1. Chemical Equilibrium Chapter 15

2. aA + bB cC + dD K C = [C] c [D] d [A] a [B] b Law of Mass Action Must be caps! Equilibrium constant Lies to the rightLies to the left

3. 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. Sample Exercise 15.3 p 636 Interpreting the Magnitude of an Equilibrium Constant (ii) < (i) < (iii).

4. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] 2NO 2 (g) N 2 O 4 (g) K = [N 2 O 4 ] [NO 2 ] 2 ' = 1 K = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

5. A + B C + D C + D E + F A + B E + F K c = ' [C][D] [A][B] K c = ″ [E][F] [C][D] [E][F] [A][B] K c = KcKc ' KcKc ″ KcKc KcKc ″ KcKc ' ∙ If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

6. Writing Equilibrium Constant Expressions Concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, concentrations can be expressed in M or in atm. Concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Summary

7. Ways of Expressing Equilibrium Constants Concentration of products and reactants may be expressed in different units, so: Heterogeneous equilibria Homogeneous equilibria K = [C] c [D] d [A] a [B] b Law of Mass Action

8. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ' [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c ∙ ' [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant

9. P CO 2 K p = CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO Fig 15.8

10. Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate K p and K c for the reaction. NH 4 HS (s) NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S ∙P= (0.265) · (0.265) = 0.0702 K p = K c (RT)  n K c = K p (RT) -  n  n = 2 – 0 = 2 T = 295 K K c = (0.0702) · (0.0821 · 295) -2 = 1.20 x 10 -4

11. Homogeneous Equilibrium CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ' [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] ≈ 55.6 M ≈ constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ' What’s the concentration of water? General practice not to include units for the equilibrium constant

12. Applications of Equilibrium Constants What does the equilibrium constant tell us? Calculating equilibrium concentrations Predicting the direction of a reaction

13. Calculating Equilibrium Concentrations 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species.

14. At 1280°C the equilibrium constant (K c ) for the reaction is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x

15. K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c = 0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M

16. Practice Exercise bottom p 647 For the equilibrium PCl 5 (g) ⇌ PCl 3 (g) + Cl 2 (g) the equilibrium constant K p is 0.497 at 500. K. A gas cylinder at 500. K is injected with PCl 5 (g) at 1.66 atm. Calculate the equilibrium pressures of all species at this temperature. Ans: P PCl5 = 0.967 atmP PCl3 = P Cl2 = 0.693 atm

17. Applications of Equilibrium Constants What does the equilibrium constant tell us? Calculating equilibrium concentrations Predicting the direction of a reaction

18. Reaction quotient (Q c ) - calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF: Q c < K c system proceeds from left to right to reach equilibrium Q c = K c the system is at equilibrium Q c > K c system proceeds from right to left to reach equilibrium Fig 15.9

19. At 448 °C the equilibrium constant K c for the reaction is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448 °C if we start with 2.0 × 10 –2 mol of HI, 1.0 × 10 –2 mol of H 2, and 3.0 × 10 –2 mol of I 2 in a 2.00-L container. Sample Exercise 15.10 Predicting the Direction of Approach to Equilibrium Solve: The initial concentrations: The reaction quotient is: