534 chapter 13 Properties of Solutions

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1 534 chapter 13 Properties of Solutions Go Figure ow does the orientation of 2 O molecules around Na + differ from that around Cl -? Solvent solute interactions between water molecules and NaCl allow solid to dissolve Crystal of NaCl in water Ions hydrated in solution ydrated Cl ion ydrated Na + ion Figure 13.3 Dissolution of the ionic solid NaCl in water. We can imagine the solution process as having three components, each with an associated enthalpy change: A cluster of n solute particles must separate from one another 1 solute 2, a cluster of m solvent particles separate from one another 1 solvent 2, and these solute and solvent particles mix 1 mix solute2 n n solute solute 2. 1solvent2 m m solvent solvent 3. n solute + m solvent solution mix 4. 1solute2 n + 1solvent2 m solution soln = solute + solvent + mix As seen above, the overall enthalpy change, soln, is the sum of the three steps: soln = solute + solvent + mix [13.1] Separation of the solute particles from one another always requires an input of energy to overcome their attractive interactions. The process is therefore endothermic 1 solute Likewise, separation of solvent molecules to accommodate the solute always requires energy 1 solvent The third component, which arises from the attractive interactions between solute particles and solvent particles, is always exothermic 1 mix The three enthalpy terms in Equation 13.1 can be added together to give either a negative or a positive sum, depending on the actual numbers for the system being considered ( Figure 13.4). Thus, the formation of a solution can be either exothermic or endothermic. For example, when magnesium sulfate 1MgSO 4 2 is added to water, the solution process is exothermic: soln = kj>mol. In contrast, the dissolution of ammonium nitrate 1N 4 NO 3 2 is endothermic: soln = 26.4 kj>mol. These particular salts are the main components in the instant heat packs and ice packs used to treat

2 section 13.1 The Solution Process 535 Go Figure ow does the magnitude of mix compare with the magnitude of solvent + solute for exothermic solution processes? Enthalpy, solvent Solvent separated Solvent aggregated solute Solute separated Solute aggregated solute + solvent Solution mix initial soln < 0 Enthalpy, solvent Solvent separated Solvent aggregated solute Solute separated Solute aggregated solute + solvent Solution mix final soln > 0 initial final Exothermic solution process Figure 13.4 Enthalpy changes accompanying the solution process. Endothermic solution process athletic injuries ( Figure 13.5). The packs consist of a pouch of water and the solid salt sealed off from the water MgSO 4 1s2 for hot packs and N 4 NO 3 1s2 for cold packs. When the pack is squeezed, the seal separating the solid from the water is broken and a solution forms, either increasing or decreasing the temperature. The enthalpy change for a process can provide insight into the extent to which the process occurs. (Section 5.4) Exothermic processes tend to proceed spontaneously. On the other hand, if soln is too endothermic, the solute might not dissolve to any significant extent in the chosen solvent. Thus, for solutions to form, the solvent solute interaction must be strong enough to make mix comparable in magnitude to solute + solvent. This fact further explains why ionic solutes do not dissolve in nonpolar solvents. The nonpolar solvent molecules experience only weak attractive interactions with the ions, and these interactions do not compensate for the energies required to separate the ions from one another. By similar reasoning, a polar liquid solute, such as water, does not dissolve in a nonpolar liquid solvent, such as octane 1C The water molecules experience strong hydrogen-bonding interactions with one another (Section 11.2) attractive forces that must be overcome if the water molecules are to be dispersed throughout the octane solvent. The energy required to separate the 2 O molecules from one another is not recovered in the form of attractive interactions between the 2 O and C 8 18 molecules. Figure 13.5 Magnesium sulfate instant hot pack. Give It Some Thought Label the following processes as exothermic or endothermic: (a) breaking solvent solvent interactions to form separated particles (b) forming solvent solute interactions from separated particles Solution Formation and Chemical Reactions In discussing solutions, we must be careful to distinguish the physical process of solution formation from chemical reactions that lead to a solution. For example, nickel metal dissolves on contact with an aqueous hydrochloric acid solution because the following reaction occurs: Ni1s2 + 2 Cl1aq2 NiCl 2 1aq g2 [13.2]

3 536 chapter 13 Properties of Solutions Go Figure What is the molar mass of nickel chloride hexahydrate, NiCl 2 # 6 2 O1s2? Nickel metal and hydrochloric acid Nickel reacts with hydrochloric acid, forming NiCl 2 (aq) and 2 (g). The solution is of NiCl 2, not Ni metal NiCl O(s) remains when solvent evaporated Figure 13.6 The reaction between nickel metal and hydrochloric acid is not a simple dissolution. The product is NiCl 2 # 6 2 O1s2, nickel(ii) chloride hexahydrate, which has exactly 6 waters of hydration in the crystal lattice for every nickel ion. In this instance, one of the resulting solutes is not Ni metal but rather its salt NiCl 2. If the solution is evaporated to dryness, NiCl 2 # 6 2 O1s2 is recovered ( Figure 13.6). Compounds such as NiCl 2 # 6 2 O1s2 that have a defined number of water molecules in the crystal lattice are known as hydrates. When NaCl(s) is dissolved in water, on the other hand, no chemical reaction occurs. If the solution is evaporated to dryness, NaCl is recovered. Our focus throughout this chapter is on solutions from which the solute can be recovered unchanged from the solution Saturated Solutions and Solubility As a solid solute begins to dissolve in a solvent, the concentration of solute particles in solution increases, increasing the chances that some solute particles will collide with the surface of the solid and reattach. This process, which is the opposite of the solution process, is called crystallization. Thus, two opposing processes occur in a solution in contact with undissolved solute. This situation is represented in this chemical equation: Solute + solvent dissolve solution [13.3] crystallize When the rates of these opposing processes become equal, a dynamic equilibrium is established, and there is no further increase in the amount of solute in solution. (Section 4.1) A solution that is in equilibrium with undissolved solute is saturated. Additional solute will not dissolve if added to a saturated solution. The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility

4 section 13.2 Saturated Solutions and Solubility 537 of that solute. That is, the solubility of a particular solute in a particular solvent is the maximum amount of the solute that can dissolve in a given amount of the solvent at a specified temperature, assuming that excess solute is present. For example, the solubility of NaCl in water at 0 C is 35.7 g per 100 ml of water. This is the maximum amount of NaCl that can be dissolved in water to give a stable equilibrium solution at that temperature. If we dissolve less solute than the amount needed to form a saturated solution, the solution is unsaturated. Thus, a solution containing 10.0 g of NaCl per 100 ml of water at 0 C is unsaturated because it has the capacity to dissolve more solute. Under suitable conditions, it is possible to form solutions that contain a greater amount of solute than needed to form a saturated solution. Such solutions are supersaturated. For example, when a saturated solution of sodium acetate is made at a high temperature and then slowly cooled, all of the solute may remain dissolved even though its solubility decreases as the temperature decreases. Because the solute in a supersaturated solution is present in a concentration higher than the equilibrium concentration, supersaturated solutions are unstable. For crystallization to occur, however, the solute particles must arrange themselves properly to form crystals. The addition of a small crystal of the solute (a seed crystal) provides a template for crystallization of the excess solute, leading to a saturated solution in contact with excess solid ( Figure 13.7). Give It Some Thought What happens if a solute is added to a saturated solution? Go Figure What is the evidence that the solution in the left photograph is supersaturated? Amount of sodium acetate dissolved is greater than its solubility at this temperature 1 Seed crystal of sodium 2 Excess sodium acetate 3 Solution arrives acetate added to crystallizes from solution at saturation supersaturated solution Figure 13.7 Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving about 170 g of the salt in 100 ml of water at 100 C and then slowly cooling it to 20 C. Because the solubility of sodium acetate in water at 20 C is 46 g per 100 ml of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.

5 538 chapter 13 Properties of Solutions Table 13.1 Solubilities of Gases in Water at 20 C, with 1 atm Gas Pressure Molar Gas Mass 1g>mol2 Solubility (M) N * 10-3 O * 10-3 Ar * 10-3 Kr * Factors Affecting Solubility The extent to which one substance dissolves in another depends on the nature of both substances. (Section 13.1) It also depends on temperature and, at least for gases, on pressure. Solute Solvent Interactions The natural tendency of substances to mix and the various interactions among solute and solvent particles are all involved in determining solubilities. Nevertheless, insight into variations in solubility can often be gained by focusing on the interaction between the solute and solvent. The data in Table 13.1 show that the solubilities of various gases in water increase with increasing molecular mass. The attractive forces between the gas molecules and solvent molecules are mainly dispersion forces, which increase with increasing size and molecular mass. (Section 11.2) Thus, the data indicate that the solubilities of gases in water increase as the attraction between solute (gas) and solvent (water) increases. In general, when other factors are comparable, the stronger the attractions between solute and solvent molecules, the greater the solubility of the solute in that solvent. Because of favorable dipole dipole attractions between solvent molecules and solute molecules, polar liquids tend to dissolve in polar solvents. Water is both polar and able to form hydrogen bonds. (Section 11.2) Thus, polar molecules, especially those that can form hydrogen bonds with water molecules, tend to be soluble in water. For example, acetone, a polar molecule with the structural formula shown below, mixes in all proportions with water. Acetone has a strongly polar C O bond and pairs of nonbonding electrons on the O atom that can form hydrogen bonds with water. O C 3 CC 3 Acetone exane Water Figure 13.8 exane is immiscible with water. exane is the top layer because it is less dense than water. Liquids that mix in all proportions, such as acetone and water, are miscible, whereas those that do not dissolve in one another are immiscible. Gasoline, which is a mixture of hydrocarbons, is immiscible with water. ydrocarbons are nonpolar substances because of several factors: The C C bonds are nonpolar, the C bonds are nearly nonpolar, and the molecules are symmetrical enough to cancel much of the weak C bond dipoles. The attraction between the polar water molecules and the nonpolar hydrocarbon molecules is not sufficiently strong to allow the formation of a solution. Nonpolar liquids tend to be insoluble in polar liquids, as Figure 13.8 shows for hexane 1C and water. Many organic compounds have polar groups attached to a nonpolar framework of carbon and hydrogen atoms. For example, the series of organic compounds in Table 13.2 all contain the polar O group. Organic compounds with this molecular feature are called alcohols. The O bond is able to form hydrogen bonds. For example, ethanol 1C 3 C 2 O2 molecules can form hydrogen bonds with water molecules as well as with each other ( Figure 13.9). As a result, the solute solute, solvent solvent, and solute solvent forces are not greatly different in a mixture of C 3 C 2 O and 2 O. No major change occurs in the environments of the molecules as they are mixed. Therefore, the increased entropy when the components mix plays a significant role in solution formation, and ethanol is completely miscible with water.

6 section 13.3 Factors Affecting Solubility 539 Table 13.2 Solubilities of Some Alcohols in Water and in exane* Solubility Solubility Alcohol in h 2 o in C 6 h 14 C 3 O 1methanol C 3 C 2 O 1ethanol2 C 3 C 2 C 2 O 1propanol2 C 3 C 2 C 2 C 2 O 1butanol C 3 C 2 C 2 C 2 C 2 O 1pentanol C 3 C 2 C 2 C 2 C 2 C 2 O 1hexanol *Expressed in mol alcohol/100 g solvent at 20 C. The infinity symbol 1 2 indicates that the alcohol is completely miscible with the solvent. ydrogen bond between two ethanol molecules Figure 13.9 ydrogen bonding involving O groups. ydrogen bond between ethanol molecule and water molecule Notice in Table 13.2 that the number of carbon atoms in an alcohol affects its solubility in water. As this number increases, the polar O group becomes an even smaller part of the molecule, and the molecule behaves more like a hydrocarbon. The solubility of the alcohol in water decreases correspondingly. On the other hand, the solubility of alcohols in a nonpolar solvent like hexane 1C increases as the nonpolar hydrocarbon chain lengthens. One way to enhance the solubility of a substance in water is to increase the number of polar groups the substance contains. For example, increasing the number of O groups in a solute increases the extent of hydrogen bonding between that solute and water, thereby increasing solubility. Glucose (C 6 12 O 6, Figure 13.10) has five O groups on a six-carbon framework, which makes the molecule very soluble in water: 830 g dissolves in 1.00 L of water at 17.5 C. In contrast, cyclohexane 1C , which has a similar structure to glucose but with all of the O groups replaced by, is essentially insoluble in water (only 55 mg of cyclohexane can dissolve in 1.00 L of water at 25 C). C C C C C C Cyclohexane, C 6 12, which has no polar O groups, is essentially insoluble in water O groups enhance the aqueous solubility because of their ability to hydrogen bond with 2 O. C 2 O O C O C C O C O O C ydrogen-bonding sites Figure The correlation of molecular structure with solubility. Chemistry and Life Fat-Soluble and Water-Soluble Vitamins Vitamins have unique chemical structures that affect their solubilities in different parts of the human body. Vitamin C and the B vitamins are soluble in water, for example, whereas vitamins A, D, E, and K are soluble in nonpolar solvents and in fatty tissue (which is nonpolar). Because of their water solubility, vitamins B and C are not stored to any appreciable extent in the body, and so foods containing these vitamins should be included in the daily diet. In contrast, the fat-soluble vitamins are stored in sufficient quantities to keep vitamin-deficiency diseases from appearing even after a person has subsisted for a long period on a vitamin-deficient diet. That some vitamins are soluble in water and others are not can be explained in terms of their structures. Notice in Figure that vitamin A (retinol) is an alcohol with a very long carbon chain. Because the O group is such a small part of the molecule, the molecule resembles the long-chain alcohols listed in Table This vitamin is nearly nonpolar. In contrast, the vitamin C molecule is smaller and has several O groups that can form hydrogen bonds with water, somewhat like glucose. Related Exercises: 13.7, 13.48

7 540 chapter 13 Properties of Solutions Most of molecule nonpolar Only one polar group to interact with water Many polar groups to interact with water 3 C 2 C 2 C C C 2 C C 3 C 3 3 C C C C C C C C C C C 3 C O O O C C O O C C C C O O Vitamin A Vitamin C Figure The molecular structures of vitamins A and C. Over years of study, examination of different solvent solute combinations has led to an important generalization: Substances with similar intermolecular attractive forces tend to be soluble in one another. This generalization is often simply stated as like dissolves like. Nonpolar substances are more likely to be soluble in nonpolar solvents; ionic and polar solutes are more likely to be soluble in polar solvents. Network solids such as diamond and quartz are not soluble in either polar or nonpolar solvents because of the strong bonding within the solid. Give It Some Thought Suppose the hydrogens on the O groups in glucose (Figure 13.10) were replaced with methyl groups, C 3. Would you expect the water solubility of the resulting molecule to be higher than, lower than, or about the same as glucose? Sample Exercise 13.1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride 1CCl 4 2 or in water: C 7 16, Na 2 SO 4, Cl, and I 2. Solution Analyze We are given two solvents, one that is nonpolar 1CCl 4 2 and the other that is polar 1 2 O2, and asked to determine which will be the better solvent for each solute listed. Plan By examining the formulas of the solutes, we can predict whether they are ionic or molecular. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be better for the nonpolar solutes, whereas the polar solvent will be better for the ionic and polar solutes. Solve C 7 16 is a hydrocarbon, so it is molecular and nonpolar. Na 2 SO 4, a compound containing a metal and nonmetals, is ionic. Cl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. I 2, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C 7 16 and I 2 (the nonpolar solutes) would be more soluble in the nonpolar CCl 4 than in polar 2 O, whereas water would be the better solvent for Na 2 SO 4 and Cl (the ionic and polar covalent solutes).

8 section 13.3 Factors Affecting Solubility 541 Practice Exercise 1 Which of the following solvents will best dissolve wax, which is a complex mixture of compounds that mostly are C 3 -C 2 -C 2 -C 2 -C 2? C C C 3 C C C C C C C 3 C C (a) exane O C 3 C C 3 (c) Acetone (b) Benzene Cl Cl C Cl Cl (d) Carbon tetrachloride O (e) Water Practice Exercise 2 Arrange the following substances in order of increasing solubility in water: (a) (c) C C C C C O C C C C C O (b) (d) C C C C C O C C C C C Cl Pressure Effects The solubilities of solids and liquids are not appreciably affected by pressure, whereas the solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent increases. We can understand the effect of pressure on gas solubility by considering Figure 13.12, which shows carbon dioxide gas distributed between the gas and solution phases. When equilibrium is established, the rate at which gas molecules enter the solution equals the rate at which solute molecules escape from the solution to enter the gas phase. The equal number of up and down arrows in the left container in Figure represent these opposing processes. Now suppose we exert greater pressure on the piston and compress the gas above the solution, as shown in the middle container in Figure If we reduce the gas volume to half its original value, the pressure of the gas increases to about twice its original value. As a result of this pressure increase, the rate at which gas molecules strike the liquid surface and enter the solution phase increases. Thus, the solubility of the gas in the solution increases until equilibrium is again established; that is, solubility increases until the rate at which gas molecules enter the solution equals the rate at which they escape from the solution. Thus,

9 542 chapter 13 Properties of Solutions Go Figure If the partial pressure of a gas over a solution is doubled, how has the concentration of gas in the solution changed after equilibrium is restored? Equilibrium Pressure is increased. More CO 2 dissolves Equilibrium restored Figure Effect of pressure on gas solubility. Go Figure ow do the slopes of the lines vary with the molecular weight of the gas? Explain the trend. the solubility of a gas in a liquid solvent increases in direct proportion to the partial pressure of the gas above the solution ( Figure 13.13). The relationship between pressure and gas solubility is expressed by enry s law: S g = kp g [13.4] Solubility (mm) O 2 CO N 2 e ere, S g is the solubility of the gas in the solvent (usually expressed as molarity), P g is the partial pressure of the gas over the solution, and k is a proportionality constant known as the enry s law constant. The value of this constant depends on the solute, solvent, and temperature. As an example, the solubility of N 2 gas in water at 25 C and 0.78 atm pressure is 4.75 * 10-4 M. The enry s law constant for N 2 in 25 C water is thus * 10-4 mol>l2>0.78 atm = 6.1 * 10-4 mol>l@atm. If the partial pressure of N 2 is doubled, enry s law predicts that the solubility in water at 25 C also doubles to 9.50 * 10-4 M. Bottlers use the effect of pressure on solubility in producing carbonated beverages, which are bottled under a carbon dioxide pressure greater than 1 atm. When the bottles are opened to the air, the partial pressure of CO 2 above the solution decreases. ence, the solubility of CO 2 decreases, and CO 2 1g2 escapes from the solution as bubbles ( Figure 13.14) Partial pressure (atm) Figure The solubility of a gas in water is directly proportional to the partial pressure of the gas. The solubilities are in millimoles of gas per liter of solution. Sample Exercise 13.2 A enry s Law Calculation Calculate the concentration of CO 2 in a soft drink that is bottled with a partial pressure of CO 2 of 4.0 atm over the liquid at 25 C. The enry s law constant for CO 2 in water at this temperature is 3.4 * 10-2 mol>l@atm. Solution Analyze We are given the partial pressure of CO 2, P CO2, and the enry s law constant, k, and asked to calculate the concentration of CO 2 in the solution. Plan With the information given, we can use enry s law, Equation 13.4, to calculate the solubility, S CO2.

10 section 13.3 Factors Affecting Solubility 543 Solve S CO2 = kp CO2 = 13.4 * 10-2 mol>l@atm214.0 atm2 = 0.14 mol>l = 0.14 M Check The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO 2 and the value of enry s constant. Practice Exercise 1 You double the partial pressure of a gas over a liquid at constant temperature. Which of these statements is then true? (a) The enry s law constant is doubled. (b) The enry s law constant is decreased by half. (c) There are half as many gas molecules in the liquid. (d) There are twice as many gas molecules in the liquid. (e) There is no change in the number of gas molecules in the liquid. Practice Exercise 2 Calculate the concentration of CO 2 in a soft drink after the bottle is opened and the solution equilibrates at 25 C under a CO 2 partial pressure of 3.0 * 10-4 atm. Temperature Effects The solubility of most solid solutes in water increases as the solution temperature increases, as Figure shows. There are exceptions to this rule, however, as seen for Ce 2 1SO 4 2 3, whose solubility curve slopes downward with increasing temperature. In contrast to solid solutes, the solubility of gases in water decreases with increasing temperature ( Figure 13.16). If a glass of cold tap water is warmed, you can see bubbles on the inside of the glass because some of the dissolved air comes out of solution. Similarly, as carbonated beverages are allowed to warm, the solubility of CO 2 decreases, and CO 2 1g2 escapes from the solution. Give It Some Thought Why do bubbles form on the inside wall of a cooking pot when water is heated on the stove, even though the water temperature is well below the boiling point of water? Figure Gas solubility decreases as pressure decreases. CO 2 bubbles out of solution when a carbonated beverage is opened because the CO 2 partial pressure above the solution is reduced. Go Figure ow does the solubility of KCl at 80 C compare with that of NaCl at the same temperature? 100 Go Figure Where would you expect N 2 to fit on this graph? C 4 Solubility (g of salt in 100 g 2 O) NaNO 3 NaCl CaCl 2 Pb(NO 3 ) 2 KNO 3 KCl KClO 3 K 2 Cr 2 O 7 Solubility (mm) O 2 CO e 10 Ce 2 (SO 4 ) Temperature ( C) Figure Solubilities of some ionic compounds in water as a function of temperature Temperature ( C) Figure Solubilities of four gases in water as a function of temperature. The solubilities are in millimoles per liter of solution, for a constant total pressure of 1 atm in the gas phase.

11 544 chapter 13 Properties of Solutions Chemistry and Life Blood Gases and Deep-Sea Diving Because gas solubility increases with increasing pressure, divers who breathe compressed air ( Figure 13.17) must be concerned about the solubility of gases in their blood. Although the gases are not very soluble at sea level, their solubilities can be appreciable at deep levels where their partial pressures are greater. Thus, divers must ascend slowly to prevent dissolved gases from being released rapidly from solution and forming bubbles in the blood and other fluids in the body. These bubbles affect nerve impulses and cause decompression sickness, or the bends, which is a painful and potentially fatal condition. Nitrogen is the main problem because it is the most abundant gas in air and because it can be removed from the body only through the respiratory system. Oxygen, in contrast, is consumed in metabolism. Deep-sea divers sometimes substitute helium for nitrogen in the air they breathe because helium has a much lower solubility in biological fluids than N 2. For example, divers working at a depth of 100 ft experience a pressure of about 4 atm. At this pressure, a mixture of 95% helium and 5% oxygen gives an oxygen partial pressure of about 0.2 atm, which is the partial pressure of oxygen in normal air at 1 atm. Figure Gas solubility increases as pressure increases. Divers who use compressed gases must be concerned about the solubility of the gases in their blood. If the oxygen partial pressure becomes too great, the urge to breathe is reduced, CO 2 is not removed from the body, and CO 2 poisoning occurs. At excessive concentrations in the body, carbon dioxide acts as a neurotoxin, interfering with nerve conduction and transmission. Related Exercises: 13.59, 13.60, expressing Solution Concentration The concentration of a solution can be expressed either qualitatively or quantitatively. The terms dilute and concentrated are used to describe a solution qualitatively. A solution with a relatively small concentration of solute is said to be dilute; one with a large concentration is said to be concentrated. Chemists use various ways to express concentration quantitatively, and we examine several of these next. Mass Percentage, ppm, and ppb One of the simplest quantitative expressions of concentration is the mass percentage of a component in a solution, given by Mass % of component = mass of component in soln total mass of soln * 100 [13.5] Because percent means per hundred, a solution of hydrochloric acid that is 36% Cl by mass contains 36 g of Cl for each 100 g of solution. We often express the concentration of very dilute solutions in parts per million (ppm) or parts per billion (ppb). These quantities are similar to mass percentage but use 10 6 (a million) or 10 9 (a billion), respectively, in place of 100, as a multiplier for the ratio of the mass of solute to the mass of solution. Thus, parts per million is defined as ppm of component = mass of component in soln total mass of soln * 10 6 [13.6] A solution whose solute concentration is 1 ppm contains 1 g of solute for each million grams of solution or, equivalently, 1 mg of solute per kilogram of solution. Because the density of water is 1 g>ml, 1 kg of a dilute aqueous solution has a volume

12 section 13.4 Expressing Solution Concentration 545 very close to 1 L. Thus, 1 ppm also corresponds to 1 mg of solute per liter of aqueous solution. The acceptable maximum concentrations of toxic or carcinogenic substances in the environment are often expressed in ppm or ppb. For example, the maximum allowable concentration of arsenic in drinking water in the United States is ppm; that is, mg of arsenic per liter of water. This concentration corresponds to 10 ppb. Give It Some Thought A solution of SO 2 in water contains g of SO 2 per liter of solution. What is the concentration of SO 2 in ppm? In ppb? Sample Exercise 13.3 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose 1C 6 12 O 6 2 in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 mg of Zn 2+. What is the concentration of Zn 2+ in parts per million? Solution (a) Analyze We are given the number of grams of solute (13.5 g) and the number of grams of solvent kg = 100 g2. From this, we must calculate the mass percentage of solute. Plan We can calculate the mass percentage by using Equation The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve Mass % of glucose = mass glucose mass soln * 100 = 13.5 g 13.5 g g * 100 = 11.9% Comment The mass percentage of water in this solution is % = 88.1%. (b) Analyze In this case we are given the number of micrograms of solute. Because 1 mg is 1 * 10-6 g, 5.4 mg = 5.4 * 10-6 g. Plan We calculate the parts per million using Equation Solve ppm = mass of solute mass of soln * 106 = 5.4 * 10-6 g 2.5 g * 10 6 = 2.2 ppm Practice Exercise 1 Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (a) %, (b) %, (c) %, (d) 2.91%, (e) 3.00%. Practice Exercise 2 A commercial bleaching solution contains 3.62% by mass of sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? Mole Fraction, Molarity, and Molality Concentration expressions are often based on the number of moles of one or more components of the solution. Recall from Section 10.6 that the mole fraction of a component of a solution is given by Mole fraction of component = moles of component total moles of all components [13.7] The symbol X is commonly used for mole fraction, with a subscript to indicate the component of interest. For example, the mole fraction of Cl in a hydrochloric acid solution is represented as X Cl. Thus, if a solution contains 1.00 mol of Cl (36.5 g) and 8.00 mol of water (144 g), the mole fraction of Cl is X Cl = mol2>11.00 mol mol2 =

13 546 chapter 13 Properties of Solutions Mole fractions have no units because the units in the numerator and the denominator cancel. The sum of the mole fractions of all components of a solution must equal 1. Thus, in the aqueous Cl solution, X 2 O = = Mole fractions are very useful when dealing with gases, as we saw in Section 10.6, but have limited use when dealing with liquid solutions. Recall from Section 4.5 that the molarity (M) of a solute in a solution is defined as Molarity = moles of solute liters of soln [13.8] For example, if you dissolve mol of Na 2 CO 3 in enough water to form L of solution, the molarity of Na 2 CO 3 in the solution is mol2> L2 = 2.00 M. Molarity is especially useful for relating the volume of a solution to the quantity of solute contained in that volume, as we saw in our discussions of titrations. (Section 4.6) The molality of a solution, denoted m, is a concentration unit that is also based on moles of solute. Molality equals the number of moles of solute per kilogram of solvent: Molality = moles of solute kilograms of solvent [13.9] Thus, if you form a solution by mixing mol of NaO (8.00 g) and kg of water (500 g), the concentration of the solution is mol2> kg2 = m (that is, molal) in NaO. The definitions of molarity and molality are similar enough that they can be easily confused. Molarity depends on the volume of solution, whereas molality depends on the mass of solvent. When water is the solvent, the molality and molarity of dilute solutions are numerically about the same because 1 kg of solvent is nearly the same as 1 kg of solution, and 1 kg of the solution has a volume of about 1 L. The molality of a given solution does not vary with temperature because masses do not vary with temperature. The molarity of the solution does change with temperature, however, because the volume of the solution expands or contracts with temperature. Thus, molality is often the concentration unit of choice when a solution is to be used over a range of temperatures. Give It Some Thought If an aqueous solution is very dilute, will its molality be greater than its molarity, nearly the same as its molarity, or smaller than its molarity? Sample Exercise 13.4 Calculation of Molality A solution is made by dissolving 4.35 g of glucose 1C 6 12 O 6 2 in 25.0 ml of water at 25 C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g>ml. Solution Analyze We are asked to calculate a solution concentration in units of molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water). Plan We use the molar mass of C 6 12 O 6 to convert grams of glucose to moles. We use the density of water to convert milliliters of water to kilograms. The molality equals the number of moles of solute (glucose) divided by the number of kilograms of solvent (water). Solve Use the molar mass of glucose, g>mol, to convert grams to moles: Mol C 6 12 O 6 = g C 6 12 O 6 2a 1 mol C 6 12 O 6 b = mol C g C 6 12 O 6 12 O 6 6 Because water has a density of 1.00 g>ml, the mass of the solvent is ml g>ml2 = 25.0 g = kg Finally, use Equation 13.9 to obtain the molality: Molality of C 6 12 O 6 = mol C 6 12 O kg 2 O = m

14 section 13.4 Expressing Solution Concentration 547 Practice Exercise 1 Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled. You take this new solution and add more solute, so that the original mass of the solute is doubled. What happens to the molality of the final solution, compared to the original molality? (a) It is doubled. (b) It is decreased by half. (c) It is unchanged. (d) It will increase or decrease depending on the molar mass of the solute. (e) There is no way to tell without knowing the molar mass of the solute. Practice Exercise 2 What is the molality of a solution made by dissolving 36.5 g of naphthalene 1C in 425 g of toluene 1C 7 8 2? Converting Concentration Units If you follow the dimensional analysis techniques you learned in Chapter 1, you can convert between concentration units, as shown in Sample Exercise To convert between molality and molarity, the density of the solution will be needed, as in Sample Exercise Sample Exercise 13.5 Calculation of Mole Fraction and Molality An aqueous solution of hydrochloric acid contains 36% Cl by mass. (a) Calculate the mole fraction of Cl in the solution. (b) Calculate the molality of Cl in the solution. Solution Analyze We are asked to calculate the concentration of the solute, Cl, in two related concentration units, given only the percentage by mass of the solute in the solution. Plan In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let s assume that there is exactly 100 g of solution. Because the solution is 36% Cl, it contains 36 g of Cl and g = 64 g of 2 O. We must convert grams of solute (Cl) to moles to calculate either mole fraction or molality. We must convert grams of solvent 1 2 O2 to moles to calculate mole fractions and to kilograms to calculate molality. Solve (a) To calculate the mole fraction of Cl, we convert the masses of Cl and 2 O to moles and then use Equation 13.7: (b) To calculate the molality of Cl in the solution, we use Equation We calculated the number of moles of Cl in part (a), and the mass of solvent is 64 g = kg: 1 mol Cl Moles Cl = 136 g Cl2a b = 0.99 mol Cl 36.5 g Cl Moles 2 O = 164 g 2 O2a 1 mol 2O 18 g 2 O b = 3.6 mol 2O moles Cl X Cl = moles 2 O + moles Cl = = = 0.22 Molality of Cl = Notice that we can t readily calculate the molarity of the solution because we don t know the volume of the 100 g of solution mol Cl kg 2 O = 15 m Practice Exercise 1 The solubility of oxygen gas in water at 40 C is 1.0 mmol per liter of solution. What is this concentration in units of mole fraction? (a) 1.00 * 10-6, (b) 1.80 * 10-5, (c) 1.00 * 10-2, (d) 1.80 * 10-2, (e) 5.55 * Practice Exercise 2 A commercial bleach solution contains 3.62% by mass of NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.

15 548 chapter 13 Properties of Solutions Sample Exercise 13.6 Calculation of Molarity Using the Density of the Solution A solution with a density of g>ml contains 5.0 g of toluene 1C and 225 g of benzene. Calculate the molarity of the solution. Solution Analyze Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution g>ml2. Plan The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute 1C is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solution = mass of solute + mass of solvent = 5.0 g g = 230 g) and its density. Solve The number of moles of solute is: Moles C 7 8 = 15.0 g C 7 8 2a 1 mol C 7 8 b = mol 92 g C 7 8 The density of the solution is used to convert the mass of the solution to its volume: Milliliters soln = 1230 g2a 1 ml b = 263 ml g Molarity is moles of solute per liter of solution: Molarity = a moles C 7 8 liter soln b = a mol C ml soln b a b = 0.21 M 263 ml soln 1 L soln Check The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25 gives a molarity of mol2>10.25 L2 = 0.2 M. The units for our answer 1mol>L2 are correct, and the answer, 0.21, has two significant figures, corresponding to the number of significant figures in the mass of solute (2). Comment Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263) are similar in magnitude, the molarity and molality are also similar in magnitude: mol C 7 8 2> kg solvent2 = 0.24 m. Practice Exercise 1 Maple syrup has a density of g>ml, and g of maple syrup contains 67 mg of calcium in the form of Ca 2+ ions. What is the molarity of calcium in maple syrup? (a) M, (b) M, (c) 0.89 M, (d) 12.6 M, (e) 45.4 M Practice Exercise 2 A solution containing equal masses of glycerol 1C 3 8 O 3 2 and water has a density of 1.10 g>ml. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution Colligative Properties Some physical properties of solutions differ in important ways from those of the pure solvent. For example, pure water freezes at 0 C, but aqueous solutions freeze at lower temperatures. We apply this behavior when we add ethylene glycol antifreeze to a car s radiator to lower the freezing point of the solution. The added solute also raises the boiling point of the solution above that of pure water, making it possible to operate the engine at a higher temperature. Lowering of the freezing point and raising of the boiling point are physical properties of solutions that depend on the quantity (concentration) but not on the kind or identity of the solute particles. Such properties are called colligative properties. (Colligative means depending on the collection ; colligative properties depend on the collective effect of the number of solute particles.) In addition to freezing-point lowering and boiling-point raising, vapor-pressure lowering and osmotic pressure are also colligative properties. As we examine each one, notice how solute concentration quantitatively affects the property. Vapor-Pressure Lowering A liquid in a closed container establishes equilibrium with its vapor. (Section 11.5) The vapor pressure is the pressure exerted by the vapor when it is at equilibrium with the liquid (that is, when the rate of vaporization equals the rate of condensation). A substance that has no measurable vapor pressure is nonvolatile, whereas one that exhibits a vapor pressure is volatile. A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In

16 section 13.5 Colligative Properties 549 Volatile solvent particles Nonvolatile solute particles Add nonvolatile solute Equilibrium Rate of vaporization reduced by presence of nonvolatile solute Equilibrium reestablished with fewer molecules in gas phase Figure Vapor-pressure lowering. The presence of nonvolatile solute particles in a liquid solvent results in a reduction of the vapor pressure above the liquid. effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in Figure Ideally, the vapor pressure of a volatile solvent above a solution containing a nonvolatile solute is proportional to the solvent s concentration in the solution. This relationship is expressed quantitatively by Raoult s law, which states that the partial pressure exerted by solvent vapor above the solution, P solution, equals the product of the mole fraction of the solvent, X solvent, times the vapor pressure of the pure solvent, P solvent : P solution = X solvent P solvent [13.10] For example, the vapor pressure of pure water at 20 C is P 2 O = 17.5 torr. Imagine holding the temperature constant while adding glucose 1C 6 12 O 6 2 to the water so that the mole fractions in the resulting solution are X 2 O = and X C6 12 O 6 = According to Equation 13.10, the vapor pressure of the water above this solution is 80.0% of that of pure water: P solution = torr2 = 14.0 torr The presence of the nonvolatile solute lowers the vapor pressure of the volatile solvent by 17.5 torr torr = 3.5 torr. The vapor-pressure lowering, P, is directly proportional to the mole fraction of the solute, X solute : P = X solute P solvent [13.11] Thus, for the example of the solution of glucose in water, we have P = X C6 12 O 6 P 2 O = torr2 = 3.50 torr The vapor-pressure lowering caused by adding a nonvolatile solute depends on the total concentration of solute particles, regardless of whether they are molecules or ions. Remember that vapor-pressure lowering is a colligative property, so its value for any solution depends on the concentration of solute particles and not on their kind or identity. Give It Some Thought Adding 1 mol of NaCl to 1 kg of water lowers the vapor pressure of water more than adding 1 mol of C 6 12 O 6. Explain. Sample Exercise 13.7 Calculation of Vapor Pressure of a Solution Glycerin 1C 3 8 O 3 2 is a nonvolatile nonelectrolyte with a density of 1.26 g>ml at 25 C. Calculate the vapor pressure at 25 C of a solution made by adding 50.0 ml of glycerin to ml of water. The vapor pressure of pure water at 25 C is 23.8 torr (Appendix B), and its density is 1.00 g>ml.

17 550 chapter 13 Properties of Solutions Solution Analyze Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan We can use Raoult s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X solvent, is the ratio of the number of moles of solvent 1 2 O2to total moles of solution 1moles C 3 8 O 3 + moles 2 O2. Solve To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 8 O 3 and 2 O: We now use Raoult s law to calculate the vapor pressure of water for the solution: Moles C 3 8 O 3 = ml C 3 8 O 3 2a 1.26 g C 3 8 O 3 1 ml C 3 8 O 3 b a 1 mol C 3 8 O g C 3 8 O 3 b = mol Moles 2 O = ml 2 O2a 1.00 g 2O 1 ml 2 O b a 1 mol 2O b = 27.8 mol 18.0 g 2 O X 2 O = mol 2 O mol 2 O + mol C 3 8 O 3 = = P 2 O = X 2 O P 2 O = torr2 = 23.2 torr Comment The vapor pressure of the solution has been lowered by 23.8 torr torr = 0.6 torr relative to that of pure water. The vapor-pressure lowering can be calculated directly using Equation together with the mole fraction of the solute, C 3 8 O 3 : Practice Exercise 1 The vapor pressure of benzene, C 6 6, is torr at 26.1 C. Assuming Raoult s law is obeyed, how many moles of a nonvolatile solute must be added to ml of benzene to decrease its vapor pressure by 10.0% at 26.1 C? The density of benzene is g>cm 3. (a) , (b) , (c) , (d) , (e) P = X C3 8 O 3 P 2 O = torr2 = 0.57 torr. Notice that the use of Equation gives one more significant figure than the number obtained by subtracting the vapor pressure of the solution from that of the pure solvent. Practice Exercise 2 The vapor pressure of pure water at 110 C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 C. Assuming that Raoult s law is obeyed, what is the mole fraction of ethylene glycol in the solution? A Closer Look Ideal Solutions with Two or More Volatile Components Solutions sometimes have two or more volatile components. Gasoline, for example, is a solution of several volatile liquids. To gain some understanding of such mixtures, consider an ideal solution of two volatile liquids, A and B. (For our purposes here, it does not matter which we call the solute and which the solvent.) The partial pressures above the solution are given by Raoult s law: P A = X A P A and P B = X B P B and the total vapor pressure above the solution is P total = P A + P B = X A P A + X B P B Figure The volatile components of organic mixtures can be separated on an industrial scale in these distillation towers. Consider a mixture of 1.0 mol of benzene 1C and 2.0 mol of toluene 1C X ben = 0.33, X tol = At 20 C, the vapor pressures of the pure substances are P ben = 75 torr and P tol = 22 torr. Thus, the partial pressures above the solution are P ben = torr2 = 25 torr P tol = torr2 = 15 torr and the total vapor pressure above the liquid is P total = P ben + P tol = 25 torr + 15 torr = 40 torr Note that the vapor is richer in benzene, the more volatile component. The mole fraction of benzene in the vapor is given by the ratio of its vapor pressure to the total pressure (Equations and 10.15): X ben in vapor = P ben P tol = 25 torr 40 torr = 0.63 Although benzene constitutes only 33% of the molecules in the solution, it makes up 63% of the molecules in the vapor. When an ideal liquid solution containing two volatile components is in equilibrium with its vapor, the more volatile component will be relatively richer in the vapor. This fact forms the basis of distillation, a technique used to separate (or partially separate) mixtures containing volatile components. (Section 1.3) Distillation is a way of purifying liquids, and is the procedure by which petrochemical plants achieve the separation of crude petroleum into gasoline, diesel fuel, lubricating oil, and other products ( Figure 13.19). Distillation is also used routinely on a small scale in the laboratory. Related Exercises: 13.67, 13.68

18 section 13.5 Colligative Properties 551 An ideal gas is defined as one that obeys the ideal-gas equation (Section 10.4), and an ideal solution is defined as one that obeys Raoult s law. Whereas ideality for a gas arises from a complete lack of intermolecular interaction, ideality for a solution implies total uniformity of interaction. The molecules in an ideal solution all influence one another in the same way in other words, solute solute, solvent solvent, and solute solvent interactions are indistinguishable from one another. Real solutions best approximate ideal behavior when the solute concentration is low and solute and solvent have similar molecular sizes and take part in similar types of intermolecular attractions. Many solutions do not obey Raoult s law exactly and so are not ideal. If, for instance, the solvent solute interactions in a solution are weaker than either the solvent solvent or solute solute interactions, the vapor pressure tends to be greater than that predicted by Raoult s law. When the solute solvent interactions in a solution are exceptionally strong, as might be the case when hydrogen bonding exists, the vapor pressure is lower than that predicted by Raoult s law. Although you should be aware that these departures from ideality occur, we will ignore them for the remainder of this chapter. Boiling-Point Elevation In Sections 11.5 and 11.6, we examined the vapor pressures of pure substances and how to use them to construct phase diagrams. ow does the phase diagram of a solution and, hence, its boiling and freezing points differ from that of the pure solvent? The addition of a nonvolatile solute lowers the vapor pressure of the solution. Thus, in Figure the vapor-pressure curve of the solution is shifted downward relative to the vapor-pressure curve of the pure solvent. Recall from Section 11.5 that the normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atm. Because the solution has a lower vapor pressure than the pure solvent, a higher temperature is required for the solution to achieve a vapor pressure of 1 atm. As a result, the boiling point of the solution is higher than that of the pure solvent. This effect is seen in Figure We find the normal boiling point of the pure solvent on the graph by locating the point where the 1-atm pressure horizontal line intersects the black vapor-pressure curve and then tracing this point down to the temperature axis. For the solution, the 1-atm line intersects the blue vapor-pressure curve at a higher temperature, indicating that the solution has a higher boiling point than the pure solvent. The increase in the boiling point of a solution, relative to the pure solvent, depends on the molality of the solute. But it is important to remember that boiling-point 1 atm Pressure Solid Liquid Vapor pressure of pure liquid solvent Gas Vapor pressure of solution Pure solvent boiling point Solution boiling point Temperature T b Figure Phase diagram illustrating boiling-point elevation. The black lines show the pure solvent s phase equilibria curves, and the blue lines show the solution s phase equilibria curves.

19 552 chapter 13 Properties of Solutions elevation is proportional to the total concentration of solute particles, regardless of whether the particles are molecules or ions. When NaCl dissolves in water, 2 mol of solute particles (1 mol of Na + and 1 mol of Cl - ) are formed for each mole of NaCl that dissolves. We take this into account by defining i, the van t off factor, as the number of fragments that a solute breaks up into for a particular solvent. The change in boiling point for a solution compared to the pure solvent is: T b = T b (solution) - T b (solvent) = ik b m [13.12] In this equation, T b 1solution2 is the boiling point of the solution, T b 1solvent2 is the boiling point of the pure solvent, m is the molality of the solute, K b is the molal boilingpoint-elevation constant for the solvent (which is a proportionality constant that is experimentally determined for each solvent), and i is the van t off factor. For a nonelectrolyte, we can always assume i = 1; for an electrolyte, i will depend on how the substance ionizes in that solvent. For instance, i = 2 for NaCl in water, assuming complete dissociation of ions. As a result, we expect the boiling-point elevation of a 1 m aqueous solution of NaCl to be twice as large as the boiling-point elevation of a 1 m solution of a nonelectrolyte such as sucrose. Thus, to properly predict the effect of a particular solute on boiling-point elevation (or any other colligative property), it is important to know whether the solute is an electrolyte or a nonelectrolyte. (Sections 4.1 and 4.3) Give It Some Thought A solute dissolved in water causes the boiling point to increase by 0.51 C. Does this necessarily mean that the concentration of the solute is 1.0 m ( Table 13.3)? Freezing-Point Depression The vapor-pressure curves for the liquid and solid phases meet at the triple point. (Section 11.6) In Figure we see that the triple-point temperature of the solution is lower than the triple-point temperature of pure liquid because the solution has a lower vapor pressure than the pure liquid. The freezing point of a solution is the temperature at which the first crystals of pure solvent form in equilibrium with the solution. Recall from Section 11.6 that the line representing the solid liquid equilibrium rises nearly vertically from the triple point. It is easy to see in Figure that the triple-point temperature of the solution is lower than that of the pure liquid, but it is also true for all points along the solid liquid equilibrium curve: the freezing point of the solution is lower than that of the pure liquid. 1 atm Liquid Vapor pressure of pure liquid solvent Pressure Vapor pressure of pure solid solvent Solution triple point Solid T f Pure solvent triple point Gas Solution freezing point Pure solvent freezing point Temperature Vapor pressure of solution Figure Phase diagram illustrating freezing-point depression. The black lines show the pure solvent s phase equilibria curves, and the blue lines show the solution s phase equilibria curves.

