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South Pasadena • AP Chemistry Name 3 ▪ Chemical Equilibrium Period 3.1 PROBLEMS 1. Write the equilibrium constant expressions for the following reactions. How are they related to one another? (a) 2 N2O (g) + 3 O2 (g) 4 NO2 (g) Keq = [NO2]4eq [N2O]2eq[O2]3eq (b) N2O (g) + 3/2 O2 (g) 2 NO2 (g) [NO2]2eq K′eq = = (Keq)1/2 [N2O] eq[O2]3/2eq (c) 4 NO2 (g) 2 N2O (g) + 3 O2 (g) K″eq = [N2O]2eq[O2]3eq = (Keq)−1 [NO2]4eq 2. Calculate the value of the equilibrium constant for the following system, given the data shown: H2 (g) + CO2 (g) H2O (g) + CO (g) Concentrations at equilibrium: [H2]eq = 1.5 M, [CO2]eq = 2.5 M [H2O]eq = 0.5 M, [CO]eq = 3.0 M Keq = [H2O]eq[CO]eq (0.5)(3.0) = = 0.40 [H2]eq[CO2]eq (1.5)(2.5) 3. Chlorine molecules will dissociate at high temperatures into chlorine atoms. At 3000C, for example, Kc for the equilibrium shown is 0.55. If the partial pressure of chlorine molecules is 1.5 atm, calculate the partial pressure of the chlorine atoms: Cl2 (g) 2 Cl (g) Kp = Kc(R·T)∆n = (0.55)((0.08206)(3273 K))2−1 = 148 Kp = (PCl)2 (PCl2) PCl = 15 atm 148 = (PCl)2 (1.5) – Date EQUILIBRIUM 4. Suppose that 0.50 moles of hydrogen gas, 0.50 moles of iodine gas, and 0.75 moles of hydrogen iodide gas are introduced into a 2.0 liter vessel and the system is allowed to reach equilibrium. H2 (g) + I2 (g) 2 HI (g) Calculate the concentrations of all three substances at equilibrium. At the temperature of the experiment, Kc equals 2.0 × 10−2. Q= [HI]2 (0.375)2 = = 2.25 > Keq [H2][I2] (0.25)(0.25) H2 (g) + I2 (g) 2 HI (g) M Initial 0.25 0.25 0.375 Change +x +x −2x Equil. 0.25+x 0.25+x 0.375−2x Keq = [HI]2eq (0.375−2x)2 = = 2.0 × 10−2 [H2]eq[I2]eq (0.25+x)(0.25+x) x = 0.16 [H2]eq = [I2]eq = 0.41 M; [HI]eq = 0.06 M 5. If the mechanism of a chemical equilibrium consists of two reversible elementary steps, each with its own equilibrium constant Kc1 and Kc2, what expression relates the equilibrium constant Kc for the overall equilibrium to the two constants Kc1 and Kc2? [B] [A] [D] Step 2: C D Kc2 = [C] Overall: A + C B + D [B][D] Kc = = Kc1 · Kc2 [A][C] Step 1: A B Kc1 = 6. When 2.0 mol of carbon disulfide and 4.0 mol of chlorine are placed in a 1.0 liter flask, the following equilibrium system results. At equilibrium, the flask is found to contain 0.30 mol of carbon tetrachloride. What quantities of the other components are present in this equilibrium mixture? CS2 (g) + 3 Cl2 (g) S2Cl2 (g) + CCl4 (g) CS2 (g) + 3 Cl2 (g) S2Cl2 (g) + CCl4 (g) M b. Is the mixture at equilibrium? No. c. In which direction will the system move to reach equilibrium? The reaction will shift to the left, because Qp > Kp. d. When the system reaches equilibrium, what will be the partial pressures of the components in the system? Initial 2.0 4.0 0.0 0.0 Change −0.30 −0.90 +0.30 +0.30 torr Equil. 1.7 3.1 0.30 0.30 Initial 675 43 23 Change +2x −2x −x Equil. 675+2x 43−2x 23−x [CS2]eq = 1.7 M, [Cl2]eq = 3.1 M, [S2Cl2]eq = 0.30 M 7. 