Partial Pressure and Initial D Equilibrium

Chemistry 102 REVIEW QUESTIONS Chapter 14

ANSWER KEY

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† 2 NO (g) + 2 H2 (g) ‡ˆˆ N2 (g) + 2 H2O (g) At equilibrium [NO] = 0.062 M. Calculate the equilibrium constant, Kc, for this reaction. ˆˆ† 2 NO + 2 H2 (g) ‡ˆˆ N2 (g) + 2 H2O (g) 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

2 2 [N 2 ][H 2 O] (0.019)(0.138) = = 650 2 [NO]2 [H 2 ]2 (0.062)2 (0.012)

2. At 700°C, Kc = 20.4 for the reaction shown below: ˆˆ† (A) SO2 (g) + ½ O2 (g) ‡ˆˆ SO3 (g) Calculate Kc and KP for the reaction shown below: (B) ˆˆ† 2 SO2 (g) + O2 (g) ‡ˆˆ 2 SO3 (g)

2 2 KC (B) = [KC (A)] = (20.4) = 416 Dn –1 KP = KC (RT) = (416) [(0.0821)(973 K)] = 5.21

3. At 100°C, Kc= 0.078 for the following reaction: ˆˆ† SO2Cl2 (g) ‡ˆˆ SO2 (g) + Cl2 (g) 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 ] 0.078 [SO 2Cl 2 ] (0.078)(0.136) = = 0.15 M [SO 2 ] 0.072

[Cl 2 ] =

1

4. At 373 K, KP= 0.416 for the equilibrium: ˆˆ† 2 NOBr (g) ‡ˆˆ 2 NO (g) + Br2 (g) If the partial pressures of NOBr and NO are equal at equilibrium, what is the partial pressure of Br2?

K p =

P 2 P 2 NO Br P 2 NOBr

= 0.416 since PNOBr = P NO

PBr 2 = 0.416 atm

5. At 250°C, the reaction ˆˆ† PCl5 (g) ‡ˆˆ PCl3 (g) + Cl2 (g) 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† PCl5 (g) ‡ˆˆ PCl3 (g) + Cl2 (g) Initial D Equilibrium 0.0200 M – x