CHEMICAL EQUILIBRIUM. Chapter 15

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1 Chapter 15 P a g e 1 CHEMICAL EQUILIBRIUM Examples of Dynamic Equilibrium Vapor above a liquid is in equilibrium with the liquid phase. rate of evaporation = rate of condensation Saturated solutions rate of ions leaving the solid surface = rate of ions returning to the solid from the liquid. Chemical Equilibrium When a reaction starts, the reactants are consumed and products are made. The reactant concentrations decrease, and the product concentrations increase. As reactant concentration decreases, the forward reaction rate decreases. Eventually, the products can react to re-form some of the reactants, assuming the products are not allowed to escape. As product concentration increases, the reverse reaction rate increases. Processes that proceed in both the forward and reverse directions are said to be reversible. reactants products As the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate. Dynamic equilibrium is the condition wherein the rates of the forward and reverse reactions are equal. Once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant because the chemicals are being consumed and made at the same rate. Consider the reaction N2O4 (g) 2 NO2(g) N2O4 gas is colorless while NO2 is brown. Suppose we obtain a sealed frozen sample of N2O4 and warm it above its boiling point (21.1 C), the solid vaporizes and the gas progressively turns darker. The N2O4 is dissociating to NO2. At some point, the color stops getting darker. This means that equilibrium has been achieved and that the rates of the forward and reverse processes are equal.

2 P a g e 2 Using kinetics, Forward reaction: N2O4(g) 2 NO2(g) Rate forward = kf[n2o4] Reverse reaction: 2 NO2 (g) N2O4(g) Rate reverse = kr[no2] 2 At equilibrium, rate forward = rate reverse kf[n2o4] = kr[no2] 2 rearranging, [NO 2] 2 [N 2 O 4 ] = k f k r = constant Dynamic Equilibrium The Equilibrium Constant It is important to note that equilibrium can be achieved from either direction of the reaction. (a) Equilibrium for the reaction N2 + 3 H2 2 NH3 is approached beginning with H2 and N2 present in the ratio 3:1 and no NH3 present. (b) Equilibrium for the same reaction is approached beginning with only NH3 in the reaction vessel.

3 P a g e 3 For the general equilibrium equation aa + bb below: cc + dd, the law of mass action gives the relationship K is known as the equilibrium constant. The expression above is the equilibrium-constant expression. The equilibrium constant is a unitless quantity. The reason is beyond the scope of our current discussion. For the reaction The value of K does not depend on the initial amounts of reactants and products. It only depends of the temperature and the particular reaction. Example 1 Express the equilibrium constant for the chemical equation: a. CH3OH(g) CO(g) + 2 H2(g) b. C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(g) c. I2(g) + Cl2(g) 2 ICl(g) d. 2 H2S(g) 2 H2(g) + S2(g) e. Ag + (aq) + 2 NH3 (aq) Ag(NH3)2 + (aq) Evaluating Kc Kc is the equilibrium constant in terms of concentrations of the reactants and products. The square brackets represent concentration. Consider the following data

4 Notice that equilibrium constant did not change as the initial concentrations were varied. In addition, equilibrium was achieved from either side of the reaction (experiments 3 and 4). P a g e 4 Evaluating Kp When the reactants and products, we can express the equilibrium constant in terms of the partial pressures instead of molar concentrations. In this case, we use Kp and parentheses. Don t forget!!! K p = (P C) c (P D ) d (P A ) a (P B ) b Example 2 Repeat examples 1 (a-d) in terms of partial pressures. Relationship between Kc and Kp If we had the molar concentrations and partial pressures of the reactants and products in examples 1 and 2, will be magnitude of Kc equal Kp? It turns out that the two are not necessarily equal. For the reaction, aa + bb cc + dd Let s begin our derivation with the ideal gas law.

