Chapter Fourteen 1 Chemical Equilibrium Prentice Hall 2005

2 Dynamic Nature of Equilibrium When a system reaches equilibrium, the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

3 Dynamic Equilibrium Illustrated Na. Cl containing radioactive Na+ is added to a saturated Na. Cl solution. After a time, the solution contains radioactive Na+ … Na. Cl dissolves and recrystallizes continuously. … and the added salt now contains some stable Na+. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

4 Concentration vs. Time If we begin with only 1 M HI, the [HI] decreases and both [H 2] and [I 2] increase. Prentice Hall © 2005 Beginning with 1 M H 2 and 1 M I 2, the [HI] increases and both [H 2] and [I 2] decrease. General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Beginning with 1 M each of H 2, I 2, and HI, the [HI] increases and both [H 2] and [I 2] decrease. Chapter Fourteen

5 Regardless of the starting concentrations; once equilibrium is reached … Prentice Hall © 2005 … the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant. General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

6 The Equilibrium Constant Expression For the general reaction: a. A + b. B g. G + h. H The equilibrium expression is: Kc = [G]g[H]h [A]a[B]b Each concentration is simply raised to the power of its coefficient Products in numerator. Reactants in denominator. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

7 The Equilibrium Constant • The equilibrium constant is constant regardless of the initial concentrations of reactants and products. • This constant is denoted by the symbol Kc and is called the concentration equilibrium constant. • Concentrations of the products appear in the numerator and concentrations of the reactants appear in the denominator. • The exponents of the concentrations are identical to the stoichiometric coefficients in the chemical equation. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

8 Example 14. 1 If the equilibrium concentrations of COCl 2 and Cl 2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl 2(g) COCl 2(g) Kc = 1. 2 x 103 at 395 °C Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

9 The Condition of Equilibrium • The kinetics view: Kc = (forward rate)/(reverse rate) = kf/kr • The thermodynamics view: – The equilibrium constant can be related to other fundamental thermodynamic properties and is called thermodynamic equilibrium constant, Keq. – The thermodynamic equilibrium constant expression uses dimensionless quantities known as activities in place of molar concentrations. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

10 Modifying the Chemical Equation Consider the reaction: 2 NO(g) + O 2(g) 2 NO 2(g) [NO 2]2 Kc = ––––– = 4. 67 x 1013 (at 298 K) [NO]2 [O 2] Now consider the reaction: 2 NO 2(g) 2 NO(g) + O 2(g) What will be the equilibrium constant K'c for the new reaction? [NO]2 [O 2] 1 1 1 K'c = ––––––––––– = 2. 14 x 10– 14 [NO 2]2 Kc 4. 67 x 1013 ––––– [NO]2 [O 2] Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

11 Modifying the Chemical Equation (cont’d) Consider the reaction: 2 NO(g) + O 2(g) 2 NO 2(g) [NO 2]2 Kc = ––––– = 4. 67 x 1013 (at 298 K) [NO]2 [O 2] Now consider the reaction: NO 2(g) NO(g) + ½ O 2(g) What will be the equilibrium constant K"c for the new reaction? [NO] [O 2]1/2 1 K"c = ––––– = ––– 2 [NO 2] Kc Prentice Hall © 2005 1/2 = 2. 14 x 10– 14 = 1. 46 x 10– 7 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

12 Modifying the Chemical Equation (cont’d) • For the reverse reaction, K is the reciprocal of K for the forward reaction. • When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. • General rule: – When the coefficients of an equation are multiplied by a common factor n to produce a new equation, we raise the original Kc value to the power n to obtain the new equilibrium constant. • It should be clear that we must write a balanced chemical equation when citing a value for Kc. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

13 Example 14. 2 The equilibrium constant for the reaction: ½ H 2(g) + ½ I 2(g) at 718 K is 7. 07. HI(g) (a) What is the value of Kc at 718 K for the reaction HI(g) ½ H 2(g) + ½ I 2(g) (b) What is the value of Kc at 718 K for the reaction H 2(g) + I 2(g) 2 HI(g) Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

14 The Equilibrium Constant for an Overall Reaction Suppose we need: N 2 O(g) + 3/2 O 2(g) 2 NO 2(g) Kc(1) = ? ? and we’re given: N 2 O(g) + ½ O 2(g) 2 NO(g) + O 2(g) 2 NO(g) Kc(2) = 1. 7 x 10– 13 2 NO 2(g) Kc(3) = 4. 67 x 1013 • Adding the given equations gives the desired equation. • Multiplying the given values of K gives the equilibrium constant for the overall reaction. • (To see why this is so, write the equilibrium constant expressions for the two given equations, and multiply them together. Examine the result …) Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

15 Equilibria Involving Gases • In reactions involving gases, it is often convenient to measure partial pressures rather than molarities. • In these cases, a partial pressure equilibrium constant, Kp, is used. g h Kp = (PG) (PH) (PA)a(PB)b Kc and Kp are related by: Kp = Kc (RT)Δn(gas) where Dn(gas) is the change in number of moles of gas as the reaction occurs in the forward direction. Dn(gas) = mol gaseous products – mol gaseous reactants Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