20 section 13.5 Colligative Properties 553 Table 13.3 Molal Boiling-Point-Elevation and Freezing-Point-Depression Constants Solvent Normal Boiling Point 1 C2 K b 1 C>m2 Normal Freezing Point 1 C2 K f 1 C>m2 Water, 2 O Benzene, C Ethanol, C 2 5 O Carbon tetrachloride, CCl Chloroform, CCl Like the boiling-point elevation, the change in freezing point T f is directly proportional to solute molality, taking into account the van t off factor i: T f = T f 1solution2 - T f 1solvent2 = -ik f m [13.13] The proportionality constant K f is the molal freezing-point-depression constant, analogous to K b for boiling point elevation. Note that because the solution freezes at a lower temperature than does the pure solvent, the value of T f is negative. Some typical values of K b and K f for several common solvents are given in Table For water, the table shows K b = 0.51 C>m, which means that the boiling point of any aqueous solution that is 1 m in nonvolatile solute particles is 0.51 C higher than the boiling point of pure water. Because solutions generally do not behave ideally, the constants listed in Table 13.3 serve well only for solutions that are rather dilute. For water, K f is 1.86 C>m. Therefore, any aqueous solution that is 1 m in nonvolatile solute particles (such as 1 m C 6 12 O 6 or 0.5 m NaCl) freezes at the temperature that is 1.86 C lower than the freezing point of pure water. The freezing-point depression caused by solutes has useful applications: it is why antifreeze works in car cooling systems, and why calcium chloride 1CaCl 2 2 promotes the melting of ice on roads during winter. Sample Exercise 13.8 Calculation of Boiling-Point Elevation and Freezing-Point Depression Automotive antifreeze contains ethylene glycol, C 2 1O2C 2 1O2, a nonvolatile nonelectrolyte, in water. Calculate the boiling point and freezing point of a 25.0% by mass solution of ethylene glycol in water. Solution Analyze We are given that a solution contains 25.0% by mass of a nonvolatile, nonelectrolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression. Plan To calculate the boiling-point elevation and the freezing-point depression using Equations and 13.13, we must express the concentration of the solution as molality. Let s assume for convenience that we have 1000 g of solution. Because the solution is 25.0% by mass ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point-depression constants (Table 13.3) to calculate T b and T f. We add T b to the boiling point and T f to the freezing point of the solvent to obtain the boiling point and freezing point of the solution. Solve The molality of the solution is calculated as follows: Molality = moles C 2 6 O 2 kilograms 2 O = a 250 g C 2 6 O g 2 O b a 1 mol C 2 6 O g C 2 6 O 2 b a 1000 g 2O 1 kg 2 O b = 5.37 m

21 554 chapter 13 Properties of Solutions We can now use Equations and to calculate the changes in the boiling and freezing points: T b = ik b m = C>m m2 = 2.7 C T f = -ik f m = C>m m2 = C ence, the boiling and freezing points of the solution are readily calculated: T b = T b 1solution2 - T b 1solvent2 2.7 C = T b 1solution C T b 1solution2 = C T f = T f 1solution2 - T f 1solvent C = T f 1solution2-0.0 C T f 1solution2 = C Comment Notice that the solution is a liquid over a larger temperature range than the pure solvent. Practice Exercise 1 Which aqueous solution will have the lowest freezing point? (a) m CaCl 2, (b) 0.15 m NaCl, (c) 0.10 m Cl, (d) m C 3 COO, (e) 0.20 m C O 11. Practice Exercise 2 Referring to Table 13.3, calculate the freezing point of a solution containing kg of CCl 3 and 42.0 g of eucalyptol 1C O2, a fragrant substance found in the leaves of eucalyptus trees. Osmosis Certain materials, including many membranes in biological systems and synthetic substances such as cellophane, are semipermeable. When in contact with a solution, these materials allow only ions or small molecules water molecules, for instance to pass through their network of tiny pores. Consider a situation in which only solvent molecules are able to pass through a semipermeable membrane placed between two solutions of different concentrations. The rate at which the solvent molecules pass from the less concentrated solution (lower solute concentration but higher solvent concentration) to the more concentrated solution (higher solute concentration but lower solvent concentration) is greater than the rate in the opposite direction. Thus, there is a net movement of solvent molecules from the solution with a lower solute concentration into the one with a higher solute concentration. In this process, called osmosis, the net movement of solvent is always toward the solution with the lower solvent (higher solute) concentration, as if the solutions were driven to attain equal concentrations. Figure shows the osmosis that occurs between an aqueous solution and pure water, separated by a semipermeable membrane. The U-tube contains water on the left and an aqueous solution on the right. Initially, there is a net movement of water through the membrane from left to right, leading to unequal liquid levels in the two arms of the U-tube. Eventually, at equilibrium (middle panel of Figure 13.22), the pressure difference resulting from the unequal liquid heights becomes so large that the net flow of water ceases. This pressure, which stops osmosis, is the osmotic pressure, Π, of the solution. If an external pressure equal to the osmotic pressure is applied to the solution, the liquid levels in the two arms can be equalized, as shown in the right panel of Figure The osmotic pressure obeys a law similar in form to the ideal-gas law, ΠV = inrt where Π is the osmotic pressure, V is the volume of the solution, i is the van t off factor, n is the number of moles of solute, R is the ideal-gas constant, and T is the absolute temperature. From this equation, we can write Π = ia n brt = imrt [13.14] V

22 section 13.5 Colligative Properties 555 Go Figure If the pure water in the left arm of the U-tube is replaced by a solution more concentrated than the one in the right arm, what will happen? Pure solvent Initially Solution At equilibrium Pressure applied to equalize fluid levels in both arms equals osmotic pressure, Solvent particle ( 2 O) Solute particle Semipermeable membrane Net movement of 2 O is from pure water side to solution side At equilibrium, the flow of 2 O is the same in both directions, so there is no net movement of 2 O Figure Osmosis is the process of a solvent moving from one compartment to another, across a semipermeable membrane, toward higher solute concentration. Osmotic pressure is generated at equilibrium due to the different heights of liquid on either side of the membrane and is equivalent to the pressure needed to equalize the fluid levels across the membrane. where M is the molarity of the solution. Because the osmotic pressure for any solution depends on the solution concentration, osmotic pressure is a colligative property. If two solutions of identical osmotic pressure are separated by a semipermeable membrane, no osmosis will occur. The two solutions are isotonic with respect to each other. If one solution is of lower osmotic pressure, it is hypotonic with respect to the more concentrated solution. The more concentrated solution is hypertonic with respect to the dilute solution. Give It Some Thought Of two KBr solutions, one 0.50 m and the other 0.20 m, which is hypotonic with respect to the other? Osmosis plays an important role in living systems. The membranes of red blood cells, for example, are semipermeable. Placing a red blood cell in a solution that is hypertonic relative to the intracellular solution (the solution inside the cells) causes water to move out of the cell (Figure 13.23). This causes the cell to shrivel, a process called crenation. Placing the cell in a solution that is hypotonic relative to the intracellular fluid causes water to move into the cell, which may cause the cell to rupture, a process called hemolysis. People who need body fluids or nutrients replaced but cannot be fed orally are given solutions by intravenous (IV) infusion, which feeds nutrients directly into the veins. To prevent crenation or hemolysis of red blood cells, the IV solutions must be isotonic with the intracellular fluids of the blood cells.

23 556 chapter 13 Properties of Solutions Go Figure If the fluid surrounding a patient s red blood cells is depleted in electrolytes, is crenation or hemolysis more likely to occur? The arrows represent the net movement of water molecules. igh solute concentration Low solute concentration Low solute concentration igh solute concentration Red blood cell in isotonic medium neither swells nor shrinks. Crenation of red blood cell placed in hypertonic environment emolysis of red blood cell placed in hypotonic environment Figure Osmosis through red blood cell walls. If water moves out of the red blood cell, it shrivels (crenation); if water moves into the red blood cell, it will swell and may burst (hemolysis). Sample Exercise 13.9 Osmotic Pressure Calculations The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose 1C 6 12 O 6 2 will be isotonic with blood? Solution Analyze We are asked to calculate the concentration of glucose in water that would be isotonic with blood, given that the osmotic pressure of blood at 25 C is 7.7 atm. Plan Because we are given the osmotic pressure and temperature, we can solve for the concentration, using Equation Because glucose is a nonelectrolyte, i = 1. Solve Π = imrt M = Π irt = 17.7 atm2 112a L@atm = 0.31 M b1298 K2 mol@k Comment In clinical situations, the concentrations of solutions are generally expressed as mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of NaCl that is isotonic with blood is 0.16 M, because i = 2 for NaCl in water (a M solution of NaCl is M in particles). A 0.16 M solution of NaCl is 0.9% mass in NaCl. This kind of solution is known as a physiological saline solution. Practice Exercise 1 Which of the following actions will raise the osmotic pressure of a solution? (a) decreasing the solute concentration, (b) decreasing the temperature, (c) adding more solvent, (d) increasing the temperature, (e) none of the above. Practice Exercise 2 What is the osmotic pressure, in atm, of a M sucrose 1C O 11 2 solution at 20.0 C?

24 section 13.5 Colligative Properties 557 There are many interesting biological examples of osmosis. A cucumber placed in concentrated brine loses water via osmosis and shrivels into a pickle. People who eat a lot of salty food retain water in tissue cells and intercellular space because of osmosis. The resultant swelling or puffiness is called edema. Water moves from soil into plant roots partly because of osmosis. Bacteria on salted meat or candied fruit lose water through osmosis, shrivel, and die thus preserving the food. Give It Some Thought Is the osmotic pressure of a 0.10 M solution of NaCl greater than, less than, or equal to that of a 0.10 M solution of KBr? Determination of Molar Mass from Colligative Properties The colligative properties of solutions provide a useful means of determining the molar mass of solutes. Any of the four colligative properties can be used, as shown in Sample Exercises and Sample Exercise Molar Mass from Freezing-Point Depression A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving g of the substance in 40.0 g of CCl 4. The boiling point of the resultant solution was C higher than that of the pure solvent. Calculate the molar mass of the solute. Solution Analyze Our goal is to calculate the molar mass of a solute based on knowledge of the boiling-point elevation of its solution in CCl 4, T b = C, and the masses of solute and solvent. Table 13.3 gives K b for the solvent 1CCl 4 2, K b = 5.02 C>m. Plan We can use Equation 13.12, T b = ik b m, to calculate the molality of the solution. Because the solute is a nonelectrolyte, i = 1. Then we can use molality and the quantity of solvent (40.0 g CCl 4 ) to calculate the number of moles of solute. Finally, the molar mass of the solute equals the number of grams per mole, so we divide the number of grams of solute (0.250 g) by the number of moles we have just calculated. Solve From Equation 13.12, we have Molality = T b ik b = C C>m = m Thus, the solution contains mol of solute per kilogram of solvent. The solution was prepared using 40.0 g = kg of solvent 1CCl 4 2. The number of moles of solute in the solution is therefore kg CCl 4 2a mol solute b = 2.84 * 10-3 mol solute kg CCl 4 The molar mass of the solute is the number of grams per mole of the substance: g Molar mass = 2.84 * 10-3 = 88.0 g>mol mol Practice Exercise 1 A mysterious white powder could be powdered sugar 1C O 11 2, cocaine 1C NO 4 2, codeine 1C NO 3 2, norfenefrine 1C 8 11 NO 2 2, or fructose 1C 6 12 O 6 2. When 80 mg of the powder is dissolved in 1.50 ml of ethanol (d = g>cm 3, normal freezing point C, K f = 1.99 C>m), the freezing point is lowered to C. What is the identity of the white powder? (a) powdered sugar, (b) cocaine, (c) codeine, (d) norfenefrine, (e) fructose. Practice Exercise 2 Camphor 1C O2 melts at C, and it has a particularly large freezing-point-depression constant, K f = 40.0 C>m. When g of an organic substance of unknown molar mass is dissolved in g of liquid camphor, the freezing point of the mixture is found to be C. What is the molar mass of the solute? Sample Exercise Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 ml of solution. The osmotic pressure of the solution at 25 C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass. Solution Analyze Our goal is to calculate the molar mass of a high-molecularmass protein, based on its osmotic pressure and a knowledge of the mass of protein and solution volume. Since the protein will be considered as a nonelectrolyte, i = 1. Plan The temperature 1T = 25 C2 and osmotic pressure 1Π = 1.54 torr2 are given, and we know the value of R so we can use Equation to calculate the molarity of the solution, M. In doing so, we must convert temperature from C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of

25 558 chapter 13 Properties of Solutions the solution (5.00 ml) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute. Solve Solving Equation for molarity gives Molarity = Π irt = torr2a 1 atm 760 torr b 112a L@atm b1298 K2 mol@k -5 mol = 8.28 * 10 L Because the volume of the solution is 5.00 ml = 5.00 * 10-3 L, the number of moles of protein must be Moles = * 10-5 mol>l * 10-3 L2 = 4.14 * 10-7 mol The molar mass is the number of grams per mole of the substance. Because we know the sample has a mass of 3.50 mg = 3.50 * 10-3 g, we can calculate the molar mass by dividing the number of grams in the sample by the number of moles we just calculated: Molar mass = grams moles = 3.50 * 10-3 g 4.14 * 10-7 mol = 8.45 * 103 g>mol Comment Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules. Practice Exercise 1 Proteins frequently form complexes in which 2, 3, 4, or even more individual proteins ( monomers ) interact specifically with each other via hydrogen bonds or electrostatic interactions. The entire assembly of proteins can act as one unit in solution, and this assembly is called the quaternary structure of the protein. Suppose you discover a new protein whose monomer molar mass is 25,000 g>mol. You measure an osmotic pressure of atm at 37 C for 7.20 g of the protein in ml of an aqueous solution. ow many protein monomers form the quaternary protein structure in solution? Treat the protein as a nonelectrolyte. (a) 1, (b) 2, (c) 3, (d) 4, (e) 8. Practice Exercise 2 A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kpa at 25 C. Calculate the molar mass of the polystyrene. A Closer Look The van t off Factor The colligative properties of solutions depend on the total concentration of solute particles, regardless of whether the particles are ions or molecules. Thus, we expect a m solution of NaCl to have a freezingpoint depression of m C>m2 = C because it is m in Na + 1aq2 and m in Cl - 1aq2. The measured freezingpoint depression is only C, however, and the situation is similar for other strong electrolytes. A m solution of KCl, for example, freezes at C. The difference between expected and observed colligative properties for strong electrolytes is due to electrostatic attractions between ions. As the ions move about in solution, ions of opposite charge collide and stick together for brief moments. While they are together, they behave as a single particle called an ion pair ( Figure 13.24). The number of independent particles is thereby reduced, causing a reduction in the freezing-point depression (as well as in boiling-point elevation, vapor-pressure reduction, and osmotic pressure). We have been assuming that the van t off factor, i, is equal to the number of ions per formula unit of the electrolyte. The true (measured) value of this factor, however, is given by the ratio of the measured value of a colligative property to the value calculated when the substance is assumed to be a nonelectrolyte. Using the freezing-point depression, for example, we have i = T f 1measured2 T f 1calculated for nonelectrolyte2 [13.15] The limiting value of i can be determined for a salt from the number of ions per formula unit. For NaCl, for example, the limiting van t off factor is 2 because NaCl consists of one Na + and one Cl - per formula unit; for K 2 SO 4 it is 3 because K 2 SO 4 consists of two K + and one SO 4 2- per formula unit. In the absence of any information about the actual value of i for a solution, we will use the limiting value in calculations. Two trends are evident in Table 13.4, which gives measured van t off factors for several substances at different Na + One ion pair Cl Figure Ion pairing and colligative properties. A solution of NaCl contains not only separated Na + 1aq2 and Cl - 1aq2 ions but ion pairs as well.

26 section 13.6 Colloids 559 dilutions. First, dilution affects the value of i for electrolytes; the more dilute the solution, the more closely i approaches the expected value based on the number of ions in the formula unit. Thus, we conclude that the extent of ion pairing in electrolyte solutions decreases upon dilution. Second, the lower the charges on the ions, the less i departs from the expected value because the extent of ion pairing decreases as the ionic charges decrease. Both trends are consistent with simple electrostatics: The force of interaction between charged particles decreases as their separation increases and as their charges decrease. Related Exercises: 13.83, 13.84, , Table 13.4 Measured and Expected van t off Factors for Several Substances at 25 C Concentration Compound m m m Expected Value Sucrose NaCl K 2 SO MgSO Colloids Some substances appear to initially dissolve in a solvent, but over time, the substance separates from the pure solvent. For example, finely divided clay particles dispersed in water eventually settle out because of gravity. Gravity affects the clay particles because they are much larger than most molecules, consisting of thousands or even millions of atoms. In contrast, the dispersed particles in a true solution (ions in a salt solution or glucose molecules in a sugar solution) are small. Between these extremes lie dispersed particles that are larger than typical molecules but not so large that the components of the mixture separate under the influence of gravity. These intermediate types of dispersions are called either colloidal dispersions or simply colloids. Colloids form the dividing line between solutions and heterogeneous mixtures. Like solutions, colloids can be gases, liquids, or solids. Examples of each are listed in Table Particle size can be used to classify a mixture as colloid or solution. Colloid particles range in diameter from 5 to 1000 nm; solute particles are smaller than 5 nm in diameter. The nanomaterials we saw in Chapter 12 (Section 12.9), when dispersed in a liquid, are colloids. A colloid particle may even consist of a single giant molecule. The hemoglobin molecule, for example, which carries oxygen in your blood, has molecular dimensions of 6.5 * 5.5 * 5.0 nm and a molar mass of 64,500 g>mol. Although colloid particles may be so small that the dispersion appears uniform even under a microscope, they are large enough to scatter light. Consequently, most colloids appear cloudy or opaque unless they are very dilute. (For example, homogenized milk is a colloid of fat and protein molecules dispersed in water.) Furthermore, because they scatter light, a light beam can be seen as it passes through a colloidal Table 13.5 Types of Colloids Phase of Colloid Dispersing (solvent-like) Substance Dispersed (solute-like) Substance Colloid Type Example Gas Gas Gas None (all are solutions) Gas Gas Liquid Aerosol Fog Gas Gas Solid Aerosol Smoke Liquid Liquid Gas Foam Whipped cream Liquid Liquid Liquid Emulsion Milk Liquid Liquid Solid Sol Paint Solid Solid Gas Solid foam Marshmallow Solid Solid Liquid Solid emulsion Butter Solid Solid Solid Solid sol Ruby glass

27 560 chapter 13 Properties of Solutions Figure Tyndall effect in the laboratory. The glass on the right contains a colloidal dispersion; that on the left contains a solution. dispersion ( Figure 13.25). This scattering of light by colloidal particles, known as the Tyndall effect, makes it possible to see the light beam of an automobile on a dusty dirt road, or the sunlight streaming through trees or clouds. Not all wavelengths are scattered to the same extent. Colors at the blue end of the visible spectrum are scattered more than those at the red end by the molecules and small dust particles in the atmosphere. As a result, our sky appears blue. At sunset, light from the sun travels through more of the atmosphere; blue light is scattered even more, allowing the reds and yellows to pass through and be seen. Go Figure What is the chemical composition of the groups that carry a negative charge? ydrophilic polar and charged groups on molecule surface help molecule remain dispersed in water and other polar solvents O C N + Figure ydrophilic colloidal particle. Examples of the hydrophilic groups that help to keep a giant molecule (macromolecule) suspended in water. ydrophilic and ydrophobic Colloids The most important colloids are those in which the dispersing medium is water. These colloids may be hydrophilic ( water loving ) or hydrophobic ( water fearing ). ydrophilic colloids are most like the solutions that we have previously examined. In the human body, the extremely large protein molecules such as enzymes and antibodies are kept in suspension by interaction with surrounding water molecules. A hydrophilic molecule folds in such a way that its hydrophobic groups are away from the water molecules, on the inside of the folded molecule, while its hydrophilic, polar groups are on the surface, interacting with the water molecules. The hydrophilic groups generally contain oxygen or nitrogen and often carry a charge ( Figure 13.26). Give it Some Thought Some proteins reside in the hydrophobic lipid bilayer of the cell membrane. Would hydrophilic groups of these proteins still be facing the lipid solvent? ydrophobic colloids can be dispersed in water only if they are stabilized in some way. Otherwise, their natural lack of affinity for water causes them to separate from the water. One method of stabilization involves adsorbing ions on the surface of the hydrophobic particles ( Figure 13.27). (Adsorption means to adhere to a surface. It differs from absorption, which means to pass into the interior, as when a

28 section 13.6 Colloids 561 Like charges keep particles from colliding, sticking together, and growing large enough to settle out of solution Adsorbed anions can interact with water + + Cations in solution + ydrophobic particle Repulsion + Water + + ydrophobic particle Figure ydrophobic colloids stabilized in water by adsorbed anions. sponge absorbs water.) The adsorbed ions can interact with water, thereby stabilizing the colloid. At the same time, the electrostatic repulsion between adsorbed ions on neighboring colloid particles keeps the particles from sticking together rather than dispersing in the water. ydrophobic colloids can also be stabilized by hydrophilic groups on their surfaces. Oil drops are hydrophobic, for example, and they do not remain suspended in water. Instead, they aggregate, forming an oil slick on the water surface. Sodium stearate ( Figure 13.28), or any similar substance having one end that is hydrophilic (either polar or charged) and one end that is hydrophobic (nonpolar), will stabilize a suspension of oil in water. Stabilization results from the interaction of the hydrophobic ends of the stearate ions with the oil drops and the hydrophilic ends with the water. Give It Some Thought Why don t oil drops stabilized by sodium stearate coagulate to form larger oil drops? Go Figure Which kind of intermolecular force attracts the stearate ion to the oil drop? Na Stearate ion + + ydrophobic end ydrophilic end + + Oil droplet + + Water Figure Stabilization of an emulsion of oil in water by stearate ions.

29 562 chapter 13 Properties of Solutions Colloid stabilization has an interesting application in the human digestive system. When fats in our diet reach the small intestine, a hormone causes the gallbladder to excrete a fluid called bile. Among the components of bile are compounds that have chemical structures similar to sodium stearate; that is, they have a hydrophilic (polar) end and a hydrophobic (nonpolar) end. These compounds emulsify the fats in the intestine and thus permit digestion and absorption of fat-soluble vitamins through the intestinal wall. The term emulsify means to form an emulsion, a suspension of one liquid in another, with milk being one example (Table 13.5). A substance that aids in the formation of an emulsion is called an emulsifying agent. If you read the labels on foods and other materials, you will find that a variety of chemicals are used as emulsifying agents. These chemicals typically have a hydrophilic end and a hydrophobic end. Chemistry and Life Sickle-Cell Anemia Our blood contains the complex protein hemoglobin, which carries oxygen from the lungs to other parts of the body. In the genetic disease sickle-cell anemia, hemoglobin molecules are abnormal and have a lower solubility in water, especially in their unoxygenated form. Consequently, as much as 85% of the hemoglobin in red blood cells crystallizes out of solution. The cause of the insolubility is a structural change in one part of an amino acid. Normal hemoglobin molecules contain an amino acid that has a C 2 C 2 COO group: This change leads to the aggregation of the defective form of hemoglobin into particles too large to remain suspended in biological fluids. It also causes the cells to distort into the sickle shape shown in Figure The sickled cells tend to clog capillaries, causing severe pain, weakness, and the gradual deterioration of vital organs. The disease is hereditary, and if both parents carry the defective genes, it is likely that their children will possess only abnormal hemoglobin. You might wonder how it is that a life-threatening disease such as sickle-cell anemia has persisted in humans through evolutionary time. The answer in part is that people with the disease are far less susceptible to malaria. Thus, in tropical climates rife with malaria, those with sickle-cell disease have lower incidence of this debilitating disease. C 2 O C O Normal Normal The polarity of the COO group contributes to the solubility of the hemoglobin molecule in water. In the hemoglobin molecules of sickle-cell anemia patients, the C 2 C 2 COO chain is absent and in its place is the nonpolar (hydrophobic) C1C group: C C 3 C 3 Abnormal Abnormal Figure A scanning electron micrograph of normal (round) and sickle (crescent-shaped) red blood cells. Normal red blood cells are about 6 * 10-3 mm in diameter. Colloidal Motion in Liquids We learned in Chapter 10 that gas molecules move at some average speed that depends inversely on their molar mass, in a straight line, until they collide with something. The mean free path is the average distance molecules travel between collisions. (Section 10.8) Recall also that the kinetic-molecular theory of gases assumes that gas molecules are in continuous, random motion. (Section 10.7)

30 section 13.6 Colloids 563 Colloidal particles in a solution undergo random motion as a result of collisions with solvent molecules. Because the colloidal particles are massive in comparison with solvent molecules, their movements from any one collision are very tiny. owever, there are many such collisions, and they cause a random motion of the entire colloidal particle, called Brownian motion. In 1905, Einstein developed an equation for the average square of the displacement of a colloidal particle, a historically very important development. As you might expect, the larger the colloidal particle, the shorter its mean free path in a given liquid ( Table 13.6). Today, the understanding of Brownian motion is applied to diverse problems in everything from cheese-making to medical imaging. Table 13.6 Calculated Mean Free Path, after One our, for Uncharged Colloidal Spheres in Water at 20 C Radius of sphere, nm Mean Free Path, mm Sample Integrative Exercise A L solution is made by dissolving g of CaCl 2 1s2 in water. (a) Calculate the osmotic pressure of this solution at 27 C, assuming that it is completely dissociated into its component ions. (b) The measured osmotic pressure of this solution is 2.56 atm at 27 C. Explain why it is less than the value calculated in (a), and calculate the van t off factor, i, for the solute in this solution. (c) The enthalpy of solution for CaCl 2 is = kj>mol. If the final temperature of the solution is 27 C, what was its initial temperature? (Assume that the density of the solution is 1.00 g>ml, that its specific heat is 4.18 J>g@K, and that the solution loses no heat to its surroundings.) Solution Putting Concepts Together (a) The osmotic pressure is given by Equation 13.14, Π = imrt. We know the temperature, T = 27 C = 300 K, and the gas constant, R = L@atm/mol@K. We can calculate the molarity of the solution from the mass of CaCl 2 and the volume of the solution: Molarity = a g CaCl 2 b a 1 mol CaCl 2 b = mol CaCl L 110 g CaCl 2 >L 2 Soluble ionic compounds are strong electrolytes. (Sections 4.1 and 4.3) Thus, CaCl 2 consists of metal cations 1Ca 2+ 2 and nonmetal anions 1Cl - 2. When completely dissociated, each CaCl 2 unit forms three ions (one Ca 2+ and two Cl - ). ence, the calculated osmotic pressure is Π = imrt = mol>l L@atm>mol@K21300 K2 = 2.93 atm (b) The actual values of colligative properties of electrolytes are less than those calculated because the electrostatic interactions between ions limit their independent movements. In this case, the van t off factor, which measures the extent to which electrolytes actually dissociate into ions, is given by Π1measured2 i = Π1calculated for nonelectrolyte atm = mol>l L@atm>mol@K21300 K2 = 2.62 Thus, the solution behaves as if the CaCl 2 has dissociated into 2.62 particles instead of the ideal 3. (c) If the solution is M in CaCl 2 and has a total volume of L, the number of moles of solute is L mol>l2 = mol. ence, the quantity of heat generated in forming the solution is mol kj>mol2 = kj. The solution

31 564 chapter 13 Properties of Solutions absorbs this heat, causing its temperature to increase. The relationship between temperature change and heat is given by Equation 5.22: q = 1specific heat21grams21 T2 The heat absorbed by the solution is q = kj = 323 J. The mass of the L of solution is 1100 ml g>ml2 = 100 g (to three significant figures). Thus, the temperature change is q T = 1specific heat of solution21grams of solution2 323 J = J>g@K21100 g2 = K A kelvin has the same size as a degree Celsius. (Section 1.4) Because the solution temperature increases by C, the initial temperature was 27.0 C C = 26.2 C. Chapter Summary and Key Terms The Solution Process (Section 13.1) Solutions form when one substance disperses uniformly throughout another. The attractive interaction of solvent molecules with solute is called solvation. When the solvent is water, the interaction is called hydration. The dissolution of ionic substances in water is promoted by hydration of the separated ions by the polar water molecules. The overall enthalpy change upon solution formation may be either positive or negative. Solution formation is favored both by a positive entropy change, corresponding to an increased dispersal of the components of the solution, and by a negative enthalpy change, indicating an exothermic process. Saturated Solutions and Solubility (Section 13.2) The equilibrium between a saturated solution and undissolved solute is dynamic; the process of solution and the reverse process, crystallization, occur simultaneously. In a solution in equilibrium with undissolved solute, the two processes occur at equal rates, giving a saturated solution. If there is less solute present than is needed to saturate the solution, the solution is unsaturated. When solute concentration is greater than the equilibrium concentration value, the solution is supersaturated. This is an unstable condition, and separation of some solute from the solution will occur if the process is initiated with a solute seed crystal. The amount of solute needed to form a saturated solution at any particular temperature is the solubility of that solute at that temperature. Factors Affecting Solubility (Section 13.3) The solubility of one substance in another depends on the tendency of systems to become more random, by becoming more dispersed in space, and on the relative intermolecular solute solute and solvent solvent energies compared with solute solvent interactions. Polar and ionic solutes tend to dissolve in polar solvents, and nonpolar solutes tend to dissolve in nonpolar solvents ( like dissolves like ). Liquids that mix in all proportions are miscible; those that do not dissolve significantly in one another are immiscible. ydrogen-bonding interactions between solute and solvent often play an important role in determining solubility; for example, ethanol and water, whose molecules form hydrogen bonds with each other, are miscible. The solubilities of gases in a liquid are generally proportional to the pressure of the gas over the solution, as expressed by enry s law: S g = kp g. The solubilities of most solid solutes in water increase as the temperature of the solution increases. In contrast, the solubilities of gases in water generally decrease with increasing temperature. Expressing Solution Concentrations (Section 13.4) Concentrations of solutions can be expressed quantitatively by several different measures, including mass percentage [(mass solute/mass solution) * 100] parts per million (ppm), parts per billion (ppb), and mole fraction. Molarity, M, is defined as moles of solute per liter of solution; molality, m, is defined as moles of solute per kilogram of solvent. Molarity can be converted to these other concentration units if the density of the solution is known. Colligative Properties (Section 13.5) A physical property of a solution that depends on the concentration of solute particles present, regardless of the nature of the solute, is a colligative property. Colligative properties include vapor-pressure lowering, freezingpoint lowering, boiling-point elevation, and osmotic pressure. Raoult s law expresses the lowering of vapor pressure. An ideal solution obeys Raoult s law. Differences in solvent solute as compared with solvent solvent and solute solute intermolecular forces cause many solutions to depart from ideal behavior. A solution containing a nonvolatile solute possesses a higher boiling point than the pure solvent. The molal boiling-point-elevation constant, K b, represents the increase in boiling point for a 1 m solution of solute particles as compared with the pure solvent. Similarly, the molal freezing-point-depression constant, K f, measures the lowering of the freezing point of a solution for a 1 m solution of solute particles. The temperature changes are given by the equations T b = ik b m and T f = -ik f m where i is the van t off factor, which represents how many particles the solute breaks up into in the solvent. When NaCl dissolves in water, two moles of solute particles are formed for each mole of dissolved salt. The boiling point or freezing point is thus elevated or depressed, respectively, approximately twice as much as that of a nonelectrolyte solution of the same concentration. Similar considerations apply to other strong electrolytes. Osmosis is the movement of solvent molecules through a semipermeable membrane from a less concentrated to a more concentrated solution. This net movement of solvent generates an osmotic pressure, Π, which can be measured in units of gas pressure, such as atm. The osmotic pressure of a solution is proportional to the solution molarity: Π = imrt. Osmosis is a very important process in living systems, in which cell walls act as semipermeable membranes, permitting the passage of water but restricting the passage of ionic and macromolecular components.

32 Key Equations 565 Colloids (Section 13.6) Particles that are large on the molecular scale but still small enough to remain suspended indefinitely in a solvent system form colloids, or colloidal dispersions. Colloids, which are intermediate between solutions and heterogeneous mixtures, have many practical applications. One useful physical property of colloids, the scattering of visible light, is referred to as the Tyndall effect. Aqueous colloids are classified as hydrophilic or hydrophobic. ydrophilic colloids are common in living organisms, in which large molecular aggregates (enzymes, antibodies) remain suspended because they have many polar, or charged, atomic groups on their surfaces that interact with water. ydrophobic colloids, such as small droplets of oil, may remain in suspension through adsorption of charged particles on their surfaces. Colloids undergo Brownian motion in liquids, analogous to the random three-dimensional motion of gas molecules. Learning Outcomes After studying this chapter, you should be able to: Describe how enthalpy and entropy changes affect solution formation. (Section 13.1) Describe the relationship between intermolecular forces and solubility, including use of the like dissolves like rule. (Sections 13.1 and 13.3) Describe the role of equilibrium in the solution process and its relationship to the solubility of a solute. (Section 13.2) Describe the effect of temperature on the solubility of solids and gases in liquids. (Section 13.3) Describe the relationship between the partial pressure of a gas and its solubility. (Section 13.3) Calculate the concentration of a solution in terms of molarity, molality, mole fraction, percent composition, and parts per million and be able to interconvert between them. (Section 13.4) Describe what a colligative property is and explain the difference between the effects of nonelectrolytes and electrolytes on colligative properties. (Section 13.5) Calculate the vapor pressure of a solvent over a solution. (Section 13.5) Calculate the boiling-point elevation and freezing-point depression of a solution. (Section 13.5) Calculate the osmotic pressure of a solution. (Section 13.5) Explain the difference between a solution and a colloid. (Section 13.6) Describe the similarities between the motions of gas molecules and the motions of colloids in a liquid. (Section 13.6) Key Equations S g = kp g [13.4] enry s law, which relates gas solubility to partial pressure Mass % of component = mass of component in soln total mass of soln * 100 [13.5] Concentration in terms of mass percent ppm of component = mass of component in soln total mass of soln * 10 6 [13.6] Concentration in terms of parts per million (ppm) Mole fraction of component = moles of component total moles of all components [13.7] Concentration in terms of mole fraction Molarity = moles of solute liters of soln [13.8] Concentration in terms of molarity Molality = moles of solute kilograms of solvent [13.9] Concentration in terms of molality P solution = X solvent P solvent [13.10] Raoult s law, calculating vapor pressure of solvent above a solution T b = ik b m [13.12] Calculating the boiling-point elevation of a solution T f = -ik f m [13.13] Calculating the freezing-point depression of a solution Π = ia n brt = imrt [13.14] Calculating the osmotic pressure of a solution V

33 566 chapter 13 Properties of Solutions Exercises Visualizing Concepts 13.1 Rank the contents of the following containers in order of increasing entropy: [Section 13.1] 13.6 The solubility of Xe in water at 1 atm pressure and 20 C is approximately 5 * 10-3 M. (a) Compare this with the solubilities of Ar and Kr in water (Table 13.1). (b) What properties of the rare gas atoms account for the variation in solubility? [Section 13.3] 13.7 The structures of vitamins E and B 6 are shown below. Predict which is more water soluble and which is more fat soluble. Explain. [Section 13.3] (a) (b) (c) 13.2 This figure shows the interaction of a cation with surrounding water molecules. + Vitamin B 6 Vitamin E (a) Which atom of water is associated with the cation? Explain. (b) Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li + than for K +? a. Li + is of lower mass than K +. b. The ionization energy of Li is higher than that for K. c. Li + has a smaller ionic radius than K +. d. Li has a lower density than K. e. Li reacts with water more slowly than K. [Section 13.1] 13.3 Consider two ionic solids, both composed of singly-charged ions, that have different lattice energies. (a) Will the solids have the same solubility in water? (b) If not, which solid will be more soluble in water, the one with the larger lattice energy or the one with the smaller lattice energy? Assume that solute-solvent interactions are the same for both solids. [Section 13.1] 13.4 Are gases always miscible with each other? Explain. [Section 13.1] 13.5 Which of the following is the best representation of a saturated solution? Explain your reasoning. [Section 13.2] 13.8 You take a sample of water that is at room temperature and in contact with air and put it under a vacuum. Right away, you see bubbles leave the water, but after a little while, the bubbles stop. As you keep applying the vacuum, more bubbles appear. A friend tells you that the first bubbles were water vapor, and the low pressure had reduced the boiling point of water, causing the water to boil. Another friend tells you that the first bubbles were gas molecules from the air (oxygen, nitrogen, and so forth) that were dissolved in the water. Which friend is mostly likely to be correct? What, then, is responsible for the second batch of bubbles? [Section 13.4] 13.9 The figure shows two identical volumetric flasks containing the same solution at two temperatures. (a) Does the molarity of the solution change with the change in temperature? Explain. (b) Does the molality of the solution change with the change in temperature? Explain. [Section 13.4] (a) (b) (c) 25 C 55 C

34 Exercises This portion of a phase diagram shows the vapor-pressure curves of a volatile solvent and of a solution of that solvent containing a nonvolatile solute. (a) Which line represents the solution? (b) What are the normal boiling points of the solvent and the solution? [Section 13.5] P (atm) T ( C) Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a 0.2 M solution of some solute and is submerged in a 0.1 M solution of the same solute: 0.1 M 0.2 M Initially, the volume of solution in the balloon is 0.25 L. Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5] The molecule n-octylglucoside, shown here, is widely used in biochemical research as a nonionic detergent for solubilizing large hydrophobic protein molecules. What characteristics of this molecule are important for its use in this way? [Section 13.6] The Solution Process (Section 13.1) Indicate whether each statement is true or false: (a) A solute will dissolve in a solvent if solute solute interactions are stronger than solute-solvent interactions. (b) In making a solution, the enthalpy of mixing is always a positive number. (c) An increase in entropy favors mixing Indicate whether each statement is true or false: (a) NaCl dissolves in water but not in benzene 1C because benzene is denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene has zero dipole moment. (c) NaCl dissolves in water but not in benzene because the water ion interactions are stronger than benzene ion interactions Indicate the type of solute solvent interaction (Section 11.2) that should be most important in each of the following solutions: (a) CCl 4 in benzene 1C 6 6 2, (b) methanol 1C 3 O2 in water, (c) KBr in water, (d) Cl in acetonitrile 1C 3 CN Indicate the principal type of solute solvent interaction in each of the following solutions and rank the solutions from weakest to strongest solute solvent interaction: (a) KCl in water, (b) C 2 Cl 2 in benzene 1C 6 6 2, (c) methanol 1C 3 O2 in water An ionic compound has a very negative soln in water. (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: solvent, solute, or mix? When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form? (a) In Equation 13.1, which of the enthalpy terms for dissolving an ionic solid would correspond to the lattice energy? (b) Which energy term in this equation is always exothermic? For the dissolution of LiCl in water, soln = -37 kj>mol. Which term would you expect to be the largest negative number: solvent, solute, or mix? Two nonpolar organic liquids, hexane 1C and heptane 1C , are mixed. (a) Do you expect soln to be a large positive number, a large negative number, or close to zero? Explain. (b) exane and heptane are miscible with each other in all proportions. In making a solution of them, is the entropy of the system increased, decreased, or close to zero, compared to the separate pure liquids? The enthalpy of solution of KBr in water is about +198 kj>mol. Nevertheless, the solubility of KBr in water is relatively high. Why does the solution process occur even though it is endothermic? Saturated Solutions; Factors Affecting Solubility (Sections 13.2 and 13.3) The solubility of Cr1NO # 9 2 O in water is 208 g per 100 g of water at 15 C. A solution of Cr1NO # 9 2 O in water at 35 C is formed by dissolving 324 g in 100 g of water. When this solution is slowly cooled to 15 C, no precipitate forms. (a) What term describes this solution? (b) What action might you take to initiate crystallization? Use molecular-level processes to explain how your suggested procedure works The solubility of MnSO 4 # 2 O in water at 20 C is 70 g per 100 ml of water. (a) Is a 1.22 M solution of MnSO 4 # 2 O in water at 20 C saturated, supersaturated, or unsaturated? (b) Given a solution of MnSO 4 # 2 O of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

35 568 chapter 13 Properties of Solutions By referring to Figure 13.15, determine whether the addition of 40.0 g of each of the following ionic solids to 100 g of water at 40 C will lead to a saturated solution: (a) NaNO 3, (b) KCl, (c) K 2 Cr 2 O 7, (d) Pb1NO By referring to Figure 13.15, determine the mass of each of the following salts required to form a saturated solution in 250 g of water at 30 C: (a) KClO 3, (b) Pb1NO 3 2 2, (c) Ce 2 1SO Consider water and glycerol, C 2 1O2C1O2C 2 O. (a) Would you expect them to be miscible in all proportions? Explain. (b) List the intermolecular attractions that occur between a water molecule and a glycerol molecule Oil and water are immiscible. Which is the most likely reason? (a) Oil molecules are denser than water. (b) Oil molecules are composed mostly of carbon and hydrogen. (c) Oil molecules have higher molar masses than water. (d) Oil molecules have higher vapor pressures than water. (e) Oil molecules have higher boiling points than water Common laboratory solvents include acetone 1C 3 COC 3 2, methanol 1C 3 O2, toluene 1C 6 5 C 3 2, and water. Which of these is the best solvent for nonpolar solutes? Would you expect alanine (an amino acid) to be more soluble in water or in hexane? Explain. 1C or glycerol, C 2 1O2C1O2C 2 O, (c) octanoic acid, C 3 C 2 C 2 C 2 C 2 C 2 C 2 COO, or acetic acid, C 3 COO? Explain your answer in each case Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane 1C or glucose 1C 6 12 O 6 2, (b) propionic acid 1C 3 C 2 COO2 or sodium propionate 1C 3 C 2 COONa2, (c) Cl or ethyl chloride 1C 3 C 2 Cl2? Explain in each case (a) Explain why carbonated beverages must be stored in sealed containers. (b) Once the beverage has been opened, why does it maintain more carbonation when refrigerated than at room temperature? Explain why pressure substantially affects the solubility of O 2 in water but has little effect on the solubility of NaCl in water The enry s law constant for helium gas in water at 30 C is 3.7 * 10-4 M>atm and the constant for N 2 at 30 C is 6.0 * 10-4 M>atm. If the two gases are each present at 1.5 atm pressure, calculate the solubility of each gas The partial pressure of O 2 in air at sea level is 0.21 atm. Using the data in Table 13.1, together with enry s law, calculate the molar concentration of O 2 in the surface water of a mountain lake saturated with air at 20 C and an atmospheric pressure of 650 torr. Concentrations of Solutions (Section 13.4) Alanine (a) Would you expect stearic acid, C 3 1C COO, to be more soluble in water or in carbon tetrachloride? Explain. (b) Which would you expect to be more soluble in water, cyclohexane or dioxane? Explain. 2 C O C 2 2 C C 2 O Dioxane 2 C C 2 C 2 2 C C 2 C 2 Cyclohexane Ibuprofen, widely used as a pain reliever, has a limited solubility in water, less than 1 mg>ml. Which part of the molecule s structure (gray, white, red) contributes to its water solubility? Which part of the molecule (gray, white, red) contributes to its water insolubility? Ibuprofen Which of the following in each pair is likely to be more soluble in hexane, C 6 14 : (a) CCl 4 or CaCl 2, (b) benzene (a) Calculate the mass percentage of Na 2 SO 4 in a solution containing 10.6 g of Na 2 SO 4 in 483 g of water. (b) An ore contains 2.86 g of silver per ton of ore. What is the concentration of silver in ppm? (a) What is the mass percentage of iodine in a solution containing mol I 2 in 125 g of CCl 4? (b) Seawater contains g of Sr 2+ per kilogram of water. What is the concentration of Sr 2+ in ppm? A solution is made containing 14.6 g of C 3 O in 184 g of 2 O. Calculate (a) the mole fraction of C 3 O, (b) the mass percent of C 3 O, (c) the molality of C 3 O A solution is made containing 20.8 g of phenol 1C 6 5 O2 in 425 g of ethanol 1C 3 C 2 O2. Calculate (a) the mole fraction of phenol, (b) the mass percent of phenol, (c) the molality of phenol Calculate the molarity of the following aqueous solutions: (a) g of Mg1NO in ml of solution, (b) 22.4 g of LiClO 4 # 3 2 O in 125 ml of solution, (c) 25.0 ml of 3.50 M NO 3 diluted to L What is the molarity of each of the following solutions: (a) 15.0 g of Al 2 1SO in ml solution, (b) 5.25 g of Mn1NO # 2 2 O in 175 ml of solution, (c) 35.0 ml of 9.00 M 2 SO 4 diluted to L? Calculate the molality of each of the following solutions: (a) 8.66 g of benzene 1C dissolved in 23.6 g of carbon tetrachloride 1CCl 4 2, (b) 4.80 g of NaCl dissolved in L of water (a) What is the molality of a solution formed by dissolving 1.12 mol of KCl in 16.0 mol of water? (b) ow many grams of sulfur 1S 8 2 must be dissolved in g of naphthalene 1C to make a 0.12 m solution? A sulfuric acid solution containing g of 2 SO 4 per liter of solution has a density of g>cm 3. Calculate

36 Exercises 569 (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of 2 SO 4 in this solution Ascorbic acid 1vitamin C, C 6 8 O 6 2 is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g>ml at 55 C. Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of ascorbic acid in this solution The density of acetonitrile 1C 3 CN2 is g>ml and the density of methanol 1C 3 O2 is g>ml. A solution is made by dissolving 22.5 ml of C 3 O in 98.7 ml of C 3 CN. (a) What is the mole fraction of methanol in the solution? (b) What is the molality of the solution? (c) Assuming that the volumes are additive, what is the molarity of C 3 O in the solution? The density of toluene 1C is g>ml, and the density of thiophene 1C 4 4 S2 is g>ml. A solution is made by dissolving 8.10 g of thiophene in ml of toluene. (a) Calculate the mole fraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution? Calculate the number of moles of solute present in each of the following aqueous solutions: (a) 600 ml of M SrBr 2, (b) 86.4 g of m KCl, (c) g of a solution that is 6.45% glucose 1C 6 12 O 6 2 by mass Calculate the number of moles of solute present in each of the following solutions: (a) 255 ml of 1.50 M NO 3 1aq2, (b) 50.0 mg of an aqueous solution that is 1.50 m NaCl, (c) 75.0 g of an aqueous solution that is 1.50% sucrose 1C O 11 2 by mass Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of 1.5 * 10-2 M KBr, (b) 125 g of m KBr, (c) 1.85 L of a solution that is 12.0% KBr by mass (the density of the solution is 1.10 g>ml), (d) a M solution of KBr that contains just enough KBr to precipitate 16.0 g of AgBr from a solution containing mol of AgNO Describe how you would prepare each of the following aqueous solutions: (a) 1.50 L of M 1N SO 4 solution, starting with solid 1N SO 4 ; (b) 225 g of a solution that is 0.65 m in Na 2 CO 3, starting with the solid solute; (c) 1.20 L of a solution that is 15.0% Pb1NO by mass (the density of the solution is 1.16 g>ml), starting with solid solute; (d) a 0.50 M solution of Cl that would just neutralize 5.5 g of Ba1O2 2 starting with 6.0 M Cl Commercial aqueous nitric acid has a density of 1.42 g>ml and is 16 M. Calculate the percent NO 3 by mass in the solution Commercial concentrated aqueous ammonia is 28% N 3 by mass and has a density of 0.90 g>ml. What is the molarity of this solution? Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of 80.0% Cu and 20.0% Zn by mass has a density of 8750 kg>m 3. (a) What is the molality of Zn in the solid solution? (b) What is the molarity of Zn in the solution? Caffeine 1C 8 10 N 4 O 2 2 is a stimulant found in coffee and tea. If a solution of caffeine in the solvent chloroform 1CCl 3 2 has a concentration of m, calculate (a) the percentage of caffeine by mass, (b) the mole fraction of caffeine in the solution. Caffeine During a person s typical breathing cycle, the CO 2 concentration in the expired air rises to a peak of 4.6% by volume. (a) Calculate the partial pressure of the CO 2 in the expired air at its peak, assuming 1 atm pressure and a body temperature of 37 C. (b) What is the molarity of the CO 2 in the expired air at its peak, assuming a body temperature of 37 C? Breathing air that contains 4.0% by volume CO 2 over time causes rapid breathing, throbbing headache, and nausea, among other symptoms. What is the concentration of CO 2 in such air in terms of (a) mol percentage, (b) molarity, assuming 1 atm pressure and a body temperature of 37 C? Colligative Properties (Section 13.5) You make a solution of a nonvolatile solute with a liquid solvent. Indicate whether each of the following statements is true or false. (a) The freezing point of the solution is higher than that of the pure solvent. (b) The freezing point of the solution is lower than that of the pure solvent. (c) The boiling point of the solution is higher than that of the pure solvent. (d) The boiling point of the solution is lower than that of the pure solvent You make a solution of a nonvolatile solute with a liquid solvent. Indicate if each of the following statements is true or false. (a) The freezing point of the solution is unchanged by addition of the solvent. (b) The solid that forms as the solution freezes is nearly pure solute. (c) The freezing point of the solution is independent of the concentration of the solute. (d) The boiling point of the solution increases in proportion to the concentration of the solute. (e) At any temperature, the vapor pressure of the solvent over the solution is lower than what it would be for the pure solvent Consider two solutions, one formed by adding 10 g of glucose 1C 6 12 O 6 2 to 1 L of water and the other formed by adding 10 g of sucrose 1C O 11 2 to 1 L of water. Calculate the vapor pressure for each solution at 20 C; the vapor pressure of pure water at this temperature is 17.5 torr (a) What is an ideal solution? (b) The vapor pressure of pure water at 60 C is 149 torr. The vapor pressure of water over a solution at 60 C containing equal numbers of moles of water and ethylene glycol (a nonvolatile solute) is 67 torr. Is the solution ideal according to Raoult s law? Explain (a) Calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose 1C O 11 2 to g of water at 338 K. (Vapor-pressure data for water are given in Appendix B.) (b) Calculate the mass of propylene glycol 1C 3 8 O 2 2 that must be added to kg of water to reduce the vapor pressure by 2.88 torr at 40 C.

