CHEMICAL EQUILIBRIUM Chapter 13

1 CHEMICAL EQUILIBRIUM Chapter 13 Pb 2+ (aq) + 2 Cl (aq) PbCl 2 (s) 1 Objectives Briefly review what we know of equilibrium Define the Equilibrium Constant (K eq ) and Reaction Quotient (Q) Determining the K eq Using K eq in calculation Equations and Manipulating K eq Disturbing a Chemical Equilibrium and Le Châtelier s Principle 2 Ch. 13.1 The Equilibrium Condition Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction Pink to blue Co(H 2 O) 6 Cl 2 Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O Co(H 2 O) 6 Cl 2 3 Chemical Equilibrium Fe 3+ + SCN - FeSCN 2+ + Fe(H 2 O) 3+ 6 + SCN - Fe(SCN)(H 2 O) 2+ 5 + H 2 O 4 Chemical Equilibrium Fe 3+ + SCN - FeSCN 2+ 5 Examples of Chemical Equilibria 6 Phase changes such as H 2 O(s) H 2 O(liq) After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.

2 7 8 Examples of Chemical Equilibria Chemical Equilibria Formation of stalactites and stalagmites CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3- (aq) CaCO 3 (s) + H 2 O(liq) + CO 2 (g) Ca 2+ (aq) + 2 HCO 3- (aq) At a given T and P of CO 2, [Ca 2+ ] and [HCO 3- ] can be found from the EQUILIBRIUM CONSTANT, K eq Ch. 13.2 The Equilibrium Constant N 2 (g) + 3H 2 (g) 2NH 3 (g) Reactant conc. declines and then becomes constant at equilibrium 9 The Equilibrium Constant Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. Rate forward = Rate backward 10 Product conc. increases and then becomes constant at equilibrium The Equilibrium Constant 11 THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT at a given T 12 If K (aka K c ) is known, then we can predict concs. of products or reactants.

3 Practice 1: Write the equilibrium expression (K) for the following reactions: 13 Practice 2 Consider an equilibrium mixture in a closed vessel reacting according to the equation: 14 a) 2O 3 (g) 3O 2 (g) b) 2NO(g) + Cl 2 (g) 2NOCl(g) c) Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) d) H 2 (g) + I 2 (g) 2HI(g) e) Cd 2+ (aq) + 4Br - (aq) CdBr 4 2- (aq) H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g)to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. Determining K When you know the equilibrium concentrations of the reactants and products, the K can be determined by substituting the values into the equilibrium constant expression. Example: In an experiment of the reaction at 600 C: 2SO 2 (g) + O 2 (g) 2SO 3 (g) The equilibrium concentrations are: [SO 2 ] = 1.50 M [O 2 ] = 1.25 M [SO 3 ] = 3.50 M 15 In an experiment of the reaction at 600 C, 2SO 2 (g) + O 2 (g) 2SO 3 (g) Determine K: Determining K.... 16 Writing and Manipulating K Expressions Changing coefficients 2SO 2 (g) + O 2 (g) 2SO 3 (g) K = 4SO 2 (g) + 2O 2 (g) 4SO 3 (g) K new = K new = K new = (4.36) 2 = 19.0 Note: K new can also be noted as K (K prime) 17 Writing and Manipulating K Expressions Changing direction 2SO 2 (g) + O 2 (g) 2SO 3 (g) 2SO 3 (g) 2SO 2 (g) + O 2 (g) K new = = K new = 1/4.36 = 0.229 K = K new = = 1/K old 18

4 Writing and Manipulating K Expressions How do these manipulations affect the equilibrium concentration values? Find the equilibrium [O 2 ] using the equation 4SO 2 (g) + 2O 2 (g) 4SO 3 (g) K= 19.0 if [SO 2 ] = 1.50 M and [SO 3 ] = 3.50 M [O 2 ] = =... = 1.25 M (Same as before!) 19 Writing and Manipulating K Expressions K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. 20 Writing and Manipulating K Expressions Equilibrium position is a set of equilibrium concentrations. 2SO 2 (g) + O 2 (g) 2SO 3 (g) Experiment 2 Equil [ ] s 0.590 M 0.0450 M 0.260 M. = 4.32.. 21 Manipulating K Practice Problem Use the data below to find the equilibrium concentration of SO 2 if the equilibrium concentrations of SO 3 and O 2 are: [SO 3 ] = 2.25 M [O 2 ] = 1.50 M Use a reaction and K from at least two different equations. See slides 16 to 18. 22 Ch. 13.3 Equilibrium Expressions Involving Pressures Concentration Units 23 Convert between K c and K p General gas reaction aa + bb cc + dd 24 We have been writing K in terms of mol/l. These are designated by K or K c But with gases, P = (n/v) RT = conc RT P is proportional to concentration, so we can write K in terms of P. These are designated by K p. K c and K p may or may not be the same. K p =K c (RT) (c+d-a-b) Also thought of as n, or change in number of moles of gas. R=0.0821L-atm/mol K K c