16 Example 14. 3 Consider the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N 2 O 4(g) 2 NO 2(g) Kp = 0. 660 at 319 K (a) What is the value of Kc for this reaction? (b) What is the value of Kp for the reaction 2 NO 2(g) N 2 O 4(g)? (c) If the equilibrium partial pressure of NO 2(g) is 0. 332 atm, what is the equilibrium partial pressure of N 2 O 4(g)? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

17 Equilibria Involving Pure Solids and Liquids • The equilibrium constant expression does not include terms for pure solid and liquid phases because their concentrations do not change in a reaction. • Although the amounts of pure solid and liquid phases change during a reaction, these phases remain pure and their concentrations do not change. Example: Ca. CO 3(s) Ca. O(s) + CO 2(g) [Ca. O] [CO 2] Kc = ––––– [Ca. CO 3] Prentice Hall © 2005 Kc = [CO 2] General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

18 Example 14. 4 The reaction of steam and coke (a form of carbon) produces a mixture of carbon monoxide and hydrogen, called watergas. This reaction has long been used to make combustible gases from coal: C(s) + H 2 O(g) CO(g) + H 2(g) Write the equilibrium constant expression for Kc for this reaction. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

19 Equilibrium Constants: When Do We Need Them and When Do We Not? • A very large numerical value of Kc or Kp signifies that a reaction goes (essentially) to completion. • A very small numerical value of Kc or Kp signifies that the forward reaction, as written, occurs only to a slight extent. • An equilibrium constant expression applies only to a reversible reaction at equilibrium. • Although a reaction may be thermodynamically favored, it may be kinetically controlled … • Thermodynamics tells us “it’s possible (or not)” • Kinetics tells us “it’s practical (or not)” Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

20 Example 14. 5 Is the reaction Ca. O(s) + CO 2(g) Ca. CO 3(s) likely to occur to any appreciable extent at 298 K? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

21 The Reaction Quotient, Q • For nonequilibrium conditions, the expression having the same form as Kc or Kp is called the reaction quotient, Qc or Qp. • The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium. • To determine the direction of net change, we compare the magnitude of Qc to that of Kc. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

22 The Reaction Quotient, Q When Q is smaller than K, the denominator of Q is too big; we have “too much reactants. ” When Q = K, equilibrium has been reached. When Q is larger than K, the numerator of Q is too big; we have “too much products. ” Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

23 Example 14. 6 Predict the direction of net change for Experiment 3 in

24 Le Châtelier’s Principle • When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. • Analogy: Begin with 100 men and 100 women at a dance. • Assume that there are 70 couples dancing, though not always the same couples (dynamic equilibrium). • If 30 more men arrive, what happens? • The equilibrium will shift, and shortly, more couples will be dancing … but probably not 30 more couples. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

25 Changing the Amounts of Reacting Species • At equilibrium, Q = Kc. • If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. • Q is now less than Kc. • This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = Kc again. • In order to do this, the concentrations of the products increase; the equilibrium is shifted to the right. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

… the acetic acid concentration first increases … 26 … then the concentrations of both reactants decrease … … and the concentrations of both products increase, until a new equilibrium is established. When acetic acid (a reactant) is added to the equilibrium mixture … Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

27 Example 14. 7 Water can be removed from an equilibrium mixture in the reaction of 1 -octanol and acetic acid, for example, by using a solid drying agent that is insoluble in the reaction mixture. Describe how the removal of a small quantity of water affects the equilibrium. CH 3(CH 2)6 CH 2 OH(soln) + CH 3 COOH(soln) H+ CH 3(CH 2)6 CH 2 OCOCH 3(soln) + H 2 O(soln) Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

28 Heterogeneous Equilibria and Le Chatelier’s Principle • Addition or removal of pure solids or pure liquids from a system at equilibrium does not affect the equilibrium. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

29 Changing External Pressure or Volume in Gaseous Equilibria • When the external pressure is increased (or system volume is reduced), an equilibrium shifts in the direction producing the smaller number of moles of gas. • When the external pressure is decreased (or the system volume is increased), an equilibrium shifts in the direction producing the larger number of moles of gas. • If there is no change in the number of moles of gas in a reaction, changes in external pressure (or system volume) have no effect on an equilibrium. Example: H 2(g) + I 2(g) 2 HI equilibrium is unaffected by pressure changes. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

30 Initial When pressure is increased … … two molecules of NO 2 combine … Prentice Hall © 2005 … to give one molecule of N 2 O 4, reducing the pressure increase. General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

31 Temperature Changes and Catalysis • Raising the temperature of an equilibrium mixture shifts equilibrium in the direction of the endothermic reaction; lowering the temperature shifts equilibrium in the direction of the exothermic reaction. – Consider heat as though it is a product of an exothermic reaction or as a reactant of an endothermic reaction, and apply Le Châtelier’s principle. • A catalyst lowers the activation energy … of both the forward and the reverse reaction. • Adding a catalyst does not affect an equilibrium state. • A catalyst merely causes equilibrium to be achieved faster. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