37 570 chapter 13 Properties of Solutions (a) Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin 1C 3 8 O 3 2 in 125 g of water at 343 K. (The vapor pressure of water is given in Appendix B.) (b) Calculate the mass of ethylene glycol 1C 2 6 O 2 2 that must be added to 1.00 kg of ethanol 1C 2 5 O2 to reduce its vapor pressure by 10.0 torr at 35 C. The vapor pressure of pure ethanol at 35 C is 1.00 * 10 2 torr. [13.67] At 63.5 C, the vapor pressure of 2 O is 175 torr, and that of ethanol 1C 2 5 O2 is 400 torr. A solution is made by mixing equal masses of 2 O and C 2 5 O. (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal-solution behavior, what is the vapor pressure of the solution at 63.5 C? (c) What is the mole fraction of ethanol in the vapor above the solution? [13.68] At 20 C, the vapor pressure of benzene 1C is 75 torr, and that of toluene 1C is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of 35 torr at 20 C? (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)? (a) Does a 0.10 m aqueous solution of NaCl have a higher boiling point, a lower boiling point, or the same boiling point as a 0.10 m aqueous solution of C 6 12 O 6? (b) The experimental boiling point of the NaCl solution is lower than that calculated assuming that NaCl is completely dissociated in solution. Why is this the case? Arrange the following aqueous solutions, each 10% by mass in solute, in order of increasing boiling point: glucose 1C 6 12 O 6 2, sucrose 1C O 11 2, sodium nitrate 1NaNO List the following aqueous solutions in order of increasing boiling point: m glucose, m LiBr, m Zn1NO List the following aqueous solutions in order of decreasing freezing point: m glycerin 1C 3 8 O 3 2, m KBr, m phenol 1C 6 5 O Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol 1C 3 8 O 3 2 in ethanol, (b) mol of naphthalene 1C in 2.45 mol of chloroform, (c) 1.50 g NaCl in kg of water, (d) 2.04 g KBr and 4.82 g glucose 1C 6 12 O 6 2 in 188 g of water Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.25 m glucose in ethanol; (b) 20.0 g of decane, C 10 22, in 50.0 g CCl 3 ; (c) 3.50 g NaO in 175 g of water, (d) 0.45 mol ethylene glycol and 0.15 mol KBr in 150 g 2 O ow many grams of ethylene glycol 1C 2 6 O 2 2 must be added to 1.00 kg of water to produce a solution that freezes at C? What is the freezing point of an aqueous solution that boils at C? What is the osmotic pressure formed by dissolving 44.2 mg of aspirin 1C 9 8 O 4 2 in L of water at 25 C? Seawater contains 3.4 g of salts for every liter of solution. Assuming that the solute consists entirely of NaCl (in fact, over 90% of the salt is indeed NaCl), calculate the osmotic pressure of seawater at 20 C Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of 0.64 g of adrenaline in 36.0 g of CCl 4 elevates the boiling point by 0.49 C. Calculate the approximate molar mass of adrenaline from this data. Adrenaline Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in kg of benzene freezes at 4.1 C. What is the molar mass of lauryl alcohol from this data? Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing g of this enzyme in 210 ml of solution has an osmotic pressure of torr at 25 C. What is the molar mass of lysozyme? A dilute aqueous solution of an organic compound soluble in water is formed by dissolving 2.35 g of the compound in water to form L of solution. The resulting solution has an osmotic pressure of atm at 25 C. Assuming that the organic compound is a nonelectrolyte, what is its molar mass? [13.83] The osmotic pressure of a M aqueous solution of CaCl 2 is found to be atm at 25 C. (a) Calculate the van t off factor, i, for the solution. (b) ow would you expect the value of i to change as the solution becomes more concentrated? Explain. [13.84] Based on the data given in Table 13.4, which solution would give the larger freezing-point lowering, a m solution of NaCl or a m solution of K 2 SO 4? ow do you explain the departure from ideal behavior and the differences observed between the two salts? Colloids (Section 13.6) (a) Do colloids made only of gases exist? Why or why not? (b) In the 1850 s, Michael Faraday prepared ruby-red colloids of gold nanoparticles in water that are still stable today. These brightly colored colloids look like solutions. What experiment(s) could you do to determine whether a given colored preparation is a solution or colloid? Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol An emulsifying agent is a compound that helps stabilize a hydrophobic colloid in a hydrophilic solvent (or a hydrophilic colloid in a hydrophobic solvent). Which of the following choices is the best emulsifying agent? (a) C 3 COO, (b) C 3 C 2 C 2 COO, (c) C 3 1C COO, (d) C 3 1C COONa Aerosols are important components of the atmosphere. Does the presence of aerosols in the atmosphere increase or decrease the amount of sunlight that arrives at the Earth s surface, compared to an aerosol-free atmosphere? Explain your reasoning. [13.89] Proteins can be precipitated out of aqueous solution by the addition of an electrolyte; this process is called salting out

38 additional Exercises 571 the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of the same electrolyte? (b) If a protein has been salted out, are the protein protein interactions stronger or weaker than they were before the electrolyte was added? (c) A friend of yours who is taking a biochemistry class says that salting out works because the waters of hydration that surround the protein prefer to surround the electrolyte as the electrolyte is added; therefore, the protein s hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the same biochemistry class says that salting out works because the incoming ions adsorb tightly to the protein, making ion pairs on the protein surface, which end up giving the protein a zero net charge in water and therefore leading to precipitation. Discuss these two hypotheses. What kind of measurements would you need to make to distinguish between these two hypotheses? Explain how (a) a soap such as sodium stearate stabilizes a colloidal dispersion of oil droplets in water; (b) milk curdles upon addition of an acid. Additional Exercises Butylated hydroxytoluene (BT) has the following molecular structure: 3 C C 3 C C 3 O BT C 3 C 3 C C 3 C 3 It is widely used as a preservative in a variety of foods, including dried cereals. Based on its structure, would you expect BT to be more soluble in water or in hexane 1C ? Explain A saturated solution of sucrose 1C O 11 2 is made by dissolving excess table sugar in a flask of water. There are 50 g of undissolved sucrose crystals at the bottom of the flask in contact with the saturated solution. The flask is stoppered and set aside. A year later a single large crystal of mass 50 g is at the bottom of the flask. Explain how this experiment provides evidence for a dynamic equilibrium between the saturated solution and the undissolved solute Most fish need at least 4 ppm dissolved O 2 in water for survival. (a) What is this concentration in mol>l? (b) What partial pressure of O 2 above water is needed to obtain 4 ppm O 2 in water at 10 C? (The enry s law constant for O 2 at this temperature is 1.71 * 10-3 mol>l@atm.) The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United States. (a) Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 C is 7.27 * 10-3 M, what is the enry s law constant for radon in water at this temperature? (b) A sample consisting of various gases contains 3.5 * 10-6 mole fraction of radon. This gas at a total pressure of 32 atm is shaken with water at 30 C. Calculate the molar concentration of radon in the water Glucose makes up about 0.10% by mass of human blood. Calculate this concentration in (a) ppm, (b) molality. (c) What further information would you need to determine the molarity of the solution? The concentration of gold in seawater has been reported to be between 5 ppt (parts per trillion) and 50 ppt. Assuming that seawater contains 13 ppt of gold, calculate the number of grams of gold contained in 1.0 * 10 3 gal of seawater The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0- ppb solution. (b) ow many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 m 3 of water? Acetonitrile 1C 3 CN2 is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1.80 M LiBr solution in acetonitrile is g>cm 3. Calculate the concentration of the solution in (a) molality, (b) mole fraction of LiBr, (c) mass percentage of C 3 CN A canned heat product used to warm buffet dishes consists of a homogeneous mixture of ethanol 1C 2 5 O2 and paraffin, which has an average formula of C What mass of C 2 5 O should be added to 620 kg of the paraffin to produce 8 torr of ethanol vapor pressure at 35 C? The vapor pressure of pure ethanol at 35 C is 100 torr A solution contains mol 2 O and an unknown number of moles of sodium chloride. The vapor pressure of the solution at 30 C is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (int: Remember that sodium chloride is a strong electrolyte.) [13.101] Two beakers are placed in a sealed box at 25 C. One beaker contains 30.0 ml of a M aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains 30.0 ml of a M aqueous solution of NaCl. The water vapor from the two solutions reaches equilibrium. (a) In which beaker does the solution level rise, and in which one does it fall? (b) What are the volumes in the two beakers when equilibrium is attained, assuming ideal behavior? A car owner who knows no chemistry has to put antifreeze in his car s radiator. The instructions recommend a mixture of 30% ethylene glycol and 70% water. Thinking he will improve his protection he uses pure ethylene glycol, which is a liquid at room temperature. e is saddened to find that the solution does not provide as much protection as he hoped. The pure ethylene glycol freezes solid in his radiator on a very cold day, while his neighbor, who did use the 30/70 mixture, has no problem. Suggest an explanation Calculate the freezing point of a m aqueous solution of K 2 SO 4, (a) ignoring interionic attractions, and (b) taking interionic attractions into consideration by using the van t off factor (Table 13.4) Carbon disulfide 1CS 2 2 boils at C and has a density of g>ml. (a) When mol of a nondissociating solute

39 572 chapter 13 Properties of Solutions is dissolved in ml of CS 2, the solution boils at C. What is the molal boiling-point-elevation constant for CS 2? (b) When 5.39 g of a nondissociating unknown is dissolved in 50.0 ml of CS 2, the solution boils at C. What is the molecular weight of the unknown? [13.105] A lithium salt used in lubricating grease has the formula LiC n 2n + 1 O 2. The salt is soluble in water to the extent of g per 100 g of water at 25 C. The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of n in the formula for the salt. Integrative Exercises Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 C, and their solubilities in water at 25 C and 1 atm fluorocarbon pressure are given as mass percentages. (a) For each fluorocarbon, calculate the molality of a saturated solution. (b) Explain why the molarity of each of the solutions should be very close numerically to the molality. (c) Based on their molecular structures, account for the differences in solubility of the four fluorocarbons. (d) Calculate the enry s law constant at 25 C for CClF 2, and compare its magnitude to that for N * 10-4 mol>l@atm2. Suggest a reason for the difference in magnitude. Fluorocarbon Solubility (mass %) CF CClF CCl 2 F CClF [13.107] At ordinary body temperature 137 C2, the solubility of N 2 in water at ordinary atmospheric pressure (1.0 atm) is g>l. Air is approximately 78 mol % N 2. (a) Calculate the number of moles of N 2 dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N 2 from air in blood at this pressure? (c) If a scuba diver suddenly surfaces from this depth, how many milliliters of N 2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood? [13.108] Consider the following values for enthalpy of vaporization 1kJ>mol2 of several organic substances: O C 3 C 30.4 Acetaldehyde O 2 C C Ethylene oxide O C 3 CC Acetone C 2 2 C C 2 Cyclopropane 24.7 (a) Account for the variations in heats of vaporization for these substances, considering their relative intermolecular forces. (b) ow would you expect the solubilities of these substances to vary in hexane as solvent? In ethanol? Use intermolecular forces, including hydrogen-bonding interactions where applicable, to explain your responses. [13.109] A textbook on chemical thermodynamics states, The heat of solution represents the difference between the lattice energy of the crystalline solid and the solvation energy of the gaseous ions. (a) Draw a simple energy diagram to illustrate this statement. (b) A salt such as NaBr is insoluble in most polar nonaqueous solvents such as acetonitrile 1C 3 CN2 or nitromethane 1C 3 NO 2 2, but salts of large cations, such as tetramethylammonium bromide 31C NBr4, are generally more soluble. Use the thermochemical cycle you drew in part (a) and the factors that determine the lattice energy (Section 8.2) to explain this fact (a) A sample of hydrogen gas is generated in a closed container by reacting g of zinc metal with 15.0 ml of 1.00 M sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is 122 ml. Calculate the partial pressure of the hydrogen gas in this volume at 25 C, ignoring any solubility of the gas in the solution. (c) The enry s law constant for hydrogen in water at 25 C is 7.8 * 10-4 mol>l@atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)? [13.111] The following table presents the solubilities of several gases in water at 25 C under a total pressure of gas and water vapor of 1 atm. (a) What volume of C 4 1g2 under standard conditions of temperature and pressure is contained in 4.0 L of a saturated solution at 25 C? (b) Explain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecular structures and intermolecular forces. (c) Compare the solubilities of O 2, N 2, and NO, and account for the variations based on molecular structures and intermolecular forces. (d) Account for the much larger values observed for 2 S and SO 2 as compared with the other gases listed. (e) Find several pairs of substances with the same or nearly the same molecular masses (for example, C 2 4 and N 2 ), and use intermolecular interactions to explain the differences in their solubilities.

40 Design an Experiment 573 Gas Solubility (mm) C 4 1methane2 1.3 C 2 6 1ethane2 1.8 C 2 4 1ethylene2 4.7 N O NO S 99 SO A small cube of lithium 1density = g/cm 3 2 measuring 1.0 mm on each edge is added to L of water. The following reaction occurs: 2 Li1s O1l2 2 LiO1aq g2 What is the freezing point of the resultant solution, assuming that the reaction goes to completion? [13.113] At 35 C the vapor pressure of acetone, 1C CO, is 360 torr, and that of chloroform, CCl 3, is 300 torr. Acetone and chloroform can form very weak hydrogen bonds between one another; the chlorines on the carbon give the carbon a sufficient partial positive charge to enable this behavior: Cl Cl C Cl C 3 C 3 A solution composed of an equal number of moles of acetone and chloroform has a vapor pressure of 250 torr at 35 C. (a) What would be the vapor pressure of the solution if it exhibited ideal behavior? (b) Use the existence of hydrogen O C bonds between acetone and chloroform molecules to explain the deviation from ideal behavior. (c) Based on the behavior of the solution, predict whether the mixing of acetone and chloroform is an exothermic 1 soln 6 02 or endothermic 1 soln 7 02 process. (d) Would you expect the same vaporpressure behavior for acetone and chloromethane (C 3 Cl)? Explain Compounds like sodium stearate, called surfactants in general, can form structures known as micelles in water, once the solution concentration reaches the value known as the critical micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cmc depends on the substance, the solvent, and the temperature. Surfactant tail Surfactant head Surfactant monomers cmc Micelle At and above the cmc, the properties of the solution vary drastically. (a) The turbidity (the amount of light scattering) of solutions increases dramatically at the cmc. Suggest an explanation. (b) The ionic conductivity of the solution dramatically changes at the cmc. Suggest an explanation. (c) Chemists have developed fluorescent dyes that glow brightly only when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would relate to the concentration of sodium stearate as the sodium stearate concentration approaches and then increases past the cmc. Design an Experiment Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is because the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a common misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents.

41 14 Chemical Kinetics Chemical reactions take time to occur. Some reactions, such as the rusting of iron or the changing of color in leaves, occur relatively slowly, requiring days, months, or years to complete. Others, such as the combustion reaction that generates the thrust for a rocket, as in the chapter-opening photograph, happen much more rapidly. As chemists, we need to be concerned about the speed with which chemical reactions occur as well as the products of the reactions. For example, the chemical reactions that govern the metabolism of food, the transport of essential nutrients, and your body s ability to adjust to temperature changes (see the Chemistry and Life box on the regulation of body temperature in Section 5.5) all require that reactions occur with the appropriate speed. Indeed, considerations of the speeds of reactions are among the most important aspects of designing new chemistry and chemical processes. The area of chemistry concerned with the speeds, or rates, of reactions is chemical kinetics. So far, we have focused on the beginning and end of chemical reactions: We start with certain reactants and see what products they yield. This view is useful but does not tell us what happens in the middle that is, which chemical bonds are broken, which are formed, and in what order these events occur. The speed at which a chemical reaction occurs is called the reaction rate. Reaction rates can occur over very different time Launch of the Juno Spacecraft at Cape Canaveral, Florida, in August The spacecraft is mounted to an Atlas V rocket, which at launch uses the very rapid combustion of kerosene and liquid oxygen to generate its thrust. What s Ahead 14.1 Factors that Affect Reaction Rates We see that four variables affect reaction rates: concentration, physical states of reactants, temperature, and presence of catalysts. These factors can be understood in terms of the collisions among reactant molecules that lead to reaction Reaction Rates We examine how to express reaction rates and how reactant disappearance rates and product appearance rates are related to the reaction stoichiometry Concentration and Rate Laws We show that the effect of concentration on rate is expressed quantitatively by rate laws and show how rate laws and rate constants are determined experimentally The Change of Concentration with Time We learn that rate equations can be written to express how concentrations change with time and look at several classifications of rate equations: zero-order, first-order, and second-order reactions.

42 14.5 Temperature and Rate We explore the effect of temperature on rate. In order to occur, most reactions require a minimum input of energy called the activation energy Reaction Mechanisms We look more closely at reaction mechanisms, the step-by-step molecular pathways leading from reactants to products Catalysis We end the chapter with a discussion of how catalysts increase reaction rates, including a discussion of biological catalysts called enzymes.

43 576 Energy chapter 14 Chemical Kinetics s 109 s (30 years) 1s 1015 s (30 million years) Time scale Go Figure If a heated steel nail were placed in pure O2, would you expect it to burn as readily as the steel wool does? Steel wool heated in air (about 20% O2) glows red-hot but oxidizes to Fe2O3 slowly Figure 14.1 Reaction rates span an enormous range of time scales. The absorption of light by an atom or a molecule is complete within one femtosecond; explosions occur within seconds; corrosion can occur over years; and the weathering of rocks can occur over millions of years. scales ( Figure 14.1). To investigate how reactions happen, we must examine the reaction rates and the factors that influence them. Experimental information on the rate of a given reaction provides important evidence that helps us formulate a reaction mechanism, which is a step-by-step, molecular-level view of the pathway from reactants to products. Our goal in this chapter is to understand how to determine reaction rates and to consider the factors that control these rates. What factors determine how rapidly food spoils, for instance? ow does one design an automotive airbag that fills extremely rapidly following a car crash? What determines the rate at which steel rusts? ow can we remove hazardous pollutants in automobile exhaust before the exhaust leaves the tailpipe? Although we will not address these specific questions, we will see that the rates of all chemical reactions are subject to the same principles Factors that Affect Reaction Rates Four factors affect the rate at which any particular reaction occurs: Red-hot steel wool in 100% O2 burns vigorously, forming Fe2O3 quickly Figure 14.2 Effect of concentration on reaction rate. The difference in behavior is due to the different concentrations of O2 in the two environments. 1. Physical state of the reactants. Reactants must come together to react. The more readily reactant molecules collide with one another, the more rapidly they react. Reactions may broadly classified as being either homogeneous, involving either all gases or all liquids, or as heterogeneous, in which reactants are in different phases. Under heterogeneous conditions, a reaction is limited by the area of contact of the reactants. Thus, heterogeneous reactions that involve solids tend to proceed more rapidly if the surface area of the solid is increased. For example, a medicine in the form of a fine powder dissolves in the stomach and enters the blood more quickly than the same medicine in the form of a tablet. 2. Reactant concentrations. Most chemical reactions proceed more quickly if the concentration of one or more reactants is increased. For example, steel wool burns only slowly in air, which contains 20% O2, but bursts into flame in pure oxygen ( Figure 14.2). As reactant concentration increases, the frequency with which the reactant molecules collide increases, leading to increased rates. 3. Reaction temperature. Reaction rates generally increase as temperature is increased. The bacterial reactions that spoil milk, for instance, proceed more rapidly at room temperature than at the lower temperature of a refrigerator. Increasing (Section 10.7) As moltemperature increases the kinetic energies of molecules. ecules move more rapidly, they collide more frequently and with higher energy, leading to increased reaction rates.

44 section 14.2 Reaction Rates The presence of a catalyst. Catalysts are agents that increase reaction rates without themselves being used up. They affect the kinds of collisions (and therefore alter the mechanism) that lead to reaction. Catalysts play many crucial roles in living organisms, including ourselves. On a molecular level, reaction rates depend on the frequency of collisions between molecules. The greater the frequency of collisions, the higher the reaction rate. For a collision to lead to a reaction, however, it must occur with sufficient energy to break bonds and with suitable orientation for new bonds to form in the proper locations. We will consider these factors as we proceed through this chapter. Give It Some Thought In a reaction involving reactants in the gas state, how does increasing the partial pressures of the gases affect the reaction rate? 14.2 Reaction Rates The speed of an event is defined as the change that occurs in a given time interval, which means that whenever we talk about speed, we necessarily bring in the notion of time. For example, the speed of a car is expressed as the change in the car s position over a certain time interval. In the United States, the speed of cars is usually measured in units of miles per hour that is, the quantity that is changing (position measured in miles) divided by a time interval (measured in hours). Similarly, the speed of a chemical reaction its reaction rate is the change in the concentration of reactants or products per unit of time. The units for reaction rate are usually molarity per second 1M>s2 that is, the change in concentration measured in molarity divided by a time interval measured in seconds. Let s consider the hypothetical reaction A B, depicted in Figure Each red sphere represents 0.01 mol of A, each blue sphere represents 0.01 mol of B, and the container has a volume of 1.00 L. At the beginning of the reaction, there is 1.00 mol A, so the concentration is 1.00 mol >L = 1.00 M. After 20 s, the concentration of A has fallen to 0.54 M and the concentration of B has risen to 0.46 M. The sum of the concentrations is still 1.00 M because 1 mol of B is produced for each mole of A that reacts. After 40 s, the concentration of A is 0.30 M and that of B is 0.70 M. Go Figure Estimate the number of moles of A in the mixture after 30 s. 0 s 20 s 40 s 1.00 mol A 0 mol B 0.54 mol A 0.46 mol B 0.30 mol A 0.70 mol B Figure 14.3 Progress of a hypothetical reaction a B. The volume of the flask is 1.0 L.

45 578 chapter 14 Chemical Kinetics The rate of this reaction can be expressed either as the rate of disappearance of reactant A or as the rate of appearance of product B. The average rate of appearance of B over a particular time interval is given by the change in concentration of B divided by the change in time: Average rate of appearance of B = change in concentration of B change in time = 3B4 at t 2-3B4 at t 1 t 2 - t 1 = 3B4 t [14.1] We use brackets around a chemical formula, as in [B], to indicate molarity. The Greek letter delta,, is read change in and is always equal to a final value minus an initial value. (Equation 5.4, Section 5.2) The average rate of appearance of B over the 20-s interval from the beginning of the reaction 1t 1 = 0 s to t 2 = 20 s2 is 0.46 M M Average rate = = 2.3 * 10-2 M>s 20 s - 0 s We could equally well express the reaction rate in term of the reactant, A. In this case, we would be describing the rate of disappearance of A, which we express as change in concentration of A Average rate of disappearance of A = - change in time = - 3A4 [14.2] t Notice the minus sign in this equation, which we use to indicate that the concentration of A decreases. By convention, rates are always expressed as positive quantities. Because [A] decreases, 3A4 is a negative number. The minus sign we put in the equation converts the negative 3A4 to a positive rate of disappearance. Because one molecule of A is consumed for every molecule of B that forms, the average rate of disappearance of A equals the average rate of appearance of B: Average rate = - 3A4 t = M M 20 s - 0 s = 2.3 * 10-2 M>s Sample Exercise 14.1 Calculating an Average Rate of Reaction From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s. Solution Analyze We are given the concentration of A at 20 s M2 and at 40 s M2 and asked to calculate the average rate of reaction over this time interval. Plan The average rate is given by the change in concentration, 3A4, divided by the change in time, t. Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity. Solve Average rate = - 3A M M = - = 1.2 * 10-2 M>s t 40 s - 20 s Practice Exercise 1 If the experiment in Figure 14.3 is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? (i) After 60 s there are 0.84 mol B in the flask. (ii) The decrease in the number of moles of A from t 1 = 0 s to t 2 = 20 s is greater than that from t 1 = 40 to t 2 = 60 s. (iii) The average rate for the reaction from t 1 = 40 s to t 2 = 60 s is 7.0 * 10-3 M>s. (a) Only one of the statements is true. (c) Statements (i) and (iii) are true. (e) All three statements are true. (b) Statements (i) and (ii) are true. (d) Statements (ii) and (iii) are true. Practice Exercise 2 Use the data in Figure 14.3 to calculate the average rate of appearance of B over the time interval from 0 s to 40 s.

46 section 14.2 Reaction Rates 579 Table 14.1 Rate Data for Reaction of C 4 h 9 Cl with Water Time, t 1s2 3C 4 h 9 Cl4 1M2 Average Rate 1M,s * * * * * * * * ,000 0 Change of Rate with Time Now let s consider the reaction between butyl chloride 1C 4 9 Cl2 and water to form butyl alcohol 1C 4 9 O2 and hydrochloric acid: C 4 9 Cl1aq2 + 2 O1l2 C 4 9 O1aq2 + Cl1aq2 [14.3] Suppose we prepare a @M aqueous solution of C 4 9 Cl and then measure the concentration of C 4 9 Cl at various times after time zero (which is the instant at which the reactants are mixed, thereby initiating the reaction). We can use the resulting data, shown in the first two columns of Table 14.1, to calculate the average rate of disappearance of C 4 9 Cl over various time intervals; these rates are given in the third column. Notice that the average rate decreases over each 50-s interval for the first several measurements and continues to decrease over even larger intervals through the remaining measurements. It is typical for rates to decrease as a reaction proceeds because the concentration of reactants decreases. The change in rate as the reaction proceeds is also seen in a graph of 3C 4 9 Cl4 versus time ( Figure 14.4). Notice how the steepness of the curve decreases with time, indicating a decreasing reaction rate. Instantaneous Rate Graphs such as Figure 14.4 that show how the concentration of a reactant or product changes with time allow us to evaluate the instantaneous rate of a reaction, which is the rate at a particular instant during the reaction. The instantaneous rate is determined from the slope of the curve at a particular point in time. We have drawn two tangent lines in Figure 14.4, a dashed line running through the point at t = 0 s and a solid line running through the point at t = 600 s. The slopes of these tangent lines give the instantaneous rates at these two time points.* To determine the instantaneous rate at 600 s, for example, we construct horizontal and vertical lines to form the blue right triangle in Figure The slope of the tangent line is the ratio of the height of the vertical side to the length of the horizontal side: Instantaneous rate = - 3C 4 9 Cl4 = - t = 6.3 * 10-5 M>s M s Go Figure ow does the instantaneous rate of reaction change as the reaction proceeds? [C 4 9 Cl] (M) C 4 9 Cl(aq) + 2 O(l) 0 Instantaneous rate at t = 0 s (initial rate) [C 4 9 Cl] C 4 9 O(aq) + Cl(aq) Instantaneous rate at time t = slope of tangent to the line at time t. Instantaneous rate at t = 600 s Time (s) Figure 14.4 Concentration of butyl chloride 1c 4 h 9 cl2 as a function of time. t *You may wish to review graphical determination of slopes in Appendix A. If you are familiar with calculus, you may recognize that the average rate approaches the instantaneous rate as the time interval approaches zero. This limit, in the notation of calculus, is the negative of the derivative of the curve at time t, -d3c 4 9 Cl4>dt.

47 580 chapter 14 Chemical Kinetics In discussions that follow, the term rate means instantaneous rate unless indicated otherwise. The instantaneous rate at t = 0 is called the initial rate of the reaction. To understand the difference between average and instantaneous rates, imagine you have just driven 98 mi in 2.0 h. Your average speed for the trip is 49 mi>hr, but your instantaneous speed at any moment during the trip is the speedometer reading at that moment. Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction Using Figure 14.4, calculate the instantaneous rate of disappearance of C 4 9 Cl at t = 0 s (the initial rate). Solution Analyze We are asked to determine an instantaneous rate from a graph of reactant concentration versus time. Plan To obtain the instantaneous rate at t = 0 s, we must determine the slope of the curve at t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this straight line equals the change in the vertical axis divided by the corresponding change in the horizontal axis (which, in the case of this example, is the change in molarity over change in time). Solve The tangent line falls from 3C 4 9 Cl4 = M to M in the time change from 0 s to 210 s. Thus, the initial rate is Rate = - 3C 4 9 Cl4 t M = - = 1.9 * 10-4 M>s s Practice Exercise 1 Which of the following could be the instantaneous rate of the reaction in Figure 14.4 at t = 1000 s? (a) 1.2 * 10-4 M>s, (b) 8.8 * 10-5 M>s, (c) 6.3 * 10-5 M>s, (d) 2.7 * 10-5 M>s, (e) More than one of these. Practice Exercise 2 Using Figure 14.4, determine the instantaneous rate of disappearance of C 4 9 Cl at t = 300 s. Give It Some Thought In Figure 14.4, order the following three rates from fastest to slowest: (i) The average rate of the reaction between 0 s and 600 s, (ii) the instantaneous rate at t = 0 s, and (iii) the instantaneous rate at t = 600 s. You should not have to do any calculations. Reaction Rates and Stoichiometry During our discussion of the hypothetical reaction A B, we saw that the stoichiometry requires that the rate of disappearance of A equal the rate of appearance of B. Likewise, the stoichiometry of Equation 14.3 indicates that 1 mol of C 4 9 O is produced for each mole of C 4 9 Cl consumed. Therefore, the rate of appearance of C 4 9 O equals the rate of disappearance of C 4 9 Cl: Rate = - 3C 4 9 Cl4 t = 3C 4 9 O4 t What happens when the stoichiometric relationships are not one-to-one? For example, consider the reaction 2 I1g2 2 1g2 + I 2 1g2. We can measure either the rate of disappearance of I or the rate of appearance of either 2 or I 2. Because 2 mol of I disappears for each mole of 2 or I 2 that forms, the rate of disappearance of I is twice the rate of appearance of either 2 or I 2. ow do we decide which number to use for the rate of the reaction? Depending on whether we monitor I, I 2, or 2, the rates can differ by a factor of 2. To fix this problem, we need to take into account the reaction stoichiometry. To arrive at a number for the reaction rate that does not

48 section 14.3 Concentration and Rate Laws 581 depend on which component we measured, we must divide the rate of disappearance of I by 2 (its coefficient in the balanced chemical equation): Rate = - 1 3I4 2 t = 3 24 t = 3I 24 t In general, for the reaction a A + b B c C + d D the rate is given by Rate = - 1 3A4 = - 1 3B4 = 1 3C4 = 1 3D4 [14.4] a t b t c t d t When we speak of the rate of a reaction without specifying a particular reactant or product, we utilize the definition in Equation 14.4.* Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear (a) ow is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 1g2 3 O 2 1g2? (b) If the rate at which O 2 appears, 3O 2 4> t, is 6.0 * 10-5 M>s at a particular instant, at what rate is O 3 disappearing at this same time, - 3O 3 4> t? Solution Analyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. Plan We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative rates of reactions. Solve (a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have: Rate = - 1 3O 3 4 = 1 3O t 3 t (b) Solving the equation from part (a) for the rate at which O 3 disappears, - 3O 3 4> t, we have: - 3O 34 t = 2 3 3O 2 4 = 2 t * 10-5 M>s2 = 4.0 * 10-5 M>s Practice Exercise 1 At a certain time in a reaction, substance A is disappearing at a rate of 4.0 * 10-2 M>s, substance B is appearing at a rate of 2.0 * 10-2 M>s, and substance C is appearing at a rate of 6.0 * 10-2 M>s. Which of the following could be the stoichiometry for the reaction being studied? (a) 2A + B 3C (b) A 2B + 3C (c) 2A B + 3C (d) 4A 2B + 3C (e) A + 2B 3C Practice Exercise 2 If the rate of decomposition of N 2 O 5 in the reaction 2 N 2 O 5 1g2 4 NO 2 1g2 + O 2 1g2 at a particular instant is 4.2 * 10-7 M>s, what is the rate of appearance of (a) NO 2 and (b) O 2 at that instant? Check We can apply a stoichiometric factor to convert the O 2 formation rate to the O 3 disappearance rate: - 3O 34 t mol O = a6.0 * >L s = 4.0 * 10-5 M>s b a 2 mol O 3 b = 4.0 * 10 mol O 3>L -5 3 mol O 2 s 14.3 Concentration and Rate Laws One way of studying the effect of concentration on reaction rate is to determine the way in which the initial rate of a reaction depends on the initial concentrations. For example, we might study the rate of the reaction N 4 + 1aq2 + NO 2-1aq2 N 2 1g O1l2 *Equation 14.4 does not hold true if substances other than C and D are formed in significant amounts. For example, sometimes intermediate substances build in concentration before forming the final products. In that case, the relationship between the rate of disappearance of reactants and the rate of appearance of products is not given by Equation All reactions whose rates we consider in this chapter obey Equation 14.4.

49 582 chapter 14 Chemical Kinetics Table 14.2 Rate Data for the Reaction of Ammonium and Nitrite Ions in Water at 25 C Experiment Number Initial nh 4 + Concentration (M) Initial no 2 Concentration (M) Observed Initial Rate 1M,s * * * * * * 10-7 by measuring the concentration of N 4 + or NO 2 - as a function of time or by measuring the volume of N 2 collected as a function of time. Because the stoichiometric coefficients on N 4 +, NO 2 -, and N 2 are the same, all of these rates are the same. Table 14.2 shows that changing the initial concentration of either reactant changes the initial reaction rate. If we double 3N while holding 3NO 2-4 constant, the rate doubles (compare experiments 1 and 2). If we increase 3N by a factor of 4 but leave 3NO 2-4 unchanged (experiments 1 and 3), the rate changes by a factor of 4, and so forth. These results indicate that the initial reaction rate is proportional to 3N When 3NO 2-4 is similarly varied while 3N is held constant, the rate is affected in the same manner. Thus, the rate is also directly proportional to the concentration of 3NO 2-4. A Closer Look Using Spectroscopic Methods to Measure Reaction Rates: Beer s Law A variety of techniques can be used to monitor reactant and product concentration during a reaction, including spectroscopic methods, which rely on the ability of substances to absorb (or emit) light. Spectroscopic kinetic studies are often performed with the reaction mixture in the sample compartment of a spectrometer, an instrument that measures the amount of light transmitted or absorbed by a sample at different wavelengths. For kinetic studies, the spectrometer is set to measure the light absorbed at a wavelength characteristic of one of the reactants or products. In the decomposition of I(g) into 2 1g2 and I 2 1g2, for example, both I and 2 are colorless, whereas I 2 is violet. During the reaction, the violet color of the reaction mixture gets more intense as I 2 forms. Thus, visible light of appropriate wavelength can be used to monitor the reaction ( Figure 14.5). Figure 14.6 shows the components of a spectrometer. The spectrometer measures the amount of light absorbed by the sample by comparing the intensity of the light emitted from the light source with the intensity of the light transmitted through the sample, for various wavelengths. As the concentration of I 2 increases and its color becomes more intense, the amount of light absorbed by the reaction mixture increases, as Figure 14.5 shows, causing less light to reach the detector. ow can we relate the amount of light detected by the spectrometer to the concentration of a species? A relationship called Beer s law gives us a direct route to the information we seek. Beer s law connects the amount of light absorbed to the concentration of the absorbing substance: A = ebc [14.5] In this equation, A is the measured absorbance, e is the extinction coefficient (a characteristic of the substance being monitored at a given wavelength of light), b is the path length through which the light Absorbance Spectrometer measures intensity of purple color as I 2 concentration increases Wavelength (nm) Figure 14.5 Visible spectra of l 2 at different concentrations. 100 mg/l 70 mg/l 40 mg/l 10 mg/l 1 mg/l passes, and c is the molar concentration of the absorbing substance. Thus, the concentration is directly proportional to absorbance. Many chemical and pharmaceutical companies routinely use Beer s law to calculate the concentration of purified solutions of the compounds that they make. In the laboratory portion of your course, you may very well perform one or more experiments in which you use Beer s law to relate absorption of light to concentration. Related Exercises: , , Design an Experiment

50 section 14.3 Concentration and Rate Laws 583 Source Lenses/slits/ collimators Monochromator (selects wavelength) Sample Detector Computer Figure 14.6 Components of a spectrometer. We express the way in which the rate depends on the reactant concentrations by the equation Rate = k3n NO [14.6] An equation such as Equation 14.6, which shows how the rate depends on reactant concentrations, is called a rate law. For the general reaction a A + b B c C + d D the rate law generally has the form Rate = k3a4 m 3B4 n [14.7] Notice that only the concentrations of the reactants generally appear in the rate law. The constant k is called the rate constant. The magnitude of k changes with temperature and therefore determines how temperature affects rate, as we will see in Section The exponents m and n are typically small whole numbers. As we will learn shortly, if we know m and n for a reaction, we can gain great insight into the individual steps that occur during the reaction. Give It Some Thought ow do reaction rate, rate law, and rate constant differ? Once we know the rate law for a reaction and the reaction rate for a set of reactant concentrations, we can calculate the value of k. For example, using the values for experiment 1 in Table 14.2, we can substitute into Equation 14.6: 5.4 * 10-7 M>s = k M M2 5.4 * 10-7 M>s k = M M2 = 2.7 * 10-4 M -1 s -1 You should verify that this same value of k is obtained using any of the other experimental results in Table Once we have both the rate law and the k value for a reaction, we can calculate the reaction rate for any set of concentrations. For example, using Equation 14.7 with k = 2.7 * 10-4 M -1 s -1, m = 1, and n = 1, we can calculate the rate for 3N = M and 3NO 2-4 = M: Rate = 12.7 * 10-4 M -1 s M M2 = 2.7 * 10-6 M>s Give It Some Thought Does the rate constant have the same units as the rate?

51 584 chapter 14 Chemical Kinetics Reaction Orders: The Exponents in the Rate Law The rate law for most reactions has the form Rate = k3reactant 14 m 3reactant 24 n c [14.8] The exponents m and n are called reaction orders. For example, consider again the + rate law for the reaction of N 4 with NO - 2 : Rate = k3n NO 2-4 Because the exponent of 3N is 1, the rate is first order in N 4 +. The rate is also first order in NO 2 -. (The exponent 1 is not shown in rate laws.) The overall reaction order is the sum of the orders with respect to each reactant represented in the rate law. Thus, for the N NO 2 - reaction, the rate law has an overall reaction order of = 2, and the reaction is second order overall. The exponents in a rate law indicate how the rate is affected by each reactant concentration. Because the rate at which N 4 + reacts with NO 2 - depends on 3N raised to the first power, the rate doubles when 3N doubles, triples when 3N triples, and so forth. Doubling or tripling 3NO 2-4 likewise doubles or triples the rate. If a rate law is second order with respect to a reactant, 3A4 2, then doubling the concentration of that substance causes the reaction rate to quadruple because = 4, whereas tripling the concentration causes the rate to increase ninefold: = 9. The following are some additional examples of experimentally determined rate laws: 2 N 2 O 5 1g2 4 NO 2 1g2 + O 2 1g2 Rate = k3n 2 O 5 4 [14.9] 2 1g2 + I 2 1g2 2 I1g2 Rate = k3 2 43I 2 4 [14.10] CCl 3 1g2 + Cl 2 1g2 CCl 4 1g2 + Cl1g2 Rate = k3ccl 3 43Cl 2 4 1>2 [14.11] Although the exponents in a rate law are sometimes the same as the coefficients in the balanced equation, this is not necessarily the case, as Equations 14.9 and show. For any reaction, the rate law must be determined experimentally. In most rate laws, reaction orders are 0, 1, or 2. owever, we also occasionally encounter rate laws in which the reaction order is fractional (as is the case with Equation 14.11) or even negative. Give It Some Thought The experimentally determined rate law for the reaction 2 NO1g g2 N 2 1g O1g2 is rate = k3no (a) What are the reaction orders in this rate law? (b) Would the reaction rate increase more if we doubled the concentration of NO or the concentration of 2? Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction A + B C for which rate = k3a43b4 2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. (1) (2) (3)

52 section 14.3 Concentration and Rate Laws 585 Solution Analyze We are given three boxes containing different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates. Plan Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate for each box. Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate: Box 1: Rate = k = 125k Box 2 contains 7 red spheres and 3 purple spheres: Box 2: Rate = k = 63k Box 3 contains 3 red spheres and 7 purple spheres: Box 3: Rate = k = 147k The slowest rate is 63k (Box 2), and the highest is 147k (Box 3). Thus, the rates vary in the order Check Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a larger reaction order. ence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order Practice Exercise 1 Suppose the rate law for the reaction in this Sample Exercise were rate = k3a4 2 3B4. What would be the ordering of the rates for the three mixtures shown above, from slowest to fastest? (a) (b) (c) (d) (e) Practice Exercise 2 Assuming that rate = k3a43b4, rank the mixtures represented in this Sample Exercise in order of increasing rate. Magnitudes and Units of Rate Constants If chemists want to compare reactions to evaluate which ones are relatively fast and which ones are relatively slow, the quantity of interest is the rate constant. A good general rule is that a large value of k ( 10 9 or higher) means a fast reaction and a small value of k (10 or lower) means a slow reaction. Give It Some Thought Suppose the reactions A B and X Y have the same value of k. When 3A4 = 3X4, will the two reactions necessarily have the same rate? The units of the rate constant depend on the overall reaction order of the rate law. In a reaction that is second order overall, for example, the units of the rate constant must satisfy the equation: Units of rate = 1units of rate constant21units of concentration2 2 ence, in our usual units of molarity for concentration and seconds for time, we have Units of rate constant = units of rate 1units of concentration2 2 = M>s M 2 = M -1 s -1 Sample Exercise 14.5 Determining Reaction Orders and Units for Rate Constants (a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11? (b) What are the units of the rate constant for the rate law in Equation 14.9? Solution Analyze We are given two rate laws and asked to express (a) the overall reaction order for each and (b) the units for the rate constant for the first reaction. Plan The overall reaction order is the sum of the exponents in the rate law. The units for the rate constant, k, are found by using the normal units for rate 1M>s2 and concentration (M) in the rate law and applying algebra to solve for k. Solve (a) The rate of the reaction in Equation 14.9 is first order in N 2 O 5 and first order overall. The reaction in Equation is first order in CCl 3 and one-half order in Cl 2. The overall reaction order is three halves. (b) For the rate law for Equation 14.9, we have Units of rate = 1units of rate constant21units of concentration2 so units of rate Units of rate constant = units of concentration = M>s M = s-1 Notice that the units of the rate constant change as the overall order of the reaction changes.

53 586 chapter 14 Chemical Kinetics Practice Exercise 1 Which of the following are the units of the rate constant for Equation 14.11? (a) M -1 2 s -1 (b) M -1 2 s -1 2 (c) M 1 2 s -1 (d) M -3 2 s -1 (e) M -3 2 s -1 2 Practice Exercise 2 (a) What is the reaction order of the reactant 2 in Equation 14.10? (b) What are the units of the rate constant for Equation 14.10? Using Initial Rates to Determine Rate Laws We have seen that the rate law for most reactions has the general form Rate = k3reactant 14 m 3reactant 24 n c Thus, the task of determining the rate law becomes one of determining the reaction orders, m and n. In most reactions, the reaction orders are 0, 1, or 2. As noted earlier in this section, we can use the response of the reaction rate to a change in initial concentration to determine the reaction order. In working with rate laws, it is important to realize that the rate of a reaction depends on concentration but the rate constant does not. As we will see later in this chapter, the rate constants (and hence the reaction rate) are affected by temperature and by the presence of a catalyst. Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows: Experiment Number [A] (M) [B] (M) Initial Rate 1M,s * * * 10-5 Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when 3A4 = 0.050M and 3B4 = M. Solution Analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction and asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table. Plan (a) We assume that the rate law has the following form: Rate = k3a4 m 3B4 n. We will use the given data to deduce the reaction orders m and n by determining how changes in the concentration change the rate. (b) Once we know m and n, we can use the Solve (a) If we compare experiments 1 and 2, we see that [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with In experiments 1 and 3, [B] is held constant, so these data show how [A] affects rate. olding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to 3A4 2 (that is, the reaction is second order in A). ence, the rate law is Rate = k3a4 2 3B4 0 = k3a4 2 (b) Using the rate law and the data from experiment 1, we have rate law and one of the sets of data to determine the rate constant k. (c) Upon determining both the rate constant and the reaction orders, we can use the rate law with the given concentrations to calculate rate. respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0). k = rate 3A4 2 = 4.0 * 10-5 M>s M2 2 = 4.0 * 10-3 M -1 s -1 (c) Using the rate law from part (a) and the rate constant from part (b), we have Rate = k3a4 2 = 14.0 * 10-3 M -1 s M2 2 = 1.0 * 10-5 M>s Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react with A.

