Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 14 1. A mixture of 0.10 mol of NO, 0.050 mol of H2 and 0.10 mol of H2O is placed in a 1.0­L flask and allowed to reach equilibrium as shown below: ˆˆ† N2 (g) + 2 H2O (g) 2 NO (g) + 2 H2 (g) ‡ˆˆ At equilibrium [NO] = 0.062 M. Calculate the equilibrium constant, Kc, for this reaction. ˆˆ† N2 (g) + 2 H2O (g) 2 NO + 2 H2 (g) ‡ˆˆ Initial D Equilibrium K C = 0.10 M – 0.038 0.062 0.050 M – 0.038 0.012 0 + 0.019 0.019 0.10 M + 0.038 0.138 [N 2 ][H 2 O] 2 (0.019)(0.138) 2 = = 650 [NO]2 [H 2 ]2 (0.062)2 (0.012) 2 2. At 700°C, Kc = 20.4 for the reaction shown below: ˆˆ† SO3 (g) (A) SO2 (g) + ½ O2 (g) ‡ˆˆ Calculate Kc and KP for the reaction shown below: ˆˆ† 2 SO3 (g) 2 SO2 (g) + O2 (g) ‡ˆˆ (B) KC (B) = [KC (A)] 2 = (20.4) 2 = 416 KP = KC (RT) Dn = (416) [(0.0821)(973 K)] –1 = 5.21 3. At 100°C, Kc= 0.078 for the following reaction: ˆˆ† SO2 (g) + Cl2 (g) SO2Cl2 (g) ‡ˆˆ In an equilibrium mixture, [SO2Cl2] = 0.136 M and [SO2] = 0.072 M. What is the concentration of Cl2 in the equilibrium mixture? K C = [SO 2 ][Cl 2 ] = 0.078 [SO 2 Cl 2 ] [Cl 2 ] = 0.078 [SO 2Cl 2 ] (0.078)(0.136) = = 0.15 M [SO 2 ]
0.072
1 4. At 373 K, KP= 0.416 for the equilibrium: ˆˆ† 2 NO (g) + Br2 (g) 2 NOBr (g) ‡ˆˆ If the partial pressures of NOBr and NO are equal at equilibrium, what is the partial pressure of Br2? 2 P NO P Br 2 K p = 2 P NOBr = 0.416 since PNOBr = P NO PBr 2 = 0.416 atm 5. At 250°C, the reaction ˆˆ† PCl3 (g) + Cl2 (g) PCl5 (g) ‡ˆˆ has an equilibrium constant Kc= 1.80. If 0.100 mol of PCl5 is added to a 5.00­L flask, what are the concentrations of PCl5, PCl3 and Cl2 at equilibrium at this temperature? [PCl 5 ] = 0.100 mol = 0.0200 M 5.00 L ˆˆ† PCl3 (g) + Cl2 (g) PCl5 (g) ‡ˆˆ Initial D Equilibrium 0.0200 M – x 0.0200 –x 0 + x x [PCl 3 ][Cl 2 ] x 2 K C = = = 1.80 [PCl 5 ]
0.0200 ­x x 2 + 1.80 x ­ 0.0360 = 0 Solving the quadratic equation, ­1.80 ± (1.80)2 ­4(­0.0360) = 0.0198 2 [PCl 3 ]=[Cl 2 ]= 0.0198 M x = [PCl 5 ]= 0.0200 ­ 0.0198 = 0.0002 M
2 0 + x x 6. When 2.00 mol each of hydrogen and iodine are mixed in a 1.00­L flask, 3.50 mol of HI is produced at equilibrium: ˆˆ† 2 HI (g) H2 (g) + I2 (g) ‡ˆˆ Calculate the equilibrium constant Kc for this reaction. ˆˆ† 2 HI (g) H2 (g) + I2 (g) ‡ˆˆ 2.00 M 2.00 M 0 – 1.75 – 1.75 + 3.50 0.25 0.25 3.50 Initial D Equilibrium [HI]2
(3.50) 2 K C = = = 196 [H 2 ] [I 2 ] (0.25) 2 7. The equilibrium constant for the reaction ˆˆ† SO3 (g) + NO (g) SO2 (g) + NO2 (g) ‡ˆˆ has a numerical value of 3.00 at a given temperature. 1.50 mol each of SO2 and NO2 are mixed in a 1.00­L flask and allowed to reach equilibrium. What percent of SO2 is converted to product? ˆˆ† SO3 (g) + NO (g) SO2 (g) + NO2 (g) ‡ˆˆ Initial D Equilibrium 1.50 M – x 1.50 –x K C = 1.50 M – x 1.50 –x [SO 3 ][NO] x 2 = = 3.00 [SO 2 ][NO 2 ] (1.50 ­ x) 2 Taking square root of each side, x = 1.73 1.50 ­ x x + 1.73 x = 2.595 x = 0.951 % = 0 + x x 0.951 x100 = 63.4% 1.50
