Chapter 15. Chemical Equilibrium 15.1 The Concept of Equilibrium Chemical equilibrium is the point at which the concentrations of all species are constant. A dynamic equilibrium exists when the rates of the forward and reverse reactions are equal. The double arrow implies that the process is dynamic. 15.2 The Equilibrium Constant Equilibrium-Constant Expression law of mass action. For a general reaction aa + bb dd + ee The equilibrium-constant expression is given by: d e [ D ][ E] K c = a b [ A] [ B] Where K c is the equilibrium constant. The subscript c indicates that molar concentrations were used to evaluate the constant. Note that the equilibrium constant expression has products in the numerator and reactants in the denominator. Equilibrium Constants in Terms of Pressure, K p When the reactants and products are gases the equilibrium constant is K p where p stands for pressure. For the reaction: aa + bb dd + ee ( PD ) ( PE ) K p = a b ( PA ) ( PB ) They can be interconverted using the ideal gas equation and our definition of molarity: PV = nrt thus P = (n/v)rt If we express volume in liters the quantity (n/v) is equivalent to molarity. Thus the partial pressure of a substance, A, is given as: P A =(n A /V)RT = [A]RT d e We can use this to obtain a general expression relating K c and K p: K p = K c (RT) n Where n = (moles of gaseous products) (moles of gaseous reactants). The numerical values of K c and K p will not differ if n = 0. (Note: See p. 635 in textbook for info about why we do not use units for K) - 1 -
Sample Exercise 15.1 (p. 634) Write the equilibrium expression for K eq for these three reactions: a) 2 O 3(g) 3 O 2(g) b) 2 NO (g) + Cl 2(g) 2 NOCl (g) c) Ag + (aq) + 2 NH 3(g) Ag(NH 3 ) 2 + (aq) Practice Exercise 1 (15.1) For the reaction 2 SO 2(g) + O 2(g) 2 SO 3(g) which of the following is the correct equilibrium-constant expression: a) K p = P SO2 2 P O2 P SO3 2 b) K p = 2 P SO2 P O2 2 P SO3 2 c) K p = P SO3 P SO2 2 P O2 d) K c = 2 P SO3 2 P SO2 P O2 Practice Exercise 2 (15.1) Write the equilibrium expression for K eq for these two reactions: a) H 2(g) + I 2(g) 2 HI (g) b) Cd 2+ (aq) + 4 Br - (aq) CdBr 4 2- (aq - 2 -
Evaluating K c The value of K eq does not depend on initial concentrations of products or reactants. The equilibrium expression depends on stoichiometry. It does not depend on the reaction mechanism. The value of K eq varies with temperature. We generally omit the units of the equilibrium constant. 15.3 Interpreting and Working with Equilibrium Constants The Magnitude of Equilibrium Constants The larger K eq the more products are present at equilibrium. The smaller K eq the more reactants are present at equilibrium. If K eq >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K eq << 1, then reactants dominate at equilibrium and the equilibrium lies to the left. - 3 -
Sample Exercise 15.3 (p. 638) 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, 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. Practice Exercise 2 (15.3) The equilibrium constant for the reaction H 2(g) + I 2(g) 2 HI (g) varies with temperature as follows: K p = 792 at 298 K; K p = 54 at 700 K. Is the formation of HI favored more at the higher or lower temperature? Relating Chemical Equations and Equilibrium Constants The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction. The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number. The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. - 4 -
Sample Exercise 15.4 (p. 640) Given the following information, HF (aq) H + (aq) + F - (aq) K c = 6.8 x 10-4 H 2 C 2 O 4(aq) 2 H + 2- (aq) + C 2 O 4 (aq) K c = 3.8 x 10-6 determine the value of K c for the following reaction: 2 HF (aq) + C 2 O 4 2- (aq) 2 F - (aq) + H 2 C 2 O 4(aq) (0.12) Practice Exercise 1 (15.4) Given the equilibrium constants for the following two reactions in aqueous solution at 25 o C, HNO 2(aq) H + - (aq) + NO 2 (aq) K c = 4.5 x 10-4 H 2 SO 3(aq) 2 H + 2- (aq) + SO 3 (aq) K c = 1.1 x 10-9 What is the value of K c for the following reaction? 2 HNO 2(aq) + SO 3 2- (aq) H 2 SO 3(aq ) + 2 NO 2 - (aq) a) 4.9 x 10-13 b) 4.1 x 10 5 c) 8.2 x 10 5 d) 1.8 x 10 2 e) 5.4 x 10-3 Practice Exercise 2 (15.4) Given the following information at 700 K, H 2(g) + I 2(g) 2HI (g) K p = 54.0 N 2(g) + 3 H 2(g) 2 NH 3(g) K p = 1.04 x 10-4 determine the value of K p (at 700 K) 2 NH 3(g) + 3 I 2(g) 6 HI (g) + N 2(g) (1.51 x 10 9 ) - 5 -