3.0 moles each of carbon monoxide, hydrogen, and carbon are placed in a 2.0 L vessel and allowed to come to equilibrium according to the equation: CO (g) + H2 (g) C (s) + H2O (g) If the equilibrium constant at the temperature of the experiment is 4.0, what is the equilibrium concentration of water vapor? CO (g) + H2 (g) C (s) + H2O (g) M Initial 1.5 1.5 0.0 Change −x −x +x Equil. 1.5−x 1.5−x x Kc = [H2O]eq x = = 4.0 [CO]eq[H2]eq (1.5−x)(1.5−x) x = 9 – 12x + 4x2 (PNO)2eq(PCl2)eq (PNOCl)2eq (43 − 2x)2(23 − x) = 0.060 (675 + 2x)2 x=3 PNOCl = 681 torr, PNO = 37 torr, PCl2 = 20 torr 9. Sulfuryl chloride decomposes at high temperatures to produce sulfur dioxide and chlorine gases: SO2Cl2 (g) SO2 (g) + Cl2 (g) At 375C, the equilibrium constant Kc is 0.045. If there are 2.0 grams of sulfuryl chloride, 0.17 gram of sulfur dioxide, and 0.19 gram of chlorine present in a 1.0 L flask, 2.0 g 1 mol = 0.015 M 1.0 L 134.97 g 0.17 g 1 mol [SO2] = = 0.0027 M 1.0 L 64.07 g 0.19 g 1 mol [Cl2] = = 0.0027 M 1.0 L 70.90 g [SO2Cl2] = [H2O]eq = 1.0 M 8. Nitrosyl chloride NOCl decomposes to nitric oxide and chlorine when heated: 2 NOCl (g) 2 NO (g) + Cl2 (g) At 600K, the equilibrium constant Kp is 0.060. In a vessel at 600K, there is a mixture of all three gases. The partial pressure of NOCl is 675 torr, the partial pressure of NO is 43 torr and the partial pressure of chlorine is 23 torr. a. What is the value of the reaction quotient? 2 (PNO) (PCl2) (43 torr) (23 torr) = (PNOCl)2 (675 torr)2 = 0.093 Qp = Kp = a. What is the value of the reaction quotient? x = 1, or 2.25 2 2 NOCl (g) 2 NO (g) + Cl2 (g) Q= [SO2][Cl2] (0.0020)(0.0027) = [SO2Cl2] (0.015) = 4.8 × 10−4 b. Is the system at equilibrium? No. c. In which direction will the system move to reach equilibrium? Q < Kc, so reaction will shift to the right. 10. Ammonium chloride is placed inside a closed vessel where it comes into equilibrium at 400C according to the equation shown. Only these three substances are present inside the vessel. If Kp for the system at 400C is 0.640, what is the pressure inside the vessel? NH4Cl (s) NH3 (g) + HCl (g) 12. When ammonia is dissolved in water, the following equilibrium is established. If the equilibrium constant is 1.8 × 10−5, calculate the hydroxide ion concentration in the solution if 0.100 mol of ammonia is dissolved in sufficient water to make 500 mL of solution. NH3 (aq) + H2O (l) NH4+ (aq) + OH− (aq) atm NH4Cl (s) NH3 (g) + HCl (g) M Initial 0 0 Initial 0.200 0 0 Change +x +x Change −x +x +x Equil. x x Equil. 0.200−x x x Kp = (PNH3)eq(PHCl)eq = (x)(x) = 0.640 Kc = x = 0.80 Ptotal = PNH3 + PHCl = 0.80 atm + 0.80 atm = 1.60 atm 11. Bromine and chlorine react to produce bromine monochloride according to the equation. Kc = 36.0 under the conditions of the experiment. Br2 (g) + Cl2 (g) 2 BrCl (g) If 0.180 moles of bromine gas and 0.180 moles of chlorine gas are introduced into a 3.0 Liter flask and allowed to come to equilibrium, what is the equilibrium concentration of the bromine monochloride? How much BrCl is produced? Br2 (g) + Cl2 (g) 2 BrCl (g) Initial 0.060 0.060 0 Change −x −x +2x Equil. 