54 section 14.4 The Change of Concentration with Time 587 Check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see if we can correctly calculate the rate. Using data from experiment 3, we have Rate = k3a4 2 = 14.0 * 10-3 M -1 s M2 2 = 1.6 * 10-4 M>s Thus, the rate law correctly reproduces the data, giving both the correct number and the correct units for the rate. Practice Exercise 1 A certain reaction X + Y Z is described as being first order in [X] and third order overall. Which of the following statements is or are true?: (i) The rate law for the reaction is: Rate = k3x43y4 2. (ii) If the concentration of X is increased by a factor of 1.5, the rate will increase by a factor of (iii) If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of (a) Only one of the statements is true. (c) Statements (i) and (iii) are true. (e) All three statements are true. (b) Statements (i) and (ii) are true. (d) Statements (ii) and (iii) are true. Practice Exercise 2 The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO1g g2 N 2 1g O1g2 Experiment Number [NO] (M) 3h 2 4 (M) Initial Rate 1M,s * * * 10-3 (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when 3NO4 = M and = M the Change of Concentration with Time The rate laws we have examined so far enable us to calculate the rate of a reaction from the rate constant and reactant concentrations. In this section, we will show that rate laws can also be converted into equations that show the relationship between concentrations of reactants or products and time. The mathematics required to accomplish this conversion involves calculus. We do not expect you to be able to perform the calculus operations, but you should be able to use the resulting equations. We will apply this conversion to three of the simplest rate laws: those that are first order overall, those that are second order overall, and those that are zero order overall. First-Order Reactions A first-order reaction is one whose rate depends on the concentration of a single reactant raised to the first power. If a reaction of the type A products is first order, the rate law is: Rate = - 3A4 = k3a4 t This form of a rate law, which expresses how rate depends on concentration, is called the differential rate law. Using the operation from calculus called integration, this relationship can be transformed into an equation known as the integrated rate law for a

55 588 chapter 14 Chemical Kinetics Sample Exercise 14.7 Using the Integrated First-Order Rate Law first-order reaction that relates the initial concentration of A, 3A4 0, to its concentration at any other time t, 3A4 t : ln3a4 t - ln3a4 0 = -kt or ln 3A4 t 3A4 0 = -kt [14.12] The function ln in Equation is the natural logarithm (Appendix A.2). Equation can also be rearranged to ln3a4 t = -kt + ln3a4 0 [14.13] Equations and can be used with any concentration units as long as the units are the same for both 3A4 t and 3A4 0. For a first-order reaction, Equation or can be used in several ways. Given any three of the following quantities, we can solve for the fourth: k, t, 3A4 0, and 3A4 t. Thus, you can use these equations to determine (1) the concentration of a reactant remaining at any time after the reaction has started, (2) the time interval required for a given fraction of a sample to react, or (3) the time interval required for a reactant concentration to fall to a certain level. The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr -1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 * 10-7 g>cm 3. Assume that the temperature of the lake is Solution Analyze We are given the rate constant for a reaction that obeys firstorder kinetics, as well as information about concentrations and times, and asked to calculate how much reactant (insecticide) remains after 1 yr. We must also determine the time interval needed to reach a particular insecticide concentration. Because the exercise gives time in (a) and asks for time in (b), we will find it most useful to use the integrated rate law, Equation constant (so that there are no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b) ow long will it take for the insecticide concentration to decrease to 3.0 * 10-7 g>cm 3? Plan (a) We are given k = 1.45 yr -1, t = 1.00 yr, and 3insecticide4 0 = 5.0 * 10-7 g>cm 3, and so Equation can be solved for 3insecticide4 t. (b) We have k = 1.45 yr -1, 3insecticide4 0 = 5.0 * 10-7 g>cm 3, and 3insecticide4 t = 3.0 * 10-7 g>cm 3, and so we can solve Equation for time, t. Solve (a) Substituting the known quantities into Equation 14.13, we have ln3insecticide4 t = 1 yr = yr yr2 + ln15.0 * We use the ln function on a calculator to evaluate the second term on the right [that is, ln15.0 * , giving ln3insecticide4 t = 1 yr = = To obtain 3insecticide4 t = 1 yr, we use the inverse natural logarithm, or e x, function on the calculator: 3insecticide4 t = 1 yr = e = 1.2 * 10-7 g>cm 3 Note that the concentration units for 3A4 t and 3A4 0 must be the same. (b) Again substituting into Equation 14.13, with 3insecticide4 t = 3.0 * 10-7 g>cm 3, gives ln13.0 * = yr -1 21t2 + ln15.0 * Solving for t gives t = -3ln13.0 * ln15.0 * >1.45 yr -1 Check In part (a) the concentration remaining after 1.00 yr (that is, 1.2 * 10-7 g>cm 3 ) is less than the original concentration 15.0 * 10-7 g>cm 3 2, as it should be. In (b) the given concentration = >1.45 yr -1 = 0.35 yr Practice Exercise 1 At 25 C, the decomposition of dinitrogen pentoxide, N 2 O 5 1g2, into NO 2 1g2 and O 2 1g2 follows first-order kinetics with k = 3.4 * 10-5 s -1. A sample of N 2 O 5 with an initial pressure of 760 torr decomposes at 25 C until its partial pressure is 650 torr. ow much time (in seconds) has elapsed? (a) 5.3 * 10-6 (b) 2000 (c) 4600 (d) 34,000 (e) 190, * 10-7 g>cm 3 2 is greater than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a reasonable answer. Practice Exercise 2 The decomposition of dimethyl ether, 1C O, at 510 C is a first-order process with a rate constant of 6.8 * 10-4 s -1 : 1C O1g2 C 4 1g g2 + CO1g2 If the initial pressure of 1C O is 135 torr, what is its pressure after 1420 s?

56 section 14.4 The Change of Concentration with Time 589 Equation can be used to verify whether a reaction is first order and to determine its rate constant. This equation has the form of the general equation for a straight line, y = mx + b, in which m is the slope and b is the y-intercept of the line (Appendix A.4): ln [A] t = kt + ln [A] 0 Methyl isonitrile y = mx + b For a first-order reaction, therefore, a graph of ln3a4 t versus time gives a straight line with a slope of -k and a y-intercept of ln3a4 0. A reaction that is not first order will not yield a straight line. As an example, consider the conversion of methyl isonitrile 1C 3 NC2 to its isomer acetonitrile 1C 3 CN2 ( Figure 14.7). Because experiments show that the reaction is first order, we can write the rate equation: ln3c 3 NC4 t = -kt + ln3c 3 NC4 0 We run the reaction at a temperature at which methyl isonitrile is a gas 1199 C2, and Figure 14.8(a) shows how the pressure of this gas varies with time. We can use pressure as a unit of concentration for a gas because we know from the ideal-gas law the pressure is directly proportional to the number of moles per unit volume. Figure 14.8(b) shows that a plot of the natural logarithm of the pressure versus time is a straight line. The slope of this line is -5.1 * 10-5 s -1. (You should verify this for yourself, remembering that your result may vary slightly from ours because of inaccuracies associated with reading the graph.) Because the slope of the line equals -k, the rate constant for this reaction equals 5.1 * 10-5 s -1. Acetonitrile Figure 14.7 The first-order reaction of ch 3 nc conversion into ch 3 cn. Second-Order Reactions A second-order reaction is one for which the rate depends either on a reactant concentration raised to the second power or on the concentrations of two reactants each raised to the first power. For simplicity, let s consider reactions of the type A products or A + B products that are second order in just one reactant, A: Rate = - 3A4 t = k3a4 2 Go Figure What can you conclude given that the plot of ln P versus t is linear? Pressure, C 3 NC (torr) ,000 20,000 30,000 Time (s) ln pressure, C 3 NC ,000 20,000 30,000 Time (s) (a) Figure 14.8 Kinetic data for conversion of methyl isonitrile into acetonitrile. (b)

57 590 chapter 14 Chemical Kinetics With the use of calculus, this differential rate law can be used to derive the integrated rate law for second-order reactions: 1 = kt + 1 [14.14] 3A4 t 3A4 0 This equation, like Equation 14.13, has four variables, k, t, 3A4 0, and 3A4 t, and any one of these can be calculated knowing the other three. Equation also has the form of a straight line 1y = mx + b2. If the reaction is second order, a plot of 1>3A4 t versus t yields a straight line with slope k and y-intercept 1>3A4 0. One way to distinguish between first- and second-order rate laws is to graph both ln3a4 t and 1>3A4 t against t. If the ln3a4 t plot is linear, the reaction is first order; if the 1>3A4 t plot is linear, the reaction is second order. Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300 C, NO 2 1g2 NO1g O 21g2. Is the reaction first or second order in NO 2? Time (s) 3no 2 4 1M Solution Analyze We are given the concentrations of a reactant at various times during a reaction and asked to determine whether the reaction is first or second order. Plan We can plot ln3no 2 4 and 1>3NO 2 4 against time. If one plot or the other is linear, we will know the reaction is either first or second order. Solve To graph ln3no 2 4 and 1>3NO 2 4 against time, we first make the following calculations from the data given: Time (s) 3no 2 4 1M2 ln 3no 2 4 1, 3no ,M ln[no 2 ] Time (s) As Figure 14.9 shows, only the plot of 1>3NO 2 4 versus time is linear. Thus, the reaction obeys a second-order rate law: Rate = k3no From the slope of this straight-line graph, we determine that k = M -1 s -1 for the disappearance of NO 2. Practice Exercise 1 For a certain reaction A products, a plot of ln[a] versus time produces a straight line with a slope of -3.0 * 10-2 s -1. Which of the following statements is or are true?: (i) The reaction follows first-order kinetics. (ii) The rate constant for the reaction is 3.0 * 10-2 s -1. (iii) The initial concentration of [A] was 1.0 M. 1/[NO 2 ] (1/M) Time (s) Figure 14.9 Kinetic data for decomposition of no 2.

58 section 14.4 The Change of Concentration with Time 591 (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 The decomposition of NO 2 discussed in the Sample Exercise is second order in NO 2 with k = M -1 s -1. If the initial concentration of NO 2 in a closed vessel is M, what is the concentration of this reactant after h? Zero-Order Reactions We have seen that in a first-order reaction the concentration of a reactant A decreases nonlinearly, as shown by the red curve in Figure As [A] declines, the rate at which it disappears declines in proportion. A zero-order reaction is one in which the rate of disappearance of A is independent of [A]. The rate law for a zero-order reaction is Rate = - 3A4 t The integrated rate law for a zero-order reaction is 3A4 t = -kt + 3A4 0 where 3A4 t is the concentration of A at time t and 3A4 0 is the initial concentration. This is the equation for a straight line with vertical intercept 3A4 0 and slope -kt, as shown in the blue curve in Figure The most common type of zero-order reaction occurs when a gas undergoes decomposition on the surface of a solid. If the surface is completely covered by decomposing molecules, the rate of reaction is constant because the number of reacting surface molecules is constant, so long as there is some gas-phase substance left. alf-life The half-life of a reaction, t 1>2, is the time required for the concentration of a reactant to reach half its initial value, 3A4 t1>2 = 1 23A4 0. alf-life is a convenient way to describe how fast a reaction occurs, especially if it is a first-order process. A fast reaction has a short half-life. We can determine the half-life of a first-order reaction by substituting 3A4 t1>2 = 1 23A4 0 for 3A4 t and t 1>2 for t in Equation 14.12: 1 23A4 0 ln = -kt 3A4 1>2 0 = k A 0 [A] Go Figure At which times during the reaction would you have trouble distinguishing a zero-order reaction from a first-order reaction? Slope = kt First-order reaction Zero-order reaction Time Figure Comparison of first-order and zero-order reactions for the disappearance of reactant A with time. ln 1 2 = -kt 1>2 t 1>2 = - ln 1 2 k = k [14.15] From Equation 14.15, we see that t 1>2 for a first-order rate law does not depend on the initial concentration of any reactant. Consequently, the half-life remains constant throughout the reaction. If, for example, the concentration of a reactant is M at some instant in the reaction, it will be M2 = M after one half-life. After one more 75 half-life passes, the concentration will drop to M, and so on. Equation also indicates that, for a first-order reaction, we can calculate t 1>2 if we know k and calculate k if we know t 1> The change in concentration over time for the first-order rearrangement of gaseous methyl isonitrile at 199 C is graphed in Figure Because the concentration of this gas is directly proportional to its pressure during the reaction, we have chosen to plot pressure rather than 0 concentration in this graph. The first half-life occurs at 13,600 s (3.78 h). At a time 13,600 s later, the methyl isonitrile pressure (and therefore, concentration) has decreased to half of one-half, or one-fourth, of the initial value. In a first-order reaction, the concentration of the reactant decreases by one-half in each of a series of regularly spaced time intervals, each interval equal to t 1>2. Pressure, C 3 NC (torr) 150 t 1/2 t 1/2 10,000 20,000 30,000 Time (s) Figure Kinetic data for the rearrangement of methyl isonitrile to acetonitrile at 199 c, showing the half-life of the reaction.

59 592 chapter 14 Chemical Kinetics Give It Some Thought If a solution containing 10.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives? Chemistry Put to Work Methyl Bromide in the Atmosphere The compounds known as chlorofluorocarbons (CFCs) are well-known agents responsible for the destruction of Earth s protective ozone layer. Another simple molecule that has the potential to destroy the stratospheric ozone layer is methyl bromide, C 3 Br ( Figure 14.12). Because this substance has a wide range of uses, including antifungal treatment of plant seeds, it has been produced in large quantities in the past (about 150 million pounds per year worldwide in 1997, at the height of its production). In the stratosphere, the C Br bond is Diffusion to stratosphere 50% decomposes in 0.8 years Methyl bromide applied as antifungal treatment Stratosphere Troposphere Lower atmosphere Figure Distribution and fate of methyl bromide in Earth s atmosphere. broken through absorption of short-wavelength radiation. The resultant Br atoms then catalyze decomposition of O 3. Methyl bromide is removed from the lower atmosphere by a variety of mechanisms, including a slow reaction with ocean water: C 3 Br1g2 + 2 O1l2 C 3 O1aq2 + Br1aq2 [14.16] To determine the potential importance of C 3 Br in destruction of the ozone layer, it is important to know how rapidly the reaction in Equation and all other reactions remove C 3 Br from the lower atmosphere before it can diffuse into the stratosphere. The average lifetime of C 3 Br in Earth s lower atmosphere is difficult to measure because the conditions that exist in the atmosphere are too complex to be simulated in the laboratory. Instead, scientists analyzed nearly 4000 atmospheric samples collected above the Pacific Ocean for the presence of several trace organic substances, including methyl bromide. From these measurements, it was possible to estimate the atmospheric residence time for C 3 Br. The atmospheric residence time is related to the half-life for C 3 Br in the lower atmosphere, assuming C 3 Br decomposes by a first-order process. From the experimental data, the half-life for methyl bromide in the lower atmosphere is estimated to be 0.8 { 0.1 yr. That is, a collection of C 3 Br molecules present at any given time will, on average, be 50% decomposed after 0.8 yr, 75% decomposed after 1.6 yr, and so on. A half-life of 0.8 yr, while comparatively short, is still sufficiently long so that C 3 Br contributes significantly to the destruction of the ozone layer. In 1997 an international agreement was reached to phase out use of methyl bromide in developed countries by owever, in recent years exemptions for critical agricultural use have been requested and granted. Nevertheless, authorized worldwide production was down to 26 million pounds in 2012, three-fourths of which is used in the United States. Related Exercise: Sample Exercise 14.9 Determining the alf-life of a First-Order Reaction The reaction of C 4 9 Cl with water is a first-order reaction. (a) Use Figure 14.4 to estimate the halflife for this reaction. (b) Use the half-life from (a) to calculate the rate constant. Solution Analyze We are asked to estimate the half-life of a reaction from a graph of concentration versus time and then to use the half-life to calculate the rate constant for the reaction. Plan (a) To estimate a half-life, we can select a concentration and then determine the time required for the concentration to decrease to half of that value. (b) Equation is used to calculate the rate constant from the half-life. Solve (a) From the graph, we see that the initial value of 3C 4 9 Cl4 is M. The half-life for this first-order reaction is the time required for 3C 4 9 Cl4 to decrease to M, which we can read off the graph. This point occurs at approximately 340 s.

60 section 14.5 Temperature and Rate 593 (b) Solving Equation for k, we have k = t 1>2 = s = 2.0 * 10-3 s -1 Check At the end of the second half-life, which should occur at 680 s, the concentration should have decreased by yet another factor of 2, to M. Inspection of the graph shows that this is indeed the case. Practice Exercise 1 We noted in an earlier Practice Exercise that at 25 C the decomposition of N 2 O 5 1g2 into NO 2 1g2 and O 2 1g2 follows first-order kinetics with k = 3.4 * 10-5 s -1. ow long will it take for a sample originally containing 2.0 atm of N 2 O 5 to reach a partial pressure of 380 torr? (a) 5.7 h (b) 8.2 h (c) 11 h (d) 16 h (e) 32 h Practice Exercise 2 (a) Using Equation 14.15, calculate t 1>2 for the decomposition of the insecticide described in Sample Exercise (b) ow long does it take for the concentration of the insecticide to reach one-quarter of the initial value? The half-life for second-order and other reactions depends on reactant concentrations and therefore changes as the reaction progresses. We obtained Equation for the half-life for a first-order reaction by substituting 3A4 t1>2 = 1 23A4 0 for 3A4 t and t 1>2 for t in Equation We find the half-life of a second-order reaction by making the same substitutions into Equation 14.14: 1 1 = kt 1> A4 0 3A = kt 3A4 0 3A4 1>2 0 t 1>2 = 1 k3a4 0 [14.17] In this case, the half-life depends on the initial concentration of reactant the lower the initial concentration, the longer the half-life. Give It Some Thought Why can we report the half-life for a first-order reaction without knowing the initial concentration, but not for a second-order reaction? 14.5 temperature and Rate The rates of most chemical reactions increase as the temperature rises. For example, dough rises faster at room temperature than when refrigerated, and plants grow more rapidly in warm weather than in cold. We can see the effect of temperature on reaction rate by observing a chemiluminescence reaction (one that produces light), such as that in Cyalume light sticks ( Figure 14.13). ow is this experimentally observed temperature effect reflected in the rate law? The faster rate at higher temperature is due to an increase in the rate constant with increasing temperature. For example, let s reconsider the first-order reaction we saw in Figure 14.7, namely C 3 NC C 3 CN. Figure shows the rate constant for this reaction as a function of temperature. The rate constant and, hence, the rate of the reaction increase rapidly with temperature, approximately doubling for each 10 C rise. The Collision Model Reaction rates are affected both by reactant concentrations and by temperature. The collision model, based on the kinetic-molecular theory (Section 10.7), accounts for both of these effects at the molecular level. The central idea of the collision model is that molecules must collide to react. The greater the number of collisions per second, the greater the reaction rate. As reactant concentration increases, therefore, the number of collisions increases, leading to an increase in reaction rate. According to the kineticmolecular theory of gases, increasing the temperature increases molecular speeds. As molecules move faster, they collide more forcefully (with more energy) and more frequently, increasing reaction rates. Go Figure Why does the light stick glow with less light in cold water than in hot water? ot water Cold water Figure Temperature affects the rate of the chemiluminescence reaction in light sticks: The chemiluminescent reaction occurs more rapidly in hot water, and more light is produced.

61 594 chapter 14 Chemical Kinetics Go Figure Would you expect this curve to eventually go back down to lower values? Why or why not? For a reaction to occur, though, more is required than simply a collision it must be the right kind of collision. For most reactions, in fact, only a tiny fraction of collisions leads to a reaction. For example, in a mixture of 2 and I 2 at ordinary temperatures and pressures, each molecule undergoes about collisions per second. If every collision between 2 and I 2 resulted in the formation of I, the reaction would be over in much less than a second. Instead, at room temperature the reaction proceeds very slowly because only about one in every collisions produces a reaction. What keeps the reaction from occurring more rapidly? k (s 1 ) C 3 NC C 3 CN Temperature ( C) The Orientation Factor In most reactions, collisions between molecules result in a chemical reaction only if the molecules are oriented in a certain way when they collide. The relative orientations of the molecules during collision determine whether the atoms are suitably positioned to form new bonds. For example, consider the reaction Cl + NOCl NO + Cl 2 which takes place if the collision brings Cl atoms together to form Cl 2, as shown in the top panel of Figure In contrast, in the collision shown in the lower panel, the two Cl atoms are not colliding directly with one another, and no products are formed. Figure Temperature dependence of the rate constant for methyl isonitrile conversion to acetonitrile. The four points indicated are used in Sample Exercise Activation Energy Molecular orientation is not the only factor influencing whether a molecular collision will produce a reaction. In 1888 the Swedish chemist Svante Arrhenius suggested that molecules must possess a certain minimum amount of energy to react. According to the collision model, this energy comes from the kinetic energies of the colliding molecules. Upon collision, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. That is, the kinetic energy is used to change the potential energy of the molecule. If molecules are moving too slowly in other words, with too little kinetic energy they merely bounce off one another without changing. The minimum energy required to initiate a chemical reaction is called the activation energy, E a, and its value varies from reaction to reaction. Effective collision, reaction occurs, Cl 2 forms Before collision Collision After collision Ineffective collision, no reaction possible, no Cl 2 Before collision Collision After collision Figure Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl.

62 section 14.5 Temperature and Rate 595 Go Figure If the barrier were lower than as shown in the figure, would the golfer have to hit the ball as hard? Figure Energy is needed to overcome a barrier between initial and final states. The situation during reactions is analogous to that shown in Figure The golfer hits the ball to make it move over the hill in the direction of the cup. The hill is a barrier between ball and cup. To reach the cup, the player must impart enough kinetic energy with the putter to move the ball to the top of the barrier. If he does not impart enough energy, the ball will roll partway up the hill and then back down toward him. In the same way, molecules require a certain minimum energy to break existing bonds during a chemical reaction. We can think of this minimum energy as an energy barrier. In the rearrangement of methyl isonitrile to acetonitrile, for example, we might imagine the reaction passing through an intermediate state in which the N C portion of the methyl isonitrile molecule is sideways: 3 C N C 3 C C N 3 C Figure shows that energy must be supplied to stretch the bond between the 3 C group and the N C group to allow the N C group to rotate. After the N C group has twisted sufficiently, the C C bond begins to form, and the energy of the molecule drops. Thus, the barrier to formation of acetonitrile represents the energy necessary to force the molecule through the relatively unstable intermediate state, analogous to forcing the ball in Figure over the hill. The difference between the energy of the starting molecule and the highest energy along the reaction pathway is the activation energy, E a. The molecule having the arrangement of atoms shown at the top of the barrier is called either the activated complex or the transition state. The conversion of 3 C N C to 3 C C N is exothermic. Figure therefore shows the product as having a lower energy than the reactant. The energy change for the reaction, E, has no effect on reaction rate, however. The rate depends on the magnitude of E a ; generally, the lower the value of E a is, the faster the reaction. Notice that the reverse reaction is endothermic. The activation energy for the reverse reaction is equal to the energy that must be overcome if approaching the barrier from the right: E + E a. Thus, to reach the activated complex for the reverse reaction requires more energy than for the forward reaction for this reaction, there is a larger barrier to overcome going from right to left than from left to right. C N Potential energy Molecule bends, C N bond begins to break 3 C N C 3 C N C Go Figure ow does the energy needed to overcome the energy barrier compare with the overall change in energy for this reaction? Activated complex forms E a E C 3 C... N Reaction progress C C bond forms 3 C 3 C Figure Energy profile for conversion of methyl isonitrile 1h 3 cnc2 to its isomer acetonitrile 1h 3 ccn2. C C N N

63 596 chapter 14 Chemical Kinetics Give It Some Thought Suppose you could measure the rates for both the forward and reverse reactions of the process in Figure In which direction would the rate be larger? Why? Any particular methyl isonitrile molecule acquires sufficient energy to overcome the energy barrier through collisions with other molecules. Recall from the kineticmolecular theory of gases that, at any instant, gas molecules are distributed in energy over a wide range. (Section 10.7) Figure shows the distribution of kinetic energies for two temperatures, comparing them with the minimum energy needed for reaction, E a. At the higher temperature a much greater fraction of the molecules have kinetic energy greater than E a, which leads to a greater rate of reaction. Give It Some Thought Suppose we have two reactions, A B and B C. You can isolate B, and it is stable. Is B the transition state for the reaction A C? Go Figure What would the curve look like for a temperature higher than that for the red curve in the figure? Fraction of molecules Lower T igher T Figure The effect of temperature on the distribution of kinetic energies of molecules in a sample. Minimum energy needed for reaction, E a Larger fraction of molecules reacts at higher temperature Kinetic energy For a collection of molecules in the gas phase, the fraction of molecules that have kinetic energy equal to or greater than E a is given by the expression f = e -E a>rt [14.18] In this equation, R is the gas constant J>mol@K2 and T is the absolute temperature. To get an idea of the magnitude of f, let s suppose that E a is 100 kj>mol, a value typical of many reactions, and that T is 300 K. The calculated value of f is 3.9 * 10-18, an extremely small number! At 320 K, f = 4.7 * Thus, only a 20 increase in temperature produces a more than tenfold increase in the fraction of molecules possessing at least 100 kj>mol of energy. The Arrhenius Equation Arrhenius noted that for most reactions the increase in rate with increasing temperature is nonlinear (Figure 14.14). e found that most reaction-rate data obeyed an equation based on (a) the fraction of molecules possessing energy E a or greater, (b) the number of collisions per second, and (c) the fraction of collisions that have the appropriate orientation. These three factors are incorporated into the Arrhenius equation: k = Ae -E a>rt [14.19] In this equation, k is the rate constant, E a is the activation energy, R is the gas constant J>mol@K2, and T is the absolute temperature. The frequency factor, A, is constant, or nearly so, as temperature is varied. This factor is related to the frequency of collisions and the probability that the collisions are favorably oriented for reaction.* As the magnitude of E a increases, k decreases because the fraction of molecules that possess the required energy is smaller. Thus, at fixed values of T and A, reaction rates decrease as E a increases. *Because collision frequency increases with temperature, A also has some temperature dependence, but this dependence is much smaller than the exponential term. Therefore, A is considered approximately constant.

64 section 14.5 Temperature and Rate 597 Sample Exercise Activation Energies and Speeds of Reaction Consider a series of reactions having these energy profiles: Potential energy 15 kj/mol 10 kj/mol 25 kj/mol 15 kj/mol 20 kj/mol 5 kj/mol Reaction progress Reaction progress Reaction progress Rank the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A. Solution The lower the activation energy, the faster the reaction. The value of E does not affect the rate. ence, the order from slowest reaction to fastest is Practice Exercise 1 Which of the following statements is or are true? (i) The activation energies for the forward and reverse directions of a reaction can be different. (ii) Assuming that A is constant, if both E a and T increase, then k will increase. (iii) For two different reactions, the one with the smaller value of E a will necessarily have the larger value for k. (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 Rank the reverse reactions from slowest to fastest. Determining the Activation Energy We can calculate the activation energy for a reaction by manipulating the Arrhenius equation. Taking the natural log of both sides of Equation 14.19, we obtain ln k = E a RT + ln A [14.20] y = mx + b which has the form of the equation for a straight line. A graph of ln k versus 1>T is a line with a slope equal to -E a >R and a y-intercept equal to ln A. Thus, the activation energy can be determined by measuring k at a series of temperatures, graphing ln k versus 1>T, and calculating E a from the slope of the resultant line. We can also use Equation to evaluate E a in a nongraphical way if we know the rate constant of a reaction at two or more temperatures. For example, suppose that at two different temperatures T 1 and T 2 a reaction has rate constants k 1 and k 2. For each condition, we have ln k 1 = - E a RT 1 Subtracting ln k 2 from ln k 1 gives + ln A and ln k 2 = - E a RT 2 + ln A E a E a ln k 1 - ln k 2 = a- + ln Ab - a- RT 1 RT 2 + ln Ab

65 598 chapter 14 Chemical Kinetics Simplifying this equation and rearranging gives ln k 1 k 2 = E a R a 1 T 2-1 T 1 b [14.21] Equation provides a convenient way to calculate a rate constant k 1 at some temperature T 1 when we know the activation energy and the rate constant k 2 at some other temperature T 2. Sample Exercise Determining the Activation Energy The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures (these are the data points in Figure 14.14): Temperature 1 C2 k 1s * * * * 10-3 (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at K? Solution Analyze We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, E a, and the rate constant, k, at a particular temperature. Plan We can obtain E a from the slope of a graph of ln k versus 1>T. Once we know E a, we can use Equation together with the given rate data to calculate the rate constant at K. Solve (a) We must first convert the temperatures from degrees T 1K2 Celsius to kelvins. We then take the inverse of each temperature, 1>T, and the natural log of each rate constant, * , T 1K 1 2 ln k ln k. This gives us the table shown at the right: * 10 A graph of ln k versus 1>T is a straight line ( Figure 14.19) * 10 The slope of the line is obtained by choosing any two well-separated points and using the coordinates of each: * Slope = y x = = -1.9 * Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the units of 1>T, namely, K - 1. Thus, the overall units for the slope are K. The slope equals -E a >R. We use the value for the gas constant R in units of J>mol@K (Table 10.2). We thus obtain Slope = - E a R E a = -1slope21R2 = * 10 4 J K2a8.314 mol@k b a 1 kj 1000 J b = 1.6 * 10 2 kj>mol = 160 kj>mol We report the activation energy to only two significant figures because we are limited by the precision with which we can read the graph in Figure ln k Slope = E a R /T (K 1 ) Figure Graphical determination of activation energy E a.

66 section 14.6 Reaction Mechanisms 599 (b) To determine the rate constant, k 1, at T 1 = K, we can use Equation with E a = 160 kj>mol and one of the rate constants and temperatures from the given data, such as k 2 = 2.52 * 10-5 s -1 and T 2 = K: ln a 2.52 * 10-5 s - 1 b = 160 kj>mol a J>mol@K b a K K b a 1000 J 1 kj b = Thus, k 1 Practice Exercise 1 Using the data in Sample Exercise 14.11, which of the following is the rate constant for the rearrangement of methyl isonitrile at 320 C? (a) 8.1 * s -1 (b) 2.2 * s -1 (c) 2.7 * 10-9 s -1 (d) 2.3 * 10-1 s -1 (e) 9.2 * 10 3 s -1 Practice Exercise 2 To one significant figure, what is the value for the frequency factor A for the data presented in Sample Exercise k * 10-5 s - 1 = e = 4.15 * 10-2 k 1 = * * 10-5 s = 1.0 * 10-6 s - 1 Note that the units of k 1 are the same as those of k Reaction Mechanisms A balanced equation for a chemical reaction indicates the substances present at the start of the reaction and those present at the end of the reaction. It provides no information, however, about the detailed steps that occur at the molecular level as the reactants are turned into products. The steps by which a reaction occurs is called the reaction mechanism. At the most sophisticated level, a reaction mechanism describes the order in which bonds are broken and formed and the changes in relative positions of the atoms in the course of the reaction. Elementary Reactions We have seen that reactions take place because of collisions between reacting molecules. For example, the collisions between molecules of methyl isonitrile 1C 3 NC2 can provide the energy to allow the C 3 NC to rearrange to acetonitrile: 3 C N C 3 C C N 3 C C N Similarly, the reaction of NO and O 3 to form NO 2 and O 2 appears to occur as a result of a single collision involving suitably oriented and sufficiently energetic NO and O 3 molecules: NO1g2 + O 3 1g2 NO 2 1g2 + O 2 1g2 [14.22] Both reactions occur in a single event or step and are called elementary reactions. The number of molecules that participate as reactants in an elementary reaction defines the molecularity of the reaction. If a single molecule is involved, the reaction is unimolecular. The rearrangement of methyl isonitrile is a unimolecular process. Elementary reactions involving the collision of two reactant molecules are bimolecular. The reaction between NO and O 3 is bimolecular. Elementary reactions involving the simultaneous collision of three molecules are termolecular. Termolecular reactions are far less probable than unimolecular or bimolecular processes and are extremely rare. The chance that four or more molecules will collide simultaneously with any regularity is even more remote; consequently, such collisions are never proposed as part of a reaction mechanism. Thus, nearly all reaction mechanisms contain only unimolecular and bimolecular elementary reactions.

67 600 chapter 14 Chemical Kinetics Give It Some Thought What is the molecularity of the elementary reaction? NO1g2 + Cl 2 1g2 NOCl1g2 + Cl1g2 Go Figure For this profile, is it easier for a molecule of the intermediate to convert to reactants or products? Potential energy Reactants Transition state Transition state Intermediate Reaction progress Figure The energy profile of a reaction, showing transition states and an intermediate. Multistep Mechanisms The net change represented by a balanced chemical equation often occurs by a multistep mechanism consisting of a sequence of elementary reactions. For example, below 225 C, the reaction NO 2 1g2 + CO1g2 NO1g2 + CO 2 1g2 [14.23] appears to proceed in two elementary reactions (or two elementary steps), each of which is bimolecular. First, two NO 2 molecules collide, and an oxygen atom is transferred from one to the other. The resultant NO 3 then collides with a CO molecule and transfers an oxygen atom to it: NO 2 1g2 + NO 2 1g2 NO 3 1g2 + NO1g2 NO 3 1g2 + CO1g2 NO 2 1g2 + CO 2 1g2 Thus, we say that the reaction occurs by a two-step mechanism. The chemical equations for the elementary reactions in a multistep mechanism must always add to give the chemical equation of the overall process. In the present example, the sum of the two elementary reactions is 2 NO 2 1g2 + NO 3 1g2 + CO1g2 NO 2 1g2 + NO 3 1g2 + NO1g2 + CO 2 1g2 Products Simplifying this equation by eliminating substances that appear on both sides gives Equation 14.23, the net equation for the process. Because NO 3 is neither a reactant nor a product of the reaction it is formed in one elementary reaction and consumed in the next it is called an intermediate. Multistep mechanisms involve one or more intermediates. Intermediates are not the same as transition states, as shown in Figure Intermediates can be stable and can therefore sometimes be identified and even isolated. Transition states, on the other hand, are always inherently unstable and as such can never be isolated. Nevertheless, the use of advanced ultrafast techniques sometimes allows us to characterize them. Sample Exercise Determining Molecularity and Identifying Intermediates It has been proposed that the conversion of ozone into O 2 proceeds by a two-step mechanism: O 3 1g2 O 2 1g2 + O1g2 O 3 1g2 + O1g2 2 O 2 1g2 (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s). Solution Analyze We are given a two-step mechanism and asked for (a) the molecularities of each of the two elementary reactions, (b) the equation for the overall process, and (c) the intermediate. Plan The molecularity of each elementary reaction depends on the number of reactant molecules in the equation for that reaction. The overall equation is the sum of the equations for the elementary reactions. The intermediate is a substance formed in one step of the mechanism and used in another and therefore not part of the equation for the overall reaction. Solve (a) The first elementary reaction involves a single reactant and is consequently unimolecular. The second reaction, which involves two reactant molecules, is bimolecular.

68 section 14.6 Reaction Mechanisms 601 (b) Adding the two elementary reactions gives 2 O 3 1g2 + O1g2 3 O 2 1g2 + O1g2 Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net equation for the chemical process: 2 O 3 1g2 3 O 2 1g2 (c) The intermediate is O(g). It is neither an original reactant nor a final product but is formed in the first step of the mechanism and consumed in the second. Practice Exercise 1 Consider the following two-step reaction mechanism: A1g2 + B1g2 X1g2 + Y1g2 X1g2 + C1g2 Y1g2 + Z1g2 Which of the following statements about this mechanism is or are true? (i) Both of the steps in this mechanism are bimolecular. (ii) The overall reaction is A1g2 + B1g2 + C1g2 Y1g2 + Z1g2. (iii) The substance X(g) is an intermediate in this mechanism. (a) Only one of the statements is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 For the reaction Mo1CO2 6 + P1C Mo1CO2 5 P1C CO the proposed mechanism is Mo1CO2 6 Mo1CO2 5 + CO Mo1CO2 5 + P1C Mo1CO2 5 P1C (a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s). Rate Laws for Elementary Reactions In Section 14.3, we stressed that rate laws must be determined experimentally; they cannot be predicted from the coefficients of balanced chemical equations. We are now in a position to understand why this is so. Every reaction is made up of a series of one or more elementary steps, and the rate laws and relative speeds of these steps dictate the overall rate law for the reaction. Indeed, the rate law for a reaction can be determined from its mechanism, as we will see shortly, and compared with the experimental rate law. Thus, our next challenge in kinetics is to arrive at reaction mechanisms that lead to rate laws consistent with those observed experimentally. We start by examining the rate laws of elementary reactions. Elementary reactions are significant in a very important way: If a reaction is elementary, its rate law is based directly on its molecularity. For example, consider the unimolecular reaction A products As the number of A molecules increases, the number that reacts in a given time interval increases proportionally. Thus, the rate of a unimolecular process is first order: Rate = k3a4 For bimolecular elementary steps, the rate law is second order, as in the reaction A + B products Rate = k3a43b4 The second-order rate law follows directly from collision theory. If we double the concentration of A, the number of collisions between the molecules of A and B doubles; likewise, if we double [B], the number of collisions between A and B doubles. Therefore, the rate law is first order in both [A] and [B] and second order overall. The rate laws for all feasible elementary reactions are given in Table Notice how each rate law follows directly from the molecularity of the reaction. It is important Table 14.3 Elementary Reactions and Their Rate Laws Molecularity Elementary Reaction Rate Law Unimolecular A products Rate = k3a4 Bimolecular A + A products Rate = k3a4 2 Bimolecular A + B products Rate = k3a43b4 Termolecular A + A + A products Rate = k3a4 3 Termolecular A + A + B products Rate = k3a4 2 3B4 Termolecular A + B + C products Rate = k3a43b43c4

69 602 chapter 14 Chemical Kinetics to remember, however, that we cannot tell by merely looking at a balanced, overall chemical equation whether the reaction involves one or several elementary steps. Sample Exercise Predicting the Rate Law for an Elementary Reaction If the following reaction occurs in a single elementary reaction, predict its rate law: 2 1g2 + Br 2 1g2 2 Br1g2 Solution Analyze We are given the equation and asked for its rate law, assuming that it is an elementary process. Plan Because we are assuming that the reaction occurs as a single elementary reaction, we are able to write the rate law using the coefficients for the reactants in the equation as the reaction orders. Solve The reaction is bimolecular, involving one molecule of 2 and one molecule of Br 2. Thus, the rate law is first order in each reactant and second order overall: Rate = k3 2 43Br 2 4 Comment Experimental studies of this reaction show that the reaction actually has a very different rate law: Rate = k3 2 43Br 2 4 1>2 Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism cannot occur by a single elementary step. It must, therefore, involve two or more elementary steps. Practice Exercise 1 Consider the following reaction: 2 A + B X + 2 Y. You are told that the first step in the mechanism of this reaction has the following rate law: Rate = k3a43b4. Which of the following could be the first step in the reaction mechanism (note that substance Z is an intermediate)? (a) A + A Y + Z (b) A X + Z (c) A + A + B X + Y + Y (d) B X + Y (e) A + B X + Z Practice Exercise 2 Consider the following reaction: 2 NO1g2 + Br 2 1g2 2 NOBr1g2. (a) Write the rate law for the reaction, assuming it involves a single elementary reaction. (b) Is a single-step mechanism likely for this reaction? The Rate-Determining Step for a Multistep Mechanism As with the reaction in Sample Exercise 14.13, most reactions occur by mechanisms that involve two or more elementary reactions. Each step of the mechanism has its own rate constant and activation energy. Often one step is much slower than the others, and the overall rate of a reaction cannot exceed the rate of the slowest elementary step. Because the slow step limits the overall reaction rate, it is called the rate-determining step (or rate-limiting step). To understand the concept of the rate-determining step for a reaction, consider a toll road with two toll plazas ( Figure 14.21). Cars enter the toll road at point 1 and pass through toll plaza A. They then pass an intermediate point 2 before passing through toll plaza B and arriving at point 3. We can envision this trip along the toll road as occurring in two elementary steps: Step 1: Step 2: Overall: Point 1 Point 2 Point 2 Point 3 Point 1 Point 3 1through toll plaza A2 1through toll plaza B2 1through both toll plazas2 Now suppose that one or more gates at toll plaza A are malfunctioning, so that traffic backs up behind the gates, as depicted in Figure 14.21(a). The rate at which cars can get to point 3 is limited by the rate at which they can get through the traffic jam at plaza A. Thus, step 1 is the rate-determining step of the journey along the toll road. If, however, all gates at A are functioning but one or more at B are not, traffic flows quickly through A but gets backed up at B, as depicted in Figure 14.21(b). In this case, step 2 is the rate-determining step. In the same way, the slowest step in a multistep reaction determines the overall rate. By analogy to Figure 14.21(a), the rate of a fast step following the rate-determining step does not speed up the overall rate. If the slow step is not the first one, as is the case in

70 section 14.6 Reaction Mechanisms 603 Go Figure For which of the two scenarios in the figure will one get from point 1 to point 3 most rapidly? Toll plaza A Toll plaza B (a) Cars slowed at toll plaza A, rate-determining step is passage through A Toll plaza A Toll plaza B (b) Cars slowed at toll plaza B, rate-determining step is passage through B Figure Rate-determining steps in traffic flow on a toll road. Figure 14.21(b), the faster preceding steps produce intermediate products that accumulate before being consumed in the slow step. In either case, the rate-determining step governs the rate law for the overall reaction. Give It Some Thought Why can t the rate law for a reaction generally be deduced from the balanced equation for the reaction? Mechanisms with a Slow Initial Step We can most easily see the relationship between the slow step in a mechanism and the rate law for the overall reaction by considering an example in which the first step in a multistep mechanism is the rate-determining step. Consider the reaction of NO 2 and CO to produce NO and CO 2 (Equation 14.23). Below 225 C, it is found experimentally that the rate law for this reaction is second order in NO 2 and zero order in CO: Rate = k3no Can we propose a reaction mechanism consistent with this rate law? Consider the two-step mechanism:* Step 1: NO 2 1g2 + NO 2 1g2 k 1 NO 3 1g2 + NO1g2 1slow2 Step 2: NO 3 1g2 + CO1g2 k 2 NO 2 1g2 + CO 2 1g2 1fast2 Overall: NO 2 1g2 + CO1g2 NO1g2 + CO 2 1g2 Step 2 is much faster than step 1; that is, k 2 77 k 1, telling us that the intermediate NO 3 1g2 is slowly produced in step 1 and immediately consumed in step 2. Because step 1 is slow and step 2 is fast, step 1 is the rate-determining step. Thus, the rate of the overall reaction depends on the rate of step 1, and the rate law of the *Note the rate constants k 1 and k 2 written above the reaction arrows. The subscript on each rate constant identifies the elementary step involved. Thus, k 1 is the rate constant for step 1, and k 2 is the rate constant for step 2. A negative subscript refers to the rate constant for the reverse of an elementary step. For example, k - 1 is the rate constant for the reverse of the first step.

