1. Chemical Equilibrium Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 1) Describe chemical equilibrium using the terms forward and reverse reactions and dynamic process. 2) Write the equilibrium constant in terms of the equilibrium concentration of products and reactants and their respective stoichiometric coefficients for both homogenous and heterogeneous equilibria. 3) Derive the relationship between K p and K c. 4) Determine equilibrium constant given equilibrium concentration data. 5) Show that 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) Relate equilibrium constant to rate constants from chemical kinetics. 7) Describe the relationship between reaction quotient and equilibrium constant and predict the direction a reaction will proceed to reach equilibrium. 8) Use the concepts of equilibrium to determine concentration of all species in a solution. 9) Use Le Chatelier’s Principle to describe how changing concentration, volume, pressure, or temperature will shift the reaction so that a equilibrium will be maintained. 10) Describe the effect of a catalyst has on equilibrium concentrations.

3. 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 the concentrations of the reactants and products remain constant Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) 14.1 H 2 O (g) 2NO 2 (g)

9. 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

10. constant

11. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N2O4][N2O4] aA + bB cC + dD K = [C] c [D] d [A] a [B] b Law of Mass Action K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will 14.1

12. 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 [N2O4][N2O4] 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)

13. 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] = 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

14. 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

15. 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

17. 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

18. 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

19. 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 x 0.265 = 0.702 K p = K c (RT)  n K c = K p (RT) -  n  n = 2 – 0 = 2 T = 295 K K c = 0.702 x (0.0821 x 295) -2 = 1.20 x 10 -3 14.2

20. 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 ‘ x 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. 14.2

21. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N2O4][N2O4] 2NO 2 (g) N 2 O 4 (g) K = [N2O4][N2O4] [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. 14.2

22. 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. 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. 14.2

24. 14.3 Chemical Kinetics and Chemical Equilibrium A + 2B AB 2 kfkf krkr rate f = k f [A][B] 2 rate r = k r [AB 2 ] Equilibrium rate f = rate r k f [A][B] 2 = k r [AB 2 ] kfkf krkr [AB 2 ] [A][B] 2 =K c =

25. 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 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 14.4

26. 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. 14.4

27. 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

28. 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 = 0.0638 M 14.4

32. 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) + H 2 (g) NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress 14.5

33. 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

34. Le Châtelier’s Principle Changes in Volume and Pressure 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

36. Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner 14.5

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

39. Le Châtelier’s Principle ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyesno Volumeyesno Temperatureyes Catalystno 14.5