0.060−x 0.060−x 2 Kc = [NH4+]eq[OH−]eq (x)(x) = = 1.8 × 10−5 [NH3]eq 0.200−x x = 0.0019 M PNH3 = PHCl = 0.80 atm M NH3 (aq) + H2O (l) NH4+ (aq) + OH− (aq) 2 2x [BrCl] eq (2x) = = 36.0 [Br2]eq[Cl2]eq (0.060−x)(0.060−x) 2x = 6.0 0.060 − x x = 0.045 M [BrCl]eq = 0.090 M, nBrCl = 0.27 mol [OH−]eq = 0.0019 M 13. The following reaction is exothermic: Ti (s) + 2 Cl2 (g) TiCl4 (g) List all the ways the yield of the product TiCl4 could be increased. Add Cl2 (g) Remove TiCl4 (g) Increase pressure Lower the temperature AP Chemistry 2004B #1 N2 (g) + 3 H2 (g) 2 NH3 (g) For the reaction represented above, the value of the equilibrium constant, Kp, is 3.1 × 10−4 at 700. K. (a) Write the expression for the equilibrium constant, Kp, for the reaction. Kp = (PNH3)2eq (PN2)eq(PH2)3eq (b) Assume the initial partial pressures of the gases are as follows: PN 2 =0.411 atm, PH 2 =0.903 atm, and PNH 3 =0.224 atm. (i) Calculate the value of the reaction quotient, Q, at these initial conditions. Qp = (PNH3)2 (0.224)2 = 0.166 3= (PN2)(PH2) (0.411)(0.903)3 (ii) Predict the direction in which the reaction will proceed at 700. K if the initial partial pressures are those given above. Justify your answer. The reaction will proceed in the reverse direction (to the left) because Q p < Kp. (c) Calculate the value of the equilibrium constant, Kc, given that the value of Kp for the reaction at 700. K is 3.1 × 10−4. Kc = Kp (3.1 × 10−4) = 1.02 ∆n = (R·T) ((0.08206)(700 K))2−4 (d) The value of Kp for the reaction represented below is 8.3 × 10−3 at 700. K. NH3 (g) + H2S (g) NH4HS (g) Calculate the value of Kp at 700. K for each of the reactions represented below. (i) NH4HS (g) NH3 (g) + H2S (g) K′p = (PNH3)(PH2S) = (Kp)−1 = 120 (PNH4HS) (ii) 2 H2S (g) + N2 (g) + 3 H2 (g) 2 NH4HS (g) (PNH4HS)2 (PNH4HS)2 (PNH3)2 = (K′p)−2·(Kp) 2 3= 2 2· (PH2S) (PN2)(PH2) (PNH3) (PH2S) (PN2)(PH2)3 = (8.3 × 10−3)−2(3.1 × 10−4) = 4.5 K″p = AP Chemistry 1995 #1 H2 (g) + CO2 (g) H2O (g) + CO (g) When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO] = 0.55 mol/L (a) What is the mole fraction of CO(g) in the equilibrium mixture? Assume 1 L XCO = 0.55 mol 0.55 mol = = 0.45 (0.20 mol + 0.30 mol + 0.55 mol + 0.55 mol) 1.60 mol (b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction. Kc = [H2O][CO] (0.55)(0.55) = = 5.0 [H2][CO2] (0.20)(0.30) (c) Determine Kp, in terms of Kc for this system. Kp = Kc(R·T)∆n = (5.0)((0.08206)(2000 K))2−2 = 5.0 (d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature. H2 (g) + CO2 (g) H2O (g) + CO (g) M Initial 0.20 Change +0.165 Equil. Kc = 0.365 0.30 0.55 0.55 +0.165 −0.165 −0.165 0.465 0.385 0.385 [H2O][CO] (0.385)(0.385) = = 0.87 [H2][CO2] (0.365)(0.465) (e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature. H2 (g) + CO2 (g) H2O (g) + CO (g) M Initial 0.17 0.17 0 0 Change −x −x +x +x Equil. 0.17−x 0.17−x x x Kc = [H2O][CO] (x)(x) = = 0.87 [H2][CO2] (0.17−x)(0.17−x) x = 0.081 [CO] = 0.081 M