5 P a g e 5 Let n = (c + d) (a + b) In calculating Kp, the partial pressures are always in atm. The values of Kp and Kc are not necessarily the same because of the difference in units. Δn is the difference between the number of moles of reactants and moles of products. Kp = Kc when Δn = 0 Example 3 In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300 C. Calculate Kp for this reaction at this temperature. Example 4 Nitrogen monoxide, a pollutant in automobile exhaust, is oxidized to nitrogen dioxide in the atmosphere according to the equation: 2 NO(g) + O2(g) 2 NO2(g) Kp = at 25 C Find Kc for this reaction.

6 P a g e 6 The Magnitude of Equilibrium Constant When the value of Keq >> 1, the reaction reaches equilibrium where there will be many more product molecules present than reactant molecules. The position of equilibrium favors products; lies to the right. When the value of Keq << 1, the reaction reaches equilibrium where there will be many more reactant molecules present than product molecules. The position of equilibrium favors reactants; lies to the left. When the value of Keq is 1, the reaction doesn t favor either side. Example 5 The following diagrams represent three different systems at equilibrium, all in the same size containers. (a) Without doing any calculations, rank the three systems in order of increasing equilibrium constant, Kc. (b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kc for each system. Relationships between the Equilibrium Constant and the Chemical Equation 1. When the reaction is written backward, the equilibrium constant is inverted. For reaction, A + 2B 3C If the equation is reversed, 3C A + 2B K forward = [C]3 [A][B] 2 K reverse = [A][B]2 [C] 3 = 1 K forward 2. When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor. For reaction, A + 2B 3C K = [C]3 [A][B] 2 If we multiply the equation by n, we get: n A + 2n B 3n C K = [C]3n [A] n [B] 2n = ( [C]3 n [A][B] 2) = K n

7 3. When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations. Consider the equations below: A 2 B K 1 = [B]2 [A] 2 B 3 C K 2 = [C]3 [A] 2 The sum of the two equations is as follows: A 2 B 2 B 3 C A 3 C K overall = [C]3 [A] P a g e 7 Notice that K overall = K 1 K 2 since [B]2 [C]3 = [C]3 [A] [B] 2 [A] Example 6 The reaction A(g) 2 B(g) has an equilibrium constant K = What is the equilibrium constant for the reaction B(g) ½ A(g)? Example 7 Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 C: N2(g) + 3 H2(g) NH3(g) K = Calculate the equilibrium constant for the following reaction at 25 C: NH3(g) 1 N2(g) + 3 H2(g) K =? 2 2

8 Example 8 Predict the equilibrium constant for the first reaction shown here given the equilibrium constants for the second and third reactions: CO2(g) + 3 H2(g) CH3OH(g) + H2O(g) K1 =? CO(g) + H2O(g) CO2(g) + H2(g) K2 = CO(g) + 2 H2(g) CH3OH(g) K3 = P a g e 8 Example 9 Given the following information, determine the value of Kc for the reaction Heterogeneous Equilibria In some equilibria, the reactants and products are in different phases, involving pure solids and liquids. Consider a saturated solution of silver chloride (AgCl): AgCl (s) Ag + (aq) + Cl - (aq) The equilibrium constant Kc = [Ag + ] [Cl - ]. When a pure solid or liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium expression. The reason is that pure liquids and solids have constant concentrations and do not change during equilibrium.

9 P a g e 9 CaCO3 (s) CaO (s) + CO2 (g) The equilibrium involving CaCO3, CaO, and CO2 is a heterogeneous equilibrium. The equilibrium pressure of CO2 is the same in the two bell jars as long as the two systems are at the same temperature, even though the relative amounts of pure CaCO3 and CaO differ greatly. The equilibrium-constant expression for the reaction is Kp = PCO2. Example 10 Write the equilibrium-constant expression for Kc for each of the following reactions: (c) 4 HCl(g) + O2(g) 2 H2O(l) + 2 Cl2(g) Example 11 Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of attaining the equilibrium CaCO3 (s) CaO (s) + CO2 (g): (a) pure CaCO3, (b) CaO and a CO2 pressure greater than the value of K p, (c) some CaCO3 and a CO2 pressure greater than the value of K p, (d) CaCO3 and CaO? Justify your answer.

