1. Chapter 15; CHEMICAL EQUILIBRIUM 14 | 1 Describing Chemical Equilibrium Chemical Equilibrium—A Dynamic Equilibrium The Equilibrium Constant Heterogeneous Equilibria; Solvents in Homogeneous Equilibria Using the Equilibrium Constant Qualitatively Interpreting the Equilibrium Constant Predicting the Direction of Reaction Calculating Equilibrium Concentrations

2. Changing Reaction Conditions: Le Châtelier’s Principle Removing Products or Adding Reactants Changing the Pressure and Temperature Effect of a Catalyst

3. 14 | 3 Chemical reactions often seem to stop before they are complete. Actually, such reactions are reversible. That is, the original reactants form products, but then the products react with themselves to give back the original reactants. When these two reactions—forward and reverse—occur at the same rate, a chemical equilibrium exists.

4. 14 | 4 The graph shows how the amounts of reactants and products change as the reaction approaches equilibrium. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g)

5. 14 | 5 This graph shows how the rates of the forward reaction and the reverse reaction change as the reaction approaches equilibrium. CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g)

6. 14 | 6 Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. We can apply stoichiometry to compute the content of the reaction mixture at equilibrium.

7. 14 | 7 When heated PCl 5, phosphorus pentachloride, forms PCl 3 and Cl 2 as follows: PCl 5 (g) PCl 3 (g) + Cl 2 (g) When 1.00 mol PCl 5 in a 1.00-L container is allowed to come to equilibrium at a given temperature, the mixture is found to contain 0.135 mol PCl 3. What is the molar composition of the mixture?

8. 14 | 8 We will organize this problem by using the chemical reaction to set up a table of initial, change, and equilibrium amounts. Initially we had 1.00 mol PCl 5 and no PCl 3 or Cl 2. The change in each is stoichiometric: If x moles of PCl 5 react, then x moles of PCl 3 and x moles of Cl 2 are produced. For reactants, this amount is subtracted from the original amount; for products, it is added to the original amount.

9. 14 | 9 We were told that the equilibrium amount of PCl 3 is 0.135 mol. That means x = 0.135 mol. We can now find the amounts of the other substances. PCl 5 (g)PCl 3 (g) +Cl 2 (g) Initial1.00 mol00 Change –x–x+x+x+x+x Equilibrium1.00 – xxx

10. 14 | 10 The Equilibrium Constant, K c The equilibrium constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration term to a power equal to its coefficient in the balanced chemical equation. The equilibrium constant, K c, is the value obtained for the K c expression when equilibrium concentrations are substituted.

11. 14 | 11 For the reaction aA + bB cC + dD The equilibrium constant expression is K c =

12. 14 | 12 An example to consider; Methanol (also called wood alcohol) is made commercially by hydrogenation of carbon monoxide at elevated temperature and pressure in the presence of a catalyst: 2H 2 (g) + CO(g) CH 3 OH(g) What is the K c expression for this reaction?

13. 14 | 13 When we are given some information about equilibrium amounts, we are able to calculate the value of K c. We need to take care to remember that the K c expression uses molar concentrations.

14. 14 | 14 Example; Carbon dioxide decomposes at elevated temperatures to carbon monoxide and oxygen: 2CO 2 (g) 2CO(g) + O 2 (g) At 3000 K, 2.00 mol CO 2 is placed into a 1.00-L container and allowed to come to equilibrium. At equilibrium, 0.90 mol CO 2 remains. What is the value of K c at this temperature?

15. 14 | 15 We can find the value of x. Equilibrium Change Initial O2(g)O2(g)2CO(g) +2CO 2 (g)

16. 14 | 16 2CO 2 (g) 2CO(g) + O 2 (g) Next we find the concentrations of all the species at equilibrium [CO 2 ]= [CO]= [O 2 ]= We then calculate Kc by substituting the various equilibrium concentrations in the Equilibrium constant expression.

17. 14 | 17 In a heterogeneous equilibrium, in the K c expression, the concentrations of solids and pure liquids are constant (due to these substances’ constant density). As a result, we incorporate those constants into the value of K c, thereby making a new constant, K c. In other words, equilibrium is not affected by solids and pure liquids as long as some of each is present. More simply, we write the K c expression by replacing the concentration of a solid or pure liquid with 1.

18. 14 | 18 Write the K c expression for the following reaction: H 2 O(g) + C(s) CO(g) + H 2 (g)

19. 14 | 19 Given: aA + bB cC + dD; K 1 When the reaction is reversed: cC + dD aA + bB; K 2 The equilibrium constant expression is inverted: K 2 =

20. 14 | 20 Given: aA + bB cC + dD; K 1 When the reaction is doubled: 2aA + 2bB 2cC + 2dD; K 2 The equilibrium constant expression, K 2, is the square of the equilibrium constant expression, K 1 : K 2 =

21. 14 | 21 For the reaction aA(g) + bB(g) cC(g) + dD(g) The equilibrium constant expressions are K c = and K p =

23. 14 | 23 How are these related? We know From the ideal gas law, we know that So,

24. 14 | 24 When you express an equilibrium constant for a gaseous reaction in terms of partial pressures, you call it the equilibrium constant, K p In general, the value of K p is different from that of K c. We will explore this relationship on the next slides. Recall the ideal gas law and the relationship between pressure and molarity of a gas:

25. 14 | 25 K p = K c (RT)  n

26. 14 | 26 For catalytic methanation, CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) the equilibrium expression in terms of partial pressures becomes and

27. 14 | 27 Example; The value of K c at 227°C is 0.0952 for the following reaction: CH 3 OH(g) CO(g) + 2H 2 (g) What is K p at this temperature? R = 0.08206 L  atm/(mol  K)

28. 14 | 28 We can use the value of the equilibrium constant in several ways. First, we can qualitatively describe the content of the reaction mixture by looking at the magnitude of K c. Second, we can determine the direction in which a reaction will proceed by comparing K c to the value of the reaction quotient, Q, which has the same expression as K c but uses nonequilibrium values. Finally, we can determine equilibrium concentrations given the initial concentrations and the value of K c.

29. 14 | 29 When K c is very large (>10 2 ), the equilibrium mixture is mostly products. When K c is very small (<10 -2 ), the equilibrium mixture is mostly reactants. When K c approaches 1, the equilibrium mixture contains appreciable amounts of both reactants and products.

30. 14 | 30 K c = 0.82 for a reaction. Describe the composition of the equilibrium mixture.

31. 14 | 31 Reaction Quotient, Q The reaction quotient has the same form as the equilibrium constant, but uses initial concentrations for its value. When K c > Q, the reaction proceeds to the right. When K c < Q, the reaction proceeds to the left. When K c = Q, the reaction is at equilibrium.

32. 14 | 32 Q c must move toward K c. Here the numerator must increase; more products must be produced. Here the denominator must increase; more reactants must be produced.