71 604 chapter 14 Chemical Kinetics overall reaction equals the rate law of step 1. Step 1 is a bimolecular process that has the rate law Rate = k 1 3NO Thus, the rate law predicted by this mechanism agrees with the one observed experimentally. The reactant CO is absent from the rate law because it reacts in a step that follows the rate-determining step. A scientist would not, at this point, say that we have proved that this mechanism is correct. All we can say is that the rate law predicted by the mechanism is consistent with experiment. We can often envision a different sequence of steps that leads to the same rate law. If, however, the predicted rate law of the proposed mechanism disagrees with experiment, we know for certain that the mechanism cannot be correct. Sample Exercise Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: N 2 O1g2 N 2 1g2 + O1g2 1slow2 N 2 O1g2 + O1g2 N 2 1g2 + O 2 1g2 1fast2 (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. Solution Analyze Given a multistep mechanism with the relative speeds of the steps, we are asked to write the overall reaction and the rate law for that overall reaction. Plan (a) Find the overall reaction by adding the elementary steps and eliminating the intermediates. (b) The rate law for the overall reaction will be that of the slow, rate-determining step. Solve (a) Adding the two elementary reactions gives 2 N 2 O1g2 + O1g2 2 N 2 1g2 + 2 O 2 1g2 + O1g2 Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction: 2 N 2 O1g2 2 N 2 1g2 + O 2 1g2 (b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order: Rate = k3n 2 O4 Practice Exercise 1 Let s consider a hypothetical reaction similar to that in Practice Exercise 1 of Sample Exercise 14.13: 2 C + D J + 2 K. You are told that the rate of this reaction is second order overall and second order in [C]. Could any of the following be a rate-determining first step in a reaction mechanism that is consistent with the observed rate law for the reaction (note that substance Z is an intermediate)? (a) C + C K + Z (b) C J + Z (c) C + D J + Z (d) D J + K (e) None of these are consistent with the observed rate law. Practice Exercise 2 Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: O 3 1g2 + 2 NO 2 1g2 N 2 O 5 1g2 + O 2 1g2 The reaction is believed to occur in two steps: O 3 1g2 + NO 2 1g2 NO 3 1g2 + O 2 1g2 NO 3 1g2 + NO 2 1g2 N 2 O 5 1g2 The experimental rate law is rate = k3o 3 43NO 2 4. What can you say about the relative rates of the two steps of the mechanism? Mechanisms with a Fast Initial Step It is possible, but not particularly straightforward, to derive the rate law for a mechanism in which an intermediate is a reactant in the rate-determining step. This situation arises in multistep mechanisms when the first step is fast and therefore not the ratedetermining step. Let s consider one example: the gas-phase reaction of nitric oxide (NO) with bromine 1Br 2 2: 2 NO1g2 + Br 2 1g2 2 NOBr1g2 [14.24]

72 section 14.6 Reaction Mechanisms 605 The experimentally determined rate law for this reaction is second order in NO and first order in Br 2 : Rate = k3no4 2 3Br 2 4 [14.25] We seek a reaction mechanism that is consistent with this rate law. One possibility is that the reaction occurs in a single termolecular step: NO1g2 + NO1g2 + Br 2 1g2 2 NOBr1g2 Rate = k3no4 2 3Br 2 4 [14.26] As noted in Practice Exercise 2 of Exercise 14.13, this does not seem likely because termolecular processes are so rare. Give It Some Thought Why are termolecular elementary steps rare in gas-phase reactions? Let s consider an alternative mechanism that does not involve a termolecular step: Step 1: NO1g2 + Br 2 1g2 k 1 k - 1 NOBr 2 1g2 1fast2 Step 2: NOBr 2 1g2 + NO1g2 k 2 2 NOBr1g2 1slow2 [14.27] In this mechanism, step 1 involves two processes: a forward reaction and its reverse. Because step 2 is the rate-determining step, the rate law for that step governs the rate of the overall reaction: Rate = k 2 3NOBr 2 43NO4 [14.28] Note that NOBr 2 is an intermediate generated in the forward reaction of step 1. Intermediates are usually unstable and have a low, unknown concentration. Thus, the rate law of Equation depends on the unknown concentration of an intermediate, which isn t desirable. We want instead to express the rate law for a reaction in terms of the reactants, or the products if necessary, of the reaction. With the aid of some assumptions, we can express the concentration of the intermediate NOBr 2 in terms of the concentrations of the starting reactants NO and Br 2. We first assume that NOBr 2 is unstable and does not accumulate to any significant extent in the reaction mixture. Once formed, NOBr 2 can be consumed either by reacting with NO to form NOBr or by falling back apart into NO and Br 2. The first of these possibilities is step 2 of our alternative mechanism, a slow process. The second is the reverse of step 1, a unimolecular process: NOBr 2 1g2 k-1 NO1g2 + Br 2 1g2 Because step 2 is slow, we assume that most of the NOBr 2 falls apart according to this reaction. Thus, we have both the forward and reverse reactions of step 1 occurring much faster than step 2. Because they occur rapidly relative to step 2, the forward and reverse reactions of step 1 establish an equilibrium. As in any other dynamic equilibrium, the rate of the forward reaction equals that of the reverse reaction: Solving for 3NOBr 2 4, we have k3no43br 2 4 = k -1 3NOBr 2 4 Rate of forward reaction Rate of reverse reaction 3NOBr 2 4 = k 1 k -1 3NO43Br 2 4 Substituting this relationship into Equation 14.28, we have Rate = k 2 k 1 k -1 3NO43Br 2 43NO4 = k3no4 2 3Br 2 4 where the experimental rate constant k equals k 2 k 1 >k -1. This expression is consistent with the experimental rate law (Equation 14.25). Thus, our alternative mechanism

73 606 chapter 14 Chemical Kinetics (Equation 14.27), which involves two steps but only unimolecular and bimolecular processes, is far more probable than the single-step termolecular mechanism of Equation In general, whenever a fast step precedes a slow one, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step. Sample Exercise Deriving the Rate Law for a Mechanism with a Fast Initial Step Show that the following mechanism for Equation also produces a rate law consistent with the experimentally observed one: Step 1: NO1g2 + NO1g2 k 1 k -1 N 2 O 2 1g2 1fast, equilibrium2 Step 2: N 2 O 2 1g2 + Br 2 1g2 k 2 2 NOBr1g2 1slow2 Solution Analyze We are given a mechanism with a fast initial step and asked to write the rate law for the overall reaction. Plan The rate law of the slow elementary step in a mechanism determines the rate law for the overall reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case, the slow step involves the intermediate N 2 O 2 as a reactant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of starting substances. Thus, we must relate the concentration of N 2 O 2 to the concentration of NO by assuming that an equilibrium is established in the first step. Solve The second step is rate determining, so the overall rate is Rate = k 2 3N 2 O 2 43Br 2 4 We solve for the concentration of the intermediate N 2 O 2 by assuming that an equilibrium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal: k 1 3NO4 2 = k -1 3N 2 O 2 4 Solving for the concentration of the intermediate, N 2 O 2, gives 3N 2 O 2 4 = k 1 k -1 3NO4 2 Substituting this expression into the rate expression gives Rate = k 2 k 1 k -1 3NO4 2 3Br 2 4 = k3no4 2 3Br 2 4 Thus, this mechanism also yields a rate law consistent with the experimental one. Remember: There may be more than one mechanism that leads to an observed experimental rate law! Practice Exercise 1 Consider the following hypothetical reaction: 2 P + Q 2 R + S. The following mechanism is proposed for this reaction: P + P T (fast) Q + T R + U 1slow2 U R + S 1fast2 Substances T and U are unstable intermediates. What rate law is predicted by this mechanism? (a) Rate = k3p4 2 (b) Rate = k3p43q4 (c) Rate = k3p4 2 3Q4 (d) Rate = k3p43q4 2 (e) Rate = k3u4 Practice Exercise 2 The first step of a mechanism involving the reaction of bromine is Br 2 1g2 k 1 2 Br1g2 1fast, equilibrium2 k -1 What is the expression relating the concentration of Br(g) to that of Br 2 1g2? So far we have considered only three reaction mechanisms: one for a reaction that occurs in a single elementary step and two for simple multistep reactions where there is one rate-determining step. There are other more complex mechanisms, however. If you take a biochemistry class, for example, you will learn about cases in which the concentration of an intermediate cannot be neglected in deriving the rate law. Furthermore, some mechanisms require a large number of steps, sometimes 35 or more, to arrive at a rate law that agrees with experimental data! 14.7 Catalysis A catalyst is a substance that changes the speed of a chemical reaction without undergoing a permanent chemical change itself. Most reactions in the body, the atmosphere, and the oceans occur with the help of catalysts. Much industrial chemical research is

74 section 14.7 Catalysis 607 devoted to the search for more effective catalysts for reactions of commercial importance. Extensive research efforts also are devoted to finding means of inhibiting or removing certain catalysts that promote undesirable reactions, such as those that corrode metals, age our bodies, and cause tooth decay. omogeneous Catalysis A catalyst that is present in the same phase as the reactants in a reaction mixture is called a homogeneous catalyst. Examples abound both in solution and in the gas phase. Consider, for example, the decomposition of aqueous hydrogen peroxide, 2 O 2 1aq2, into water and oxygen: 2 2 O 2 1aq2 2 2 O1l2 + O 2 1g2 [14.29] In the absence of a catalyst, this reaction occurs extremely slowly. Many substances are capable of catalyzing the reaction, however, including bromide ion, which reacts with hydrogen peroxide in acidic solution, forming aqueous bromine and water ( Figure 14.22). 2 Br - 1aq2 + 2 O 2 1aq Br 2 1aq O1l2 [14.30] Go Figure Why does the solution in the middle cylinder have a brownish color? NaBr catalyst about to be added to reaction mixture 2 O 2 (aq) in acidic solution 2 Br (aq) + 2 O 2 (aq) (aq) Br 2 (aq) O(l) brown Figure omogeneous catalysis. Effect of catalyst on the speed of hydrogen peroxide decomposition to water and oxygen gas. Br 2 (aq) + 2 O 2 (aq) 2 Br (aq) (aq) + O 2 (g) colorless bubbles

75 608 chapter 14 Chemical Kinetics Go Figure Where are the intermediates and transition states in this diagram? Potential energy 2 2 O O Br Catalyzed reaction Reaction progress Figure Energy profiles for the uncatalyzed and bromide-catalyzed decomposition of h 2 o 2. If this were the complete reaction, bromide ion would not be a catalyst because it undergoes chemical change during the reaction. owever, hydrogen peroxide also reacts with the Br 2 1aq2 generated in Equation 14.30: Br 2 1aq2 + 2 O 2 1aq2 2 Br - 1aq aq2 + O 2 1g2 [14.31] The sum of Equations and is just Equation 14.29, a result which you can check for yourself. When the 2 O 2 has been completely decomposed, we are left with a colorless solution of Br - 1aq2, which means that this ion is indeed a catalyst of the reaction because it speeds up the reaction without itself undergoing any net change. In contrast, Br 2 is an intermediate because it is first formed (Equation 14.30) and then consumed (Equation 14.31). Neither the catalyst nor the intermediate appears in the equation for the overall reaction. Notice, however, that the catalyst is there at the start of the reaction, whereas the intermediate is formed during the course of the reaction. ow does a catalyst work? If we think about the general form of rate laws (Equation 14.7, rate = k3a4 m 3B4 n ), we must Uncatalyzed reaction 2 2 O + O O + O Br conclude that the catalyst must affect the numerical value of k, the rate constant. On the basis of the Arrhenius equation (Equation 14.19, k = Ae -E a>rt ), k is determined by the activation energy 1E a 2 and the frequency factor (A). A catalyst may affect the rate of reaction by altering the value of either E a or A. We can envision this happening in two ways: The catalyst could provide a new mechanism for the reaction that has an E a value lower than the E a value for the uncatalyzed reaction, or the catalyst could assist in the orientation of reactants and so increase A. The most dramatic catalytic effects come from lowering E a. As a general rule, a catalyst lowers the overall activation energy for a chemical reaction. A catalyst can lower the activation energy for a reaction by providing a different mechanism for the reaction. In the decomposition of hydrogen peroxide, for example, two successive reactions of 2 O 2, first with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decomposition, both of them must have significantly lower activation energies than the uncatalyzed decomposition ( Figure 14.23). eterogeneous Catalysis A heterogeneous catalyst is one that exists in a phase different from the phase of the reactant molecules, usually as a solid in contact with either gaseous reactants or reactants in a liquid solution. Many industrially important reactions are catalyzed by the surfaces of solids. For example, raw petroleum is transformed into smaller hydrocarbon molecules by using what are called cracking catalysts. eterogeneous catalysts are often composed of metals or metal oxides. The initial step in heterogeneous catalysis is usually adsorption of reactants. Adsorption refers to the binding of molecules to a surface, whereas absorption refers to the uptake of molecules into the interior of a substance. (Section 13.6) Adsorption occurs because the atoms or ions at the surface of a solid are extremely reactive. Because the catalyzed reaction occurs on the surface, special methods are often used to prepare catalysts so that they have very large surface areas. Unlike their counterparts in the interior of the substance, surface atoms and ions have unused bonding capacity that can be used to bond molecules from the gas or solution phase to the surface of the solid.

76 section 14.7 Catalysis 609 ydrogen Carbon 3 One free has attached to C 2 4 to form C 2 5 (ethyl group) intermediate 1 2 and C 2 4 adsorbed 2 After bond breaks, atoms on metal surface migrate along metal surface 4 Second free about to attach to C 2 5 intermediate to form C Ethane, C 2 6, desorbs from metal surface Figure eterogeneous catalysis. Mechanism for reaction of ethylene with hydrogen on a catalytic surface. The reaction of hydrogen gas with ethylene gas to form ethane gas provides an example of heterogeneous catalysis: C 2 4 1g g2 C 2 6 1g2 = -137 kj>mol [14.32] Ethylene Ethane Even though this reaction is exothermic, it occurs very slowly in the absence of a catalyst. In the presence of a finely powdered metal, however, such as nickel, palladium, or platinum, the reaction occurs easily at room temperature via the mechanism diagrammed in Figure Both ethylene and hydrogen are adsorbed on the metal surface. Upon adsorption, the bond of 2 breaks, leaving two atoms initially bonded to the metal surface but relatively free to move. When a hydrogen encounters an adsorbed ethylene molecule, it can form a s bond to one of the carbon atoms, effectively destroying the C C p bond and leaving an ethyl group 1C bonded to the surface via a metal-to-carbon s bond. This s bond is relatively weak, so when the other carbon atom also encounters a hydrogen atom, a sixth C s bond is readily formed, and an ethane molecule 1C is released from the metal surface. Give It Some Thought ow does a homogeneous catalyst compare with a heterogeneous one regarding the ease of recovery of the catalyst from the reaction mixture? Enzymes The human body is characterized by an extremely complex system of interrelated chemical reactions, all of which must occur at carefully controlled rates to maintain life. A large number of marvelously efficient biological catalysts known as enzymes are necessary for many of these reactions to occur at suitable rates. Most enzymes

77 610 chapter 14 Chemical Kinetics Chemistry Put to Work Catalytic Converters eterogeneous catalysis plays a major role in the fight against urban air pollution. Two components of automobile exhausts that help form photochemical smog are nitrogen oxides and unburned hydrocarbons. In addition, automobile exhaust may contain considerable quantities of carbon monoxide. Even with the most careful attention to engine design, it is impossible under normal driving conditions to reduce the quantity of these pollutants to an acceptable level in the exhaust gases. It is therefore necessary to remove them from the exhaust before they are vented to the air. This removal is accomplished in the catalytic converter. The catalytic converter, which is part of an automobile s exhaust system, must perform two functions: (1) oxidation of CO and unburned hydrocarbons 1C x y 2 to carbon dioxide and water, and (2) reduction of nitrogen oxides to nitrogen gas: CO, C x y O 2 CO O NO, NO 2 N 2 These two functions require different catalysts, so the development of a successful catalyst system is a difficult challenge. The catalysts must be effective over a wide range of operating temperatures. They must continue to be active despite the fact that various components of the exhaust can block the active sites of the catalyst. And the catalysts must be sufficiently rugged to withstand exhaust gas turbulence and the mechanical shocks of driving under various conditions for thousands of miles. Catalysts that promote the combustion of CO and hydrocarbons are, in general, the transition-metal oxides and the noble metals. These materials are supported on a structure ( Figure 14.25) that allows the best possible contact between the flowing exhaust gas and the catalyst surface. A honeycomb structure made from alumina 1Al 2 O 3 2 and impregnated with the catalyst is employed. Such catalysts operate by first adsorbing oxygen gas present in the exhaust gas. This adsorption weakens the O O bond in O 2, so that oxygen atoms are available for reaction with adsorbed CO to form CO 2. ydrocarbon oxidation probably proceeds somewhat similarly, with the hydrocarbons first being adsorbed followed by rupture of a C bond. Transition-metal oxides and noble metals are also the most effective catalysts for reduction of NO to N 2 and O 2. The catalysts that are most effective in one reaction, however, are usually much less effective in the other. It is therefore necessary to have two catalytic components. Catalytic converters contain remarkably efficient heterogeneous catalysts. The automotive exhaust gases are in contact with the catalyst for only 100 to 400 ms, but in this very short time, 96% of the hydrocarbons and CO is converted to CO 2 and 2 O, and the emission of nitrogen oxides is reduced by 76%. There are costs as well as benefits associated with the use of catalytic converters, one being that some of the metals are very expensive. Catalytic converters currently account for about 35% of the platinum, 65% of the palladium, and 95% of the rhodium used annually. All of these metals, which come mainly from Russia and South Africa, can be far more expensive than gold. Related Exercises: 14.62, 14.81, 14.82, Figure Cross section of a catalytic converter. Metal catalyst impregnated on high surface area alumina honeycomb Go Figure Why is the reaction faster when the liver is ground up? Catalase present in beef liver rapidly converts 2 O 2 to water and O 2 Ground beef liver Figure Enzymes speed up reactions. O 2 gas 2 O 2 and 2 O are large protein molecules with molecular weights ranging from about 10,000 to about 1 million amu. They are very selective in the reactions they catalyze, and some are absolutely specific, operating for only one substance in only one reaction. The decomposition of hydrogen peroxide, for example, is an important biological process. Because hydrogen peroxide is strongly oxidizing, it can be physiologically harmful. For this reason, the blood and liver of mammals contain an enzyme, catalase, that catalyzes the decomposition of hydrogen peroxide into water and oxygen (Equation 14.29). Figure shows the dramatic acceleration of this chemical reaction by the catalase in beef liver. The reaction any given enzyme catalyzes takes place at a specific location in the enzyme called the active site. The substances that react at this site are called substrates. The lock-and-key model provides a simple explanation for the specificity of an enzyme ( Figure 14.27). The substrate is pictured as fitting neatly into the active site, much like a key fits into a lock. Lysozyme is an enzyme that is important to the functioning of our immune system because it accelerates reactions that damage (or lyse ) bacterial cell walls. Figure shows a model of the enzyme lysozyme without and with a bound substrate molecule.

78 section 14.7 Catalysis 611 Go Figure Which molecules must bind more tightly to the active site, substrates or products? Substrate Products Enzyme Enzyme substrate complex Enzyme Figure Lock-and-key model for enzyme action. The combination of enzyme and substrate is called the enzyme substrate complex. Although Figure shows both the active site and its substrate as having a fixed shape, the active site is often fairly flexible and so may change shape as it binds the substrate. The binding between substrate and active site involves dipole dipole attractions, hydrogen bonds, and dispersion forces. (Section 11.2) As substrate molecules enter the active site, they are somehow activated so that they are capable of reacting rapidly. This activation process may occur, for example, by the withdrawal or donation of electron density from a particular bond or group of atoms in the enzyme s active site. In addition, the substrate may become distorted in the process of fitting into the active site and made more reactive. Once the reaction occurs, the products depart from the active site, allowing another substrate molecule to enter. The activity of an enzyme is destroyed if some molecule other than the substrate specific to that enzyme binds to the active site and blocks entry of the substrate. Such substances are called enzyme inhibitors. Nerve poisons and certain toxic metal ions, such as lead and mercury, are believed to act in this way to inhibit enzyme activity. Some other poisons act by attaching elsewhere on the enzyme, thereby distorting the active site so that the substrate no longer fits. Enzymes are enormously more efficient than nonbiochemical catalysts. The number of individual catalyzed reaction events occurring at a particular active site, called the turnover number, is generally in the range of 10 3 to 10 7 per second. Such large turnover numbers correspond to very low activation energies. Compared with a simple chemical catalyst, enzymes can increase the rate constant for a given reaction by a millionfold or more. Figure Lysozyme was one of the first enzymes for which a structure function relationship was described. This model shows how the substrate (yellow) fits into the active site of the enzyme. Give It Some Thought Is it reasonable to say that enzymes lower the energy of the transition state for a reaction?

79 612 chapter 14 Chemical Kinetics Chemistry and Life Nitrogen Fixation and Nitrogenase Nitrogen is one of the most essential elements in living organisms, found in many compounds vital to life, including proteins, nucleic acids, vitamins, and hormones. Nitrogen is continually cycling through the biosphere in Ammonia (N 3 ) Nitrogen in atmosphere (N 2 ) Groundwater Nitrate (NO 3 ) Figure Simplified picture of the nitrogen cycle. Plant and animal waste Fe S various forms, as shown in Figure For example, certain microorganisms convert the nitrogen in animal waste and dead plants and animals into N 2 1g2, which then returns to the atmosphere. For the food chain to be sustained, there must be a means of converting atmospheric N 2 1g2 into a form plants can use. For this reason, if a chemist were asked to name the most important chemical reaction in the world, she might easily say nitrogen fixation, the process by which atmospheric N 2 1g2 is converted into compounds suitable for plant use. Some fixed nitrogen results from the action of lightning on the atmosphere, and some is produced industrially using a process we will discuss in Nitrite (NO 2 ) Nitrogen fixation caused by lightning Chapter 15. About 60% of fixed nitrogen, however, is a consequence of the action of the remarkable and complex enzyme nitrogenase. This enzyme is not present in humans or other animals; rather, it is found in bacteria that live in the root nodules of certain plants, such as the legumes clover and alfalfa. Nitrogenase converts N 2 into N 3, a process that, in the absence of a catalyst, has a very large activation energy. This process is a reduction reaction in which the oxidation state of N is reduced from 0 in N 2 to -3 in N 3. The mechanism by which nitrogenase reduces N 2 is not fully understood. Like many other enzymes, including catalase, the active site of nitrogenase contains transition-metal atoms; such enzymes are called metalloenzymes. Because transition metals can readily change oxidation state, metalloenzymes are especially useful for effecting transformations in which substrates are either oxidized or reduced. It has been known for nearly 40 years that a portion of nitrogenase contains iron and molybdenum atoms. This portion, called the FeMo-cofactor, is thought to serve as the active site of the enzyme. The FeMo-cofactor of nitrogenase is a cluster of seven Fe atoms and one Mo atom, all linked by sulfur atoms ( Figure 14.30). It is one of the wonders of life that simple bacteria can contain beautifully complex and vitally important enzymes such as nitrogenase. Because of this enzyme, nitrogen is continually cycled between its comparatively inert role in the atmosphere and its critical role in living organisms. Without nitrogenase, life as we know it could not exist on Earth. Mo Related Exercises: 14.86, , Figure The FeMo-cofactor of nitrogenase. Nitrogenase is found in nodules in the roots of certain plants, such as the white clover roots shown at the left. The cofactor, which is thought to be the active site of the enzyme, contains seven Fe atoms and one Mo atom, linked by sulfur atoms. The molecules on the outside of the cofactor connect it to the rest of the protein.

80 section 14.7 Catalysis 613 Sample Integrative Exercise Putting Concepts Together Formic acid (COO) decomposes in the gas phase at elevated temperatures as follows: COO1g2 CO 2 1g g2 The uncatalyzed decomposition reaction is determined to be first order. A graph of the partial pressure of COO versus time for decomposition at 838 K is shown as the red curve in Figure When a small amount of solid ZnO is added to the reaction chamber, the partial pressure of acid versus time varies as shown by the blue curve in Figure (a) Estimate the half-life and first-order rate constant for formic acid decomposition. (b) What can you conclude from the effect of added ZnO on the decomposition of formic acid? (c) The progress of the reaction was followed by measuring the partial pressure of formic acid vapor at selected times. Suppose that, instead, we had plotted the concentration of formic acid in units of mol>l. What effect would this have had on the calculated value of k? (d) The pressure of formic acid vapor at the start of the reaction is 3.00 * 10 2 torr. Assuming constant temperature and ideal-gas behavior, what is the pressure in the system at the end of the reaction? If the volume of the reaction chamber is 436 cm 3, how many moles of gas occupy the reaction chamber at the end of the reaction? (e) The standard heat of formation of formic acid vapor is f = kj>mol. Calculate for the overall reaction. If the activation energy 1E a 2 for the reaction is 184 kj>mol, sketch an approximate energy profile for the reaction, and label E a,, and the transition state. Solution (a) The initial pressure of COO is 3.00 * 10 2 torr. On the graph we move to the level at which the partial pressure of COO is 1.50 * 10 2 torr, half the initial value. This corresponds to a time of about 6.60 * 10 2 s, which is therefore the half-life. The first-order rate constant is given by Equation 14.15: k = 0.693>t 1>2 = 0.693>660 s = 1.05 * 10-3 s -1. (b) The reaction proceeds much more rapidly in the presence of solid ZnO, so the surface of the oxide must be acting as a catalyst for the decomposition of the acid. This is an example of heterogeneous catalysis. (c) If we had graphed the concentration of formic acid in units of moles per liter, we would still have determined that the half-life for decomposition is 660 s, and we would have computed the same value for k. Because the units for k are s -1, the value for k is independent of the units used for concentration. (d) According to the stoichiometry of the reaction, two moles of product are formed for each mole of reactant. When reaction is completed, therefore, the pressure will be 600 torr, just twice the initial pressure, assuming ideal-gas behavior. (Because we are working at quite high temperature and fairly low gas pressure, assuming ideal-gas behavior is reasonable.) The number of moles of gas present can be calculated using the ideal-gas equation (Section 10.4): n = PV RT = 1600>760 atm L L@atm>mol@K21838 K2 = 5.00 * 10-3 mol Pressure, COO (torr) COO(g) alone COO(g) with ZnO(s) Time (s) Figure Variation in pressure of COO(g) as a function of time at 838 K. (e) We first calculate the overall change in energy, (Section 5.7 and Appendix C), as in = f 1CO 2 1g22 + f 1 2 1g22 - f 1COO1g22 = kj>mol kj>mol2 = kj>mol From this and the given value for E a, we can draw an approximate energy profile for the reaction, in analogy to Figure Transition state Potential energy E a Reaction progress

81 614 chapter 14 Chemical Kinetics Chapter Summary and Key Terms Introduction to Kinetics (Introduction and Section 14.1) Chemical kinetics is the area of chemistry in which reaction rates are studied. Factors that affect reaction rate are the physical state of the reactants; concentration; temperature; and the presence of catalysts. Reaction Rates (Section 14.2) Reaction rates are usually expressed as changes in concentration per unit time: Typically, for reactions in solution, rates are given in units of molarity per second 1M>s2. For most reactions, a plot of molarity versus time shows that the rate slows down as the reaction proceeds. The instantaneous rate is the slope of a line drawn tangent to the concentration-versus-time curve at a specific time. Rates can be written in terms of the appearance of products or the disappearance of reactants; the stoichiometry of the reaction dictates the relationship between rates of appearance and disappearance. Rates and Concentration (Section 14.3) The quantitative relationship between rate and concentration is expressed by a rate law, which usually has the following form: Rate = k3reactant 14 m 3reactant 24 n c The constant k in the rate law is called the rate constant; the exponents m, n, and so forth are called reaction orders for the reactants. The sum of the reaction orders gives the overall reaction order. Reaction orders must be determined experimentally. The units of the rate constant depend on the overall reaction order. For a reaction in which the overall reaction order is 1, k has units of s -1 ; for one in which the overall reaction order is 2, k has units of M -1 s -1. Spectroscopy is one technique that can be used to monitor the course of a reaction. According to Beer s law, the absorption of electromagnetic radiation by a substance at a particular wavelength is directly proportional to its concentration. Concentration and Time (Section 14.4) Rate laws can be used to determine the concentrations of reactants or products at any time during a reaction. In a first-order reaction the rate is proportional to the concentration of a single reactant raised to the first power: Rate = k3a4. In such cases the integrated form of the rate law is ln3a4 t = -kt + ln3a4 0, where 3A4 t is the concentration of reactant A at time t, k is the rate constant, and 3A4 0 is the initial concentration of A. Thus, for a first-order reaction, a graph of ln[a] versus time yields a straight line of slope -k. A second-order reaction is one for which the overall reaction order is 2. If a second-order rate law depends on the concentration of only one reactant, then rate = k3a4 2, and the time dependence of [A] is given by the integrated form of the rate law: 1>3A4 t = 1>3A4 0 + kt. In this case, a graph of 1>3A4 t versus time yields a straight line. A zeroorder reaction is one for which the overall reaction order is 0. Rate = k if the reaction is zero order. The half-life of a reaction, t 1>2, is the time required for the concentration of a reactant to drop to one-half of its original value. For a first-order reaction, the half-life depends only on the rate constant and not on the initial concentration: t 1>2 = 0.693>k. The half-life of a second-order reaction depends on both the rate constant and the initial concentration of A: t 1>2 = 1>k3A4 0. The Effect of Temperature on Rates (Section 14.5) The collision model, which assumes that reactions occur as a result of collisions between molecules, helps explain why the magnitudes of rate constants increase with increasing temperature. The greater the kinetic energy of the colliding molecules, the greater is the energy of collision. The minimum energy required for a reaction to occur is called the activation energy, E a. A collision with energy E a or greater can cause the atoms of the colliding molecules to reach the activated complex (or transition state), which is the highest energy arrangement in the pathway from reactants to products. Even if a collision is energetic enough, it may not lead to reaction; the reactants must also be correctly oriented relative to one another in order for a collision to be effective. Because the kinetic energy of molecules depends on temperature, the rate constant of a reaction is very dependent on temperature. The relationship between k and temperature is given by the Arrhenius equation: k = Ae -E a>rt. The term A is called the frequency factor; it relates to the number of collisions that are favorably oriented for reaction. The Arrhenius equation is often used in logarithmic form: ln k = ln A - E a >RT. Thus, a graph of ln k versus 1>T yields a straight line with slope -E a >R. Reaction Mechanisms (Section 14.6) A reaction mechanism details the individual steps that occur in the course of a reaction. Each of these steps, called elementary reactions, has a well-defined rate law that depends on the number of molecules (the molecularity) of the step. Elementary reactions are defined as either unimolecular, bimolecular, or termolecular, depending on whether one, two, or three reactant molecules are involved, respectively. Termolecular elementary reactions are very rare. Unimolecular, bimolecular, and termolecular reactions follow rate laws that are first order overall, second order overall, and third order overall, respectively. Many reactions occur by a multistep mechanism, involving two or more elementary reactions, or steps. An intermediate is produced in one elementary step, is consumed in a later elementary step, and therefore does not appear in the overall equation for the reaction. When a mechanism has several elementary steps, the overall rate is limited by the slowest elementary step, called the rate-determining step. A fast elementary step that follows the rate-determining step will have no effect on the rate law of the reaction. A fast step that precedes the rate-determining step often creates an equilibrium that involves an intermediate. For a mechanism to be valid, the rate law predicted by the mechanism must be the same as that observed experimentally. Catalysts (Section 14.7) A catalyst is a substance that increases the rate of a reaction without undergoing a net chemical change itself. It does so by providing a different mechanism for the reaction, one that has a lower activation energy. A homogeneous catalyst is one that is in the same phase as the reactants. A heterogeneous catalyst has a different phase from the reactants. Finely divided metals are often used as heterogeneous catalysts for solution- and gas-phase reactions. Reacting molecules can undergo binding, or adsorption, at the surface of the catalyst. The adsorption of a reactant at specific sites on the surface makes bond breaking easier, lowering the activation energy. Catalysis in living organisms is achieved by enzymes, large protein molecules that usually catalyze a very specific reaction. The specific reactant molecules involved in an enzymatic reaction are called substrates. The site of the enzyme where the catalysis occurs is called the active site. In the lock-and-key model for enzyme catalysis, substrate molecules bind very specifically to the active site of the enzyme, after which they can undergo reaction. Learning Outcomes After studying this chapter, you should be able to: List the factors that affect the rate of chemical reactions. (Section 14.1) Determine the rate of a reaction given time and concentration. (Section 14.2) Relate the rate of formation of products and the rate of disappearance of reactants given the balanced chemical equation for the reaction. (Section 14.2)

82 Exercises 615 Explain the form and meaning of a rate law, including the ideas of reaction order and rate constant. (Section 14.3) Determine the rate law and rate constant for a reaction from a series of experiments given the measured rates for various concentrations of reactants. (Section 14.3) Apply the integrated form of a rate law to determine the concentration of a reactant at a given time. (Section 14.4) Apply the relationship between the rate constant of a first-order reaction and its half-life. (Section 14.4) Explain how the activation energy affects a rate and be able to use the Arrhenius equation. (Section 14.5) Predict a rate law for a reaction having a multistep mechanism given the individual steps in the mechanism. (Section 14.6) Explain the principles underlying catalysis. (Section 14.7) Key Equations Rate = - 1 a 3A4 t = - 1 b 3B4 t = 1 c 3C4 t = 1 d 3D4 t [14.4] Definition of reaction rate in terms of the components of the balanced chemical equation a A + b B c C + d D Rate = k3a4 m 3B4 n [14.7] General form of a rate law for the reaction A + B products ln3a4 t - ln3a4 0 = -kt or ln 3A4 t 3A4 0 = -kt [14.12] The integrated form of a first-order rate law for the reaction A products 1 = kt + 1 [14.14] The integrated form of the second-order rate law for the reaction 3A4 t 3A4 0 A products t 1>2 = k k = Ae -E a>rt [14.15] Relating the half-life and rate constant for a first-order reaction [14.19] The Arrhenius equation, which expresses how the rate constant depends on temperature E a ln k = - RT + ln A [14.20] Logarithmic form of the Arrhenius equation Exercises Visualizing Concepts 14.1 An automotive fuel injector dispenses a fine spray of gasoline into the automobile cylinder, as shown in the bottom drawing here. When an injector gets clogged, as shown in the top drawing, the spray is not as fine or even and the performance of the car declines. ow is this observation related to chemical kinetics? [Section 14.1] 14.2 Consider the following graph of the concentration of a substance X over time. Is each of the following statements true or false? (a) X is a product of the reaction. (b) The rate of the reaction remains the same as time progresses. (c) The average rate between points 1 and 2 is greater than the average rate between points 1 and 3. (d) As time progresses, the curve will eventually turn downward toward the x-axis. [Section 14.2] 3 [X] 2 1 Time

83 616 chapter 14 Chemical Kinetics 14.3 You study the rate of a reaction, measuring both the concentration of the reactant and the concentration of the product as a function of time, and obtain the following results: of the following mixtures will have the fastest initial rate? [Section 14.3] A Concentration (1) (2) (3) Time (a) Which chemical equation is consistent with these data: (i) A B, (ii) B A, (iii) A 2 B, (iv) B 2 A? (b) Write equivalent expressions for the rate of the reaction in terms of the appearance or disappearance of the two substances. [Section 14.2] 14.4 Suppose that for the reaction K + L M, you monitor the production of M over time, and then plot the following graph from your data: B 14.6 A friend studies a first-order reaction and obtains the following three graphs for experiments done at two different temperatures. (a) Which two graphs represent experiments done at the same temperature? What accounts for the difference in these two graphs? In what way are they the same? (b) Which two graphs represent experiments done with the same starting concentration but at different temperatures? Which graph probably represents the lower temperature? ow do you know? [Section 14.4] ln [A] 3 2 [M] 1 Time t, min 14.7 (a) Given the following diagrams at t = 0 min and t = 30 min, what is the half-life of the reaction if it follows first-order kinetics? (a) Is the reaction occurring at a constant rate from t = 0 to t = 15 min? (b) Is the reaction completed at t = 15 min? (c) Suppose the reaction as plotted here were started with 0.20 mol K and 0.40 mol L. After 30 min, an additional 0.20 mol K are added to the reaction mixture. Which of the following correctly describes how the plot would look from t = 30 min to t = 60 min? (i) [M] would remain at the same constant value it has at t = 30 min, (ii) [M] would increase with the same slope as t = 0 to 15 min, until t = 45 min at which point the plot becomes horizontal again, or (iii) [M] decreases and reaches 0 at t = 45 min. [Section 14.2] 14.5 The following diagrams represent mixtures of NO(g) and O 2 1g2. These two substances react as follows: 2 NO1g2 + O 2 1g2 2 NO 2 1g2 It has been determined experimentally that the rate is second order in NO and first order in O 2. Based on this fact, which t = 0 min t = 30 min (b) After four half-life periods for a first-order reaction, what fraction of reactant remains? [Section 14.4] 14.8 Which of the following linear plots do you expect for a reaction A products if the kinetics are (a) zero order, (b) first order, or (c) second order? [Section 14.4]

84 Exercises 617 (i) (iii) (v) ln[a] 1/[A] [A] Time Time Time (ii) [A] 1/[A] ln[a] (iv) (vi) Time Time Time 14.9 The following diagram shows a reaction profile. Label the components indicated by the boxes. [Section 14.5] The following graph shows two different reaction pathways for the same overall reaction at the same temperature. Is each of the following statements true or false? (a) The rate is faster for the red path than for the blue path. (b) For both paths, the rate of the reverse reaction is slower than the rate of the forward reaction. (c) The energy change E is the same for both paths. [Section 14.6] Potential energy Reaction progress Consider the diagram that follows, which represents two steps in an overall reaction. The red spheres are oxygen, the blue ones nitrogen, and the green ones fluorine. (a) Write the chemical equation for each step in the reaction. (b) Write the equation for the overall reaction. (c) Identify the intermediate in the mechanism. (d) Write the rate law for the overall reaction if the first step is the slow, rate-determining step. [Section 14.6] (2) + + Energy (1) (3) (4) + Reaction progress The accompanying graph shows plots of ln k versus 1>T for two different reactions. The plots have been extrapolated to the y-intercepts. Which reaction (red or blue) has (a) the larger value for E a, and (b) the larger value for the frequency factor, A? [Section 14.5] Based on the following reaction profile, how many intermediates are formed in the reaction A C? ow many transition states are there? Which step, A B or B C, is the faster? For the reaction A C, is E positive, negative, or zero? [Section 14.6] ln k Potential energy A B C 1/T Reaction progress

85 618 chapter 14 Chemical Kinetics Draw a possible transition state for the bimolecular reaction depicted here. (The blue spheres are nitrogen atoms, and the red ones are oxygen atoms.) Use dashed lines to represent the bonds that are in the process of being broken or made in the transition state. [Section 14.6] The following diagram represents an imaginary two-step mechanism. Let the red spheres represent element A, the green ones element B, and the blue ones element C. (a) Write the equation for the net reaction that is occurring. (b) Identify the intermediate. (c) Identify the catalyst. [Sections 14.6 and 14.7] Draw a graph showing the reaction pathway for an overall exothermic reaction with two intermediates that are produced at different rates. On your graph indicate the reactants, products, intermediates, transition states, and activation energies. [Sections 14.6 and 14.7] Reaction Rates (Sections 14.1 and 14.2) (a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products? (a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate? Consider the following hypothetical aqueous reaction: A1aq2 S B1aq2. A flask is charged with mol of A in a total volume of ml. The following data are collected: Time (min) Moles of A (a) Calculate the number of moles of B at each time in the table, assuming that there are no molecules of B at time zero, and that A cleanly converts to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 10-min interval in units of M>s. (c) Between t = 10 min and t = 30 min, what is the average rate of appearance of B in units of M>s? Assume that the volume of the solution is constant A flask is charged with mol of A and allowed to react to form B according to the hypothetical gas-phase reaction A1g2 B1g2. The following data are collected: Time (s) Moles of A (a) Calculate the number of moles of B at each time in the table, assuming that A is cleanly converted to B with no intermediates. (b) Calculate the average rate of disappearance of A for each 40-s interval in units of mol>s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A? The isomerization of methyl isonitrile 1C 3 NC2 to acetonitrile 1C 3 CN2 was studied in the gas phase at 215 C, and the following data were obtained: Time (s) 3Ch 3 nc4 1M , , (a) Calculate the average rate of reaction, in M>s, for the time interval between each measurement. (b) Calculate the average rate of reaction over the entire time of the data from t = 0 to t = 15,000 s. (c) Which is greater, the average rate between t = 2000 and t = 12,000 s, or between t = 8000 and t = 15,000 s? (d) Graph 3C 3 NC4 versus time and determine the instantaneous rates in M>s at t = 5000 s and t = 8000 s The rate of disappearance of Cl was measured for the following reaction: C 3 O1aq2 + Cl1aq2 C 3 Cl1aq2 + 2 O1l2 The following data were collected: Time (min) [Cl] (M) (a) Calculate the average rate of reaction, in M>s, for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from t = 0.0 min to t = min. (c) Which is greater, the average rate between t = 54.0 and t = min, or between t = and t = min? (d) Graph [Cl] versus time and determine the instantaneous rates in M>min and M>s at t = 75.0 min and t = 250 min For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) 2 O 2 1g2 2 1g2 + O 2 1g2 (b) 2 N 2 O1g2 2 N 2 1g2 + O 2 1g2 (c) N 2 1g g2 2 N 3 1g2 (d) C 2 5 N 2 1g2 C 2 4 1g2 + N 3 1g2

86 Exercises For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) 2 2 O1g g2 + O 2 1g2 (b) 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2 (c) 2 NO1g g2 N 2 1g O1g2 (d) N 2 1g g2 N 2 4 1g ( a ) Consider the combustion of 2 1g2: 2 2 1g2 + O 2 1g2 2 2 O1g2. If hydrogen is burning at the rate of 0.48 mol>s, what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of 56 torr/min, what is the rate of change of the total pressure of the vessel? (a) Consider the combustion of ethylene, C 2 4 1g2 + 3 O 2 1g2 2 CO 2 1g O1g2. If the concentration of C 2 4 is decreasing at the rate of M>s, what are the rates of change in the concentrations of CO 2 and 2 O? (b) The rate of decrease in N 2 4 partial pressure in a closed reaction vessel from the reaction N 2 4 1g g2 2 N 3 1g2 is 74 torr per hour. What are the rates of change of N 3 partial pressure and total pressure in the vessel? Rate Laws (Section 14.3) A reaction A + B C obeys the following rate law: Rate = k3b4 2. (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for A and B? What is the overall reaction order? (c) What are the units of the rate constant? Consider a hypothetical reaction between A, B, and C that is first order in A, zero order in B, and second order in C. (a) Write the rate law for the reaction. (b) ow does the rate change when [A] is doubled and the other reactant concentrations are held constant? (c) ow does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) ow does the rate change when [C] is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half? The decomposition reaction of N 2 O 5 in carbon tetrachloride is 2 N 2 O 5 4 NO 2 + O 2. The rate law is first order in N 2 O 5. At 64 C the rate constant is 4.82 * 10-3 s -1. (a) Write the rate law for the reaction. (b) What is the rate of reaction when 3N 2 O 5 4 = M? (c) What happens to the rate when the concentration of N 2 O 5 is doubled to M? (d) What happens to the rate when the concentration of N 2 O 5 is halved to M? Consider the following reaction: 2 NO1g g2 N 2 1g O1g2 (a) The rate law for this reaction is first order in 2 and second order in NO. Write the rate law. (b) If the rate constant for this reaction at 1000 K is 6.0 * 10 4 M -2 s -1, what is the reaction rate when 3NO4 = M and = M? (c) What is the reaction rate at 1000 K when the concentration of NO is increased to 0.10 M, while the concentration of 2 is M? (d) What is the reaction rate at 1000 K if [NO] is decreased to M and is increased to M? Consider the following reaction: C 3 Br1aq2 + O - 1aq2 C 3 O1aq2 + Br - 1aq2 The rate law for this reaction is first order in C 3 Br and first order in O -. When 3C 3 Br4 is 5.0 * 10-3 M and 3O - 4 is M, the reaction rate at 298 K is M>s. (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of O - were tripled? (d) What would happen to the rate if the concentration of both reactants were tripled? The reaction between ethyl bromide 1C 2 5 Br2 and hydroxide ion in ethyl alcohol at 330 K, C 2 5 Br1alc2 + O - 1alc2 C 2 5 O1l2 + Br - 1alc2, is first order each in ethyl bromide and hydroxide ion. When 3C 2 5 Br4 is M and 3O - 4 is M, the rate of disappearance of ethyl bromide is 1.7 * 10-7 M>s. (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) ow would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution? The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl - + I - OI - + Cl -. This rapid reaction gives the following rate data: 3oCl 4 1M2 3i 4 1M2 initial Rate 1M,s2 1.5 * * * * * * * * * 10-4 (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when 3OCl - 4 = 2.0 * 10-3 M and 3I - 4 = 5.0 * 10-4 M The reaction 2 ClO 2 1aq2 + 2 O - 1aq2 ClO - 3 1aq2 + ClO - 2 1aq2 + 2 O1l2 was studied with the following results: Experiment 1Clo 2 2 1M2 3oh 4 1M2 initial Rate 1M,s (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when 3ClO 2 4 = M and 3O - 4 = M The following data were measured for the reaction BF 3 1g2 + N 3 1g2 F 3 BN 3 1g2: Experiment 3BF 3 4 1m2 3nh 3 4 1M2 initial Rate 1M,s (a) What is the rate law for the reaction? (b) What is the overall order of the reaction? (c) Calculate the rate constant with proper units? (d) What is the rate when 3BF 3 4 = M and 3N 3 4 = M?

87 620 chapter 14 Chemical Kinetics The following data were collected for the rate of disappearance of NO in the reaction 2 NO1g2 + O 2 1g2 2 NO 2 1g2: Experiment 3no4 1M2 3o 2 4 1M2 initial Rate 1M,s * * * 10-1 (a) What is the rate law for the reaction? (b) What are the units of the rate constant? (c) What is the average value of the rate constant calculated from the three data sets? (d) What is the rate of disappearance of NO when 3NO4 = M and 3O 2 4 = M? (e) What is the rate of disappearance of O 2 at the concentrations given in part (d)? [14.37] Consider the gas-phase reaction between nitric oxide and bromine at 273 C: 2 NO1g2 + Br 2 1g2 2 NOBr1g2. The following data for the initial rate of appearance of NOBr were obtained: Experiment 3no4 1m2 3Br 2 4 1M2 initial Rate 1M,s (a) Determine the rate law. (b) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. (c) ow is the rate of appearance of NOBr related to the rate of disappearance of Br 2? (d) What is the rate of disappearance of Br 2 when 3NO4 = M and 3Br 2 4 = 0.25 M? [14.38] Consider the reaction of peroxydisulfate ion 1S 2 O with iodide ion 1I - 2 in aqueous solution: S 2 O 8 2-1aq2 + 3 I - 1aq2 2 SO 4 2-1aq2 + I 3-1aq2 At a particular temperature the initial rate of disappearance of S 2 O varies with reactant concentrations in the following manner: Experiment 3s 2 o M2 3i 4 1M2 initial Rate 1M,s * * * * 10-5 (a) Determine the rate law for the reaction and state the units of the rate constant. (b) What is the average value of the rate constant for the disappearance of S 2 O based on the four sets of data? (c) ow is the rate of disappearance of S 2 O related to the rate of disappearance of I -? (d) What is the rate of disappearance of I - when 3S 2 O = M and 3I - 4 = M? Change of Concentration with Time (Section 14.4) (a) Define the following symbols that are encountered in rate equations for the generic reaction A S B: 3A4 0, t 1>2 3A4 t, k. (b) What quantity, when graphed versus time, will yield a straight line for a first-order reaction? (c) ow can you calculate the rate constant for a first-order reaction from the graph you made in part (b)? (a) For a generic second-order reaction A B, what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) ow do the half-lives of first-order and second-order reactions differ? (a) The gas-phase decomposition of SO 2 Cl 2, SO 2 Cl 2 1g2 SO 2 1g2 + Cl 2 1g2, is first order in SO 2 Cl 2. At 600 K the halflife for this process is 2.3 * 10 5 s. What is the rate constant at this temperature? (b) At 320 C the rate constant is 2.2 * 10-5 s -1. What is the half-life at this temperature? Molecular iodine, I 2 1g2, dissociates into iodine atoms at 625 K with a first-order rate constant of s -1. (a) What is the half-life for this reaction? (b) If you start with M I 2 at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form I 2? As described in Exercise 14.41, the decomposition of sulfuryl chloride 1SO 2 Cl 2 2 is a first-order process. The rate constant for the decomposition at 660 K is 4.5 * 10-2 s -1. (a) If we begin with an initial SO 2 Cl 2 pressure of 450 torr, what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of SO 2 Cl 2 decline to one-tenth its initial value? The first-order rate constant for the decomposition of N 2 O 5, 2 N 2 O 5 1g2 4 NO 2 1g2 + O 2 1g2, at 70 C is 6.82 * 10-3 s -1. Suppose we start with mol of N 2 O 5 1g2 in a volume of 2.0 L. (a) ow many moles of N 2 O 5 will remain after 5.0 min? (b) ow many minutes will it take for the quantity of N 2 O 5 to drop to mol? (c) What is the half-life of N 2 O 5 at 70 C? The reaction SO 2 Cl 2 1g2 SO 2 1g2 + Cl 2 1g2 is first order in SO 2 Cl 2. Using the following kinetic data, determine the magnitude and units of the first-order rate constant: Time (s) Pressure so 2 Cl 2 (atm) , From the following data for the first-order gas-phase isomerization of C 3 NC at 215 C, calculate the first-order rate constant and half-life for the reaction: Time (s) Pressure Ch 3 nc (torr) , , Consider the data presented in Exercise (a) By using appropriate graphs, determine whether the reaction is first order or second order. (b) What is the rate constant for the reaction? (c) What is the half-life for the reaction?

