Chapter 9. Chemical Equilibrium

9.1 The Nature of Chemical Equilibrium -Approach to Equilibrium [Co(H 2 O) 6 ] 2+ + 4 Cl- [CoCl 4 ] 2- + 6 H 2 O

Characteristics of the Equilibrium State example) H 2 O(l) H 2 O(g) 1. They display no macroscopic evidence of change. 2. They are reached through spontaneous processes. 3. They show a dynamic balance of forward and reverse processes. 4. They are the same regardless of direction of approach. Steady states: macroscopic concentrations of species are not changing with time, even though the system in not at equilibrium.

9.2 The Empirical Law of Mass Action aa + bb K C and K P : empirical equilibrium constant Law of mass action: cc + dd 1) The numerical value of K C or K P is an inherent property of the chemical reaction itself and does not depend on the specific initial concentrations of reactants and products. 2) The magnitude gives direct information about the nature of the equilibrium state or position of the reaction.

Law of Mass Action for Gas-Phase Reactions A deeper study shows that instead of inserting just the partial pressure for each reactant or product, we must insert the value of the partial pressure relative to a specified reference pressure P ref.

Example 9.1 Write equilibrium expressions for the following gasphase chemical equilibria. (a) 2 NOCl(g) 2 NO(g) + Cl 2 (g) (b) CO(g) + 0.5 O 2 (g) CO 2 (g)

Law of Mass Action for Reactions in Solution

Example 9.2 Household laundry bleach is a solution of sodium hypochlorite (NaOCl) prepared by adding gaseous Cl 2 to a solution of sodium hydroxide: Cl 2 (aq) + 2 OH - (aq) ClO - (aq) + Cl - (aq) + H 2 O(l) The active bleaching agent is the hypochlorite ion, which can decompose to chloride and chlorate ions in a side reaction that competes with bleaching: 3 ClO - (aq) 2 Cl - (aq) + ClO 3- (aq) Write the equilibrium expression for the decomposition reaction.

Law of Mass Action for Reactions Involving Pure Substances and Multiple Phases 1. H 2 O(l) H 2 O(g) PH 2 O = K Experiments show that as long as some liquid qater is in the container, the pressure of water vapor at 25oC is 0.03126 atm. 2. I 2 (s) I 2 (aq) [I 2 ] = K Experiments show that the position of the equilibrium (given by the concentration of I 2 dissolved at a given temperature) is independent of the amount of solid present, as long as there is some. 3. CaCO 3 (s) CaO(s) + CaCO 3 (aq) PCO 2 = K Experiments show that the pressure of CO 2 is constant, independent of the amount of solid present, as long as there is some.

General Procedure for writing the mass action law for these more complex reactions 1. Gases enter the equilibrium expression as partial pressures, measured in atmospheres. 2. Dissolved species enter as concentrations, in moles per liter. 3. Pure solids and pure liquids do not appear in equilibrium expressions; neither does a solvent taking part in a chemical reaction, provided the solution is dilute. 4. Partial pressures and concentrations of products appear in the numerator, and those of reactants in the denominator; each is raised to a power equal to its coefficient in the balanced chemical equation for the reaction. This procedure gives the dimensionless thermodynamic equilibrium constant K because each species has entered the equilibrium expression relative to its standard reference state.

Example 9.3 Hypochlorous acid (HOCl) is produced by bubbling chlorine through an agitated suspension of mercury(ii) oxide in water. The chemical equation for this process is 2 Cl 2 (g) + 2 HgO(s) + H 2 O(l) HgO HgCl 2 (s) + 2 HOCl(aq) Write the equilibrium expression for this reaction.

The preceding discussion has shown the procedures for setting up the mass action law for broad classes of chemical reactions. However, numerous fundamental questions about chemical equilibrium are not answered by these procedures. Why should the law of mass action exist in the first place, and why should it take the particular mathematical form shown here? Why should the equilibrium constant take a unique value for each individual chemical reaction? What factors determine that value? Why does the value of the equilibrium constant change slightly when studied over broad ranges of concentration? Why should the equilibrium constant depend on temperature, and can a quantitative explanation be provided for this dependence?

9.3 Thermodynamic Description of the Equilibrium State In this section we use thermodynamics to demonstrate why the mass action law takes its special mathematical form and why the thermodynamic equilibrium constant K is a dimensionless quantity. Thermodynamics views a chemical reaction as a process in which atoms flow from reactants to products. If the reaction is spontaneous and is carried out at constant T and P, thermodynamics requires that ΔG < 0 for the process. When a chemical reaction has reached equilibrium, ΔG = 0.