5 Writing and Manipulating K Expressions K using concentration and pressure units K p = K c (RT) n For S(s) + O 2 (g) SO 2 (g) n = 0 and K p = K c For SO 2 (g) + 1/2 O 2 (g) SO 3 (g) n = 1/2 and K p = K c (RT) 1/2 25 Example with K p At 500 K, the following equilibrium is established: 2NO(g) + Cl 2 (g) 2NOCl(g). 0.095 0.171 0.28 An equilibrium mixture of the three gases has a partial pressures of 0.095 atm, 0.171 atm and 0.28 atm for NO, Cl 2 and NOCl, respectively. Calculate the K p for this reaction at 500 K. K p = =... = 51 26 Practice with K p A mixture of NO, H 2, and H 2 O are placed in a 1.0 L vessel at 400K. The following equilibrium pressures are established: 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g). Equil (atm) 1.53 0.300 0.465 3.39 Find Kp. Convert pressures to molarity and determine Kc. Compare with Kp to Kc conversion 27 Ch. 13.4 Heterogeneous Equilibria Pure solids and liquids NEVER appear in equilibrium expressions. S(s) + O 2 (g) SO 2 (g) 28 Heterogeneous Equilibria 29 What Are the Equilibrium Expressions for These Reactions? 30 Pure solids and liquids NEVER appear in equilibrium expressions. NH 3 (aq) + H 2 O(liq) NH 4+ (aq) + OH - (aq) SnO 2 (s) + 2CO(g) Sn(s) + 2CO 2 (g) CaCO 3 (s) CaO(s) + CO 2 (g) Zn(s) + Cu 2+ (aq) Cu(s) + Zn 2+ (aq)

6 Writing and Manipulating K Expressions When adding equations for reactions S(s) + O 2 (g) SO 2 (g) SO 2 (g) + 1/2 O 2 (g) SO 3 (g) The net equation is S(s) + 3/2 O 2 (g) SO 3 (g) And then multiply the individual K s: 31 Manipulating Equilibrium Constants Practice Determine the K eq for the following reaction: 2NH 3 (g) + 3I 2 (g) 6HI(g) + N 2 (g) 3*(H 2 (g) + I 2 (g) 2HI(g)) K 1 = (5.40 x 10 1 ) 3-1(N 2 (g) + 3H 2 (g) 2NH 3 (g)) K 2 = (1.04 x 10-4 ) -1 3H 2 (g) + 3I 2 (g) 6HI(g) K 1 = 1.57 x 10 5 2NH 3 (g) N 2 (g) + 3H 2 (g) K 2 = 9.62 x 10 3 2NH 3 (g) + 3I 2 (g) 6HI(g) + N 2 (g) Keq = K 1 * K 2 = 1.57x10 5 (9.62x10 3 ) = 1.51 x 10 9 32 Ch. 13.5 Applications of the Equilibrium Constant 1. Can tell if a reaction is product-favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 33 The Meaning of K For AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10-5 34 A large value of K means the concentration of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored. Conc. of products is much less than that of reactants at equilibrium. The reaction with small K is strongly reactant-favored. Ag + (aq) + Cl - (aq) AgCl(s) is product-favored. Product- or Reactant Favored 35 The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION 36 QUOTIENT, Q. a A + b B c C + d D Product-favored K > 1 Reactant-favored K < 1 If Q = K, then system is at equilibrium.

7 Reaction Quotient & Equilibrium Constant 37 Reaction Quotient & Equilibrium Constant 38 Equilibrium achieved At any point in the reaction H 2 + I 2 2 HI In the equilibrium region The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. 39 The Meaning of K 40 If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction shift to approach equilibrium? If Q = K, 41 If Q > K, 42 the system is at equilibrium. there is too much product and the equilibrium shifts to the left.

