Chemical Equilibria Lecture Notes Examples

Chemical EquilibriaLecture Notes ExamplesEx 1Write the equilibrium expression (K c ) for each of the following:(a) 2 H 2(g) + O 2(g) ↔ 2 H 2 O (g)K C = [H 2 O] 2 / [H 2 ] 2 [O 2 ](b) 2 C 6 H 6(g) + 15 O 2(g) ↔ 12 CO 2(g) + 6 H 2 O (g)K C = [CO 2 ] 12 [H 2 O] 6 / [C 6 H 6 ] 2 [O 2 ] 15Ex 2 For the equilibrium 2 SO 3(g) ↔ 2 SO 2(g) + O 2(g) at a temperature of 1000 K,K c has the value 4.08 × 10 –3 . Calculate the value for K p .K P = K C (RT) ∆n = (4.08 × 10 –3 ) (RT) 3–2 = 0.335Ex 3The equilibrium constant for the reaction H 2(g) + I 2(g) ↔ 2 HI (g) varies with temperature inthe following way: K c = 794 @ 298 K; K c = 54 @ 700 K. Is the formation of HI favoredmore at higher or lower temperatures?K C decreases as temperature increases. Since K C is the ratio of products toreactants, when K C gets smaller the amount of product is getting smaller. So theformation of HI is favored at lower temperatures.Ex 4Write the equilibrium expressions for K c and K p for the reactionK C = [H 2 ] 4 / [H 2 O] 43 Fe (s) + 4 H 2 O (g) ↔ Fe 3 O 4(s) + 4 H 2(g)K P = K C (RT) 4-4 = K CEx 5An experiment is run at 425°C to determine the equilibrium constant for the reactionH 2(g) + I 2(g) ↔ 2 HI (g)at that temperature. H 2(g) and I 2(g), both at the same initial partial pressure of5.73 atm, are mixed. When the reaction comes to equilibrium, the partial pressure ofHI is measured to be 9.00 atm. Compute the equilibrium constant, assuming that noside reactions occur.H 2(g) + I 2(g) ↔ 2 HI (g)Initial 5.73 5.73 0Change –x –x +2xEquilibrium 5.73 –x 5.73–x 9.00x = 4.50 atm, so P H2 = P I2 = 5.73 atm – 4.50 atm = 1.23 atmK P = (P HI ) 2 / (P H2 )(P I2 ) = (9.00) 2 / (1.23) 2 = 53.5

Ex 6 The reaction 2 NO (g) + Br 2(g) ↔ 2 NOBr (g)has an equilibrium constant of 116.6 at 25°C. In each of the following mixtures,compute the reaction quotient Q and use it to decide the direction the reaction takes(right to left) to come to equilibrium. The initial partial pressures (all in atm) in thegas mixture are(a) P NO = 0.188; P Br 2 = 0.300; P NOBr = 1.203Q = (P NOBr ) 2 / (P NO ) 2 (P Br2 ) = (1.203) 2 / (0.188) 2 (0.300) = 136Q > K so the reaction will proceed to the left (toward reactants) to reachequilibrium.(b) P NO = 0.300; P Br 2 = 0.188; P NOBr = 1.203Q = (P NOBr ) 2 / (P NO ) 2 (P Br2 ) = (1.203) 2 / (0.300) 2 (0.188) = 85.5Q < K so the reaction will proceed to the right (toward products) to reachequilibrium.Ex 7 The progress of the reaction H 2(g) + Br 2(g) ↔ 2 HBr (g)can be monitored visually by following changes in the color of the reaction mixture(Br 2 is reddish brown, and H 2 and HBr are colorless). A gas mixture is prepared at700 K, in which 0.40 atm is the initial partial pressure of both H 2 and Br 2 and0.90 atm is the initial partial pressure of HBr. The color of this mixture fades as thereaction progresses toward equilibrium. What condition must K satisfy (for example,it must be greater than or smaller than a given number)?Q = (P HBr ) 2 / (P H2 ) (P Br2 ) = (0.90) 2 / (0.40) (0.40) = 5.1As the reaction moves toward equilibrium the color of the mixture fades.This tells us that the reaction is moving to the right toward the formation ofmore colorless product. If the reaction is not yet at equilibrium and weneed to move to the right to make more products, then Q < K. Therefore, Kmust be greater than 5.1.Ex 8Consider the following equilibrium2 BrCl (g) ↔ Br 2(g) + Cl 2(g) K = 32 @ 500KCalculate the partial pressure of BrCl that is in equilibrium with a mixture of0.182 atm of bromine and 0.0876 atm of chlorine at 500K.K P = 32 = (P Br2 ) (P Cl2 ) / (P BrCl ) 2 = (0.182) (0.0876) / (P BrCl ) 2(P BrCl ) 2 = 4.98 x 10 –4P BrCl = 0.022 atmEx 9Given the following exothermic reaction:2 SO 2(g) + O 2(g) ↔ 2 SO 3(g)Which direction will the equilibrium shift when the following changes are made?(a) Add more O 2 ? Right(b) Remove some SO 3 ? Right(c) Remove some SO 2 ? Left(d) Increase the temperature? Left(e) Decrease the volume (constant T)? Right