15.4 Heterogeneous Equilibria Equilibria in which one or more reactants or products are present in a different phase are called heterogeneous equilibria. If a pure solid or pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium constant expression. Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established! systems where the solvent is involved as a reactant or product and the solutes are present at low concentrations. Example: H 2 O(l) + CO 3 2 (aq) OH (aq) + HCO 3 (aq) K c = [OH ][HCO 3 ] / [CO 3 2 ] Here the concentration of water is essentially constant and we can think of it as a pure liquid. Sample Exercise 15.5 (p. 643) Write the equilibrium-constant K eq for each of the following reactions: a) CO 2(g) + H 2(g) CO (g) + H 2 O (l) K eq = b) SnO 2(s) + 2 CO (g) Sn (s) + 2 CO 2(g) K eq = Practice Exercise 1 (15.5) Consider the equilibrium that is established in a saturated solution of silver chloride, Ag + (aq) + Cl - (aq) AgCl(s). If more solid AgCl is added to this solution, what will happen to the concentration of Ag + and Cl - ions in solution? a) [Ag + ] and [Cl - ] will both increase. b) [Ag + ] and [Cl - ] will both decrease. c) [Ag + ] will increase and [Cl - ] will decrease. d) [Ag + ] will decrease and [Cl - ] will increase. e) Neither [Ag + ] nor [Cl - ] will change. Practice Exercise 2 (15.5) Write the equilibrium-constant expressions for each of the following reactions: a) Cr (s) + 3 Ag + (aq) Cr 3+ (aq) + 3 Ag (s) K c = b) 3 Fe (s) + 4 H 2 O (g) Fe 3 O 4(s) + 4 H 2(g) K p = - 6 -
Sample Exercise 15.6 (p. 643) Each of the following mixtures was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equilibrium CaCO 3(s) CaO (s) + CO 2(g) a) pure CaCO 3 b) CaO and a pressure of CO 2 greater than the value of K p c) Some CaCO 3 and a pressure of CO 2 greater than the value of K p d) CaCO 3 and CaO Practice Exercise 1 (15.6) If 8.0 g of NH 4 HS (s) is placed in a sealed vessel with a volume of 1.0 L and heated to 200 o C the reaction NH 4 HS (s) NH 3(g) + H 2 S (g) will occur. When the system comes to equilibrium, some NH 4 HS (s) is still present. Which of the following changes will lead to a reduction in the amount of NH 4 HS (s) that is present, assuming in all cases that equilibrium is re-established following the change? a) Adding more NH 3(g) to the vessel b) Adding more H 2 S (g) to the vessel c) Adding more NH 4 HS (s) to the vessel d) Increasing the volume of the vessel e) Decreasing the volume of the vessel Practice Exercise 2 (15.6) When added to Fe 3 O 4(s) in a closed container, which one of the following substances H 2(g), H 2 O (g), O 2(g) - will allow equilibrium to be established in the reaction 3 Fe (s) + 4 H 2 O (g) Fe 3 O 4(s) + 4 H 2(g) - 7 -
15.5 Calculating Equilibrium Constants Sample Exercise 15.7 (p. 644) A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472 o C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2, 2.46 atm N 2, and 0.166 atm NH 3. From these data calculate the equilibrium constant, K p, for N 2(g) + 3 H 2(g) 2 NH 3(g) (2.79 x 10-5 ) Practice Exercise 1 (15.7) 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(g). If the vessel contains 0.669 atm SO 2(g), 0.395 atm O 2(g), and 0.0851 atm SO 3(g) after the system has reached equilibrium, what is the equilibrium constant K p for the reaction 2 SO 2(g) + O 2(g) 2 SO 3(g)? a) 0.0410 b) 0.322 c) 24.4 d) 3.36 e) 3.11 Practice Exercise 2 (15.7) An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25 o C: [CH 3 COOH] = 1.65 x 10-2 M; [H + ] = 5.44 x 10-4 M; and [CH 3 COO - ] = 5.44 x 10-4 M. Calculate the equilibrium constant, K c, for the ionization of acetic acid at 25 o C. The reaction is (1.79 x 10-5 ) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) - 8 -
If we do not know the equilibrium concentrations of all species involved in the reaction, proceed as follows: 1. Tabulate all the known initial and equilibrium concentrations of the species that appear in the equilibrium-constant expression. 2. For those species for which both the 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, use the coefficients in the balanced chemical equation) to calculate the changes in concentration of all the other species in the equilibrium. 4. From the initial concentrations and the changes in concentration, calculate the equilibrium concentrations. These are then used to evaluate the equilibrium constant. Sample Exercise 15.8 (p. 645) A closed system initially containing 1.000 x 10-3 M H 2 and 2.000 x 10-3 M I 2 at 448 o C is allowed to reach equilibrium, and at equilibrium the concentration of HI is 1.87 x 10-3 M. Calculate K c at 448 o C for the reaction taking place, which is (1.81 x 10-5 ) - 9 -
Practice Exercise 1 (15.8) In Section 15.1, we discussed the equilibrium between N 2 O 4(g) and NO 2(g). Let s return to that equation in a quantitative example. When 9.2 g of frozen N 2 O 4(g) is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibrium, the concentration of N 2 O 4 is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction N 2 O 4(g) 2 NO 2(g) at 400 K? a) 0.23 b) 0.36 c) 0.13 d) 1.4 e) 2.5 Practice Exercise 2 (15.8) Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO 3(g) 2 SO 2(g) + O 2(g) Initially the vessel is charged at 1000 K with SO 3(g) at a partial pressure of 0.500 atm. At equilibrium the SO 3 partial pressure is 0.200 atm. Calculate the value of K p at 1000 K. (0.338) - 10 -