8 If Q < K, there is too much reactant, and the equilibrium shifts to the right. 43 The Meaning of K All reacting chemical systems are characterized by their REACTION QUOTIENT, Q. 44 If Q = K, then system is at equilibrium. Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become and [n] must. Practice Problem Ethyl acetate is synthesized in a nonreacting solvent (not water) according to the reaction: CH 3 CO 2 H + C 2 H 5 OH CH 3 CO 2 C 2 H 5 + H 2 O K = 2.2 acetic acid ethanol ethyl acetate In the following mixtures, will the concentration of H 2 O increase, decrease or remain the same as equilibrium is established? a) [CH 3 CO 2 C 2 H 5 ]=0.22 M; [H 2 O]=0.10 M, [CH 3 CO 2 H]= 0.010 M, [C 2 H 5 OH]=0.010 M b) [CH 3 CO 2 C 2 H 5 ]=0.22 M; [H 2 O]=0.0020 M, [CH 3 CO 2 H]= 0.0020 M, [C 2 H 5 OH]=0.10 M c) [CH 3 CO 2 C 2 H 5 ]=0.88 M; [H 2 O]=0.12 M, [CH 3 CO 2 H]= 0.044 M, [C 2 H 5 OH]=6.0 M 45 Determining K More typically, you will have to determine the equilibrium concentrations: 2 NOCl(g) 2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. Calculate K. Solution Set of an ICE table of concentrations [NOCl] [NO] [Cl 2 ] Initial 2.00 0 0 Change Equilibrium 0.66 46 Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/l of NO. Calculate K. Solution Set of an ICE table of concentrations [NOCl] [NO] [Cl 2 ] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 47 Determining K 2 NOCl(g) 2 NO(g) + Cl 2 (g) [NOCl] [NO] [Cl 2 ] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33 48

9 Determine K Practice A solution of 0.050 M diiodocyclohexane, C 6 H 10 I 2 is prepared in CCl 4. When the solution has come to equilibrium, the concentration of I 2 is 0.035 M. C 6 H 10 I 2 (sol) C 6 H 10 (sol) + I 2 (sol) 1) Determine the equilibrium concentrations of C 6 H 10 I 2 and C 6 H 10 at equilibrium. 2) Calculate the equilibrium constant, K c. 49 Ch. 13.6 Solving Equilibrium Problems PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g) 2 HI(g) 50 H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. 51 H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. 52 [H 2 ] [I 2 ] [HI] [H 2 ] [I 2 ] [HI] Initial 1.00 1.00 0 Initial 1.00 1.00 0 Change Change -x -x +2x Equilib Equilib 1.00-x 1.00-x 2x where x is defined as am t of H 2 and I 2 consumed on approaching equilibrium. H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 2. Put equilibrium concentrations into K c expression. 53 H 2 (g) + I 2 (g) 2 HI(g) K c = 55.3 Step 3. Solve K c expression - take square root of both sides. 54 x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

10 Practice Problem Consider the following reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 moles of SO 3 were found to be present. Calculate the equilibrium concentrations of O 2 and SO 2 and K c. 55 Practice Problem #1 2SO 2 (g) + O 2 (g) 2SO 3 (g) [SO 2 ] [O 2 ] [SO 3 ] Initial 2.00 1.50 3.00 Change -0.50-0.25 +0.50 Equilibrium 1.50 1.25 3.50 K 3.50 2 c = =4.36 1.25(1.50) 2 56 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) 57 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ] [NO 2 ] Initial 0.50 0 Change Equilib 58 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) 59 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Step 2. Substitute into K c expression and solve. 60 If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N 2 O 4 ] [NO 2 ] Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029-0.0059x = 4x 2 4x 2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029

11 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4 b = 0.0059 c = -0.0029 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 61 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion: x = 0.026 M [N 2 O 4 ] = 0.50 - x = 0.47 M [NO 2 ] = 2x = 0.052 M 62 Solving Quadratic Equations Recommend you solve the equation exactly on a calculator or use the quadratic equation when Kc<<1. Otherwise, assume x is small relative to initial conc (C>100*K) 63 Solving Quadratic Equations Molecular Iodine can dissociate into atomic iodine as shown by the reaction: I 2 (g) 2I(g) The Keq for this reaction is: K eq = [I] 2 = 5.6 x 10-12 [I 2 ] If the initial concentration of I 2 is 0.45 M, what are the equilibrium concentrations? 64 Example I 2 (g) 2I(g) [I 2 ] [I] Initial 0.45 0 Change -x 2x Equilibrium 0.45 x 2x 65 Example I 2 (g) 2I(g) [I 2 ] [I] Initial 0.45 0 Change -x 2x Equilibrium 0.45 x 2x 66 (2x) 2 K eq = =5.6 x 10-12 (0.45 - x) X is going to be very small as compared to 0.45 (0.45>>100*5.6x10-12 ) Which means we can ignore x in the denominator and rewrite as: K (2x) 2 eq = =5.6 x 10 (0.45) -12 x = 7.9 x 10-7 So [I] = 2x = 2(7.9x10-7 ) or 1.6x10-6 M