88 Exercises Consider the data presented in Exercise (a) Determine whether the reaction is first order or second order. (b) What is the rate constant? (c) What is the half-life? The gas-phase decomposition of NO 2, 2 NO 2 1g2 2 NO1g2 + O 2 1g2, is studied at 383 C, giving the following data: Time (s) 3no 2 4 (M) (a) Is the reaction first order or second order with respect to the concentration of NO 2? (b) What is the rate constant? (c) Predict the reaction rates at the beginning of the reaction for initial concentrations of M, M, and M NO Sucrose 1C O 11 2, commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula C 6 22 O 6. At 23 C and in 0.5 M Cl, the following data were obtained for the disappearance of sucrose: Time (min) 3C 12 h 22 o M (a) Is the reaction first order or second order with respect to 3C O 11 4? (b) What is the rate constant? (c) Using this rate constant, calculate the concentration of sucrose at 39, 80, 140, and 210 min if the initial sucrose concentration was M and the reaction were zero order in sucrose. Temperature and Rate (Section 14.5) (a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant? (c) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature? (a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction: NO + O NO 2 or + Cl Cl? (b) ow does the kinetic-molecular theory help us understand the temperature dependence of chemical reactions? Calculate the fraction of atoms in a sample of argon gas at 400 K that has an energy of 10.0 kj or greater (a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is 160 kj>mol. Calculate the fraction of methyl isonitrile molecules that has an energy of kj or greater at 500 K. (b) Calculate this fraction for a temperature of 520 K. What is the ratio of the fraction at 520 K to that at 500 K? The gas-phase reaction Cl1g2 + Br1g2 Cl1g2 + Br1g2 has an overall energy change of -66 kj. The activation energy for the reaction is 7 kj. (a) Sketch the energy profile for the reaction, and label E a and E. (b) What is the activation energy for the reverse reaction? For the elementary process N 2 O 5 1g2 NO 2 1g2 + NO 3 1g2 the activation energy 1E a 2 and overall E are 154 kj>mol and 136 kj>mol, respectively. (a) Sketch the energy profile for this reaction, and label E a and E. (b) What is the activation energy for the reverse reaction? Indicate whether each statement is true or false. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants Indicate whether each statement is true or false. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? (a) E a = 45 kj>mol; E = -25 kj>mol (b) E a = 35 kj>mol; E = -10 kj>mol (c) E a = 55 kj>mol; E = 10 kj>mol Which of the reactions in Exercise will be fastest in the reverse direction? Which will be slowest? (a) A certain first-order reaction has a rate constant of 2.75 * 10-2 s -1 at 20 C. What is the value of k at 60 C if E a = 75.5 kj>mol? (b) Another first-order reaction also has a rate constant of 2.75 * 10-2 s -1 at 20 C What is the value of k at 60 C if E a = 125 kj>mol? (c) What assumptions do you need to make in order to calculate answers for parts (a) and (b)? Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide (NO) to N 2 and O 2 is second order with a rate constant of M -1 s -1 at 737 C and M -1 s -1 at 947 C. Calculate the activation energy for the reaction The rate of the reaction C 3 COOC 2 5 1aq2 + O - 1aq2 C 3 COO - 1aq2 + C 2 5 O1aq2 was measured at several temperatures, and the following data were collected: Temperature 1 C2 k 1M 1 s

89 622 chapter 14 Chemical Kinetics Calculate the value of E a by constructing an appropriate graph The temperature dependence of the rate constant for a reaction is tabulated as follows: Calculate E a and A. Temperature (K) k 1M 1 s Reaction Mechanisms (Section 14.6) (a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism? (d) What is meant by the term rate-determining step? (a) What is meant by the term molecularity? (b) Why are termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism? (d) What are the differences between an intermediate and a transition state? What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) Cl 2 1g2 2 Cl1g2 (b) OCl - 1aq2 + 2 O1l2 OCl1aq2 + O - 1aq2 (c) NO1g2 + Cl 2 1g2 NOCl 2 1g What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) 2 NO1g2 N 2 O 2 1g2 C 2 (b) 2 C C 2 1g2 C 2 C C 3 1g2 (c) SO 3 1g2 SO 2 1g2 + O1g (a) Based on the following reaction profile, how many intermediates are formed in the reaction A D? (b) ow many transition states are there? (c) Which step is the fastest? (d) For the reaction A D, is E positive, negative, or zero? Consider the following energy profile. Potential energy Reaction progess (a) ow many elementary reactions are in the reaction mechanism? (b) ow many intermediates are formed in the reaction? (c) Which step is rate limiting? (d) For the overall reaction, is E positive, negative, or zero? The following mechanism has been proposed for the gasphase reaction of 2 with ICl: 2 1g2 + ICl1g2 I1g2 + Cl1g2 I1g2 + ICl1g2 I 2 1g2 + Cl1g2 (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction? The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: 2 O 2 1aq2 + I - 1aq2 2 O1l2 + IO - 1aq2 1slow2 IO - 1aq2 + 2 O 2 1aq2 2 O1l2 + O 2 1g2 + I - 1aq2 1fast2 (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process. [14.73] The reaction 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 was performed and the following data obtained under conditions of constant 3Cl 2 4: 1/[NO] (under conditions of constant [Cl 2 ]) Time Potential energy A B C D (a) Is the following mechanism consistent with the data? NO1g2 + Cl 2 1g2 NOCl 2 1g2 1fast2 NOCl 2 1g2 + NO1g2 2 NOCl1g2 1slow2 (b) Does the linear plot guarantee that the overall rate law is second order? You have studied the gas-phase oxidation of Br by O 2 : Reaction progress 4 Br1g2 + O 2 1g2 2 2 O1g2 + 2 Br 2 1g2

90 Exercises 623 You find the reaction to be first order with respect to Br and first order with respect to O 2. You propose the following mechanism: Br1g2 + O 2 1g2 OOBr1g2 OOBr1g2 + Br1g2 2 OBr1g2 OBr1g2 + Br1g2 2 O1g2 + Br 2 1g2 (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect OBr or OOBr among the products, does this disprove your mechanism? Catalysis (Section 14.7) (a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both? (a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst? In Figure 14.22, we saw that Br - 1aq2 catalyzes the decomposition of 2 O 2 1aq2 into 2 O1l2 and O 2 1g2. Suppose that some KBr1s2 is added to an aqueous solution of hydrogen peroxide. Make a sketch of 3Br - 1aq24 versus time from the addition of the solid to the end of the reaction In solution, chemical species as simple as + and O - can serve as catalysts for reactions. Imagine you could measure the of a solution containing an acid-catalyzed reaction as it occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the concentration profile you would measure as a function of time for the reaction, assuming t = 0 is when you add a drop of acid to the reaction The oxidation of SO 2 to SO 3 is accelerated by NO 2. The reaction proceeds according to: NO 2 1g2 + SO 2 1g2 NO1g2 + SO 3 1g2 2 NO1g2 + O 2 1g2 2 NO 2 1g2 (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of SO 2 by O 2 to give SO 3. (b) Do we consider NO 2 a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis? The addition of NO accelerates the decomposition of N 2 O, possibly by the following mechanism: NO1g2 + N 2 O1g2 N 2 1g2 + NO 2 1g2 2 NO 2 1g2 2 NO1g2 + O 2 1g2 (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of N 2 O, NO 2 does not accumulate in measurable quantities, does this rule out the proposed mechanism? Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina 1Al 2 O 3 2 or silica 1SiO 2 2. (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) ow does the surface area affect the rate of reaction? (a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important? When D 2 reacts with ethylene 1C in the presence of a finely divided catalyst, ethane with two deuteriums, C 2 D C 2 D, is formed. (Deuterium, D, is an isotope of hydrogen of mass 2). Very little ethane forms in which two deuteriums are bound to one carbon (for example, C 3 CD 2 ). Use the sequence of steps involved in the reaction (Figure 14.24) to explain why this is so eterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to poisoning, which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons. [14.85] The enzyme carbonic anhydrase catalyzes the reaction CO 2 1g2 + 2 O1l2 CO - 3 1aq aq2. In water, without the enzyme, the reaction proceeds with a rate constant of s -1 at 25 C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 1.0 * 10 6 s -1 at 25 C. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction. [14.86] The enzyme urease catalyzes the reaction of urea, 1N 2 CON 2 2, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 * 10-5 s -1 at 100 C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 * 10 4 s -1 at 21 C. (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at 100 C as it is at 21 C, what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at 100 C as compared to that at 21 C? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions? [14.87] The activation energy of an uncatalyzed reaction is 95 kj>mol. The addition of a catalyst lowers the activation energy to 55 kj>mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25 C, (b) 125 C? [14.88] Suppose that a certain biologically important reaction is quite slow at physiological temperature 137 C2 in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a 1 * 10 increase in the reaction rate?

91 624 chapter 14 Chemical Kinetics Additional Exercises Consider the reaction A + B C + D. Is each of the following statements true or false? (a) The rate law for the reaction must be Rate = k3a43b4. (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction ydrogen sulfide 1 2 S2 is a common and troublesome pollutant in industrial wastewaters. One way to remove 2 S is to treat the water with chlorine, in which case the following reaction occurs: 2 S1aq2 + Cl 2 1aq2 S1s aq2 + 2 Cl - 1aq2 The rate of this reaction is first order in each reactant. The rate constant for the disappearance of 2 S at 28 C is 3.5 * 10-2 M -1 s -1. If at a given time the concentration of 2 S is 2.0 * 10-4 M and that of Cl 2 s M, what is the rate of formation of Cl -? The reaction 2 NO1g2 + O 2 1g2 2 NO 2 1g2 is second order in NO and first order in O 2. When 3NO4 = M, and 3O 2 4 = M, the observed rate of disappearance of NO is 9.3 * 10-5 M>s. (a) What is the rate of disappearance of O 2 at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8? You perform a series of experiments for the reaction A B + C and find that the rate law has the form rate = k3a4 x. Determine the value of x in each of the following cases: (a) There is no rate change when 3A4 0 is tripled. (b) The rate increases by a factor of 9 when 3A4 0 is tripled. (c) When 3A4 0 is doubled, the rate increases by a factor of Consider the following reaction between mercury(ii) chloride and oxalate ion: 2 gcl 2 1aq2 + C 2 O 4 2-1aq2 2 Cl - 1aq2 + 2 CO 2 1g2 + g 2 Cl 2 1s2 The initial rate of this reaction was determined for several concentrations of gcl 2 and C 2 O 4 2 -, and the following rate data were obtained for the rate of disappearance of C 2 O : Experiment 3hgCl 2 4 1M2 3C 2 o M2 Rate 1m,s * * * * 10-5 (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of gcl 2 is M and that of 1C 2 O is 0.25 M if the temperature is the same as that used to obtain the data shown? The following kinetic data are collected for the initial rates of a reaction 2 X + Z products: Experiment 3x4 0 1M2 3Z4 0 1M2 Rate 1M,s * * * 10 2 (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of X is 0.75 M and that of Z is 1.25 M? The reaction 2 NO 2 2 NO + O 2 has the rate constant k = 0.63 M - 1 s - 1. (a) Based on the units for k, is the reaction first or second order in NO 2? (b) If the initial concentration of NO 2 is M, how would you determine how long it would take for the concentration to decrease to M? Consider two reactions. Reaction (1) has a constant half-life, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations? [14.97] A first-order reaction A B has the rate constant k = 3.2 * 10-3 s -1. If the initial concentration of A is 2.5 * 10-2 M, what is the rate of the reaction at t = 660 s? (a) The reaction 2 O 2 1aq2 2 O1l O 2 1g2 is first order. Near room temperature, the rate constant equals 7.0 * 10-4 s -1. Calculate the half-life at this temperature. (b) At 415 C, 1C O decomposes in the gas phase, 1C O1g2 C 4 1g2 + CO1g2. If the reaction is first order with a half-life of 56.3 min at this temperature, calculate the rate constant in s Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of k = 1.6 * 10-3 yr -1. By contrast, iodine-125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of k = day -1. (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) ow much of a 1.00-mg sample of each isotope remains after 3 half-lives? (d) ow much of a 1.00-mg sample of each isotope remains after 4 days? Urea 1N 2 CON 2 2 is the end product in protein metabolism in animals. The decomposition of urea in 0.1 M Cl occurs according to the reaction N 2 CON 2 1aq aq O1l2 2 N 4 + 1aq2 + CO - 3 1aq2 The reaction is first order in urea and first order overall. When 3N 2 CON 2 4 = M, the rate at C is 8.56 * 10-5 M>s. (a) What is the rate constant, k? (b) What is the concentration of urea in this solution after 4.00 * 10 3 s if the starting concentration is M? (c) What is the halflife for this reaction at C? [14.101] The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at 520 nm. The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5.60 * 10 3 M -1 cm -1 at 520 nm. (a) Calculate the initial concentration of the colored reactant if the absorbance is at the beginning of the reaction. (b) The absorbance falls to at 30.0 min. Calculate the rate constant in units of s - 1. (c) Calculate the half-life of the reaction. (d) ow long does it take for the absorbance to fall to 0.100? [14.102] A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an extinction coefficient of 4.7 * 10 4 M -1 cm -1 at that wavelength. You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data:

92 additional Exercises 625 Time (min) Absorbance at 608 nm From these data, determine the rate law for the reaction dye product and determine the rate constant Cyclopentadiene 1C reacts with itself to form dicyclopentadiene 1C A M solution of C 5 6 was monitored as a function of time as the reaction 2 C 5 6 C proceeded. The following data were collected: Time (s) 3C 5 h 6 4 1M Plot 3C versus time, ln3c versus time, and 1>3C versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant? The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: Temperature (K) Rate Constant 1s * * * * 10-7 From these data, calculate the activation energy in units of kj>mol At 28 C, raw milk sours in 4.0 h but takes 48 h to sour in a refrigerator at 5 C. Estimate the activation energy in kj>mol for the reaction that leads to the souring of milk? [14.106] The following is a quote from an article in the August 18, 1998, issue of The New York Times about the breakdown of cellulose and starch: A drop of 18 degrees Fahrenheit [from 77 F to 59 F4 lowers the reaction rate six times; a 36-degree drop [from 77 F to 41 F] produces a fortyfold decrease in the rate. (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of E a calculated from the 36 drop and that the rate of breakdown is first order with a half-life at 25 C of 2.7 yr, calculate the half-life for breakdown at a temperature of -15 C The following mechanism has been proposed for the reaction of NO with 2 to form N 2 O and 2 O: NO1g2 + NO1g2 N 2 O 2 1g2 N 2 O 2 1g g2 N 2 O1g2 + 2 O1g2 (a) Show that the elementary reactions of the proposed mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the mechanism. (c) Identify any intermediates in the mechanism. (d) The observed rate law is rate = k3no If the proposed mechanism is correct, what can we conclude about the relative speeds of the first and second reactions? Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: Cl1g2 + O 3 1g2 ClO1g2 + O 2 1g2 ClO1g2 + O1g2 Cl1g2 + O 2 1g2 (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction? Using Figure as your basis, draw the energy profile for the bromide-catalyzed decomposition of hydrogen peroxide. (a) Label the curve with the activation energies for reactions [14.30] and [14.31]. (b) Notice from Figure that when Br - 1aq2 is first added, Br 2 accumulates to some extent during the reaction and the solution turns brown. What does this tell us about the relative rates of the reactions represented by Equations and 14.31? [14.110] The following mechanism has been proposed for the gasphase reaction of chloroform 1CCl 3 2 and chlorine: Step 1: Cl 2 1g2 k 1 k Cl1g2 1fast2 Step 2: Cl1g2 + CCl 3 1g2 k 2 Step 3: Cl1g2 + CCl 3 1g2 k 3 Cl1g2 + CCl 3 1g2 1slow2 CCl 4 1fast2 (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (int: The overall reaction order is not an integer.) [14.111] Consider the hypothetical reaction 2 A + B 2 C + D. The following two-step mechanism is proposed for the reaction: Step 1: A + B C + X Step 2: A + X C + D X is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii). [14.112] In a hydrocarbon solution, the gold compound 1C AuP 3 decomposes into ethane 1C and a different gold compound, 1C 3 2AuP 3. The following mechanism has been proposed for the decomposition of 1C AuP 3 : Step 1: 1C AuP 3 k 1 Step 2: 1C Au k 2 k -1 Step 3: 1C 3 2Au + P 3 k 3 1C Au + P 3 1fast2 C C 3 2Au 1slow2 1C 3 2AuP 3 1fast2

93 626 chapter 14 Chemical Kinetics (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding P 3 to the solution of 1C AuP 3? [14.113] Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of Å. (a) Estimate how many platinum atoms would fit into a 2.0-nm sphere; the volume of a sphere is 14>32pr 3. Recall that 1 A = 1 * m and 1 nm = 1 * 10-9 m. (b) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere 14pr 2 2 and assuming that the footprint of one Pt atom can be estimated from its atomic diameter of 2.8 A. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0-nm nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why? One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself Integrative Exercises [14.117] Dinitrogen pentoxide 1N 2 O 5 2 decomposes in chloroform as a solvent to yield NO 2 and O 2. The decomposition is first order with a rate constant at 45 C of 1.0 * 10-5 s -1. Calculate the partial pressure of O 2 produced from 1.00 L of M N 2 O 5 solution at 45 C over a period of 20.0 h if the gas is collected in a 10.0-L container. (Assume that the products do not dissolve in chloroform.) [14.118] The reaction between ethyl iodide and hydroxide ion in ethanol 1C 2 5 O2 solution, C 2 5 I1alc2 + O - 1alc2 C 2 5 O1l2 + I - 1alc2, has an activation energy of 86.8 kj>mol and a frequency factor of 2.10 * M -1 s -1. (a) Predict the rate constant for the reaction at 35 C. (b) A solution of KO in ethanol is made up by dissolving g KO in ethanol to form ml of solution. Similarly, g of C 2 5 I is dissolved in ethanol to form ml of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at 35 C? (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 C. [14.119] You obtain kinetic data for a reaction at a set of different temperatures. You plot ln k versus 1>T and obtain the following graph: ln k 1/T rapidly enough of the CO 2 accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to 10 7 CO 2 molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number? Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature 137 C2. In order to be physiologically useful, the reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kj>mol must an enzyme lower the activation energy of the reaction to make it useful? Enzymes are often described as following the two-step mechanism: E + S ES 1fast2 ES E + P 1slow2 where E = enzyme, S = substrate, ES = enzyme9substrate complex, and P = product. (a) If an enzyme follows this mechanism, what rate law is expected for the reaction? (b) Molecules that can bind to the active site of an enzyme but are not converted into product are called enzyme inhibitors. Write an additional elementary step to add into the preceding mechanism to account for the reaction of E with I, an inhibitor. Suggest a molecular-level interpretation of these unusual data The gas-phase reaction of NO with F 2 to form NOF and F has an activation energy of E a = 6.3 kj>mol. and a frequency factor of A = 6.0 * 10 8 M -1 s -1. The reaction is believed to be bimolecular: NO1g2 + F 2 1g2 NOF1g2 + F1g2 (a) Calculate the rate constant at 100 C. (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction The mechanism for the oxidation of Br by O 2 to form 2 2 O and Br 2 is shown in Exercise (a) Calculate the overall standard enthalpy change for the reaction process. (b) Br does not react with O 2 at a measurable rate at room temperature under ordinary conditions. What can you infer from this about the magnitude of the activation energy for the rate-determining step? (c) Draw a plausible Lewis structure for the intermediate OOBr. To what familiar compound of hydrogen and oxygen does it appear similar? [14.122] The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: C 4 1g2 + Cl 2 1g2 C 3 Cl1g2 + Cl1g2

94 Design an Experiment 627 This reaction is very slow in the absence of light. owever, Cl 2 1g2 can absorb light to form Cl atoms: Reaction 2: Cl 2 1g2 + hv 2 Cl1g2 Once the Cl atoms are generated, they can catalyze the reaction of C 4 and Cl 2, according to the following proposed mechanism: Reaction 3: C 4 1g2 + Cl1g2 C 3 1g2 + Cl1g2 Reaction 4: C 3 1g2 + Cl 2 1g2 C 3 Cl1g2 + Cl1g2 The enthalpy changes and activation energies for these two reactions are tabulated as follows: Reaction h 1kJ,mol2 E a 1kJ,mol (a) By using the bond enthalpy for Cl 2 (Table 8.4), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, C 4 1g2 + Cl 2 1g2, should be placed on your diagram in part (b). Use this result to estimate the value of E a for the reaction C 4 1g2 + Cl 2 1g2 C 3 1g2 + Cl1g2 + Cl1g2. (d) The species Cl(g) and C 3 1g2 in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of C 3, and verify that it is a radical. (e) The sequence of reactions 3 and 4 comprises a radical chain mechanism. Why do you think this is called a chain reaction? Propose a reaction that will terminate the chain reaction. [14.123] Many primary amines, RN 2, where R is a carbon-containing fragment such as C 3, C 3 C 2, and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a C atom for R ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate? [14.124] The NO x waste stream from automobile exhaust includes species such as NO and NO 2. Catalysts that convert these species to N 2. are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of NO, NO 2, and N 2. (b) Using a resource such as Table 8.4, look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of NO x into N 2, describing what wavelengths of light need to be monitored as a function of time. [14.125] As shown in Figure 14.24, the first step in the heterogeneous hydrogenation of ethylene is adsorption of the ethylene molecule on a metal surface. One proposed explanation for the sticking of ethylene to a metal surface is the interaction of the electrons in the C C p bond with vacant orbitals on the metal surface. (a) If this notion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared to ethylene? (b) Based on its Lewis structure, would you expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene? Design an Experiment Let s explore the chemical kinetics of our favorite hypothetical reaction: a A + b B c C + d D. We shall assume that all the substances are soluble in water and that we carry out the reaction in aqueous solution. Substances A and C both absorb visible light, and the absorption maxima are 510 nm for A and for 640 nm for C. Substances B and D are colorless. You are provided with pure samples of all four substances and you know their chemical formulas. You are also provided appropriate instrumentation to obtain visible absorption spectra (see the Closer Look box on using spectroscopic methods in Section 14.3). Let s design an experiment to ascertain the kinetics of our reaction. (a) What experiments could you design to determine the rate law and the rate constant for the reaction at room temperature? Would you need to know the values of the stoichiometric constants a and c in order to find the rate law? (b) Design an experiment to determine the activation energy for the reaction. What challenges might you face in actually carrying out this experiment? (c) You now want to test whether a particular water-soluble substance Q is a homogeneous catalyst for the reaction. What experiments can you carry out to test this notion? (d) If Q does indeed catalyze the reaction, what follow-up experiments might you undertake to learn more about the reaction profile for the reaction?

95 15 Chemical Equilibrium To be in equilibrium is to be in a state of balance. A tug of war in which the two sides pull with equal force so that the rope does not move is an example of a static equilibrium, one in which an object is at rest. Equilibria can also be dynamic, whereby a forward process and the reverse process take place at the same rate so that no net change occurs. A saturated solution (Section 13.2) of an ionic compound in contact with undissolved crystals of the same compound is a good example of dynamic equilibrium. The rate at which ions leave the surface of the crystals and enter the solution (dissolution) is equal to the rate at which ions leave the solution and become part of the solid (crystallization). ence the concentration of ions in solution and the amount of undissolved solid do not change with time. If some of the solvent is lost due to evaporation, the solution becomes more concentrated, which increases the rate of crystallization. This change leads to a net migration of ions from the solution into the solid until the solution concentration is reduced so that the rate of crystallization and dissolution are once again equal and equilibrium is reestablished. A striking example of this effect is the formation of intricate salt formations in the Dead Sea, as illustrated in the chapter-opening photograph. The Dead Sea, which borders Jordan and Israel, is the lowest point on the surface of the Earth, and has a salt concentration that is almost nine times higher than the ocean. An extended period of hot weather leads to extensive evaporation of the Dead Sea, resulting in an increase in the salt concentration and subsequent crystallization and growth of intricate salt formations. A saturated solution is one of many instances of dynamic equilibrium that we have already encountered. Vapor pressure (Section 11.5) is another example of dynamic equilibrium. The pressure of a vapor above a liquid in a closed container reaches equilibrium with the liquid phase, and therefore stops changing when the rate at which molecules escape from the liquid into the gas phase equals the rate at which molecules SALT PILLARS in the Dead Sea. These pillars are formed in shallow bays where evaporation can lead to salt concentrations that exceed equilibrium values. What s Ahead 15.1 The Concept of Equilibrium We begin by examining reversible reactions and the concept of equilibrium The Equilibrium Constant We define the equilibrium constant based on rates of forward and reverse reactions, and learn how to write equilibrium-constant expressions for homogeneous reactions Understanding and Working with Equilibrium Constants We learn to interpret the magnitude of an equilibrium constant and how its value depends on the way the corresponding chemical equation is expressed eterogeneous Equilibria We learn how to write equilibrium-constant expressions for heterogeneous reactions Calculating Equilibrium Constants We see that the value of an equilibrium constant can be calculated from equilibrium concentrations of reactants and products.

96 15.6 Applications of Equilibrium Constants We learn that equilibrium constants can be used to predict equilibrium concentrations of reactants and products and to determine the direction in which a reaction mixture must proceed to achieve equilibrium Le Châtelier s Principle We discuss Le Châtelier s principle, which predicts how a system at equilibrium responds to changes in concentration, volume, pressure, and temperature.

97 630 chapter 15 Chemical Equilibrium in the gas phase become part of the liquid. enry s law (Section 13.3), which governs the solubility of gases in a solvent, is yet another example of dynamic equilibrium. In this chapter, we consider dynamic equilibria in chemical reactions. Chemical equilibrium occurs when opposing reactions proceed at equal rates: The rate at which the products form from the reactants equals the rate at which the reactants form from the products. As a result, concentrations cease to change, making the reaction appear to be stopped. In this and the next two chapters, we will explore chemical equilibrium in some detail. Later, in Chapter 19, we will learn how to relate chemical equilibria to thermodynamics. ere, we learn how to express the equilibrium state of a reaction in quantitative terms and study the factors that determine the relative concentrations of reactants and products in equilibrium mixtures The Concept of Equilibrium Let s examine a simple chemical reaction to see how it reaches an equilibrium state a mixture of reactants and products whose concentrations no longer change with time. We begin with N 2 O 4, a colorless substance that dissociates to form brown NO 2. Figure 15.1 shows a sample of frozen N 2 O 4 inside a sealed tube. The solid N 2 O 4 becomes a gas as it is warmed above its boiling point C2, and the gas turns darker as the colorless N 2 O 4 gas dissociates into brown NO 2 gas. Eventually, even though there is still N 2 O 4 in the tube, the color stops getting darker because the system reaches equilibrium. We are left with an Go Figure If you were to let the tube on the right sit overnight and then take another picture would the brown color look darker, lighter, or the same? N 2 O 4 NO 2 Frozen N 2 O 4 sample is nearly colorless On warming, the N 2 O 4 becomes a gas and partially dissociates to form brown NO 2 (g) Colors stop changing, equilibrium reached: rate of reaction N 2 O 4 (g) 2 NO 2 (g) = rate of reaction 2 NO 2 (g) N 2 O 4 (g) Figure 15.1 The equilibrium between NO 2 and N 2 O 4.

98 section 15.1 The Concept of Equilibrium 631 equilibrium mixture of N 2 O 4 and NO 2 in which the concentrations of the gases no longer change as time passes. Because the reaction is in a closed system, where no gases can escape, equilibrium will eventually be reached. The equilibrium mixture results because the reaction is reversible: N 2 O 4 can form NO 2, and NO 2 can form N 2 O 4. Dynamic equilibrium is represented by writing the equation for the reaction with two half arrows pointing in opposite directions (Section 4.1): N 2 O 4 1g2 2 NO 2 1g2 [15.1] Colorless Brown We can analyze this equilibrium using our knowledge of kinetics. Let s call the decomposition of N 2 O 4 the forward reaction and the formation of N 2 O 4 the reverse reaction. In this case, both the forward reaction and the reverse reaction are elementary reactions. As we learned in Section 14.6, the rate laws for elementary reactions can be written from their chemical equations: Forward reaction: N 2 O 4 1g2 2 NO 2 1g2 Rate f = k f 3N 2 O 4 4 [15.2] Reverse reaction: 2 NO 2 1g2 N 2 O 4 1g2 Rate r = k r 3NO [15.3] At equilibrium, the rate at which NO 2 forms in the forward reaction equals the rate at which N 2 O 4 forms in the reverse reaction: k f 3N 2O 4 4 = k r 3NO [15.4] Forward reaction Reverse reaction Rearranging this equation gives 3NO N 2 O 4 4 = k f = a constant [15.5] k r From Equation 15.5, we see that the quotient of two rate constants is another constant. We also see that, at equilibrium, the ratio of the concentration terms equals this same constant. (We consider this constant, called the equilibrium constant, in Section 15.2.) It makes no difference whether we start with N 2 O 4 or with NO 2, or even with some mixture of the two. At equilibrium, at a given temperature, the ratio equals a specific value. Thus, there is an important constraint on the proportions of N 2 O 4 and NO 2 at equilibrium. Once equilibrium is established, the concentrations of N 2 O 4 and NO 2 no longer change, as shown in Figure 15.2(a). owever, the fact that the composition of the equilibrium mixture remains constant with time does not mean that N 2 O 4 and NO 2 stop reacting. On the contrary, the equilibrium is dynamic which means some N 2 O 4 is always converting to NO 2 and some NO 2 is always converting to N 2 O 4. At equilibrium, however, the two processes occur at the same rate, as shown in Figure 15.2(b). Go Figure At equilibrium, is the ratio 3NO 2 4>3N 2 O 4 4 less than, greater to, or equal to 1? Concentration 0 Time (a) NO 2 N 2 O 4 Equilibrium achieved Rate 0 k f [N 2 O 4 ] k r [NO 2 ] 2 Time (b) Equilibrium achieved (rates are equal) Figure 15.2 Achieving chemical equilibrium in the N 2 O 4 1g2 2 NO 2 1g2 reaction. Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.

99 632 chapter 15 Chemical Equilibrium We learn several important lessons about equilibrium from this example: At equilibrium, the concentrations of reactants and products no longer change with time. For equilibrium to occur, neither reactants nor products can escape from the system. At equilibrium, a particular ratio of concentration terms equals a constant. Give It Some Thought (a) Which quantities are equal in a dynamic equilibrium? (b) If the rate constant for the forward reaction in Equation 15.1 is larger than the rate constant for the reverse reaction, will the constant in Equation 15.5 be greater than 1 or smaller than 1? 15.2 The Equilibrium Constant A reaction in which reactants convert to products and products convert to reactants in the same reaction vessel naturally leads to an equilibrium, regardless of how complicated the reaction is and regardless of the nature of the kinetic processes for the forward and reverse reactions. Consider the synthesis of ammonia from nitrogen and hydrogen: N 2 1g g2 2 N 3 1g2 [15.6] This reaction is the basis for the aber process, which is critical for the production of fertilizers and therefore critical to the world s food supply. In the aber process, N 2 and 2 react at high pressure and temperature in the presence of a catalyst to form ammonia. In a closed system, however, the reaction does not lead to complete consumption of the N 2 and 2. Rather, at some point the reaction appears to stop with all three components of the reaction mixture present at the same time. ow the concentrations of 2, N 2, and N 3 vary with time is shown in Figure Notice that an equilibrium mixture is obtained regardless of whether we begin with N 2 and 2 or with N 3. The equilibrium condition is reached from either direction. Starting with only reactants Starting with only products Concentration Equilibrium achieved Time 2 N 3 N 2 Figure 15.3 The same equilibrium is reached whether we start with only reactants (N 2 and 2 ) or with only product 1N 3 2. Concentration Equilibrium achieved Time 2 N 3 N 2 Give It Some Thought ow do we know when equilibrium has been reached in a chemical reaction? An expression similar to Equation 15.5 governs the concentrations of N 2, 2, and N 3 at equilibrium. If we were to systematically change the relative amounts of the three gases in the starting mixture and then analyze each equilibrium mixture, we could determine the relationship among the equilibrium concentrations.

100 section 15.2 The Equilibrium Constant 633 Chemists carried out studies of this kind on other chemical systems in the nineteenth century before aber s work. In 1864, Cato Maximilian Guldberg ( ) and Peter Waage ( ) postulated their law of mass action, which expresses, for any reaction, the relationship between the concentrations of the reactants and products present at equilibrium. Suppose we have the general equilibrium equation a A + b B d D + e E [15.7] where A, B, D, and E are the chemical species involved and a, b, d, and e are their coefficients in the balanced chemical equation. According to the law of mass action, the equilibrium condition is described by the expression K c = 3D4d 3E4 e 3A4 a 3B4 b products reactants [15.8] We call this relationship the equilibrium-constant expression (or merely the equilibrium expression) for the reaction. The constant K c, the equilibrium constant, is the numerical value obtained when we substitute molar equilibrium concentrations into the equilibriumconstant expression. The subscript c on the K indicates that concentrations expressed in molarity are used to evaluate the constant. Chemistry Put to Work The aber Process The quantity of food required to feed the ever-increasing human population far exceeds that provided by nitrogen-fixing plants. (Section 14.7, Nitrogen Fixation and Nitrogenase ) Therefore, human agriculture requires substantial amounts of ammonia-based fertilizers for croplands. Of all the chemical reactions that humans have learned to control for their own purposes, the synthesis of ammonia from hydrogen and atmospheric nitrogen is one of the most important. In 1912, the German chemist Fritz aber ( ) developed the aber process (Equation 15.6). The process is sometimes also called the aber Bosch process to honor Karl Bosch, the engineer who developed the industrial process on a large scale ( Figure 15.4). The engineering needed to implement the aber process requires the use of temperatures and pressures (approximately 500 C and 200 to 600 atm) that were difficult to achieve at that time. The aber process provides a historically interesting example of the complex impact of chemistry on our lives. At the start of World War I, in 1914, Germany depended on nitrate deposits in Chile for the nitrogencontaining compounds needed to manufacture explosives. During the war, the Allied naval blockade of South America cut off this supply. owever, by using the aber reaction to fix nitrogen from air, Germany was able to continue to produce explosives. Experts have estimated that World War I would have ended before 1918 had it not been for the aber process. From these unhappy beginnings as a major factor in international warfare, the aber process has become the world s principal source of fixed nitrogen. The same process that prolonged World War I has enabled the manufacture of fertilizers that have increased crop yields, thereby saving millions of people from starvation. About 40 billion pounds of ammonia are manufactured annually in the United States, mostly by the aber process. The ammonia can be applied directly to the soil, or it can be converted into ammonium salts that are also used as fertilizers. aber was a patriotic German who gave enthusiastic support to his nation s war effort. e served as chief of Germany s Chemical Warfare Service during World War I and developed the use of chlorine as a poison-gas weapon. Consequently, the decision to award him the Nobel Prize in Chemistry in 1918 was the subject of considerable controversy and criticism. The ultimate irony, however, came in 1933 when aber was expelled from Germany because he was Jewish. Related Exercises: 15.44, 15.76, Figure 15.4 A high-pressure steel reactor used in the aber process is on display at Karlsruhe Institute of Technology in Germany where the aber process was developed.

101 634 chapter 15 Chemical Equilibrium Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: (a) 2 O 3 1g2 3 O 2 1g2 (b) 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 (c) Ag + 1aq2 + 2 N 3 1aq2 Ag1N aq2 The numerator of the equilibrium-constant expression is the product of the concentrations of all substances on the product side of the equilibrium equation, each raised to a power equal to its coefficient in the balanced equation. The denominator is similarly derived from the reactant side of the equilibrium equation. Thus, for the aber process, N 2 1g g2 2 N 3 1g2, the equilibrium-constant expression is K c = 3N N [15.9] Once we know the balanced chemical equation for a reaction that reaches equilibrium, we can write the equilibrium-constant expression even if we do not know the reaction mechanism. The equilibrium-constant expression depends only on the stoichiometry of the reaction, not on its mechanism. The value of the equilibrium constant at any given temperature does not depend on the initial amounts of reactants and products. It also does not matter whether other substances are present, as long as they do not react with a reactant or a product. The value of K c depends only on the particular reaction and on the temperature. Solution Analyze We are given three equations and are asked to write an equilibrium-constant expression for each. Plan Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced chemical equation. Solve (a) K c = 3O O (b) K c = 3NOCl42 3NO4 2 3Cl 2 4 (c) K c = 3Ag1N Ag + 43N Practice Exercise 1 For the reaction 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2 which of the following is the correct equilibrium-constant expression? (a) K C = 3SO O 2 4 3SO (b) K C = 23SO 243O SO 3 4 3SO SO 3 4 (c) K C = 3SO (d) K C = 3O SO 2 43O 2 4 Practice Exercise 2 Write the equilibrium-constant expression K c for (a) 2 1g2 + I 2 1g2 2 I1g2, (b) Cd 2+ 1aq2 + 4 Br - 1aq2 CdBr 4 2-1aq2. Evaluating K c We can illustrate how the law of mass action was discovered empirically and demonstrate that the equilibrium constant is independent of starting concentrations by examining a series of experiments involving dinitrogen tetroxide and nitrogen dioxide: N 2 O 4 1g2 2 NO 2 1g2 K c = 3NO 24 2 [15.10] 3N 2 O 4 4 We start with several sealed tubes containing different concentrations of NO 2 and N 2 O 4. The tubes are kept at 100 C until equilibrium is reached. We then analyze the mixtures and determine the equilibrium concentrations of NO 2 and N 2 O 4, which are shown in Table To evaluate K c, we insert the equilibrium concentrations into the equilibriumconstant expression. For example, using Experiment 1 data, 3NO 2 4 = M and 3N 2 O 4 4 = M, we find K c = 3NO N 2 O 4 4 = = 0.211

102 section 15.2 The Equilibrium Constant 635 Table 15.1 Initial and Equilibrium Concentrations of n 2 o 4 1g2 and no 2 1g2 at 100 C Experiment Initial 3n 2 o 4 4 1M2 Initial 3no 2 4 1M2 Equilibrium 3n 2 o 4 4 1M2 Equilibrium 3no 2 4 1M2 K c Proceeding in the same way, the values of K c for the other samples are calculated. Note from Table 15.1 that the value for K c is constant (within the limits of experimental error) even though the initial concentrations vary. Furthermore, Experiment 4 shows that equilibrium can be achieved beginning with N 2 O 4 rather than with NO 2. That is, equilibrium can be approached from either direction. Figure 15.5 shows how Experiments 3 and 4 result in the same equilibrium mixture even though the two experiments start with very different NO 2 concentrations. Notice that no units are given for K c either in Table 15.1 or in the calculation we just did using Experiment 1 data. It is common practice to write equilibrium constants without units for reasons that we address later in this section. Give It Some Thought ow does the value of K c in Equation depend on the starting concentrations of NO 2 and N 2 O 4? [NO 2 ] (M) Experiment 3 Experiment 4 Equilibrium Constants in Terms of Pressure, K p When the reactants and products in a chemical reaction are gases, we can formulate the equilibrium-constant expression in terms of partial pressures. When partial pressures in atmospheres are used in the expression, we denote the equilibrium constant K p (where the subscript p stands for pressure). For the general reaction in Equation 15.7, we have K p = 1P D2 d 1P E 2 e 1P A 2 a 1P B 2 b [15.11] where P A is the partial pressure of A in atmospheres, P B is the partial pressure of B in atmospheres, and so forth. For example, for our N 2 O 4 >NO 2 reaction, we have K p = 1P NO P N2 O 4 0 Time Figure 15.5 The same equilibrium mixture is produced regardless of the initial NO 2 concentration. The concentration of NO 2 either increases or decreases until equilibrium is reached. Give It Some Thought What is the difference between the equilibrium constant K c and the equilibrium constant K p? For a given reaction, the numerical value of K c is generally different from the numerical value of K p. We must therefore take care to indicate, via subscript c or p, which constant we are using. It is possible, however, to calculate one from the other using the ideal-gas equation (Section 10.4): PV = nrt, so P = n RT [15.12] V The usual units for n>v are mol>l, which equals molarity, M. For substance A in our generic reaction, we therefore see that P A = n A RT = 3A4RT [15.13] V

103 636 chapter 15 Chemical Equilibrium When we substitute Equation and like expressions for the other gaseous components of the reaction into Equation 15.11, we obtain a general expression relating K p and K c : K p = K c 1RT2 n [15.14] The quantity n is the change in the number of moles of gas in the balanced chemical equation. It equals the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants: n = 1moles of gaseous product2-1moles of gaseous reactant2 [15.15] For example, in the N 2 O 4 1g2 2 NO 2 1g2 reaction, there are two moles of product NO 2 and one mole of reactant N 2 O 4. Therefore, n = 2-1 = 1, and K p = K c 1RT2, for this reaction. Give It Some Thought Is it possible to have a reaction where K c = K p? If so, under what conditions would this relationship hold? Sample Exercise 15.2 Converting between K c and K p For the aber process, N 2 1g g2 2 N 3 1g2 K c = 9.60 at 300 C. Calculate K p for this reaction at this temperature. Solution Analyze We are given K c for a reaction and asked to calculate K p. Plan The relationship between K c and K p is given by Equation To apply that equation, we must determine n by comparing the number of moles of product with the number of moles of reactants (Equation 15.15). Solve With 2 mol of gaseous products 12 N 3 2 and 4 mol of gaseous reactants 11 N , n = 2-4 = -2. (Remember that functions are always based on products minus reactants.) The temperature is = 573 K. The value for the ideal-gas constant, R, is L@atm>mol@K. Using K c = 9.60, we therefore have K p = K c 1RT2 n = * = = 4.34 * * 5732 Practice Exercise 1 For which of the following reactions is the ratio K p >K c largest at 300 K? (a) N 2 1g2 + O 2 1g2 2 NO1g2 (b) CaCO 3 1s2 CaO1s2 + CO 2 1g2 (c) Ni1CO2 4 1g2 Ni1s2 + 4 CO1g2 (d) C1s g2 C 4 1g2 Practice Exercise 2 For the equilibrium 2 SO 3 1g2 2 SO 2 1g2 + O 2 1g2, K c is 4.08 * 10-3 at 1000 K. Calculate the value for K p. Equilibrium Constants and Units You may wonder why equilibrium constants are reported without units. The equilibrium constant is related to the kinetics of a reaction as well as to the thermodynamics. (We explore this latter connection in Chapter 19.) Equilibrium constants derived from thermodynamic measurements are defined in terms of activities rather than concentrations or partial pressures.

104 section 15.3 Understanding and Working with Equilibrium Constants 637 The activity of any substance in an ideal mixture is the ratio of the concentration or pressure of the substance either to a reference concentration 11 M2 or to a reference pressure (1 atm). For example, if the concentration of a substance in an equilibrium mixture is M, its activity is M>1 M = The units of such ratios always cancel and, consequently, activities have no units. Furthermore, the numerical value of the activity equals the concentration. For pure solids and pure liquids, the situation is even simpler because the activities then merely equal 1 (again with no units). In real systems, activities are also ratios that have no units. Even though these activities may not be exactly numerically equal to concentrations, we will ignore the differences. All we need to know at this point is that activities have no units. As a result, the thermodynamic equilibrium constants derived from them also have no units. It is therefore common practice to write all types of equilibrium constants without units, a practice that we adhere to in this text. In more advanced chemistry courses, you may make more rigorous distinctions between concentrations and activities. Give It Some Thought If the concentration of N 2 O 4 in an equilibrium mixture is M, what is its activity? (Assume the solution is ideal.) 15.3 understanding and Working with Equilibrium Constants Before doing calculations with equilibrium constants, it is valuable to understand what the magnitude of an equilibrium constant can tell us about the relative concentrations of reactants and products in an equilibrium mixture. It is also useful to consider how the magnitude of any equilibrium constant depends on how the chemical equation is expressed. The Magnitude of Equilibrium Constants The magnitude of the equilibrium constant for a reaction gives us important information about the composition of the equilibrium mixture. For example, consider the experimental data for the reaction of carbon monoxide gas and chlorine gas at 100 C to form phosgene 1COCl 2 2, a toxic gas used in the manufacture of certain polymers and insecticides: CO1g2 + Cl 2 1g2 COCl 2 1g2 K c = 3COCl 24 = 4.56 * 109 3CO43Cl 2 4 For the equilibrium constant to be so large, the numerator of the equilibrium-constant expression must be approximately a billion times larger than the denominator. Thus, the equilibrium concentration of COCl 2 must be much greater than that of CO or Cl 2, and in fact, this is just what we find experimentally. We say that this equilibrium lies to the right (that is, toward the product side). Likewise, a very small equilibrium constant indicates that the equilibrium mixture contains mostly reactants. We then say that the equilibrium lies to the left. In general, If K W 1 (large K): Equilibrium lies to right, products predominate If K V 1 (small K): Equilibrium lies to left, reactants predominate These situations are summarized in Figure Remember, it is forward and reverse reaction rates, not reactant and product concentrations, that are equal at equilibrium. Go Figure What would this figure look like for a reaction in which K 1? Reactants Products K >> 1, equilibrium lies to the right Reactants Products K << 1, equilibrium lies to the left Figure 15.6 Relationship between magnitude of K and composition of an equilibrium mixture.

105 638 chapter 15 Chemical Equilibrium Sample Exercise 15.3 interpreting the Magnitude of an Equilibrium Constant The following diagrams represent three systems at equilibrium, all in the same-size containers. (a) Without doing any calculations, rank the systems in order of increasing K c. (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate K c for each system. Solution Analyze We are asked to judge the relative magnitudes of three equilibrium constants and then to calculate them. Plan (a) The more product present at equilibrium, relative to reactant, the larger the equilibrium constant. (b) The equilibrium constant is given by Equation Solve (i) (ii) (a) Each box contains 10 spheres. The amount of product in each varies as follows: (i) 6, (ii) 1, (iii) 8. Therefore, the equilibrium constant varies in the order (ii) 6 (i) 6 (iii), from smallest (most reactant) to largest (most products). (b) In (i), we have 0.60 mol>l product and 0.40 mol>l reactant, giving K c = 0.60>0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres>4 spheres = 1.5.) In (ii), we have 0.10 mol>l product and 0.90 mol>l reactant, giving K c = 0.10>0.90 = 0.11 (or 1 sphere>9 spheres = 0.11). In (iii), we have 0.80 mol>l product and 0.20 mol>l reactant, giving K c = 0.80>0.20 = 4.0 (or 8 spheres>2 spheres = 4.0). These calculations verify the order in (a). Comment Imagine a drawing that represents a reaction with a very small or very large value of K c. For example, what would the drawing look like if K c = 1 * 10-5? In that case there would need to be 100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw. (iii) Practice Exercise 1 The equilibrium constant for the reaction N 2 O 4 1g2 2 NO 2 1g2 at 2 C is K c = 2.0. If each yellow sphere represents 1 mol of N 2 O 4 and each brown sphere 1 mol of NO 2 which of the following 1.0 L containers represents the equilibrium mixture at 2 C? 2 (a) (b) (c) (d) (e) Practice Exercise 2 For the reaction 2 1g2 + I 2 1g2 2 I1g2, K p = 794 at 298 K and K p = 55 at 700 K. Is the formation of I favored more at the higher or lower temperature?

