1. © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium.

2. © 2009, Prentice-Hall, Inc. What Does the Value of K Mean? If K>>1, the reaction is product-favored; product predominates at equilibrium. If K<<1, the reaction is reactant-favored; reactant predominates at equilibrium.

3. Sample Exercise 15.3 Interpreting the Magnitude of an Equilibrium Constant Solution Analyze: We are first asked to judge the relative magnitudes of three equilibrium constants and then to calculate them. Plan: (a) The more product that is present at equilibrium, relative to the reactant, the greater the equilibrium constant. (b) The equilibrium constant is given by the concentrations of products over reactants. 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.

4. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N 2 O 4 (g)   2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)  

5. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N 2 O 4(g)   2 NO 2(g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4(g)   4 NO 2(g)

6. Sample Exercise 15.3 Interpreting the Magnitude of an Equilibrium Constant Solution (continued) Solve: (a) Each box contains 10 spheres. The amount of product in each varies as follows: (i) 6, (ii) 1, (iii) 8. Thus, the equilibrium constant varies in the order (ii) < (i) < (iii). (b) In (i) we have 0.60 mol/L product and 0.40 mol/L reactant, giving K c = 0.60/0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres/4 spheres = 1.5.) In (ii) we have 0.10 mol/L product and 0.90 mol/L reactant, giving K c = 0.10/0.90 = 0.11 (or 1 sphere/9 spheres = 0.11). In (ii) we have 0.80 mol/L product and 0.29 mol/L reactant, giving K c = 0.80/0.20 = 4.0 (or 8 spheres/2 spheres = 4.0). These calculations verify the order in (a). Comment: Imagine that there was a drawing, like those above, that represents a reaction with a very small or very large value of K c. For example, what would the drawing look like if K c = 1 × 10 –5 ? In that case there would need to be 100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw.

7. Sample Exercise 15.4 Evaluating an Equilibrium Constant When an Equation is Reversed The equilibrium constant for the reaction of N 2 with O 2 to form NO equals K c = 1 × 10 –30 at 25 °C: Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction: Solution

8. © 2009, Prentice-Hall, Inc. Manipulating Equilibrium Constants 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.

9. Sample Exercise 15.5 Combining Equilibrium Expressions Given the following information, determine the value of K c for the reaction Solution

10. © 2009, Prentice-Hall, Inc. Heterogeneous Equilibrium

11. © 2009, Prentice-Hall, Inc. The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by multiplying the density of the substance by its molar mass — and both of these are constants at constant temperature.

12. © 2009, Prentice-Hall, Inc. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression. K c = [Pb 2+ ] [Cl - ] 2 PbCl 2 (s)   Pb 2+ (aq) + 2 Cl - (aq)

13. © 2009, Prentice-Hall, Inc. As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s)   CO 2 (g) + CaO (s)

14. Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Write the equilibrium-constant expression for K c for each of the following reactions: Solution Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the corresponding equilibrium-constant expressions. Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the expressions. Solve: (a) The equilibrium-constant expression is Because H 2 O appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-constant expression. (b) The equilibrium-constant expression is Because SnO 2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant expression.

15. Sample Exercise 15.6 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions Practice Exercise