10 P a g e 10 Example 12 When added to Fe3O4 (s) in a closed container, which one of the following substances H2(g), H2O(g), O2(g) will allow equilibrium to be established in the reaction: 3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)? Calculating Equilibrium Constants Case 1: all equilibrium concentrations are known. Example 13 A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and atm NH3. From these data, calculate the equilibrium constant Kp for the reaction: N2(g) + 3 H2(g) 2NH3(g) Example 14 An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 C: [HC2H3O2] = M; [H + ] = M; and [C2H3O2 ] = M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 C. The reaction is HC2H3O2 (aq) H + (aq) + C2H3O2 - (aq) Example 15 Consider the following reaction: 2 COF2(g) CO2(g) + CF4(g) Kc = 2.00 at 1000 C In an equilibrium mixture, the concentration of COF2 is M and the concentration of CF4 is M. What is the equilibrium concentration of CO2?

11 P a g e 11 Case 2: finding K from initial and equilibrium concentrations. Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration. Set up an ICE table based on the information provided. Use the change in the concentration of the substance you know to determine the change in the other chemicals in the reaction. Consider the equilibrium reaction aa + bb cc + dd K = [C]c [D] d [A] a [B] b ICE Initial, Change, and Equilibrium. ICE Table [A] [B] [C] [D] Initial Change -ax -bx +cx +dx Equilibrium Example 16 Consider the following reaction: CO(g) + 2 H2(g) CH3OH(g) A reaction mixture at 780 C initially contains [CO] = M and [H2] = 1.00 M. At equilibrium, the CO concentration is found to be 0.15 M. What is the value of the equilibrium constant? Example 17 A closed system initially containing M H2 and M I2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is M. Calculate Kc at 448 C for the reaction taking place, which is H2(g) + I2(g) 2 HI(g)

12 P a g e 12 Example 18 Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3 (g) 2SO2 (g) + O2 (g) Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of atm. At equilibrium, the SO3 partial pressure is atm. Calculate the value of Kp at 1000 K. Case 3: finding equilibrium concentrations from initial concentrations. Example 19 A L flask is filled with mol of H2 and mol of I2 at 448 C. The value of the equilibrium constant Kc for the reaction H2(g) + I2(g) 2 HI(g) at 448 C is What are the equilibrium concentrations of H2, I2, and HI in moles per liter?

13 Example 20 P a g e 13 For the equilibrium PCl5(g) PCl3(g) + Cl2(g), the equilibrium constant K p has the value at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature? The Reaction Quotient If a reaction mixture containing both reactants and products is not at equilibrium, how can we determine from which direction the reaction will reach equilibrium? The answer is to compare the current concentration ratios to the equilibrium constant. The non-equilibrium concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. For the gas phase reaction a A + b B c C + d D the reaction quotient is as follows: To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. The relative magnitudes of the reaction quotient Q and the equilibrium constant K indicate how the reaction mixture changes as it moves toward equilibrium. If Q is smaller than K, the reaction proceeds from left to right until Q = K. When Q = K, the reaction is at equilibrium and has no tendency to change. If Q is larger than K, the reaction proceeds from right to left until Q = K.

14 P a g e 14 If a reaction mixture contains just reactants, then Q = 0, and the reaction will proceed in the forward direction. If a reaction mixture contains just products, then Q =, and the reaction will proceed in the reverse direction. Example 21 Consider the reaction and its equilibrium constant at 25.0 C: I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9 A reaction mixture contains PI2 = atm, PCl2 = atm, and PICl = atm. Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed? Example 22 Consider the reaction and its equilibrium constant: N2O4(g) 2 NO2(g) Kc = (at some temperature) A reaction mixture contains [NO2] = M and [N2O4] = M. Calculate Qc and determine the direction in which the reaction will proceed.