3 0 + x x 8. The following equilibrium exists at 1000 ºC with KC = 2.00. ˆˆ† CO2 (g) + CF4 (g) 2 COF2 (g) ‡ˆˆ If a 5.00­L mixture contains 0.145 mol COF2 , 0.262 mol of CO2 and 0.074 mol of CF4 at 1000 ºC, in which direction will the mixture proceed to reach equilibrium? 0.145 mol
0.262 mol = 0.0290 M [CO 2 ] = = 0.0524 M 5.00 L
5.00 L 0.074 mol [CF4 ] = = 0.0148 M 5.00 L [CO 2 ][CF4 ] (0.0524)(0.0148) Q c = = = 0.922 [COF2 ]2
(0.0290) 2 [COF2 ] = Since Q c < K c , reaction will proceed in the forward direction 9. Formamide decomposes at high temperatures according to the equation shown below: ˆˆ† NH3 (g) + CO (g) HCONH2 (g) ‡ˆˆ KC = 4.84 at 400 K If 0.186 mol of formamide is placed in a 2.16­L flask and allowed to decompose at 400 K, what will be the total pressure at equilibrium? PHCONH 2 = nRT (0.186)(0.0821)(400 K) = = 2.83 atm V
2.16 L Kp = Kc (RT) Dn = 4.84 [(0.0821)(400)] 1 = 159 ˆˆ† NH3 (g) + CO (g) HCONH2 (g) ‡ˆˆ Initial 2.83 atm 0 0 – x + x + x D Equilibrium 2.83 – x x x K p = PNH 3 P CO PHCONH 2 = (x) 2 = 159 (2.83 ­ x) Solving the quadratic equation, x = 2.78 PHCONH 2 = 2.83 ­ 2.78 = 0.05 atm PNH 3 = PCO = 2.78 atm PTotal = 2.78 + 2.78 + 0.05 = 5.61 atm
4 10. Predict how each of the following changes affect the amount of H2 present in an equilibrium mixture in the reaction ˆˆ† Fe3O4 (s) + 4 H2 (g) 3 Fe (s) + 4 H2O (g) ‡ˆˆ ∆H = −150 kJ a) Raising the temperature of the mixture. Since reaction is exothermic, raising temperature will shift the equilibrium to the left (¬) and reduce amount of hydrogen. b) Adding more H2O (g). Adding more water, will shift the equilibrium to the right (®) and increase the amount of hydrogen. c) Doubling the volume of the container holding the mixture. Increasing the volume of the container will reduce the pressure but the equilibrium will not be affected, and the amount of hydrogen will not change. d) Adding a catalyst. Adding catalyst does not alter the equilibrium and the amount of hydrogen produced. 11. At 2000 °C the equilibrium constant for the reaction below is Kc= 2.4x10 3 . If the initial concentration of NO is 0.500 M, what are the equilibrium concentrations of each substance? ¾¾
® N2 (g) + O2 (g) 2 NO (g) ¬¾
¾ Initial D Equilibrium ˆˆ† N2 (g) + O2 (g) 2 NO (g) ‡ˆˆ 0.500 M 0 0 – 2x + x + x 0.500 –x x x [N 2 ][O 2 ] x 2 = = 2.4x10 3 2
2 [NO]
(0.500 ­ 2x) Taking square root of each side, x = 49 [N 2 ] = [O 2 ] = x = 0.25 M 0.500 ­ 2x x + 98 x = 24.5 [NO] = 0.500 ­ 2x = 0.500 ­ 0.494 K C = x = 0.247 [NO] = 6.0 x 10­3 M
5 12. The reaction below has an equilibrium constant Kc = 6.90. If 0.100 mol of BrCl is placed in a 500­mL flask and allowed to come to equilibrium, what are the equilibrium concentrations of each substance? ¾¾
® 2 BrCl (g) Br2 (g) + Cl2 (g) ¬¾
¾ [BrCl] = 0.100 mol = 0.200 M 0.500 L Br2 (g) Initial D Equilibrium 0 +x x + ˆˆ† 2 BrCl (g) Cl2 (g) ‡ˆˆ 0 0.200 M + x –2x x 0.200 –2x [BrCl]2
(0.200 ­ 2x) 2 K C = = = 6.90 [Br2 ][Cl 2 ]
(x) 2 Taking square root of each side, 0.200 ­2x = 2.63 [B 2 ] = [C 2 ] = x = 0.0432 M x 2.63 x + 2 x = 0.200 [BrCl] = 0.200 ­ 2x = 0.200 ­ 0.0864 x = 0.0432 [BrCl] = 0.114 M
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