12 Practice Problem #2 At 35 C, K = 1.6 x 10-5 for the reaction 2NOCl(g) 2NO(g) + Cl 2 (g) Calculate the concentration of all species at equilibrium for each of the following initial mixtures: a) 2.0 mol pure NOCl in a 2.0 L flask b) 1.0 mol of NOCl and 1.0 mol NO in a 1.0 L flask c) 2.0 mol NOCl and 1.0 mol Cl 2 in a 1.0 L flask. 67 Ch. 13.7 Le Châtelier s Principle Temperature, catalysts, and changes in concentration affect equilibria. The outcome is governed by LE CHATELIER S PRINCIPLE...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance. 68 EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Châtelier 1850-1936 Studied mining engineering. Interested in glass and ceramics. 69 EQUILIBRIUM AND EXTERNAL EFFECTS Temperature change change in K Consider the fizz in a soft drink CO 2 (aq) + HEAT CO 2 (g) + H 2 O(liq) K = P (CO 2 ) / [CO 2 ] Increase T. What happens to equilibrium position? To value of K? K increases as T goes up because P(CO 2 ) increases and [CO 2 ] decreases. Decrease T. Now what? Equilibrium shifts left and K decreases. 70 71 72 Temperature Effects on Equilibrium Temperature Effects on Equilibrium N 2 O 4 (colorless) + heat 2 NO 2 (brown) H o = + 57.2 kj K c (273 K) = 0.00077 K c (298 K) = 0.0059

13 EQUILIBRIUM AND EXTERNAL EFFECTS 73 EQUILIBRIUM AND EXTERNAL EFFECTS 74 Concentration changes no change in K only the equilibrium composition changes. Catalytic exhaust system Add catalyst no change in K A catalyst only affects the RATE of approach to equilibrium. 75 76 Le Chatelier s Principle Le Chatelier s Principle Adding a reactant to a chemical system. Removing a reactant from a chemical system. 77 78 Le Chatelier s Principle Le Chatelier s Principle Adding a product to a chemical system. Removing a product from a chemical system.

14 butane Butane- Isobutane Equilibrium 79 Butane Isobutane 80 isobutane At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5. Add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? Butane Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 Solution Calculate Q immediately after adding more butane and compare with K. Q is LESS THAN K. Therefore, the reaction will shift to the. 81 Butane Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right away from butane and toward isobutane. Set up ICE table [butane] [isobutane] Initial 0.50 + 1.50 1.25 Change -x + x Equilibrium 2.00 - x 1.25 + x 82 Butane Isobutane You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [isobutane] = 2.32 M. Equilibrium has shifted toward isobutane. 83 Le Chatelier s Principle Change T change in K therefore change in P or concentrations at equilibrium Use a catalyst: reaction goes to equilibrium more quickly. K not changed. Add or take away reactant or product: K does not change Reaction adjusts to new equilibrium position 84

15 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) 85 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) 86 Increase P in the system by reducing the volume (at constant T). Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO 2 decreases and P of N 2 O 4 increases. 87 The Meaning of K K comes from thermodynamics. (See Chapter 17.7 17.9) G < 0: reaction is product favored G > 0: reaction is reactant-favored 88 If K > 1, then ΔG is negative If K < 1, then ΔG is positive Thermodynamics and K eq (Ch. 17.7 17.9) Thermodynamics and K eq 89 90 FACT: r G o is the change in free energy when pure reactants convert COMPLETELY to pure products. The equilibrium point occurs at the lowest value of free energy available to the reaction system. FACT: Product-favored systems have K eq > 1. Therefore, both r G and K eq are related to reaction favorability. ΔG = 0 = ΔG + RT ln(k) ΔG = RT ln(k) Copyright 2009 Brooks/Cole Cengage Learning. - Cengage All rights reserved 90

16 Change in Free Energy to Reach Equilibrium 91 Thermodynamics and K eq K eq is related to reaction favorability and so to r G o. The larger the value of K the more negative the value of r G o r G o = - RT ln(k) where R = 8.31 J/K mol 92 Copyright 2009 Brooks/Cole Cengage Learning. - Cengage All rights reserved 91 Calculate K for the reaction N 2 O 4 2 NO 2 r G o = +4.8 kj r G o = +4800 J = - (8.31 J/K)(298 K) ln K Thermodynamics and K eq r G o = - RT ln(k) K = 0.14 When r G o > 0, then K < 1 93 ΔG and K eq Practice Determine the value of r G and K eq for the reaction at 25.0 C given the f G for the reactants and product. Is reaction product or reactant favored? C(s) + CO 2 (g) 2CO(g) f G (kj/mol) 0-394 -137 r G = Σ f G (product) - Σ f G (reactants) r G = 2(-137) (-394) = 120 kj 120,000 J = -8.314 J/(mol*K)*298 ln K Ln K = 120,000/(-8.314*298) = -48.4 K = 9.23 x 10-22 not spont; reactant fav d 94 AP Exam Practice 2010B AP Exam #1 2008 AP Exam #1 2007B AP Exam #1 2004B AP Exam #1 2003B AP Exam #1 95