15 P a g e 15 Example 23 Consider the reaction: N2(g) + O2(g) 2 NO(g) Kc = 0.10 (at 2000 C) A reaction mixture at 2000 C initially contains [N2] = M and [O2] = M. Find the equilibrium concentrations of the reactants and product at this temperature. Example 24 The reaction in Example 23 is carried out at a different temperature at which Kc = This time, however, the reaction mixture starts with only the product, [NO] = M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium. Example 25 Consider the reaction: I2(g) + Cl2(g) 2 ICl(g) Kp = 81.9 (at 25 C) A reaction mixture at 25 C initially contains PI2 = atm, PCl2 = atm, and PICl = atm. Find the equilibrium partial pressures of I2, Cl2, and ICl at this temperature.

16 P a g e 16 Example 26 Consider the reaction for the decomposition of hydrogen disulfide: 2 H2S(g) 2 H2(g) + S2(g) Kc = at 800 C A L reaction vessel initially contains mol of H2S at 800 C. Find the equilibrium concentrations of H2 and S2. Le Châtelier's Principle French industrial chemist, Henri-Louis Le Châtelier. The principle states that if a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position to counteract/ minimize/ annul the effect of the disturbance. We will make qualitative predictions for the following: a. Adding or removing a reactant or product b. Changing the pressure by changing the volume c. Changing the temperature. Change in Reactant or Product Concentration If a chemical system is at equilibrium and we increase the concentration of a substance (either reactant or product), the system reacts to consume some of the substance. Conversely, if we decrease the concentration of a substance, the system reacts to produce some of the substance. o Remember, adding more of a solid or liquid does not change its concentration; therefore, it has no effect on the equilibrium.

17 P a g e 17

18 P a g e 18 Effects of Volume and Pressure Changes I. Decreasing the volume of the container increases the concentration of all the gases in the container. It increases their partial pressures. If their partial pressures increase, then the total pressure in the container will increase. According to Le Châtelier s principle, the equilibrium should shift to remove that pressure. The way the system reduces the pressure is to reduce the number of gas molecules in the container. When the volume decreases, the equilibrium shifts to the side with fewer gas molecules. II. Adding a gaseous reactant increases its partial pressure, causing the equilibrium to shift to the right. Increasing its partial pressure increases its concentration. III. Adding an inert gas (noble gases) to the mixture has no effect on the position of equilibrium. It does not affect the partial pressures of the gases in the reaction. IV. If the number of moles of gas are equal on both sides of the reaction, then changing the pressure will not influence the position of the equilibrium. Effect of Temperature Changes Recall that Exothermic reactions release energy, and endothermic reactions absorb energy o Exothermic reaction For an exothermic reaction, heat is a product. Increasing the temperature is like adding heat. Adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants.

19 P a g e 19 o Adding heat to an exothermic reaction will decrease the value of K. Endothermic reaction For an endothermic reaction, heat is a reactant. Increasing the temperature is like adding heat. Adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products. Adding heat to an endothermic reaction will increase the value of K.

20 P a g e 20 Example 27 Consider the equilibrium N2O4 (g) 2NO2 (g) ΔH = 58.0 kj In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased? The Effects of Catalysts Catalysts provide an alternative, more efficient, mechanism. Catalysts work for both forward and reverse reactions. Catalysts affect the rate of the forward and reverse reactions by the same factor. Therefore, catalysts do not affect the position of equilibrium. Application of Le Châtelier s principle: The Haber Process Industrial synthesis of ammonia according to the reaction N2 (g) + 3H2(g) 2NH3(g) Incoming N2 and H2 gases are heated to approximately 500 C and passed over a catalyst. The resultant gas mixture is allowed to expand and cool, causing NH3 to liquefy. Unreacted N2 and H2 gases are recycled. The NH3 is continuously removed by selectively liquefying it; the boiling point of NH3 (-33 C) is much higher than that of N2 (-196 C) and H2 (-253 C).

21 P a g e 21 Example 28 [Integrative] At temperatures near 800 C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H2: C(s) + H2O(g) CO(g) + H2(g) 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 Kp = What are the equilibrium partial pressures of H2O, CO, and H2 in the equilibrium mixture at this temperature if we start with solid carbon and mol of H2O 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 Kp for this reaction is Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at equilibrium, should the pressure of the system be increased or decreased?

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