Predicting the Direction of Reaction 15.6 Applications of Equilibrium Constants For a general reaction: aa + bb cc + dd We define Q, the reaction quotient, as: c [ ] [ ] C D Q = [ A] a [ B] b Where [A], [B], [C], and [D] are molarities (for substances in solution) or partial pressures (for gases) at any given time. Note: Q = K eq only at equilibrium. If Q < K eq then the forward reaction must occur to reach equilibrium. If Q > K eq then the reverse reaction must occur to reach equilibrium. Products are consumed, reactants are formed. Q decreases until it equals K eq. Calculating Equilibrium Concentrations The same steps used to calculate equilibrium constants are used to calculate equilibrium concentrations. Generally, we do not have a number for the change in concentration. Therefore, we need to assume that x mol/l of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions. d Sample Exercise 15.9 (p. 647) At 448 o C the equilibrium constant, K c, for the reaction H 2(g) + I 2(g) 2 HI (g) is 51. Predict how the reaction will proceed to reach equilibrium at 448 o C if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mole of H 2, and 3.0 x 10-2 mol of I 2 in a 2.00-L container. (Q = 1.3, so reaction must proceed from left to right) - 11 -
Practice Exercise 1 (15.9) 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 (15.9) At 1000 K the value of K p for the reaction 2 SO 3(g) 2 SO 2(g) + O 2(g) is 0.338. 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. (Q p = 16; Q p > K p, so reaction will proceed from right to left) Sample Exercise 15.10 (p. 648) For the Haber process, N 2(g) + 3 H 2(g) 2 NH 3(g), K p = 1.45 x 10-5 at 500 o C. In an equilibrium mixture of the three gases at 500 o C, the partial pressure of H 2 is 0.928 atm and that of N 2 is 0.432 atm. What is the partial pressure of NH 3 in this equilibrium mixture? (2.24 x 10-3 atm) - 12 -
Practice Exercise 1 (15.10) At 500 K, the reaction 2 NO (g) + Cl 2(g) 2 NOCl (g) has K p = 51. In an equilibrium mixture at 500 K, the partial pressure of NO is 0.125 atm and that of PCl 3 is 0.165 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 x 10-5 atm e) 0.125 atm Practice Exercise 2 (15.10) At 500 K the reaction PCl 5(g) PCl 3(g) + Cl 2(g) has K p = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl 5 is 0.860 atm and that of PCl 3 is 0.350 atm. What is the partial pressure of Cl 2 in the equilibrium mixture? (1.22 atm) - 13 -
15.7 Le Châtelier s Principle If a system at equilibrium is disturbed by a change in temperature, a change in pressure, or a change in the concentration of one or more components, the system will shift its equilibrium position in such a way as to counteract the effects of the disturbance. Change in Reactant or Product Concentration If a chemical system is at equilibrium and we add or remove a product or reactant, the reaction will shift so as to reestablish equilibrium. Effects of Volume and Pressure Changes An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of moles of gas in the products and reactants, changing the pressure has no effect on the equilibrium. In addition, no change will occur if we increase the total gas pressure by the addition of a gas that is not involved in the reaction. Effect of Temperature Changes The equilibrium constant is temperature dependent, and dependent on the particular reaction. Sample Exercise 15.13 (p. 656) Consider the following equilibrium: N 2 O 4(g) 2 NO 2(g) H o = 58.0 kj In what direction will the equilibrium shift when a) N 2 O 4 is added, b) NO 2 is removed, c) the pressure is increased by adding N 2(g), d) the volume is increased, e) the temperature is decreased? Practice Exercise 1 (15.12) For the reaction 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) H o = -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 Ne (g) to the reaction vessel. - 14 -
Practice Exercise 2 (15.12) For the reaction PCl 5(g) PCl 3(g) + Cl 2(g) in which direction will the equilibrium shift when H o = 87.9 kj a) Cl 2(g) is removed, b) the temperature is decreased, c) the volume of the reaction system is increased, d) PCl 3(g) is added? The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not affect the composition of the equilibrium mixture. Sample Integrative Exercise 15 (p. 658) At temperatures near 800 o C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H 2 : C (s) + H 2 O (g) CO (g) + H 2(g) The mixture of gases that results is an important industrial fuel called water gas. a) At 800 o C the equilibrium constant for this reaction is K p = 14.1. What are the equilibrium partial pressures of H 2 O, CO, and H 2 in the equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H 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 o C the value of K eq for this reaction is 1.7 x 10-21. Is the reaction exothermic or endothermic? e) To produce the maximum amount of CO and H 2 at equilibrium, should the pressure of the system be increased or decreased? - 15 -