106 The Direction of the Chemical Equation and K We have seen that we can represent the N 2 O 4 >NO 2 equilibrium as N 2 O 4 1g2 2 NO 2 1g2 K c = 3NO N 2 O 4 4 section 15.3 Understanding and Working with Equilibrium Constants 639 = at 100 C2 [15.16] We could equally well consider this equilibrium in terms of the reverse reaction: The equilibrium expression is then 2 NO 2 1g2 N 2 O 4 1g2 K c = 3N 2O 4 4 3NO = = at 100 C2 [15.17] Equation is the reciprocal of the expression in Equation The equilibrium-constant expression for a reaction written in one direction is the reciprocal of the expression for the reaction written in the reverse direction. Consequently, the numerical value of the equilibrium constant for the reaction written in one direction is the reciprocal of that for the reverse reaction. Both expressions are equally valid, but it is meaningless to say that the equilibrium constant for the equilibrium between NO 2 and N 2 O 4 is or 4.72 unless we indicate how the equilibrium reaction is written and specify the temperature. Therefore, whenever you are using an equilibrium constant, you should always write the associated balanced chemical equation. Give It Some Thought For the reaction PCl 5 1g2 PCl 3 1g2 + Cl 2 1g2, the equilibrium constant K C = 1.1 * 10-2 at 400 K. What is the equilibrium constant for the reaction PCl 3 1g2 + Cl 2 1g2 PCl 5 1g2 at 400 K? Relating Chemical Equation Stoichiometry and Equilibrium Constants There are many ways to write a balanced chemical equation for a given reaction. For example, if we multiply Equation 15.1, N 2 O 4 1g2 2 NO 2 1g2 by 2, we have 2 N 2 O 4 1g2 4 NO 2 1g2 This chemical equation is balanced and might be written this way in some contexts. Therefore, the equilibrium-constant expression for this equation is K c = 3NO N 2 O which is the square of the equilibrium-constant expression given in Equation for the reaction as written in Equation 15.1: 3NO >3N 2 O 4 4. Because the new equilibrium-constant expression equals the original expression squared, the new equilibrium constant K c equals the original constant squared: = at 100 C2. Once again, it is important to remember that you must relate each equilibrium constant you work with to a specific balanced chemical equation. The concentrations of the substances in the equilibrium mixture will be the same no matter how you write the chemical equation, but the value of K c you calculate depends completely on how you write the reaction. Give It Some Thought ow does the magnitude of K p for the reaction 2 I1g2 2 1g2 + I 2 1g2 change if the equilibrium is written 6 I1g g2 + 3 I 2 1g2?

107 640 chapter 15 Chemical Equilibrium It is also possible to calculate the equilibrium constant for a reaction if we know the equilibrium constants for other reactions that add up to give us the one we want, similar to ess s law. (Section 5.6) For example, consider the following two reactions, their equilibrium-constant expressions, and their equilibrium constants at 100 C: 1. 2 NOBr1g2 2 NO1g2 + Br 2 1g2 K c1 = 3NO42 3Br 2 4 3NOBr4 2 = Br 2 1g2 + Cl 2 1g2 2 BrCl1g2 K c2 = 3BrCl42 3Br 2 43Cl 2 4 = 7.2 The net sum of these two equations is: 3. 2 NOBr1g2 + Cl 2 1g2 2 NO1g2 + 2 BrCl1g2 You can prove algebraically that the equilibrium-constant expression for reaction 3 is the product of the expressions for reactions 1 and 2: Thus, To summarize: K c3 = 3NO42 3BrCl4 2 3NOBr4 2 3Cl 2 4 = 3NO42 3Br 2 4 3NOBr4 2 * 3BrCl42 3Br 2 43Cl 2 4 K c3 = 1K c1 21K c2 2 = = The equilibrium constant of a reaction in the reverse direction is the inverse (or reciprocal) of the equilibrium constant of the reaction in the forward direction: A + B C + D K 1 C + D A + B K = 1>K 1 2. The equilibrium constant of a reaction that has been multiplied by a number is equal to the original equilibrium constant raised to a power equal to that number. A + B C + D K 1 na + nb nc + nd K = K 1 n 3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the equilibrium constants for the individual reactions: A + B C + D K 1 C + F G + A K 2 B + F D + G K 3 = 1K 1 21K 2 2 Sample Exercise 15.4 Combining Equilibrium Expressions Given the reactions F1aq2 + 1aq2 + F - 1aq2 K c = 6.8 * C 2 O 4 1aq aq2 + C 2 O 4 2-1aq2 K c = 3.8 * 10-6 determine the value of K c for the reaction 2 F1aq2 + C 2 O 4 2-1aq2 2 F - 1aq2 + 2 C 2 O 4 1aq2 Solution Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two. Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate these equations to come up with equations that we can add to give us the desired equation.

108 section 15.4 eterogeneous Equilibria 641 Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get 2 F1aq aq2 + 2 F - 1aq2 K c = 16.8 * = 4.6 * 10-7 Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives aq2 + C 2 O - 4 1aq2 2 C 2 O 4 1aq2 K c = 1 = 2.6 * * 10 Now, we have two equations that sum to give the net equation, and we can multiply the individual K c values to get the desired equilibrium constant. 2 F1aq aq2 + 2 F - 1aq aq2 + C 2 O 4 2-1aq2 2 C 2 O 4 1aq2 2 F1aq2 + C 2 O 4 2-1aq2 2 F - 1aq2 + 2 C 2 O 4 1aq2 K c = 4.6 * 10-7 K c = 2.5 * 10 5 K c = 14.6 * * = 0.12 Practice Exercise 1 Given the equilibrium constants for the following two reactions in aqueous solution at 25 C, NO 2 1aq2 + 1aq2 + NO - 2 1aq2 K c = 4.5 * SO 3 1aq aq2 + SO - 3 1aq2 K c = 1.1 * 10-9 what is the value of K c for the reaction? 2 NO 2 1aq2 + SO3 2-1aq2 2 SO 3 1aq2 + 2 NO - 2 1aq2 (a) 4.9 * 10-13, (b) 4.1 * 10 5, (c) 8.2 * 10 5, (d) 1.8 * 10 2, (e) 5.4 * Practice Exercise 2 Given that, at 700 K, K p = 54.0 for the reaction 2 1g2 + I 2 1g2 2 I1g2 and K p = 1.04 * 10-4 for the reaction N 2 1g g2 2 N 3 1g2, determine the value of K p for the reaction 2 N 3 1g2 + 3 I 2 1g2 6 I1g2 + N 2 1g2 at 700 K eterogeneous Equilibria Many equilibria involve substances that are all in the same phase. Such equilibria are called homogeneous equilibria. In some cases, however, the substances in equilibrium are in different phases, giving rise to heterogeneous equilibria. As an example of the latter, consider the equilibrium that occurs when solid lead(ii) chloride dissolves in water to form a saturated solution: PbCl 2 1s2 Pb 2+ 1aq2 + 2 Cl - 1aq2 [15.18] This system consists of a solid in equilibrium with two aqueous species. If we want to write the equilibrium-constant expression for this process, we encounter a problem we have not encountered previously: ow do we express the concentration of a solid? If we were to carry out experiments starting with varying amounts of products and reactants we would find that the equilibrium-constant expression for the reaction of Equation is K c = 3Pb 2+ 43Cl [15.19] Thus, our problem of how to express the concentration of a solid is not relevant in the end, because PbCl 2 1s2 does not show up in the equilibrium-constant expression. More generally, we can state that whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression. The fact that pure solids and pure liquids are excluded from equilibrium-constant expressions can be explained in two ways. First, the concentration of a pure solid or liquid has a constant value. If the mass of a solid is doubled, its volume also doubles. Thus, its concentration, which relates to the ratio of mass to volume, stays the same. Because

109 642 chapter 15 Chemical Equilibrium equilibrium-constant expressions include terms only for reactants and products whose concentrations can change during a chemical reaction, the concentrations of pure solids and pure liquids are omitted. The omission can also be rationalized in a second way. Recall from Section 15.2 that what is substituted into a thermodynamic equilibrium expression is the activity of each substance, which is a ratio of the concentration to a reference value. For a pure substance, the reference value is the concentration of the pure substance, so that the activity of any pure solid or liquid is always 1. Give It Some Thought Write the equilibrium-constant expression for the evaporation of water, 2O1l 2 2O1g2, in terms of partial pressures. Decomposition of calcium carbonate is another example of a heterogeneous reaction: CaCO31s2 CaO1s2 + CO21g2 Omitting the concentrations of the solids from the equilibrium-constant expression gives Kc = 3CO24 and Kp = PCO2 These equations tell us that at a given temperature, an equilibrium among CaCO3, CaO, and CO2 always leads to the same CO2 partial pressure as long as all three components are present. As shown in Figure 15.7, we have the same CO2 pressure regardless of the relative amounts of CaO and CaCO3. Go Figure If some of the CO21g2 were released from the upper bell jar and the seal then restored and the system allowed to return to equilibrium, would the amount of CaCO31s2 increase, decrease or remain the same? CaCO3(s) CO2 (g) CaCO3 CaO Large amount of CaCO3, small amount of CaO, gas pressure P CaO(s) + CO2(g) CO2 (g) CaCO3 CaO Small amount of CaCO3, large amount of CaO, gas pressure still P Figure 15.7 At a given temperature, the equilibrium pressure of CO2 in the bell jars is the same no matter how much of each solid is present.

110 section 15.4 eterogeneous Equilibria 643 Sample Exercise 15.5 Writing Equilibrium-Constant Expressions for eterogeneous Reactions Write the equilibrium-constant expression K c for (a) CO 2 1g g2 CO1g2 + 2 O1l2 (b) SnO 2 1s2 + 2 CO1g2 Sn1s2 + 2 CO 2 1g2 Solution Analyze We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions. Plan We use the law of mass action, remembering to omit any pure solids and pure liquids from the expressions. Solve (a) The equilibrium-constant expression is 3CO4 K c = 3CO Because 2 O appears in the reaction as a liquid, its concentration does not appear in the equilibrium-constant expression. (b) The equilibrium-constant expression is K c = 3CO CO4 2 Because SnO 2 and Sn are pure solids, their concentrations do not appear in the equilibrium-constant expression. Practice Exercise 1 Consider the equilibrium that is established in a saturated solution of silver chloride, Ag + 1aq2 + Cl - 1aq2 AgCl1s2. If solid AgCl is added to this solution, what will happen to the concentration of Ag + and Cl - ions in solution? (a) 3Ag + 4 and 3Cl - 4 will both increase, (b) 3Ag + 4 and 3Cl - 4 will both decrease, (c) 3Ag + 4 will increase and 3Cl - 4 will decrease, (d) 3Ag + 4 will decrease and 3Cl - 4 will increase, (e) neither 3Ag + 4 nor 3Cl - 4 will change. Practice Exercise 2 Write the following equilibrium-constant expressions: (a) K c for Cr1s2 + 3 Ag + 1aq2 Cr 3+ 1aq2 + 3 Ag1s2, (b) K p for 3 Fe1s O1g2 Fe 3 O 4 1s g2. Sample Exercise 15.6 Analyzing a eterogeneous Equilibrium Each of these mixtures was placed in a closed container and allowed to stand: (a) CaCO 3 1s2 (b) CaO(s) and CO 2 1g2 at a pressure greater than the value of K p (c) CaCO 3 1s2 and CO 2 1g2 at a pressure greater than the value of K p (d) CaCO 3 1s2 and CaO1s2 Determine whether or not each mixture can attain the equilibrium CaCO 3 1s2 CaO1s2 + CO 2 1g2 Solution Analyze We are asked which of several combinations of species can establish an equilibrium between calcium carbonate and its decomposition products, calcium oxide and carbon dioxide. Plan For equilibrium to be achieved, it must be possible for both the forward process and the reverse process to occur. For the forward process to occur, some calcium carbonate must be present. For the reverse process to occur, both calcium oxide and carbon dioxide must be present. In both cases, either the necessary compounds may be present initially or they may be formed by reaction of the other species. Solve Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present. (a) CaCO 3 simply decomposes, forming CaO(s) and CO 2 1g2 until the equilibrium pressure of CO 2 is attained. There must be enough CaCO 3, however, to allow the CO 2 pressure to reach equilibrium. (b) CO 2 continues to combine with CaO until the partial pressure of the CO 2 decreases to the equilibrium value. (c) Because there is no CaO present, equilibrium cannot be attained; there is no way the CO 2 pressure can decrease to its equilibrium value (which would require some CO 2 to react with CaO). (d) The situation is essentially the same as in (a): CaCO 3 decomposes until equilibrium is attained. The presence of CaO initially makes no difference. Practice Exercise 1 If 8.0 g of N 4 S1s2 is placed in a sealed vessel with a volume of 1.0 L and heated to 200 C the reaction N 4 S1s2 N 3 1g2 + 2 S1g2 will occur. When the system comes to equilibrium, some N 4 S1s2 is still present. Which of the following changes will lead to a reduction in the amount of N 4 S1s2 that is present, assuming in all cases that equilibrium is re-established following the change? (a) Adding more N 3 1g2 to the vessel, (b) Adding more 2 S1g2 to the vessel, (c) Adding more N 4 S1s2 to the vessel, (d) Increasing the volume of the vessel, (e) decreasing the volume of the vessel. Practice Exercise 2 When added to Fe 3 O 4 1s2 in a closed container, which one of the following substances 2 1g2, 2 O1g2, O 2 1g2 allows equilibrium to be established in the reaction 3 Fe1s O1g2 Fe 3 O 4 1s g2?

111 644 chapter 15 Chemical Equilibrium When a solvent is a reactant or product in an equilibrium, its concentration is omitted from the equilibrium-constant expression, provided the concentrations of reactants and products are low, so that the solvent is essentially a pure substance. Applying this guideline to an equilibrium involving water as a solvent, 2 O1l2 + CO 2-3 1aq2 O - 1aq2 + CO - 3 1aq2 [15.20] gives an equilibrium-constant expression that does not contain 3 2 O4: K c = 3O- 43CO 3-4 3CO [15.21] Give It Some Thought Write the equilibrium-constant expression for the reaction N 3 1aq2 + 2 O1l 2 N 4 + 1aq2 + O - 1aq Calculating Equilibrium Constants If we can measure the equilibrium concentrations of all the reactants and products in a chemical reaction, as we did with the data in Table 15.1, calculating the value of the equilibrium constant is straightforward. We simply insert all the equilibrium concentrations into the equilibrium-constant expression for the reaction. Sample Exercise 15.7 Calculating K When All Equilibrium Concentrations Are Known After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium at 472 C, it is found to contain 7.38 atm 2, 2.46 atm N 2, and atm N 3. From these data, calculate the equilibrium constant K p for the reaction N 2 1g g2 2 N 3 1g2 Solution Analyze We are given a balanced equation and equilibrium partial pressures and are asked to calculate the value of the equilibrium constant. Plan Using the balanced equation, we write the equilibrium-constant expression. We then substitute the equilibrium partial pressures into the expression and solve for K p. Solve K p = 1P N P N2 1P = = 2.79 * Practice Exercise 1 A mixture of gaseous sulfur dioxide and oxygen are added to a reaction vessel and heated to 1000 K where they react to form SO 3 1g2. If the vessel contains atm SO 2 1g2, atm O 2 1g2, and atm SO 3 1g2 after the system has reached equilibrium, what is the equilibrium constant K p for the reaction 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2? (a) , (b) 0.322, (c) 24.4, (d) 3.36, (e) Practice Exercise 2 An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C: 3C 3 COO4 = 1.65 * 10-2 M; = 5.44 * 10-4 M; and 3C 3 COO - 4 = 5.44 * 10-4 M. Calculate the equilibrium constant K c for the ionization of acetic acid at 25 C. The reaction is C 3 COO1aq2 + 1aq2 + C 3 COO - 1aq2 Often, we do not know the equilibrium concentrations of all species in an equilibrium mixture. If we know the initial concentrations and the equilibrium concentration of at least one species, however, we can generally use the stoichiometry of the reaction

112 section 15.5 Calculating Equilibrium Constants 645 to deduce the equilibrium concentrations of the others. The following steps outline the procedure: 1. Tabulate all known initial and equilibrium concentrations of the species that appear in the equilibrium-constant expression. 2. For those species for which initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium. 3. Use the stoichiometry of the reaction (that is, the coefficients in the balanced chemical equation) to calculate the changes in concentration for all other species in the equilibrium-constant expression. 4. Use initial concentrations from step 1 and changes in concentration from step 3 to calculate any equilibrium concentrations not tabulated in step Determine the value of the equilibrium constant. The best way to illustrate how to do this type of calculation is by example, as we do in Sample Exercise Sample Exercise 15.8 Calculating K from Initial and Equilibrium Concentrations A closed system initially containing * 10-3 M 2 and * 10-3 M I 2 at 448 C is allowed to reach equilibrium, and at equilibrium the I concentration is 1.87 * 10-3 M. Calculate K c at 448 C for the reaction taking place, which is 2 1g2 + I 2 1g2 2 I1g2 Solution Analyze We are given the initial concentrations of 2 and I 2 and the equilibrium concentration of I. We are asked to calculate the equilibrium constant K c for 2 1g2 + I 2 1g2 2 I1g2. Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant. Solve (1) We tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table. (2) We calculate the change in I concentration, which is the difference between the equilibrium and initial values: (3) We use the coefficients in the balanced equation to relate the change in [I] to the changes in and 3I 2 4: (4) We calculate the equilibrium concentrations of 2 and I 2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed: (5) Our table now is complete (with equilibrium concentrations in blue for emphasis): 2 1g2 + I 2 1g2 2 I1g2 Initial concentration (M) * * Change in concentration (M) Equilibrium concentration (M) 1.87 * 10-3 Change in 3I4 = 1.87 * 10-3 M - 0 = 1.87 * 10-3 M -3 mol I a1.87 * 10 b a 1 mol 2 L 2 mol I b = * mol L -3 mol I a1.87 * 10 b a 1 mol I 2 L 2 mol I b = * mol I L = * 10-3 M * 10-3 M = * 10-3 M 3I 2 4 = * 10-3 M * 10-3 M = * 10-3 M 2 1g2 + I 2 1g2 2 I1g2 Initial concentration (M) * * Change in concentration (M) * * * 10-3 Equilibrium concentration (M) * * * 10-3

113 646 chapter 15 Chemical Equilibrium Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. Finally, we use the equilibrium-constant expression to calculate the equilibrium constant: K c = 3I I 2 4 = * * * = 51 Comment The same method can be applied to gaseous equilibrium problems to calculate K p, in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for Initial Change Equilibrium. Practice Exercise 1 In Section 15.1, we discussed the equilibrium between N 2 O 4 1g2 and NO 2 1g2. Let s return to that equation in a quantitative example. When 9.2 g of frozen N 2 O 4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilbirum, the concentration of N 2 O 4 is determined to be M. Given this information, what is the value of K c for the reaction N 2 O 4 1g2 2 NO 2 1g2 at 400 K? (a) 0.23, (b) 0.36, (c) 0.13, (d) 1.4, (e) 2.5. Practice Exercise 2 The gaseous compound BrCl decomposes at high temperature in a sealed container: 2 BrCl1g2 Br 2 1g2 + Cl 2 1g2. Initially, the vessel is charged at 500 K with BrCl(g) at a partial pressure of atm. At equilibrium, the BrCl(g) partial pressure is atm. Calculate the value of K p at 500 K applications of Equilibrium Constants We have seen that the magnitude of K indicates the extent to which a reaction proceeds. If K is very large, the equilibrium mixture contains mostly substances on the product side of the equation for the reaction. That is, the reaction proceeds far to the right. If K is very small (that is, much less than 1), the equilibrium mixture contains mainly substances on the reactant side of the equation. The equilibrium constant also allows us to (1) predict the direction in which a reaction mixture achieves equilibrium and (2) calculate equilibrium concentrations of reactants and products. Predicting the Direction of Reaction For the formation of N 3 from N 2 and 2 (Equation 15.6), K c = at 472 C. Suppose we place 2.00 mol of 2, 1.00 mol of N 2, and 2.00 mol of N 3 in a 1.00-L container at 472 C. ow will the mixture react to reach equilibrium? Will N 2 and 2 react to form more N 3, or will N 3 decompose to N 2 and 2? To answer this question, we substitute the starting concentrations of N 2, 2, and N 3 into the equilibrium-constant expression and compare its value to the equilibrium constant: 3N N = = whereas K c = [15.22] To reach equilibrium, the quotient 3N >3N must decrease from the starting value of to the equilibrium value of Because the system is closed, this change can happen only if 3N 3 4 decreases and 3N 2 4 and increase. Thus, the reaction proceeds toward equilibrium by forming N 2 and 2 from N 3 ; that is, the reaction as written in Equation 15.6 proceeds from right to left. This approach can be formalized by defining a quantity called the reaction quotient. The reaction quotient, Q, is a number obtained by substituting reactant and product concentrations or partial pressures at any point during a reaction into an equilibrium-constant expression. Therefore, for the general reaction a A + b B d D + e E

114 section 15.6 Applications of Equilibrium Constants 647 the reaction quotient in terms of molar concentrations is Q c = 3D4d 3E4 e 3A4 a 3B4 b [15.23] (A related quantity Q p can be written for any reaction that involves gases by using partial pressures instead of concentrations.) Although we use what looks like the equilibrium-constant expression to calculate the reaction quotient, the concentrations we use may or may not be the equilibrium concentrations. For example, when we substituted the starting concentrations into the equilibrium-constant expression of Equation 15.22, we obtained Q c = whereas K c = The equilibrium constant has only one value at each temperature. The reaction quotient, however, varies as the reaction proceeds. Of what use is Q? One practical thing we can do with Q is tell whether our reaction really is at equilibrium, which is an especially valuable option when a reaction is very slow. We can take samples of our reaction mixture as the reaction proceeds, separate the components, and measure their concentrations. Then we insert these numbers into Equation for our reaction. To determine whether we are at equilibrium, or in which direction the reaction proceeds to achieve equilibrium, we compare the values of Q c and K c or Q p and K p. Three possible situations arise: Q 6 K: The concentration of products is too small and that of reactants too large. The reaction achieves equilibrium by forming more products; it proceeds from left to right. Q = K: The reaction quotient equals the equilibrium constant only if the system is at equilibrium. Q 7 K: The concentration of products is too large and that of reactants too small. The reaction achieves equilibrium by forming more reactants; it proceeds from right to left. These relationships are summarized in Figure Q At equilibrium K Q K K Q < K Reaction proceeds to form more products Q = K Q > K Figure 15.8 Predicting the direction of a reaction by comparing Q and K at a given temperature. Q Equilibrium Reaction proceeds to form more reactants Sample Exercise 15.9 Predicting the Direction of Approach to Equilibrium At 448 C, the equilibrium constant K c for the reaction 2 1g2 + I 2 1g2 2 I1g2 is Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 * 10-2 mol of I, 1.0 * 10-2 mol of 2, and 3.0 * 10-2 mol of I 2 in a 2.00-L container. Solution Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium. Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Q c. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed. Solve The initial concentrations are 3I4 = 2.0 * 10-2 mol>2.00 L = 1.0 * 10-2 M = 1.0 * 10-2 mol>2.00 L = 5.0 * 10-3 M 3I 2 4 = 3.0 * 10-2 mol>2.00 L = 1.5 * 10-2 M The reaction quotient is therefore Q c = 3I I 2 4 = 11.0 * * * = 1.3 Because Q c 6 K c, the concentration of I must increase and the concentrations of 2 and I 2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium. Practice Exercise 1 Which of the following statements accurately describes what would happen to the direction of the reaction described in the sample exercise above, if the size of the container were different from 2.00 L? (a) The reaction would proceed in the opposite direction (from right to left) if the container volume were reduced sufficiently. (b) The reaction would proceed in the opposite direction if the container volume were expanded sufficiently. (c) The direction of this reaction does not depend on the volume of the container. Practice Exercise 2 At 1000 K, the value of K p for the reaction 2 SO 3 1g2 2 SO 2 1g2 + O 2 1g2 is Calculate the value for Q p, and predict the direction in which the reaction proceeds toward equilibrium if the initial partial pressures are P SO3 = 0.16 atm; P SO2 = 0.41 atm; P O2 = 2.5 atm.

115 648 chapter 15 Chemical Equilibrium Calculating Equilibrium Concentrations Sample Exercise Calculating Equilibrium Concentrations Chemists frequently need to calculate the amounts of reactants and products present at equilibrium in a reaction for which they know the equilibrium constant. The approach in solving problems of this type is similar to the one we used for evaluating equilibrium constants: We tabulate initial concentrations or partial pressures, changes in those concentrations or pressures, and final equilibrium concentrations or partial pressures. Usually, we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise For the aber process, N 2 1g g2 2 N 3 1g2, K p = 1.45 * 10-5, at 500 C. In an equilibrium mixture of the three gases at 500 C, the partial pressure of 2 is atm and that of N 2 is atm. What is the partial pressure of N 3 in this equilibrium mixture? Solution Analyze We are given an equilibrium constant, K p, and the equilibrium partial pressures of two of the three substances in the equation 1N 2 and 2 2, and we are asked to calculate the equilibrium partial pressure for the third substance 1N 3 2. Plan We can set K p equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation. Solve We tabulate the equilibrium pressures: Equilibrium pressure 1atm N 2 1g g2 2 N 3 1g2 Because we do not know the equilibrium pressure of N 3, we represent it with x. At equilibrium, the pressures must satisfy the equilibriumconstant expression: K p = 1P N P N2 1P = x 2 = 1.45 * x We now rearrange the equation to solve for x: x 2 = * = 5.01 * 10-6 x = * 10-6 = 2.24 * 10-3 atm = P N3 Check We can always check our answer by using it to recalculate the value of the equilibrium constant: K p = * = 1.45 * Practice Exercise 1 At 500 K, the reaction 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 has K p = 51. In an equilibrium mixture at 500 K, the partial pressure of NO is atm and Cl 2 is atm. What is the partial pressure of NOCl in the equilibrium mixture? (a) 0.13 atm, (b) 0.36 atm, (c) 1.0 atm, (d) 5.1 * 10-5 atm, (e) atm. Practice Exercise 2 At 500 K, the reaction PCl 5 1g2 PCl 3 1g2 + Cl 2 1g2 has K p = In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is atm and that of PCl 3 is atm. What is the partial pressure of Cl 2 in the equilibrium mixture? In many situations, we know the value of the equilibrium constant and the initial amounts of all species. We must then solve for the equilibrium amounts. Solving this type of problem usually entails treating the change in concentration as a variable. The stoichiometry of the reaction gives us the relationship between the changes in the amounts of all the reactants and products, as illustrated in Sample Exercise The calculations frequently involve the quadratic formula, as you will see in this exercise. Sample Exercise Calculating Equilibrium Concentrations from Initial Concentrations A L flask is filled with mol of 2 1g2 and mol of I 2 1g2 at 448 C. The value of the equilibrium constant K c for the reaction 2 1g2 + I 2 1g2 2 I1g2 at 448 C is What are the equilibrium concentrations of 2, I 2, and I in moles per liter? Solution Analyze We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species. Plan In this case, we are not given any of the equilibrium concentrations. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.8, where we calculated an equilibrium constant using initial concentrations.

116 section 15.6 Applications of Equilibrium Constants 649 Solve (1) We note the initial concentrations of 2 and I 2 : (2) We construct a table in which we tabulate the initial concentrations: = M and 3I 2 4 = M 2 1g2 + I 2 1g2 2 I1g2 Initial concentration (M) Change in concentration (M) Equilibrium concentration (M) (3) We use the stoichiometry of the reaction to determine the changes in concentration that occur as the reaction proceeds to equilibrium. The 2 and I 2 concentrations will decrease as equilibrium is established and that of I will increase. Let s represent the change in concentration of 2 by x. The balanced chemical equation tells us the relationship between the changes in the concentrations of the three gases. For each x mol of 2 that reacts, x mol of I 2 are consumed and 2x mol of I are produced: 2 1g2 + I 2 1g2 2 I1g2 Initial concentration (M) Change in concentration (M) -x -x +2x Equilibrium concentration (M) (4) We use initial concentrations and changes in concentrations, as dictated by stoichiometry, to express the equilibrium concentrations. With all our entries, our table now looks like this: 2 1g2 + I 2 1g2 2 I1g2 Initial concentration (M) Change in concentration (M) -x -x +2x Equilibrium concentration (M) x x 2x (5) We substitute the equilibrium concentrations into the equilibrium-constant expression and solve for x: K c = 3I I 2 4 = 12x x x2 = 50.5 If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this expression to obtain a quadratic equation in x: 4x 2 = 50.51x x x x = 0 Solving the quadratic equation (Appendix A.3) leads to two solutions for x: When we substitute x = into the expressions for the equilibrium concentrations, we find negative concentrations of 2 and I 2. Because a negative concentration is not chemically meaningful, we reject this solution. We then use x = to find the equilibrium concentrations: x = { = x = M 3I 2 4 = x = M 3I4 = 2x = 1.87 M = or Check We can check our solution by putting these numbers into the equilibrium-constant expression to assure that we correctly calculate the equilibrium constant: K c = 3I I 2 4 = = 51 Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not chemically meaningful. Reject this solution to the quadratic equation.

117 650 chapter 15 Chemical Equilibrium Practice Exercise 1 For the equilibrium Br 2 1g2 + Cl 2 1g2 2 BrCl1g2, the equilibrium constant K p is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl? (a) 0.57 atm, (b) 0.22 atm, (c) 0.45 atm, (d) 0.15 atm, (e) 0.31 atm. Practice Exercise 2 For the equilibrium PCl 5 1g2 PCl 3 1g2 + Cl 2 1g2, the equilibrium constant K p is at 500 K. A gas cylinder at 500 K is charged with PCl 5 1g2 at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl 5, PCl 3, and Cl 2 at this temperature? 15.7 Le Châtelier s Principle Many of the products we use in everyday life are obtained from the chemical industry. Chemists and chemical engineers in industry spend a great deal of time and effort to maximize the yield of valuable products and minimize waste. For example, when aber developed his process for making ammonia from N 2 and 2, he examined how reaction conditions might be varied to increase yield. Using the values of the equilibrium constant at various temperatures, he calculated the equilibrium amounts of N 3 formed under a variety of conditions. Some of aber s results are shown in Figure Notice that the percent of N 3 present at equilibrium decreases with increasing temperature and increases with increasing pressure. We can understand these effects in terms of a principle first put forward by enri-louis Le Châtelier* ( ), a French industrial chemist: If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position so as to counteract the effect of the disturbance. Go Figure At what combination of pressure and temperature should you run the reaction to maximize N 3 yield? Percent N 3 produced Percent of N 3 increases with increasing pressure % 27.4% 47.8% 18.9% 35.9% Total pressure (atm) % 54.9% 26.0% % 42.9% 60.6% 32.2% 16.9% 48.8% % % 500 Percent of N 3 decreases with increasing temperature 550 Temperature ( C) Figure 15.9 Effect of temperature and pressure on N 3 yield in the aber process. Each mixture was produced by starting with a 3 : 1 molar mixture of 2 and N 2. *Pronounced le-sot-lee-ay.

118 section 15.7 Le Châtelier s Principle 651 Le Châtelier s Principle If a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, the system will shift its equilibrium position so as to counter the effect of the disturbance. Concentration: adding or removing a reactant or product If a substance is added to a system at equilibrium, the system reacts to consume some of the substance. If a substance is removed from a system, the system reacts to produce more of substance. Initial equilibrium Substance added Equilibrium reestablished Substances react Pressure: changing the pressure by changing the volume At constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. Initial volume Pressure System shifts to direction of fewer moles of gas Temperature: If the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the excess reactant, namely heat. Endothermic Exothermic Increasing T Reaction shifts right Increasing T Reaction shifts left Decreasing T Reaction shifts left Decreasing T Reaction shifts right In this section, we use Le Châtelier s principle to make qualitative predictions about how a system at equilibrium responds to various changes in external conditions. We consider three ways in which a chemical equilibrium can be disturbed: (1) adding or removing a reactant or product, (2) changing the pressure by changing the volume, and (3) changing the temperature. Change in Reactant or Product Concentration A system at dynamic equilibrium is in a state of balance. When the concentrations of species in the reaction are altered, the equilibrium shifts until a new state of balance is attained. What does shift mean? It means that reactant and product concentrations change over time to accommodate the new situation. Shift does not mean that the equilibrium constant itself is altered; the equilibrium constant remains the same. Le Châtelier s principle states that the shift is in the direction that minimizes or reduces the effect of the change. Therefore, if a chemical system is already at equilibrium and the concentration of any substance in the mixture is increased (either reactant or product), the system reacts to consume some of that substance. Conversely, if the concentration of a substance is decreased, the system reacts to produce some of that substance. There is no change in the equilibrium constant when we change the concentrations of reactants or products. As an example, consider our familiar equilibrium mixture of N 2, 2, and N 3 : N 2 1g g2 2 N 3 1g2

119 652 chapter 15 Chemical Equilibrium Go Figure Why does the nitrogen concentration decrease after hydrogen is added? N 2 (g) (g) 2 N 3 (g) Initial equilibrium 2 added Equilibrium reestablished 2 Partial pressure N 3 N 2 Time Figure Effect of adding 2 to an equilibrium mixture of N 2, 2, and N 3. Adding 2 causes the reaction as written to shift to the right, consuming some N 2 to produce more N 3. Adding 2 causes the system to shift so as to reduce the increased concentration of 2 ( Figure 15.10). This change can occur only if the reaction consumes 2 and simultaneously consumes N 2 to form more N 3. Adding N 2 to the equilibrium mixture likewise causes the reaction to shift toward forming more N 3. Removing N 3 also causes a shift toward producing more N 3, whereas adding N 3 to the system at equilibrium causes the reaction to shift in the direction that reduces the increased N 3 concentration: Some of the added ammonia decomposes to form N 2 and 2. All of these shifts are entirely consistent with predictions that we would make by comparing the reaction quotient Q with the equilibrium constant K. In the aber reaction, therefore, removing N 3 from an equilibrium mixture of N 2, 2, and N 3 causes the reaction to shift right to form more N 3. If the N 3 can be removed continuously as it is produced, the yield can be increased dramatically. In the industrial production of ammonia, the N 3 is continuously removed by selectively liquefying it ( Figure 15.11). (The boiling point of N 3, -33 C, is much higher than those of N 2, -196 C, and 2, -253 C.) The liquid N 3 is removed, and the N 2 and 2 are recycled to form more N 3. As a result of the product being continuously removed, the reaction is driven essentially to completion. Give It Some Thought Does the equilibrium 2 NO1g2 + O 2 1g2 2 NO 2 1g2 shift to the right (more products) or left (more reactants) if (a) O 2 is added to the system? (b) NO is removed? Effects of Volume and Pressure Changes If a system containing one or more gases is at equilibrium and its volume is decreased, thereby increasing its total pressure, Le Châtelier s principle indicates that the system responds by shifting its equilibrium position to reduce the pressure. A system can reduce its pressure by reducing the total number of gas molecules (fewer molecules of gas

120 section 15.7 Le Châtelier s Principle 653 ot gases N Incoming N 2 and 2 gases eat exchanger eat exchanger 5 Unreacted N 2 and 2 recycled 3 eated gases pass over catalyst, N 3 forms 4 As gas mixture cools, N 3 (g) liquifies 2 N 2 and 2 gases heated to approximately 500 C Liquid N 3 outlet Figure Diagram of the industrial production of ammonia. Incoming N 2 1g2 and 2 1g2 are heated to approximately 500 C and passed over a catalyst. When the resultant N 2, 2, and N 3 mixture is cooled, the N 3 liquefies and is removed from the mixture, shifting the reaction to produce more N 3. exert a lower pressure). Thus, at constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. Increasing the volume causes a shift in the direction that produces more gas molecules ( Figure 15.12). 2 A(g) B(g) Decrease volume, increase pressure Increase volume, decrease pressure New equilibrium favors products to reduce total moles of gas Initial volume New equilibrium favors reactants to increase total moles of gas Figure Pressure and Le Châtelier s principle.

121 654 chapter 15 Chemical Equilibrium Give It Some Thought What happens to the equilibrium 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2, if the volume of the system is increased? In the reaction N 2 1g g2 2 N 3 1g2, four molecules of reactant are consumed for every two molecules of product produced. Consequently, an increase in pressure (caused by a decrease in volume) shifts the reaction in the direction that produces fewer gas molecules, which leads to the formation of more N 3, as indicated in Figure In the reaction 2 1g2 + I 2 1g2 2 I1g2, the number of molecules of gaseous products (two) equals the number of molecules of gaseous reactants; therefore, changing the pressure does not influence the position of equilibrium. Keep in mind that, as long as temperature remains constant, pressure volume changes do not change the value of K. Rather, these changes alter the partial pressures of the gaseous substances. In Sample Exercise 15.7, we calculated K p = 2.79 * 10-5 for the aber reaction, N 2 1g g2 2 N 3 1g2, in an equilibrium mixture at 472 C containing 7.38 atm 2, 2.46 atm N 2, and atm N 3. Consider what happens when we suddenly reduce the volume of the system by one-half. If there were no shift in equilibrium, this volume change would cause the partial pressures of all substances to double, giving P 2 = atm, P N2 = 4.92 atm, and P N3 = atm. The reaction quotient would then no longer equal the equilibrium constant: Q p = 1P N P N2 1P = = 6.97 * 10-6 K p Because Q p 6 K p, the system would no longer be at equilibrium. Equilibrium would be reestablished by increasing P N3 and/or decreasing P N2 and P 2 until Q p = K p = 2.79 * Therefore, the equilibrium shifts to the right in the reaction as written, as Le Châtelier s principle predicts. It is possible to change the pressure of a system in which a chemical reaction is running without changing its volume. For example, pressure increases if additional amounts of any reacting components are added to the system. We have already seen how to deal with a change in concentration of a reactant or product. owever, the total pressure in the reaction vessel might also be increased by adding a gas that is not involved in the equilibrium. For example, argon might be added to the ammonia equilibrium system. The argon would not alter the partial pressures of any of the reacting components and therefore would not cause a shift in equilibrium. Effect of Temperature Changes Changes in concentrations or partial pressures shift equilibria without changing the value of the equilibrium constant. In contrast, almost every equilibrium constant changes as the temperature changes. For example, consider the equilibrium established when cobalt(ii) chloride 1CoCl 2 2 is dissolved in hydrochloric acid, Cl(aq), in the endothermic reaction Co1 2 O aq2 + 4 Cl - 1aq2 CoCl 2-4 1aq O1l2 7 0 [15.24] Pale pink Deep blue Because Co1 2 O2 6 is pink and CoCl 4 is blue, the position of this equilibrium is readily apparent from the color of the solution ( Figure 15.13). When the solution is heated it turns blue, indicating that the equilibrium has shifted to form more CoCl 2-4. Cooling the solution leads to a pink solution, indicating that the equilibrium has shifted to produce more Co1 2 O We can monitor this reaction by spectroscopic methods, measuring the concentration of all species at the different temperatures. (Section 14.2) We can then calculate the equilibrium constant at each temperature. ow do we explain why the equilibrium constants and therefore the position of equilibrium both depend on temperature?

122 section 15.7 Le Châtelier s Principle 655 > 0, endothermic reaction eat + Co( 2 O) 2+ 6 (aq) + 4 Cl (aq) Pink CoCl 2 4 (aq) O(l) Blue CoCl 4 2+ Co( 2 O) 6 2 Cool eat Solution appears pink because lowering the temperature shifts the equilibrium to favor formation of the pink Co( 2 O) 6 2+ ion. Solution appears violet because appreciable amounts of both pink Co( 2 O) 6 2+ and blue CoCl 4 2 are present. Solution appears blue because raising the temperature shifts the equilibrium to favor formation of the blue CoCl 4 2 ion. Figure Temperature and Le Châtelier s principle. In the molecular level views, only the CoCl and Co1 2 O ions are shown for clarity. We can deduce the rules for the relationship between K and temperature from Le Châtelier s principle. We do this by treating heat as a chemical reagent. In an endothermic (heat-absorbing) reaction, we consider heat a reactant, and in an exothermic (heat-releasing) reaction, we consider heat a product: Endothermic: Exothermic: Reactants + heat products Reactants products + heat When the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the excess reactant (or product), namely heat. Give It Some Thought Use Le Châtelier s principle to explain why the equilibrium vapor pressure of a liquid increases with increasing temperature. In an endothermic reaction, such as Equation 15.24, heat is absorbed as reactants are converted to products. Thus, increasing the temperature causes the equilibrium to shift to the right, in the direction of making more products, and K increases. In an exothermic reaction, the opposite occurs: eat is produced as reactants are converted to

123 656 chapter 15 Chemical Equilibrium products. Thus, increasing the temperature in this case causes the equilibrium to shift to the left, in the direction of making more reactants, and K decreases. Endothermic: Exothermic: Increasing T results in higher K value Increasing T results in lower K value Cooling a reaction has the opposite effect. As we lower the temperature, the equilibrium shifts in the direction that produces heat. Thus, cooling an endothermic reaction shifts the equilibrium to the left, decreasing K, as shown in Figure 15.13, and cooling an exothermic reaction shifts the equilibrium to the right, increasing K. Sample Exercise Using Le Châtelier s Principle to Predict Shifts in Equilibrium Consider the equilibrium N 2 O 4 1g2 2 NO 2 1g2 = 58.0 kj In which direction will the equilibrium shift when (a) N 2 O 4 is added, (b) NO 2 is removed, (c) the pressure is increased by addition of N 2 1g2, (d) the volume is increased, (e) the temperature is decreased? Solution Analyze We are given a series of changes to be made to a system at equilibrium and are asked to predict what effect each change will have on the position of the equilibrium. Plan Le Châtelier s principle can be used to determine the effects of each of these changes. Solve (a) The system will adjust to decrease the concentration of the added N 2 O 4, so the equilibrium shifts to the right, in the direction of product. (b) The system will adjust to the removal of NO 2 by shifting to the side that produces more NO 2 ; thus, the equilibrium shifts to the right. (c) Adding N 2 will increase the total pressure of the system, but N 2 is not involved in the reaction. The partial pressures of NO 2 and N 2 O 4 are therefore unchanged, and there is no shift in the position of the equilibrium. (d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right. (e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium shifts to the left, toward the formation of more N 2 O 4. Note that only this last change also affects the value of the equilibrium constant, K. Practice Exercise 1 For the reaction 4 N 3 1g2 + 5 O 2 1g2 4 NO1g O1g2 = -904 kj which of the following changes will shift the equilibrium to the right, toward the formation of more products? (a) Adding more water vapor, (b) Increasing the temperature, (c) Increasing the volume of the reaction vessel, (d) Removing O 2 (g), (e) Adding 1 atm of Ne(g) to the reaction vessel. Practice Exercise 2 For the reaction PCl 5 1g2 PCl 3 1g2 + Cl 2 1g2 = 87.9 kj in which direction will the equilibrium shift when (a) Cl 2 1g2 is removed, (b) the temperature is decreased, (c) the volume of the reaction system is increased, (d) PCl 3 1g2 is added? Sample Exercise Predicting the Effect of Temperature on K (a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the reaction N 2 1g g2 2 N 3 1g2 (b) Determine how the equilibrium constant for this reaction should change with temperature. Solution Analyze We are asked to determine the standard enthalpy change of a reaction and how the equilibrium constant for the reaction varies with temperature. Plan (a) We can use standard enthalpies of formation to calculate for the reaction. (b) We can then use Le Châtelier s principle to determine what effect temperature will have on the equilibrium constant. Solve (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation, minus the same quantities for the reactants. (Section 5.7) At 25 C, f for N 3 1g2 is kj>mol. The f values for 2 1g2 and N 2 1g2 are zero by definition because the enthalpies of formation of the elements in their normal

124 section 15.7 Le Châtelier s Principle 657 states at 25 C are defined as zero. (Section 5.7) Because 2 mol of N 3 is formed, the total enthalpy change is 12 mol kj>mol2-0 = kj (b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less N 3 and more N 2 and 2. This effect is seen in the values for K p presented in Table Notice that K p changes markedly with changes in temperature and that it is larger at lower temperatures. Comment The fact that K p for the formation of N 3 from N 2 and 2 decreases with increasing temperature is a matter of great practical importance. To form N 3 at a reasonable rate requires higher temperatures. At higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to N 3 is smaller. To compensate for this, higher pressures are needed because high pressure favors N 3 formation. Table 15.2 Variation in K p with Temperature for n 2 + 3h 2 2 nh 3 Temperature 1 C * * * * * * 10-6 K p Practice Exercise 1 The standard enthalpy of formation of Cl(g) is kj>mol. Given only this information, in which direction would you expect the equilibrium for the reaction 2 1g2 + Cl 2 1g2 2 Cl1g2 to shift as the temperature increases: (a) to the left, (b) to the right, (c) no shift in equilibrium? Practice Exercise 2 Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2 POCl 3 1g2 2 PCl 3 1g2 + O 2 1g2 Use this result to determine how the equilibrium constant for the reaction should change with temperature. The Effect of Catalysts What happens if we add a catalyst to a chemical system that is at equilibrium? As shown in Figure 15.14, a catalyst lowers the activation barrier between reactants and products. The activation energies for both the forward and reverse reactions are lowered. The catalyst thereby increases the rates of both forward and reverse reactions. Since K is the ratio of the forward and reverse rate constants for a reaction, you can predict, correctly, that the presence of a catalyst, even though it changes the reaction rate, does not affect the numeric value of K (Figure 15.14). As a result, a catalyst increases the rate at which equilibrium is achieved but does not change the composition of the equilibrium mixture. The rate at which a reaction approaches equilibrium is an important practical consideration. As an example, let s again consider the synthesis of ammonia from Go Figure What quantity dictates the speed of a reaction: (a) the energy difference between the initial state and the transition state or (b) the energy difference between the initial state and the final state? Transition states Catalyzed reaction proceeds more rapidly [B] eq Energy Catalyzed reaction has lower activation energy A [B] The two reactions reach the same equilibrium mixture, but the catalyzed reaction achieves equilibrium faster Reaction pathway B Time Figure An energy profile for the reaction A B (left), and the change in concentration of B as a function of time (right), with and without a catalyst. Green curves show the reaction with a catalyst; black curves show the reaction without a catalyst.

