1. Chemical Equilibrium Chapter 15 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 15; Equilibrium I.Dynamic Equilibrium II.Equilibrium Constant; Kc and Kp III.Homogenous vs. Heterogeneous Equilibrium IV.Calculation of K from Equilibrium Concentrations

3. Chapter 15; Equilibrium V.Predicting Reaction Direction for Nonequilibria Mixtures Reaction Quotient, Q VI.Le Chatlier’s Principle VII.Calculating Equilibrium Concentrations from Initial Concentrations and K Square Root Quadratic Formula

4. Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and they are not zero. the concentrations of the reactants and products remain constant Chemical equilibrium N2O4 (g)N2O4 (g) 14.1 2NO 2 (g)

5. N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium 14.1

6. constant

7. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will 14.1

8. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P In most cases K c  K p aA (g) + bB (g) cC (g) + dD (g) 14.2 K p = K c (RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

9. 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 x ‘ [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. 14.2

10. 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 14.2

11. Ways Different States of Matter Can Appear in the Equilibrium Constant, K MolarityPartial Pressure gas, (g) YES aqueous, (aq)YES--- liquid, (l)--- solid, (s)---

12. Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The 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. 14.2

13. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7 14.2

14. The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

15. 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] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ‘ General practice not to include units for the equilibrium constant. 14.2

16. At 300 o C Keq = 27.5 for the for the reaction of NO 2(g) and O 2(g). Find Q and predict how the reaction will proceed if [NO 2 ] = 0.15M, [O 2 ] = 0.25M, [NO 3 ] = 0.42M 2NO 2(g) + O 2(g) 2NO 3(g)

17. Predicting Reaction Direction from a Nonequilibrium Mixture Calculate Reaction Quotient, Q The expression for Q is the same as K BUT nonequilibrium concentrations rather than equilibrium concentrations are put into the expression Compare the value of Q to K.

18. The reaction quotient (Q c ) is 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 right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

19. Le Châtelier’s Principle 1.System starts at equilibrium. 2.A change/stress is then made to system at equilibrium. Change in concentration Change in volume Change in pressure Change in Temperature Add Catalyst 3.System responds by shifting to reactant or product side to restore equilibrium.

20. If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5

21. Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

22. Le Chatlier’s Principle Removing SO 2, O 2

23. Le Chatlier’s Principle (cont)

26. Le Châtelier’s Principle Changes in Volume and Pressure (Only a factor with gases) A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5

27. Le Châtelier’s Principle Chemistry; The Science in Context;by Thomas R. Gilbert, Rein V. Kirss, and Geoffrey Davies, Norton Publisher, 2004, p 752

28. Le Châtelier’s Principle Changes in Temperature  Only factor that can change value of K ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases 14.5

29. Le Châtelier’s Principle; Temperature Effect Chemistry; The Science in Context;by Thomas R. Gilbert, Rein V. Kirss, and Geoffrey Davies, Norton Publisher, 2004, p 764

30. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

31. uncatalyzedcatalyzed 14.5 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

32. Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5 Animation of Le Chatelier’s Principle

33. Calculating Equilibrium Concentrations from K and Initial Concentrations 1.Balance chemical equation. 2.Determine Q to see which way reaction will proceed to get to equilibrium 3.Define changes needed to get to equilibrium. Use ‘x’ to represent the change in concentration to get to equilibrium. Determine expression for equilibrium concentration. 4.Write the expression for equilibrium constant, K. 5.Substitute expressions defined in step 3 into the equilibrium constant (step 4). 6.Solve for x and plug value back to table made in step 3. 14.4

34. At 1280 0 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 14.4

35. 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 14.4