Reactions among Ideal Gases Dependence of Gibbs Free Energy of a Gas on Pressure

The Equilibrium Expression for Reactions in the Gas Phase Prove that for the general reaction aa + bb cc + dd

Example 9.4 The ΔG o of the chemical reaction 3 NO(g) N 2 O(g) + NO 2 (g) was calculated in Example 8.10. Now calculate the equilibrium constant of this reaction at 25 o C.

Reactions in Ideal Solutions

Example 9.5 Calculate ΔG o and equilibrium constant at 25 o C for the chemical reaction 3 ClO - (aq) 2 Cl - (aq) + ClO 3- (aq) whose equilibrium expression we developed in Example 9.2.

Reactions Involving Pure Solids and Liquids and Multiple Phases: The Concept of Activity The mass action law for homogeneous reactions in ideal gases and ideal solutions was written in Section 9.2 by straightforward inspection of the balanced equation for the reaction under study. If one or more of the reactants was a solid or liquid in its pure state, the procedure was less obvious, because concentration has no meaning for a pure species. This apparent difficulty is resolved by the concept of activity, which is a convenient means for comparing the properties of a substance in a general thermodynamic state with its properties in a specially selected reference state.

Example 9.6 The compound urea, important in biochemistry, can be prepared in aqueous solution by the following reaction: CO 2 (g) + 2 NH 3 (g) CO(NH 2 ) 2 (aq) + H 2 O(l) (a) Write the mass action law for this reaction. (b) Calculate ΔG o for this reaction at 25 o C. (c) Calculate K for this reaction at 25 o C.

9.4 The Law of Mass Action for Related and Simultaneous Equilibria Relationships among Equilibrium Expressions

Relationships among Equilibrium Expressions (continued)

Example 9.7 The concentration of the oxides of nitrogen are monitored in airpollution reports. At 25 C, the equilibrium constant for the reaction NO(g) + ½ O 2 (g) NO 2 (g) is K 1 = 1.3 x 10 6 and that for ½ N 2 (g) + ½ O 2 (g) NO(g) is K 2 = 6.5 x 10-16 Find the equilibrium constant K 3 for the reaction N 2 (g) + 2 O 2 (g) NO 2 (g)

Consecutive Equilibria; Hemoglobin and Oxygen Transport

9.5 Equilibrium Calculations for Gas-Phase and Heterogeneous Reactions Problem Solving Technique for two classes of problems. - Evaluating Equilibrium Constants from Reaction Data - Calculating Equilibrium Compositions When K is Known Evaluating Equilibrium Constants from Reaction Data Example 9.8: Phosgene, COCl 2, forms from CO and Cl 2 according to the equilibrium CO(g) + Cl 2 (g) COCl 2 (g) At 600 C, a gas mixture of CO and Cl 2 is prepared that has initial partial pressures (before reaction) of 0.60 atm for CO and 1.10 atm for Cl 2. After the reaction mixture has reached equilibrium, the partial pressure of COCl 2 (g) at this temperature is measured to be 0.10 atm. Calculate the equilibrium constant for this reaction. The reaction is carried out ina vessel of fixed volume.

For a more complex reaction, 2 C 2 H 6 (g) + 7 O 2 (g) 4 CO 2 (g) + 6 H 2 O(g) Example 9.9 Graphite (a form of solid carbon) is added to a vessel that contains CO 2 (g) at a pressure of 0.824 atm at a certain high temperature. The pressure rises due to a reaction that produces CO(g). The total pressure reaches an equilibrium value of 1.366 atm. (a) Write a balanced equation for the process. (b) Calculate the equilibrium constant.

Calculating Equilibrium Composition When K is Known Example 9.10 Suppose H 2 (g) and I 2 (g) are sealed in a flask at T = 400 K with partial pressures PH 2 = 1.320 atm and PI 2 = 1.140. At this temperature H 2 and I 2 do not react rapidly to form HI(g), although after a long enough time they would produce HI(g) at its equilibrium partial pressure. Suppose, instead, that the gases are heated in the sealed flask to 600 K, a temperature at which they quickly reach equilibrium: H 2 (g) + I 2 (g) 2 HI(g) The equilibrium constant for the reaction is 92.6 at 600 K: (a) What are the equilibrium values of PH 2, PI 2, and PHI at 600 K? (b) What percentage of the I 2 originally present has reacted when equilibrium is reached?

Example 9.11 Hydrogen is made from natural gas (methane) for immediate consumption in industrial processes, such as ammonia production. The first step is called the steam reforming of methane CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) The equilibrium constant for this reaction is 1.8 x 10-7 at 600 K. Gaseous CH 4, H 2 O, and CO are introduced into an evacuated container at 600 K, and their initial partial pressures (before reaction) are 1.40 atm, 2.30 atm, and 1.60 atm, respectively. Determine the partial pressure of H 2 (g) that will result at equilibrium.

Suppose the data for a gas-phase equilibrium are given in terms of concentrations rather than partial pressures. In such cases all concentrations can be converted to partial pressures before the calculations are carried out, or the equilibrium expression can be rewritten in terms of concentration variables.

Example 9.12 At elevated temperatures, PCl 5 dissociates extensively according to PCl 5 (g) PCl 3 (g) + Cl 2 (g) At 300 C, the equilibrium constant for this reaction is K = 11.5. The concentrations of PCl 3 and Cl 2 at equilibrium in a container at 300 C are both 0.01100 mol L -1. Calculate [PCl 5 ].

9.6 The Direction of Change in Chemical Reactions: Empirical Description The Reaction Quotient

Example 9.13 The reaction between nitrogen and hydrogen to produce ammonia N 2 (g) + 3 H 2 (g) 2 NH 3 (g) is essential in making nitrogen-containing fertilizers. This reaction has an equilibrium constant equal to 1.9 x 10-4 at 400 C. Suppose that 0.10 mol of N 2, 0.040 mol of H 2, and 0.020 mol of NH 3 are sealed in a 1.00 L vessel at 400 C. In which direction will the reaction process?

Example 9.14 Solid ammonium chloride is in equilibrium with ammonia and HCl gases: NH 4 Cl(s) NH 3 (g) + HCl(g) The equilibrium constant at 275 C is 1.04 x 10-2. We replace 0.980 g of solid NH 4 Cl into a closed vessel with volume 1.00 L and heat to 275 C. (a) In what direction does the reaction proceed? (b) What is the partial pressure of each gas at equilibrium? (c) What is the mass of solid NH 4 Cl at equilibrium?

External Effects on K: Principle of Le Chatelier A principle stated by Henri Le Chatelier in 1884 A system in equilibrium that is subjected to a stress will react in a way that tends to counteract the stress. Le Chatelier s principle provides a way to predict qualitatively the direction of change of a system under an external perturbation. It relies heavily on Q as a predictive tool.

Effect of Changing the Concentration of a Reactant or Product Example 9.15 An equilibrium gas mixture of H 2 (g), I 2 (g), and HI(g) at 600 K has partial pressures of 0.4756 atm, 0.2056 atm, and 3.009 atm, respectively. This is essentially the final equilibrium state of Example 9.10. Enough H 2 is added to increase its partial pressure to 2.000 atm at 600 K before any reaction take place. This mixture then once again reaches equilibrium at 600 K. What are the final partial pressures of the three gases?

Effect of Changing the Volume

Effect of Changing the Temperature

Maximizing the yield of a reaction

9.7 The Direction of Change in Chemical Reactions: Thermodynamic Explanation Here, we identify thermodynamic factors that determine the magnitude of K. We also provide a thermodynamic criterion for predicting the direction in which a reaction proceeds from a given initial condition.

The Magnitude of the Equilibrium Constant

Free Energy Changes and the Reaction Quotient

Temperature Dependence of Equilibrium Constants

Example 9.16 Calculate K for the equilibrium of Example 9.4 at T = 400 K, assuming ΔH o to be approximately independent of temperature over the range from 298 to 400 K. Exmple 9.4 The ΔG o of the chemical reaction 3 NO(g) N 2 O(g) + NO 2 (g) was calculated in Example 8.10. Now calculate the equilibrium constant of this reaction at 25 o C.

Temperature Dependence of Vapor Pressure

Example 9.17 The ΔH vap for water is 40.66 kj mol -1 at the normal boiling point, Tb = 373 K. Assuming ΔH vap and ΔS vap are approximately independent of temperature from 50 C to 100 C, estimate the vapor pressure of water at 50 C (323 K).

9.8 Distribution of a Single Species between Immiscible Phases: Extraction and Separation Processes

Extraction Processes Example 9.18 An aqueous solution has an iodine concentration of 2.00 x 10-3 M. Calculate the percentage of iodine remaining in the aqueous phase after extraction of 0.100 L of this aqueous solution with 0.050 L of CCl 4 at 25 C.

Chromatography Chromatographic Separations

Column Chromatography Gas-Liquid Chromatography