125 658 chapter 15 Chemical Equilibrium N 2 and 2. In designing his process, aber had to deal with a rapid decrease in the equilibrium constant with increasing temperature (Table 15.2). At temperatures sufficiently high to give a satisfactory reaction rate, the amount of ammonia formed was too small. The solution to this dilemma was to develop a catalyst that would produce a reasonably rapid approach to equilibrium at a sufficiently low temperature, so that the equilibrium constant remained reasonably large. The development of a suitable catalyst thus became the focus of aber s research efforts. After trying different substances to see which would be most effective, Carl Bosch settled on iron mixed with metal oxides, and variants of this catalyst formulation are still used today. (Section 15.2, The aber Process ) These catalysts make it possible to obtain a reasonably rapid approach to equilibrium at around 400 to 500 C and 200 to 600 atm. The high pressures are needed to obtain a satisfactory equilibrium amount of N 3. If a catalyst could be found that leads to sufficiently rapid reaction at temperatures lower than 400 C, it would be possible to obtain the same extent of equilibrium conversion at pressures much lower than 200 to 600 atm. This would result in great savings in both the cost of the high-pressure equipment and the energy consumed in the production of ammonia. It is estimated that the aber process consumes approximately 1% of the energy generated in the world each year. Not surprisingly chemists and chemical engineers are actively searching for improved catalysts for the aber process. A breakthrough in this field would not only increase the supply of ammonia for fertilizers, it would also reduce the global consumption of fossil fuels in a significant way. Give It Some Thought Can a catalyst be used to increase the amount of product produced for a reaction that reaches equilibrium quickly without a catalyst? Sample Integrative Exercise Putting Concepts Together At temperatures near 800 C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and 2 : C1s2 + 2 O1g2 CO1g g2 The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 C the equilibrium constant for this reaction is K p = What are the equilibrium partial pressures of 2 O, CO, and 2 in the equilibrium mixture at this temperature if we start with solid carbon and mol of 2 O in a 1.00-L vessel? (b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25 C the value of K p for this reaction is 1.7 * Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and 2 at equilibrium, should the pressure of the system be increased or decreased? Solution (a) To determine the equilibrium partial pressures, we use the ideal-gas equation, first determining the starting partial pressure of water. P 2 O = n 2 ORT mol L@atm>mol@K K2 = = 8.81 atm V 1.00 L We then construct a table of initial partial pressures and their changes as equilibrium is achieved: C1s2 + 2 O1g2 CO1g g2 Initial partial pressure (atm) Change in partial pressure (atm) -x +x +x Equilibrium partial pressure (atm) x x x There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the equilibrium-constant expression. Substituting the equilibrium partial pressures of the other species into the equilibrium-constant expression for the reaction gives K p = P COP 2 P 2 O = 1x21x x2 = 14.1

126 section 15.7 Le Châtelier s Principle 659 Multiplying through by the denominator gives a quadratic equation in x: x 2 = x2 x x = 0 Solving this equation for x using the quadratic formula yields x = 6.14 atm. ence, the equilibrium partial pressures are P CO = x = 6.14 atm, P 2 = x = 6.14 atm, and P 2 O = x2 = 2.67 atm. (b) Part (a) shows that x = 6.14 atm of 2 O must react for the system to achieve equilibrium. We can use the ideal-gas equation to convert this partial pressure into a mole amount. n = PV RT = atm L2 = mol L@atm>mol@K K2 Thus, mol of 2 O and the same amount of C must react to achieve equilibrium. As a result, there must be at least mol of C (0.836 g C) present among the reactants at the start of the reaction. (c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures: P total = P 2 O + P CO + P 2 = 2.67 atm atm atm = atm (d) In discussing Le Châtelier s principle, we saw that endothermic reactions exhibit an increase in K p with increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases, the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our prediction by calculating the enthalpy change for the reaction, = f 1CO1g22 + f 1 2 1g22 - f 1C(s, graphite2 - f 1 2 O1g22 = kj The positive sign for indicates that the reaction is endothermic. (e) According to Le Châtelier s principle, a decrease in the pressure causes a gaseous equilibrium to shift toward the side of the equation with the greater number of moles of gas. In this case, there are 2 mol of gas on the product side and only one on the reactant side. Therefore, the pressure should be decreased to maximize the yield of the CO and 2. Chemistry Put to Work Controlling Nitric Oxide Emissions The formation of NO from N 2 and O 2, 1 2 N 2 1g O 21g2 NO1g2 = 90.4 kj [15.25] provides an interesting example of the practical importance of the fact that equilibrium constants and reaction rates change with temperature. By applying Le Châtelier s principle to this endothermic reaction and treating heat as a reactant, we deduce that an increase in temperature shifts the equilibrium in the direction of more NO. The equilibrium constant K p for formation of 1 mol of NO from its elements at 300 K is only about 1 * ( Figure 15.15). At 2400 K, however, the equilibrium constant is about 0.05, which is times larger than the 300 K value. Figure helps explain why NO is a pollution problem. In the cylinder of a modern high-compression automobile engine, the temperature during the fuel-burning part of the cycle is approximately 2400 K. Also, there is a fairly large excess of air in the cylinder. These conditions favor the formation of NO. After combustion, however, the gases cool quickly. As the temperature drops, the equilibrium in Equation shifts to the left (because the reactant heat is being removed). owever, the lower temperature also means that the reaction rate decreases, so the NO formed at 2400 K is essentially trapped in that form as the gas cools. The gases exhausting from the cylinder are still quite hot, perhaps 1200 K. At this temperature, as shown in Figure 15.15, the equilibrium constant for formation of NO is about 5 * 10-4, much smaller than the Go Figure Estimate the value of K p at 1200 K, the exhaust gas temperature. K p N 2 (g) + 2 O 2 (g) NO(g) Exhaust gas temperature Temperature (K) Cylinder temperature during combustion Figure Equilibrium and temperature. The equilibrium constant increases with increasing temperature because the reaction is endothermic. It is necessary to use a log scale for K p because the values vary over such a large range.

127 660 chapter 15 Chemical Equilibrium value at 2400 K. owever, the rate of conversion of NO to N 2 and O 2 is too slow to permit much loss of NO before the gases are cooled further. As discussed in the Chemistry Put to Work box in Section 14.7, one of the goals of automotive catalytic converters is to achieve rapid conversion of NO to N 2 and O 2 at the temperature of the exhaust gas. Some catalysts developed for this reaction are reasonably effective under the grueling conditions in automotive exhaust systems. Nevertheless, scientists and engineers are continuously searching for new materials that provide even more effective catalysis of the decomposition of nitrogen oxides. Chapter Summary and Key Terms The Concept of Equilibrium (Section 15.1) A chemical reaction can achieve a state in which the forward and reverse processes are occurring at the same rate. This condition is called chemical equilibrium, and it results in the formation of an equilibrium mixture of the reactants and products of the reaction. The composition of an equilibrium mixture does not change with time if temperature is held constant. The Equilibrium Constant (Section 15.2) An equilibrium that is used throughout this chapter is the reaction N 2 1g g2 2 N 3 1g2. This reaction is the basis of the aber process for the production of ammonia. The relationship between the concentrations of the reactants and products of a system at equilibrium is given by the law of mass action. For an equilibrium equation of the form a A + b B d D + e E, the equilibriumconstant expression is written as K c = 3D4d 3E4 e 3A4 a 3B4 b where K c is a dimensionless constant called the equilibrium constant. When the equilibrium system of interest consists of gases, it is often convenient to express the concentrations of reactants and products in terms of gas pressures: K p = 1P D2 d 1P E 2 e 1P A 2 a 1P B 2 b K c and K p are related by the expression K p = K c 1RT2 n. To do this conversion properly, use R = L@atm>mol@K and temperature in kelvins. Understanding and Working with Equilibrium Constants (Section 15.3) The value of the equilibrium constant changes with temperature. A large value of K c indicates that the equilibrium mixture contains more products than reactants and therefore lies toward the product side of the equation. A small value for the equilibrium constant means that the equilibrium mixture contains less products than reactants and therefore lies toward the reactant side. The equilibrium-constant expression and the equilibrium constant of the reverse of a reaction are the reciprocals of those of the forward reaction. If a reaction is the sum of two or more reactions, its equilibrium constant will be the product of the equilibrium constants for the individual reactions. eterogeneous Equilibria (Section 15.4) Equilibria for which all substances are in the same phase are called homogeneous equilibria; in heterogeneous equilibria, two or more phases are present. Because their activities are exactly 1 the concentrations of pure solids and liquids are left out of the equilibrium-constant expression for a heterogeneous equilibrium. Calculating Equilibrium Constants (Section 15.5) If the concentrations of all species in an equilibrium are known, the equilibrium-constant expression can be used to calculate the equilibrium constant. The changes in the concentrations of reactants and products on the way to achieving equilibrium are governed by the stoichiometry of the reaction. Applications of Equilibrium Constants (Section 15.6) The reaction quotient, Q, is found by substituting reactant and product concentrations or partial pressures at any point during a reaction into the equilibrium-constant expression. If the system is at equilibrium, Q = K. If Q K, however, the system is not at equilibrium. When Q 6 K, the reaction will move toward equilibrium by forming more products (the reaction proceeds from left to right); when Q 7 K, the reaction will move toward equilibrium by forming more reactants (the reaction proceeds from right to left). Knowing the value of K makes it possible to calculate the equilibrium amounts of reactants and products, often by the solution of an equation in which the unknown is the change in a partial pressure or concentration. Le Châtelier s Principle (Section 15.7) Le Châtelier s principle states that if a system at equilibrium is disturbed, the equilibrium will shift to minimize the disturbing influence. Therefore, if a reactant or product is added to a system at equilibrium, the equilibrium will shift to consume the added substance. The effects of removing reactants or products and of changing the pressure or volume of a reaction can be similarly deduced. For example, if the volume of the system is reduced, the equilibrium will shift in the direction that decreases the number of gas molecules. While changes in concentration or pressure lead to shifts in the equilibrium concentrations they do not change the value of the equilibrium constant, K. Changes in temperature affect both the equilibrium concentrations and the equilibrium constant. We can use the enthalpy change for a reaction to determine how an increase in temperature affects the equilibrium: For an endothermic reaction, an increase in temperature shifts the equilibrium to the right; for an exothermic reaction, a temperature increase shifts the equilibrium to the left. Catalysts affect the speed at which equilibrium is reached but do not affect the magnitude of K. Learning Outcomes After studying this chapter, you should be able to: Explain what is meant by chemical equilibrium and how it relates to reaction rates. (Section 15.1) Write the equilibrium-constant expression for any reaction. (Section 15.2) Given the value of K c convert to K p and vice versa. (Section 15.2) Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products present in an equilibrium mixture. (Section 15.3) Manipulate the equilibrium constant to reflect changes in the chemical equation. (Section 15.3)

128 Exercises 661 Write the equilibrium-constant expression for a heterogeneous reaction. (Section 15.4) Calculate an equilibrium constant from concentration measurements. (Section 15.5) Predict the direction of a reaction given the equilibrium constant and the concentrations of reactants and products. (Section 15.6) Calculate equilibrium concentrations given the equilibrium constant and all but one equilibrium concentration. (Section 15.6) Calculate equilibrium concentrations, given the equilibrium constant and the starting concentrations. (Section 15.6) Use Le Châtelier s principle to predict how changing the concentrations, volume, or temperature of a system at equilibrium affects the equilibrium position. (Section 15.7) Key Equations K c = 3D4d 3E4 e 3A4 a 3B4 b [15.8] The equilibrium-constant expression for a general reaction of the type a A + b B d D + e E, the concentrations are equilibrium concentrations only K p = 1P D2 d 1P E 2 e 1P A 2 a 1P B 2 b [15.11] The equilibrium-constant expression in terms of equilibrium partial pressures K p = K c 1RT2 n [15.14] Relating the equilibrium constant based on pressures to the equilibrium constant based on concentration Q c = 3D4d 3E4 e [15.23] The reaction quotient. The concentrations are for any time during 3A4 a 3B4 b a reaction. If the concentrations are equilibrium concentrations, then Q c = K c. Exercises Visualizing Concepts 15.1 (a) Based on the following energy profile, predict whether k f 7 k r or k f 6 k r. (b) Using Equation 15.5, predict whether the equilibrium constant for the process is greater than 1 or less than 1. [Section 15.1] greater or smaller than 1 if the volume is 1 L and each atom/ molecule in the diagram represents 1 mol? [Section 15.2] Potential energy Reactants Products Reaction progress 15.2 The following diagrams represent a hypothetical reaction A B, with A represented by red spheres and B represented by blue spheres. The sequence from left to right represents the system as time passes. Does the system reach equilibrium? If so, in which diagram is the system in equilibrium? [Sections 15.1 and 15.2] 15.4 The following diagram represents a reaction shown going to completion. Each molecule in the diagram represents 0.1 mol and the volume of the box is 1.0 L. (a) Letting A = red spheres and B = blue spheres, write a balanced equation for the reaction. (b) Write the equilibrium-constant expression for the reaction. (c) Calculate the value of K c. (d) Assuming that all of the molecules are in the gas phase, calculate n, the change in the number of gas molecules that accompanies the reaction. (e) Calculate the value of K p. [Section 15.2] The following diagram represents an equilibrium mixture produced for a reaction of the type A + X AX. Is K 15.5 Snapshots of two hypothetical reactions, A1g2 + B1g2 AB1g2 and X1g2 + Y1g2 XY1g2 at five different times

129 662 chapter 15 Chemical Equilibrium are shown here. Which reaction has a larger equilibrium constant? [Sections 15.1 and 15.2] t = 10 s 20 s 30 s 40 s 50 s A(g) + B(g) AB(g) t = 10 s 20 s 30 s 40 s 50 s 5.0 g PbO 2 (g) in both vessels X(g) + Y(g) XY(g) Vessel A V = 50 ml Vessel B V = 100 ml 15.6 Ethene 1C reacts with halogens 1X 2 2 by the following reaction: C 2 4 1g2 + X 2 1g2 C 2 4 X 2 1g2 The following figures represent the concentrations at equilibrium at the same temperature when X 2 is Cl 2 (green), Br 2 (brown), and I 2 (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3] 15.8 The reaction A 2 + B 2 2 AB has an equilibrium constant K c = 1.5. The following diagrams represent reaction mixtures containing A 2 molecules (red), B 2 molecules (blue), and AB molecules. (a) Which reaction mixture is at equilibrium? (b) For those mixtures that are not at equilibrium, how will the reaction proceed to reach equilibrium? [Sections 15.5 and 15.6] (a) (b) (i) (ii) (iii) 15.9 The reaction A 2 1g2 + B1g2 A1g2 + AB1g2 has an equilibrium constant of K p = 2. The accompanying diagram shows a mixture containing A atoms (red), A 2 molecules, and AB molecules (red and blue). ow many B atoms should be added to the diagram to illustrate an equilibrium mixture? [Section 15.6] (c) 15.7 When lead (IV) oxide is heated above 300 C it decomposes according to the following reaction PbO 2 1s2 PbO1s2 + O 2 1g2. Consider the two sealed vessels of PbO 2 shown here. If both vessels are heated to 400 C and allowed to come to equilibitum which of the following statements is true? (a) There will be less PbO 2 remaining in vessel A, (b) There will be less PbO 2 remaining in vessel B, (c) The amount of PbO 2 remaining in each vessel will be the same. [Section 15.4] The diagram shown here represents the equilibrium state for the reaction A 2 1g2 + 2 B1g2 2AB1g2. (a) Assuming the volume is 2 L, calculate the equilibrium constant K c for the reaction. (b) If the volume of the equilibrium mixture is decreased, will the number of AB molecules increase or decrease? [Sections 15.5 and 15.7]

130 Exercises The following diagrams represent equilibrium mixtures for the reaction A 2 + B A + AB at 300 K and 500 K. The A atoms are red, and the B atoms are blue. Is the reaction exothermic or endothermic? [Section 15.7] 300 K 500 K The following graph represents the yield of the compound AB at equilibrium in the reaction A1g2 + B1g2 AB1g2 at two different pressures, x and y, as a function of temperature. [AB] P=x Temperature P=y (a) Is this reaction exothermic or endothermic? (b) Is P = x greater or smaller than P = y? [Section 15.7] Equilibrium; The Equilibrium Constant (Sections ) Suppose that the gas-phase reactions A B and B A are both elementary processes with rate constants of 4.7 * 10-3 s -1 and 5.8 * 10-1 s -1, respectively. (a) What is the value of the equilibrium constant for the equilibrium A1g2 B1g2? (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Consider the reaction A + B C + D. Assume that both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse? Write the expression for K c for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 3 NO1g2 N 2 O1g2 + NO 2 1g2 (b) C 4 1g S1g2 CS 2 1g g2 (c) Ni1CO2 4 1g2 Ni1s2 + 4 CO1g2 (d) F1aq2 + 1aq2 + F - 1aq2 (e) 2 Ag1s2 + Zn 2+ 1aq2 2 Ag + 1aq2 + Zn1s2 (f) 2 O1l2 + 1aq2 + O - 1aq2 (g) 2 2 O1l aq2 + 2 O - 1aq Write the expressions for K c for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 2 O 3 1g2 3 O 2 1g2 (b) Ti1s2 + 2 Cl 2 1g2 TiCl 4 1l2 (c) 2 C 2 4 1g O1g2 2 C 2 6 1g2 + O 2 1g2 (d) C1s g2 C 4 1g2 (e) 4 Cl1aq2 + O 2 1g2 2 2 O1l2 + 2 Cl 2 1g2 (f) 2 C l O 2 1g2 16 CO 2 1g O1g2 (g) 2 C l O 2 1g2 16 CO 2 1g O1l When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) N 2 1g2 + O 2 1g2 2 NO1g2; K c = 1.5 * (b) 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2; K p = 2.5 * Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) 2 NO1g2 + O 2 1g2 2 NO 2 1g2; K p = 5.0 * (b) 2 Br1g2 2 1g2 + Br 2 1g2; K c = 5.8 * Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases the speed at which a reaction reaches equilibrium increases Which of the following statements are true and which are false? (a) For the reaction 2 A1g2 + B1g2 A 2 B1g2 K c and K p are numerically the same. (b) It is possible to distinguish K c from K p by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of K c increases with increasing pressure If K c = for PCl 3 1g2 + Cl 2 1g2 PCl 5 1g2 at 500 K, what is the value of K p for this reaction at this temperature? Calculate K c at 303 K for SO 2 1g2 + Cl 2 1g2 SO 2 Cl 2 1g2 if K p = 34.5 at this temperature The equilibrium constant for the reaction 2 NO1g2 + Br 2 1g2 2 NOBr1g2 is K c = 1.3 * 10-2 at 1000 K. (a) At this temperature does the equilibrium favor NO and Br 2, or does it favor NOBr? (b) Calculate K c for 2 NOBr1g2 2 NO1g2 + Br 2 1g2. (c) Calculate K c for NOBr1g2 NO1g Br 2 1g Consider the following equilibrium: 2 2 1g2 + S 2 1g2 2 2 S1g2 K c = 1.08 * 10 7 at 700 C

131 664 chapter 15 Chemical Equilibrium (a) Calculate K p. (b) Does the equilibrium mixture contain mostly 2 and S 2 or mostly 2 S? (c) Calculate the value of K c if you rewrote the equation 2 1g S 21g2 2 S1g At 1000 K, K p = 1.85 for the reaction SO 2 1g O 2 1g2 SO 3 1g2 (a) What is the value of K p for the reaction SO 3 1g2 SO 2 1g O 21g2? (b) What is the value of K p for the reaction 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2? (c) What is the value of K c for the reaction in part (b)? Consider the following equilibrium, for which K p = at 480 C: 2 Cl 2 1g O1g2 4 Cl1g2 + O 2 1g2 (a) What is the value of K p for the reaction 4 Cl1g2 + O 2 1g2 2 Cl 2 1g O1g2? (b) What is the value of K p for the reaction Cl 2 1g2 + 2 O1g2 2 Cl1g O 21g2? (c) What is the value of K c for the reaction in part (b)? The following equilibria were attained at 823 K: CoO1s g2 Co1s2 + 2 O1g2 K c = 67 CoO1s2 + CO1g2 Co1s2 + CO 2 1g2 K c = 490 Based on these equilibria, calculate the equilibrium constant for 2 1g2 + CO 2 1g2 CO1g2 + 2 O1g2 at 823 K Consider the equilibrium N 2 1g2 + O 2 1g2 + Br 2 1g2 2 NOBr1g2 Calculate the equilibrium constant K p for this reaction, given the following information (at 298 K): 2 NO1g2 + Br 2 1g2 2 NOBr1g2 K c = NO1g2 N 2 1g2 + O 2 1g2 K c = 2.1 * Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: 2 g 2 O1s2 4 g1l2 + O 2 1g2. (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium-constant expression in terms of molarities for the reaction, using (solv) to indicate solvation Consider the equilibrium Na 2 O1s2 + SO 2 1g2 Na 2 SO 3 1s2. (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibriumconstant expression in terms of molarities for the aqueous reaction. Calculating Equilibrium Constants (Section 15.5) Methanol 1C 3 O2 is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: CO1g g2 C 3 O1g2. An equilibrium mixture in a 2.00-L vessel is found to contain mol C 3 O, mol CO, and mol 2 at 500 K. Calculate K c at this temperature Gaseous hydrogen iodide is placed in a closed container at 425 C, where it partially decomposes to hydrogen and iodine: 2 I1g2 2 1g2 + I 2 1g2. At equilibrium it is found that 3I4 = 3.53 * 10-3 M, = 4.79 * 10-4 M, and 3I 2 4 = 4.79 * 10-4 M. What is the value of K c at this temperature? The equilibrium 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 is established at 500 K. An equilibrium mixture of the three gases has partial pressures of atm, atm, and 0.28 atm for NO, Cl 2, and NOCl, respectively. (a) Calculate K p for this reaction at K. (b) If the vessel has a volume of 5.00 L, calculate K c at this temperature Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl 3 1g2 + Cl 2 1g2 PCl 5 1g2. A 7.5-L gas vessel is charged with a mixture of PCl 3 1g2 and Cl 2 1g2, which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are P PCl3 = atm, P Cl2 = atm, and P PCl5 = 1.30 atm. (a) What is the value of K p at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate K c for this reaction at 450 K A mixture of 0.10 mol of NO, mol of 2, and 0.10 mol of 2 O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2 NO1g g2 N 2 1g O1g2 At equilibrium 3NO4 = M. (a) Calculate the equilibrium concentrations of 2, N 2, and 2 O. (b) Calculate K c A mixture of g of 2 and g of Br 2 is heated in a 2.00-L vessel at 700 K. These substances react according to 2 1g2 + Br 2 1g2 2 Br1g2 At equilibrium, the vessel is found to contain g of 2. (a) Calculate the equilibrium concentrations of 2, Br 2, and Br. (b) Calculate K c A mixture of mol of CO 2, mol of 2, and mol of 2 O is placed in a L vessel. The following equilibrium is established at 500 K: CO 2 1g g2 CO1g2 + 2 O1g2 (a) Calculate the initial partial pressures of CO 2, 2, and 2 O. (b) At equilibrium P 2 O = 3.51 atm. Calculate the equilibrium partial pressures of CO 2, 2, and CO. (c) Calculate K p for the reaction. (d) Calculate K c for the reaction A flask is charged with atm of N 2 O 4 1g2 and 1.00 atm NO 2 1g2 at 25 C, and the following equilibrium is achieved: N 2 O 4 1g2 2 NO 2 1g2 After equilibrium is reached, the partial pressure of NO 2 is atm. (a) What is the equilibrium partial pressure of N 2 O 4? (b) Calculate the value of K p for the reaction. (c) Calculate K c for the reaction Two different proteins X and Y are dissolved in aqueous solution at 37 C. The proteins bind in a 1:1 ratio to form XY. A solution that is initially 1.00 mm in each protein is allowed to reach equilibrium. At equilibrium, 0.20 mm of free X and 0.20 mm of free Y remain. What is K c for the reaction? A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate molecules bind to a protein involved in cancer. The drug

132 Exercises 665 molecules bind the protein in a 1:1 ratio to form a drug protein complex. The protein concentration in aqueous solution at 25 C is 1.50 * 10-6 M. Drug A is introduced into the protein solution at an initial concentration of 2.00 * 10-6 M. Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00 * 10-6 M. At equilibrium, the drug A-protein solution has an A-protein complex concentration of 1.00 * 10-6 M, and the drug B solution has a B-protein complex concentration of 1.40 * 10-6 M. Calculate the K c value for the A-protein binding reaction and for the B- protein binding reaction. Assuming that the drug that binds more strongly will be more effective, which drug is the better choice for further research? Applications of Equilibrium Constants (Section 15.6) (a) If Q c 6 K c, in which direction will a reaction proceed in order to reach equilibrium? (b) What condition must be satisfied so that Q c = K c? (a) If Q c 7 K c, how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of Q c at this point in the reaction? At 100 C, the equilibrium constant for the reaction COCl 2 1g2 CO1g2 + Cl 2 1g2 has the value K c = 2.19 * Are the following mixtures of COCl 2, CO, and Cl 2 at 100 C at equilibrium? If not, indicate the direction that the reaction must proceed to achieve equilibrium. (a) 3COCl 2 4 = 2.00 * 10-3 M, 3CO4 = 3.3 * 10-6 M, 3Cl 2 4 = 6.62 * 10-6 M (b) 3COCl 2 4 = 4.50 * 10-2 M, 3CO4 = 1.1 * 10-7 M, 3Cl 2 4 = 2.25 * 10-6 M (c) 3COCl 2 4 = M, 3CO4 = 3Cl 2 4 = 1.48 * 10-6 M As shown in Table 15.2, K p for the equilibrium N 2 1g g2 2 N 3 1g2 is 4.51 * 10-5 at 450 C. For each of the mixtures listed here, indicate whether the mixture is at equilibrium at 450 C. If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm N 3, 45 atm N 2, 55 atm 2 (b) 57 atm N 3, 143 atm N 2, no 2 (c) 13 atm N 3, 27 atm N 2, 82 atm At 100 C, K c = for the reaction SO 2 Cl 2 1g2 SO 2 1g2 + Cl 2 1g2 In an equilibrium mixture of the three gases, the concentrations of SO 2 Cl 2 and SO 2 are M and M, respectively. What is the partial pressure of Cl 2 in the equilibrium mixture? At 900 K, the following reaction has K p = 0.345: 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2 In an equilibrium mixture the partial pressures of SO 2 and O 2 are atm and atm, respectively. What is the equilibrium partial pressure of SO 3 in the mixture? At 1285 C, the equilibrium constant for the reaction Br 2 1g2 2 Br1g2 is K c = 1.04 * A L vessel containing an equilibrium mixture of the gases has g Br 2 1g2 in it. What is the mass of Br1g2 in the vessel? For the reaction 2 1g2 + I 2 1g2 2 I1g2, K c = 55.3 at 700 K. In a 2.00-L flask containing an equilibrium mixture of the three gases, there are g 2 and 4.36 g I 2. What is the mass of I in the flask? At 800 K, the equilibrium constant for I 2 1g2 2 I1g2 is K c = 3.1 * If an equilibrium mixture in a 10.0-L vessel contains 2.67 * 10-2 g of I(g), how many grams of I 2 are in the mixture? For 2 SO 2 1g2 + O 2 1g2 2 SO 3 1g2, K p = 3.0 * 10 4 at 700 K. In a 2.00-L vessel, the equilibrium mixture contains 1.17 g of SO 3 and g of O 2. ow many grams of SO 2 are in the vessel? At 2000 C, the equilibrium constant for the reaction 2 NO1g2 N 2 1g2 + O 2 1g2 is K c = 2.4 * If the initial concentration of NO is M, what are the equilibrium concentrations of NO, N 2, and O 2? For the equilibrium Br 2 1g2 + Cl 2 1g2 2 BrCl1g2 at 400 K, K c = 7.0. If 0.25 mol of Br 2 and 0.55 mol of Cl 2 are introduced into a 3.0-L container at 400 K, what will be the equilibrium concentrations of Br 2, Cl 2, and BrCl? At 373 K, K p = for the equilibrium 2 NOBr1g2 2 NO1g2 + Br 2 1g2 If the pressures of NOBr(g) and NO(g) are equal, what is the equilibrium pressure of Br 2 1g2? At 218 C, K c = 1.2 * 10-4 for the equilibrium N 4 S1s2 N 3 1g2 + 2 S1g2 Calculate the equilibrium concentrations of N 3 and 2 S if a sample of solid N 4 S is placed in a closed vessel at 218 C and decomposes until equilibrium is reached Consider the reaction CaSO 4 1s2 Ca 2+ 1aq2 + SO 4 2-1aq2 At 25 C, the equilibrium constant is K c = 2.4 * 10-5 for this reaction. (a) If excess CaSO 4 1s2 is mixed with water at 25 C to produce a saturated solution of CaSO 4, what are the equilibrium concentrations of Ca 2+ and SO 2-4? (b) If the resulting solution has a volume of 1.4 L, what is the minimum mass of CaSO 4 1s2 needed to achieve equilibrium? At 80 C, K c = 1.87 * 10-3 for the reaction P 3 BCl 3 1s2 P 3 1g2 + BCl 3 1g2 (a) Calculate the equilibrium concentrations of P 3 and BCl 3 if a solid sample of P 3 BCl 3 is placed in a closed vessel at 80 C and decomposes until equilibrium is reached. (b) If the flask has a volume of L, what is the minimum mass of P 3 BCl 3 1s2 that must be added to the flask to achieve equilibrium? For the reaction I 2 + Br 2 1g2 2 IBr1g2, K c = 280 at 150 C. Suppose that mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150 C. What are the equilibrium concentrations of IBr, I 2, and Br 2?

133 666 chapter 15 Chemical Equilibrium At 25 C, the reaction CaCrO 4 1s2 Ca 2+ 1aq2 + CrO 4 2-1aq2 has an equilibrium constant K c = 7.1 * What are the equilibrium concentrations of Ca and CrO 4 in a saturated solution of CaCrO 4? Methane, C 4, reacts with I 2 according to the reaction C 4 1g2 + l 2 1g2 C 3 l1g2 + I1g2. At 630 K, K p for this reaction is 2.26 * A reaction was set up at 630 K with initial partial pressures of methane of torr and of 7.96 torr for I 2. Calculate the pressures, in torr, of all reactants and products at equilibrium The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually 2 SO 4 ). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: C 3 COO1solv2 + C 3 C 2 O1solv2 C 3 COOC 2 C 3 1solv2 + 2 O1solv2 where (solv) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at 55 C is A pharmaceutical chemist makes up 15.0 L of a solution that is initially M in acetic acid and 3.85 M in ethanol. At equilibrium, how many grams of ethyl acetate are formed? Le Châtelier s Principle (Section 15.7) Consider the following equilibrium for which SO 2 1g2 + O 2 1g2 2 SO 3 1g2 ow will each of the following changes affect an equilibrium mixture of the three gases: (a) O 2 1g2 is added to the system; (b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture; (e) the total pressure of the system is increased by adding a noble gas; (f) SO 3 1g2 is removed from the system? Consider the reaction 4 N 3 1g2 + 5 O 2 1g2 4 NO1g O1g2, = kj Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (a) increase 3N 3 4; (b) increase 3 2 O4; (c) decrease 3O 2 4; (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature ow do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst? For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side? Consider the following equilibrium between oxides of nitrogen 3 NO1g2 NO 2 1g2 + N 2 O1g2 (a) Use data in Appendix C to calculate for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture? Methanol 1C 3 O2 can be made by the reaction of CO with 2 : CO1g g2 C 3 O1g2 (a) Use thermochemical data in Appendix C to calculate for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure? Ozone, O 3, decomposes to molecular oxygen in the stratosphere according to the reaction 2 O 3 1g2 3 O 2 1g2. Would an increase in pressure favor the formation of ozone or of oxygen? The water gas shift reaction CO1g2 + 2 O1g2 CO 2 1g g2 is used industrially to produce hydrogen. The reaction enthalpy is = -41 kj. (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of 2 1g2? Additional Exercises Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: CO1g2 + Cl 2 1g2 COCl1g2 + Cl1g2 At 25 C, the rate constants for the forward and reverse reactions are 1.4 * M -1 s -1 and 9.3 * M -1 s -1, respectively. (a) What is the value for the equilibrium constant at 25 C? (b) Are reactants or products more plentiful at equilibrium? If K c = 1 for the equilibrium 2 A1g2 B1g2, what is the relationship between [A] and [B] at equilibrium? A mixture of C 4 and 2 O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L flask and is found to contain 8.62 g of CO, 2.60 g of 2, 43.0 g of C 4, and 48.4 g of 2 O. Assuming that equilibrium has been reached, calculate K c and K p for the reaction C 4 1g2 + 2 O1g2 CO1g g When 2.00 mol of SO 2 Cl 2 is placed in a 2.00-L flask at 303 K, 56% of the SO 2 Cl 2 decomposes to SO 2 and Cl 2 : SO 2 Cl 2 1g2 SO 2 1g2 + Cl 2 1g2 (a) Calculate K c for this reaction at this temperature. (b) Calculate K p for this reaction at 303 K. (c) According to Le Châtelier s principle, would the percent of SO 2 Cl 2 that decomposes increase, decrease or stay the same if the mixture

134 additional Exercises 667 were transferred to a L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of SO 2 Cl 2 that decomposes when 2.00 mol of SO 2 Cl 2 is placed in a L vessel at 303 K A mixture of 2, S, and 2 S is held in a 1.0-L vessel at 90 C and reacts according to the equation: 2 1g2 + S1s2 2 S1g2 At equilibrium, the mixture contains 0.46 g of 2 S and 0.40 g 2. (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of K c for the reaction at this temperature? A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2 NOBr1g2 2 NO1g2 + Br 2 1g2 An equilibrium mixture in a 5.00-L vessel at 100 C contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br 2. (a) Calculate K c. (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr? Consider the hypothetical reaction A1g2 2 B1g2. A flask is charged with 0.75 atm of pure A, after which it is allowed to reach equilibrium at 0 C. At equilibrium, the partial pressure of A is 0.36 atm. (a) What is the total pressure in the flask at equilibrium? (b) What is the value of K p? (c) What could we do to maximize the yield of B? As shown in Table 15.2, the equilibrium constant for the reaction N 2 1g g2 2 N 3 1g2 is K p = 4.34 * 10-3 at 300 C. Pure N 3 is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are 1.05 g N 3 in the equilibrium mixture. (a) What are the masses of N 2 and 2 in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel? For the equilibrium 2 IBr1g2 I 2 1g2 + Br 2 1g2 K p = 8.5 * 10-3 at 150 C. If atm of IBr is placed in a 2.0-L container, what is the partial pressure of all substances after equilibrium is reached? For the equilibrium P 3 BCl 3 1s2 P 3 1g2 + BCl 3 1g2 K p = at 60 C. (a) Calculate K c. (b) After 3.00 g of solid P 3 BCl 3 is added to a closed L vessel at 60 C, the vessel is charged with g of BCl 3 1g2. What is the equilibrium concentration of P 3? [15.79] Solid N 4 S is introduced into an evacuated flask at 24 C. The following reaction takes place: N 4 S1s2 N 3 1g2 + 2 S1g2 At equilibrium, the total pressure (for N 3 and 2 S taken together) is atm. What is K p for this equilibrium at 24 C? [15.80] A g sample of SO 3 is placed in a 1.00-L container and heated to 1100 K. The SO 3 decomposes to SO 2 and O 2 : 2 SO 3 1g2 2 SO 2 1g2 + O 2 1g2 At equilibrium, the total pressure in the container is atm. Find the values of K p and K c for this reaction at 1100 K Nitric oxide (NO) reacts readily with chlorine gas as follows: 2 NO1g2 + Cl 2 1g2 2 NOCl1g2 At 700 K, the equilibrium constant K p for this reaction is Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) P NO = 0.15 atm, P Cl2 = 0.31 atm, P NOCl = 0.11 atm (b) P NO = 0.12 atm, P Cl2 = 0.10 atm, P NOCl = atm (c) P NO = 0.15 atm, P Cl2 = 0.20 atm, P NOCl = 5.10 * 10-3 atm At 900 C, K c = for the reaction CaCO 3 1s2 CaO1s2 + CO 2 1g2 A mixture of CaCO 3, CaO, and CO 2 is placed in a 10.0-L vessel at 900 C. For the following mixtures, will the amount of CaCO 3 increase, decrease, or remain the same as the system approaches equilibrium? (a) 15.0 g CaCO 3, 15.0 g CaO, and 4.25 g CO 2 (b) 2.50 g CaCO 3, 25.0 g CaO, and 5.66 g CO 2 (c) 30.5 g CaCO 3, 25.5 g CaO, and 6.48 g CO When 1.50 mol CO 2 and 1.50 mol 2 are placed in a 3.00-L container at 395 C, the following reaction occurs: CO 2 1g g2 CO1g2 + 2 O1g2. If K c = 0.802, what are the concentrations of each substance in the equilibrium mixture? The equilibrium constant K c for C1s2 + CO 2 1g2 2 CO1g2 is 1.9 at 1000 K and at 298 K. (a) If excess C is allowed to react with 25.0 g of CO 2 in a 3.00-L vessel at 1000 K, how many grams of CO are produced? (b) ow many grams of C are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic? NiO is to be reduced to nickel metal in an industrial process by use of the reaction NiO1s2 + CO1g2 Ni1s2 + CO 2 1g2 At 1600 K, the equilibrium constant for the reaction is K p = 6.0 * If a CO pressure of 150 torr is to be employed in the furnace and total pressure never exceeds 760 torr, will reduction occur? Le Châtelier noted that many industrial processes of his time could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and CO 2 according to the reaction Fe 2 O 3 1s2 + 3 CO1g2 2 Fe1s2 + 3 CO 2 1g2 Even in Le Châtelier s time, it was noted that a great deal of CO was wasted, expelled through the chimneys over the furnaces. Le Châtelier wrote, Because this incomplete reaction was thought to be due to an insufficiently prolonged contact between carbon monoxide and the iron ore [oxide], the dimensions of the furnaces have been increased. In England, they have been made as high as 30 m. But the proportion of carbon monoxide escaping has not diminished, thus demonstrating, by an experiment costing several hundred thousand francs, that the reduction of iron oxide by carbon monoxide is a limited reaction. Acquaintance with the laws of chemical equilibrium would have permitted the same conclusion to be reached more rapidly and far more economically. What does this anecdote tell us about the equilibrium constant for this reaction?

135 668 chapter 15 Chemical Equilibrium [15.87] At 700 K, the equilibrium constant for the reaction CCl 4 1g2 C1s2 + 2 Cl 2 1g2 is K p = A flask is charged with 2.00 atm of CCl 4, which then reaches equilibrium at 700 K. (a) What fraction of the CCl 4 is converted into C and Cl 2? (b) What are the partial pressures of CCl 4 and Cl 2 at equilibrium? [15.88] The reaction PCl 3 1g2 + Cl 2 1g2 PCl 5 1g2 has K p = at 300 C. A flask is charged with 0.50 atm PCl 3, 0.50 atm Cl 2, and 0.20 atm PCl 5 at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of Cl 2 in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of Cl 2 in the equilibrium mixture? [15.89] An equilibrium mixture of 2, I 2, and I at 458 C contains mol 2, mol I 2, and mol I in a 5.00-L vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of mol of I? [15.90] Consider the hypothetical reaction A1g2 + 2 B1g2 2 C1g2, for which K c = 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol>l of compound A present at equilibrium. (a) In terms of x, what are the equilibrium concentrations of compounds B and C? (b) What limits must be placed on the value of x so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of x) into the equilibriumconstant expression, derive an equation that can be solved for x. (d) The equation from part (c) is a cubic equation (one that has the form ax 3 + bx 2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. owever, you can estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution. (e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (int: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.) At 1200 K, the approximate temperature of automobile exhaust gases (Figure 15.15), K p for the reaction 2 CO 2 1g2 2 CO1g2 + O 2 1g2 is about 1 * Assuming that the exhaust gas (total pressure 1 atm) contains 0.2% CO, 12% CO 2, and 3% O 2 by volume, is the system at equilibrium with respect to the CO 2 reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the CO 2 reaction? Recall that at a fixed pressure and temperature, volume % = mol % Suppose that you worked at the U.S. Patent Office and a patent application came across your desk claiming that a newly developed catalyst was much superior to the aber catalyst for ammonia synthesis because the catalyst led to much greater equilibrium conversion of N 2 and 2 into N 3 than the aber catalyst under the same conditions. What would be your response? Integrative Exercises Consider the reaction IO - 4 1aq O1l2 4 IO - 6 1aq2; K c = 3.5 * If you start with 25.0 ml of a M solution of NaIO 4, and then dilute it with water to ml, what - is the concentration of 4 IO 6 at equilibrium? [15.94] Silver chloride, AgCl(s), is an insoluble strong electrolyte. (a) Write the equation for the dissolution of AgCl(s) in 2 O1l2. (b) Write the expression for K c for the reaction in part (a). (c) Based on the thermochemical data in Appendix C and Le Châtelier s principle, predict whether the solubility of AgCl in 2 O increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of AgCl in water is 1.6 * at 25 C. In addition, Ag + 1aq2 can react with Cl - 1aq2 according to the reaction Ag + 1aq2 + 2 Cl - 1aq2 AgCl 2-1aq2 where K c = 1.8 * 10 5 at 25 C. Although AgCl is not soluble in water, the complex AgCl 2 - is soluble. At 25 C, is the solubility of AgCl in a M NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble AgCl 2 - ions? Or is the AgCl solubility in M NaCl less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations. (int: Any form in which silver is in solution counts as solubility. ) [15.95] Consider the equilibrium A B in which both the forward and reverse reactions are elementary (single-step) reactions. Assume that the only effect of a catalyst on the reaction is to lower the activation energies of the forward and reverse reactions, as shown in Figure Using the Arrhenius equation (Section 14.5), prove that the equilibrium constant is the same for the catalyzed reaction as for the uncatalyzed one. [15.96] The phase diagram for SO 2 is shown here. (a) What does this diagram tell you about the enthalpy change in the reaction SO 2 1l2 SO 2 1g2? (b) Calculate the equilibrium constant for this reaction at 100 C and at 0 C. (c) Why is it not possible to calculate an equilibrium constant between the gas and liquid phases in the supercritical region? (d) At which of the three points marked in red does SO 2 1g2 most closely approach ideal-gas behavior? (e) At which of the three red points does SO 2 1g2 behave least ideally? Pressure (atm) Liquid Critical point Gas Supercritical region Temperature ( C)

136 Design an Experiment 669 [15.97] Write the equilibrium-constant expression for the equilibrium C1s2 + CO 2 1g2 2 CO1g2 The table that follows shows the relative mole percentages of CO 2 1g2 and CO(g) at a total pressure of 1 atm for several temperatures. Calculate the value of K p at each temperature. Is the reaction exothermic or endothermic? Temperature 1 C2 Co 2 1mol %2 Co 1mol % In Section 11.5, we defined the vapor pressure of a liquid in terms of an equilibrium. (a) Write the equation representing the equilibrium between liquid water and water vapor and the corresponding expression for K p. (b) By using data in Appendix B, give the value of K p for this reaction at 30 C. (c) What is the value of K p for any liquid in equilibrium with its vapor at the normal boiling point of the liquid? Water molecules in the atmosphere can form hydrogenbonded dimers, 1 2 O2 2. The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The K p for water dimer formation in the gas phase is at 300 K and at 350 K. Is water dimer formation endothermic or exothermic? The protein hemoglobin (b) transports O 2 in mammalian blood. Each b can bind 4 O 2 molecules. The equilibrium constant for the O 2 binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the P50 value, defined as the partial pressure of oxygen at which 50% of the protein is saturated. Fetal hemoglobin has a P50 value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger K c is for the aqueous reaction 4 O 2 1g2 + b1aq2 3b1O aq24. Design an Experiment The reaction between hydrogen and iodine to form hydrogen iodide was used to illustrate Beer s law in Chapter 14 (Figure 14.5). The reaction can be monitored using visible-light spectroscopy because I 2 has a violet color while 2 and I are colorless. At 300 K, the equilibrium constant for the reaction 2 1g2 + I 2 1g2 2 I1g2 is K c = 794. To answer the following questions assume you have access to hydrogen, iodine, hydrogen iodide, a transparent reaction vessel, a visible-light spectrometer, and a means for changing the temperature. (a) Which gas or gases concentration could you readily monitor with the spectrometer? (b) To use Beer s law (Equation 14.5) you need to determine the extinction coefficient, e, for the substance in question. ow would you determine e? (c) Describe an experiment for determining the equilibrium constant at 600 K. (d) Use the bond enthalpies in Table 8.4 to estimate the enthalpy of this reaction. (e) Based on your answer to part (d), would you expect K c at 600 K to be larger or smaller than at 300 K?

137 16 Acid Base Equilibria The acids and bases that you have used so far in the laboratory are probably solutions of relatively simple inorganic substances, such as hydrochloric acid, sulfuric acid, sodium hydroxide, and the like. But acids and bases are important even when we are not in the lab. They are ubiquitous, including in the foods we eat. The characteristic flavor of the grapes shown in the opening photograph is largely due to tartaric acid 1 2 C 4 4 O 6 2 and malic acid 1 2 C 4 4 O 5 2 (Figure 16.1), two closely related (they differ by only one O atom) organic acids that are found in biological systems. Fermentation of the sugars in the grapes ultimately forms vinegar, the tangy, sour flavor of which is due to acetic acid 1C 3 COO2, a substance we discussed in Section 4.3. The sour taste of oranges, lemons, and other citrus fruits is due to citric acid 1 3 C 6 5 O 7 2, and, to a lesser extent, ascorbic acid 1 2 C 6 6 O 6 2, better known as Vitamin C. Acids and bases are among the most important substances in chemistry, and they affect our daily lives in innumerable ways. Not only are they present in our foods, but acids and bases are also crucial components of living systems, such as the amino acids that are used to synthesize proteins and the nucleic acids that code genetic information. Both citric and malic acids are among several acids involved in the Krebs cycle (also called the citric acid cycle) that is used to generate energy in aerobic organisms. The application of acid base chemistry has also had critical roles in shaping modern society, including such human-driven activities as industrial manufacturing, the creation of advanced pharmaceuticals, and many aspects of the environment. The impact of acids and bases depends not only on the type of acid or base, but also on how much is present. The time required for a metal object immersed in water to corrode, the ability of an aquatic environment to support fish and plant life, the fate of pollutants washed out of the air by rain, and even the rates of reactions that maintain Clusters of grapes and balsamic vinegar. Grapes contain several acids that contribute to their characteristic flavor. The distinctive flavor of all vinegars is due to acetic acid. Balsamic vinegar is obtained by fermenting grapes. What s Ahead 16.1 Acids and Bases: A Brief Review We begin by reviewing the Arrhenius definition of acids and bases Brønsted Lowry Acids and Bases We learn that a Brønsted Lowry acid is a proton donor and a Brønsted Lowry base is a proton acceptor. Two species that differ by the presence or absence of a proton are known as a conjugate acid base pair The Autoionization of Water We see that the autoionization of water produces small quantities of 3 O + and O ions. The equilibrium constant for autoionization, K w = [ 3 O + ][O ] defines the relationship between 3 O + and O concentrations in aqueous solutions The p Scale We use the p scale to describe the acidity or basicity of an aqueous solution. Neutral solutions have a p = 7, acidic solutions have p below 7, and basic solutions have p above Strong Acids and Bases We categorize acids and bases as being either strong or weak electrolytes. Strong acids and bases are strong electrolytes, ionizing or dissociating completely in aqueous solution. Weak acids and bases are weak electrolytes and ionize only partially.