32 Example 14. 8 An equilibrium mixture of O 2(g), SO 2(g), and SO 3(g) is transferred from a 1. 00 -L flask to a 2. 00 -L flask. In which direction does a net reaction proceed to restore equilibrium? The balanced equation for the reaction is 2 SO 3(g) 2 SO 2(g) + O 2(g) Example 14. 9 Is the amount of NO(g) formed from given amounts of N 2(g) and O 2(g), N 2(g) + O 2(g) 2 NO(g) ΔH° = +180. 5 k. J greater at high or low temperatures? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

33 Example 14. 10 A Conceptual Example Flask A, pictured below, initially contains an equilibrium mixture of the reactants and products of the reaction CO(g) + H 2 O(g) CO 2(g) + H 2(g) ΔH = – 41 k. J; Kc = 9. 03 at 698 K It is isolated from flask B by a closed valve. When the valve is opened, a new equilibrium is established as the contents of the two flasks mix. Describe, qualitatively, how the amounts of CO, H 2, CO 2, and H 2 O in the new equilibrium compare with the amounts in the initial equilibrium if (a) flask B initially contains Ar(g) at 1 atm pressure; (b) flask B initially contains 1. 0 mol CO 2; (c) flask B initially contains 1. 0 mol CO and the temperature of the A–B mixture is raised by 100 °C. If you are uncertain of the result in any of the three cases, explain why. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

34 Determining Values of Equilibrium Constants Experimentally • When initial amounts of one or more species, and equilibrium amounts of one or more species, are given, the amounts of the remaining species in the equilibrium state and, therefore, the equilibrium concentrations often can be established. • A useful general approach is to tabulate under the chemical equation: – the concentrations of substances present initially – changes in these concentrations that occur in reaching equilibrium – the equilibrium concentrations. • This sort of table is sometimes called an “ICE” table: Initial/Change/Equilibrium. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

35 Example 14. 11 In a 10. 0 -L vessel at 1000 K, 0. 250 mol SO 2 and 0. 200 mol O 2 react to form 0. 162 mol SO 3 at equilibrium. What is Kc, at 1000 K, for the reaction that is shown here? 2 SO 2(g) + O 2(g) 2 SO 3(g) Example 14. 12 Consider the reaction H 2(g) + I 2(g) 2 HI(g) Kc = 54. 3 at 698 K If we start with 0. 500 mol I 2(g) and 0. 500 mol H 2(g) in a 5. 25 -L vessel at 698 K, how many moles of each gas will be present at equilibrium? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

36 Calculating Equilibrium Quantities from Kc and Kp Values • When starting with initial reactants and no products and with the known value of the equilibrium constant, these data are used to calculate the amount of substances present at equilibrium. • Typically, an ICE table is constructed, and the symbol x is used to identify one of the changes in concentration that occurs in establishing equilibrium. • Then, all the other concentration changes are related to x, the appropriate terms are substituted into the equilibrium constant expression, and the equation solved for x. Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

37 Example 14. 13 Suppose that in the reaction of Example 14. 12, the initial amounts are 0. 800 mol H 2 and 0. 500 mol I 2. What will be the amounts of reactants and products when equilibrium is attained? Example 14. 14 Carbon monoxide and chlorine react to form phosgene, COCl 2, which is used in the manufacture of pesticides, herbicides, and plastics: COCl 2(g) CO(g) + Cl 2(g) Kc = 1. 2 x 103 at 668 K How much of each substance, in moles, will there be at equilibrium in a reaction mixture that initially has 0. 0100 mol CO, 0. 0100 mol Cl 2, and 0. 100 mol COCl 2 in a 10. 0 -L flask? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

38 Example 14. 15 A sample of phosgene, COCl 2(g), is introduced into a constant-volume vessel at 395 °C and observed to exert an initial pressure of 0. 351 atm. When equilibrium is established for the reaction CO(g) + Cl 2(g) COCl 2(g) Kp = 22. 5 what will be the partial pressure of each gas and the total gas pressure? Prentice Hall © 2005 General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen

39 Cumulative Example A mixture of H 2 S(g) and CH 4(g) in the mole ratio 2: 1 was brought to equilibrium at 700 °C and a total pressure of 1. 00 atm. The equilibrium mixture was analyzed and found to contain 9. 54 x 10– 3 mol H 2 S. The CS 2 present at equilibrium was converted, first to H 2 SO 4 and then to Ba. SO 4, with 1. 42 x 10– 3 mol Ba. SO 4 being obtained. Use these data to determine Kp at 700 °C for the reaction 2 H 2 S(g) + CH 4(g) Prentice Hall © 2005 CS 2(g) + 4 H 2(g) General Chemistry 4 th edition, Hill, Petrucci, Mc. Creary, Perry Chapter Fourteen