Some problems to try on your own:1. The value of the equilibrium constant for the following reaction is K 1 .3 H 2(g) + N 2(g) ↔ 2 NH 3(g)How is K 1 related to the equilibrium constant K 2 for the related equilibrium?2 NH 3(g) ↔ 3 H 2(g) + N 2(g)K 2 = 1/K 12. The equilibrium constant (K p ) for the reaction3/2 Fe (s) + 2 H 2 O (g) ↔ 1/2 Fe 3 O 4(s) + 2 H 2(g)is 3.51 at 527°C. 5.42 g of Fe and 7.88 g of H 2 O are placed in a closed, 5.0-L container andallowed to reach equilibrium. Calculate the pressure of H 2(g) at equilibrium.mol H 2 O = (7.88 g)(1 mol / 18.0152 g) = 0.4374 molP H2O = (0.4374 mol)R(800 K)/5.0 L = 5.74 atmIgnore the solids3/2 Fe (s) + 2 H 2 O (g) ↔ 1/2 Fe 3 O 4(s) + 2 H 2(g)Initial --- 5.74 --- 0Change --- –2x --- +2xEquilibrium --- 5.74–2x --- 2xK P = 3.51 = (P H2 ) 2 / (P H2O ) 2 = (2x) 2 / (5.74–2x) 23.51 = 2x/(5.74 – 2x)x = 2.23P H2 = 2x = 2(2.23) = 4.46 atm3. Find the equilibrium partial pressure of N 2 O 4(g) and NO 2(g) in the reactionN 2 O 4(g) ↔ 2 NO 2(g) K = 0.98 @ 298Kif a flask is charged with enough N 2 O 4 to exert an initial partial pressure of 0.050 atm.N 2 O 4(g) ↔ 2 NO 2(g)Initial 0.050 0Change –x +2xEquilibrium 0.050 – x 2xK P = 0.98 = (P NO2 ) 2 / (P N2O4 ) = (2x) 2 / (0.050–x)0.98 = (2x) 2 / (0.050–x)4x 2 + 0.98x – 0.049 = 0x = –0.288, 0.0426 The negative root is impossible so x = 0.0426P N2O4 = 0.050 – 0.0426 = 0.007 atmP NO2 = 2(0.0426) = 0.085 atm

4. For the reaction(a)SO 2 Cl 2 (g) ↔ SO 2(g) + Cl 2(g) K = 2.40 @ 100°CInitial partial pressure of SO 2 Cl 2 in a tank with rigid walls is 1.398 atm, no other species arepresent. Calculate the Q. Does total pressure increase or decrease as the rxn →equilibrium?Q = 0 The total pressure will increase as the reaction approaches equilibrium.(b)Calculate the partial pressures of SO 2 Cl 2 , SO 2 , and Cl 2 after equilibrium is reached.SO 2 Cl 2 (g) ↔ SO 2(g) + Cl 2(g)Initial 1.398 0 0Change –x +x +xEquilibrium 1.398 – x x xK P = 2.40 = (P SO2 )(P Cl2 ) / (P SO2Cl2 ) = (x) 2 / (1.398 – x)2.40 = (x) 2 / (1.398 – x)x 2 + 2.40x – 3.3552 = 0x = –3.39, 0.990 Only the positive will work.P SO2Cl2 = 1.398 – 0.990 = 0.408 atmP SO2 = P SO2 = 0.990 atm5. 0.500 atm of H 2 and 0.500 atm of I 2 are placed in a container. The reaction between the two toform HI is allowed to come to equilibrium. What are the equilibrium partial pressures of allcompounds? H 2(g) + I 2(g) ↔ 2 HI (g) K = 54 @ 700KH 2 (g) + I 2(g) ↔ 2 HI (g)Initial 0.500 0.500 0Change –x –x +2xEquilibrium 0.500 – x 0.500 – x 2xK P = 54 = (P HI ) 2 / (P H2 )(P I2 ) = (2x) 2 / (0.500 – x) 254 = 4x 2 / (0.500 – x) 2x 2 + 2.40x – 3.3552 = 0x = 0.687, 0.393 Only the smaller number will work.P HI = 2(0.393) = 0.786 atmP H2 = P I2 = 0.500 – 0.393 = 0.107 atm

6. Given the following data:N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) K @25°C = 3.5 × 10 8N 2(g) + O 2(g) ↔ 2 NO (g) K @25°C = 1.7 × 10 –32 NO 2(g) ↔ N 2 O 4(g) K @25°C = 1.7 × 10 2Arrange the following reactions in order from most reactant-favored to most product-favored.(1) 2 NH 3(g) ↔ N 2(g) + 3 H 2(g) K = 2.9 × 10 –101, 3, 2(2) 2 NO (g) ↔ N 2(g) + O 2(g) K = 588(3) 2 NO 2(g) ↔ N 2 O 4(g) K = 1.7 × 10 2