The document contains 6 chemistry problems involving gas phase chemical equilibria calculations: 1) Calculating the equilibrium partial pressure of N2O4 and the equilibrium constant Kp for the reaction N2O4 ⇌ 2NO2. 2) Calculating the equilibrium concentrations of H2, Br2, and HBr and the equilibrium constant Kc for the reaction H2 + Br2 ⇌ 2HBr. 3) Calculating the equilibrium constant Kc for the reaction SO2 + Cl2 ⇌ SO2Cl2 given the value of Kp. 4) Calculating the value of Kp for the reverse reaction of 2NO2 ⇌

1. KEY
GENERAL CHEMISTRY-II (1412)
S.I. # 18
1. A flask is charged with 1.500 atm of N2O4(g) and 1.00 atm NO2(g) at 25°C, and
the following equilibrium is achieved:
N2O4 (g)   2 NO2 (g)
After equilibrium is reached, the partial pressure of NO2 is 0.512 atm
(a) What is the equilibrium partial pressure of N2O4?
(b) Calculate the value of Kp for the reaction
N2O4  NO2
I 1.5 1.0
C (1/2)X=(1/2) 1.0-0.512=
(+0.488)= -0.488
+0.244
E 1.5+0.244= 1.744 0.512
Kp= PNO22 = (0.512)2 = 0.1503 = 0.150
PN2O4 (1.744)
2. A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00L vessel at 700K.
These substances react as follows:
H2(g) + Br2(g)  2 HBr(g)
At equilibrium the vessel is found to contain 0.566 g of H2
(a) calculate the equilibrium concentrations of H2, Br2, and HBr.
(b) Calculate Kc
[H2]initial = 1.374g H2 x 1mol H2 x 1 = 0.34079 = 0.341 M
2.0159g H2 2.00L
[Br2] = 70.31 g Br2 x 1 mol Br2 x 1 = 0.21998 = 0.220 M
159.81g Br2 2.00L
[H2]equil = 0.566 g H2 x 1 mol H2 x 1 =0.14038 = 0.140 M
2.0159 g H2 2.00 L
H2 Br2  HBr
I 0.34079 0.21998 0
C X = -0.20041 -0.020041 (2)X
+0.04008
E 0.14038 0.01957 0.40082
Kc = [HBr]2 = (0.40082)2 = (0.401)2 = 58.48 = 58
[H2][Br2] (0.14038)(0.01957) (1.40)(0.020)

2. KEY
3. Calculate Kc at 303K for SO2 (g) + Cl2(g) SO2Cl2(g) if kp = 34.5 at this
temperature.
Kp = Kc(RT)∆n and ∆n = 1-2 = -1 so
34.5 = Kc(RT)-1  34.5 = Kc / RT  Kc = (RT)(34.5)
Kc = (0.0821)(303) = 857.81 = 858
4. Kp = 1.48x104 at 184°C for 2NO(g) + O2(g)  2NO2 (g)
(a) calculate Kp for 2NO2 (g)  2NO(g) + O2 (g)
(b) Doest the equilibrium favor the products or reactants at this temperature?
Kp (forward) = P2NO2 = 1.48x104
P2NOPO2
Kp (reverse) = P2NOPO2 = 1 = 6.67x10-5
P2NO2 1.48x104
Equilibrium will favor the reverse reaction because it takes less to make it happen.
5. An equilibrium mix of CO(g) + 2H2 (g)  CH3OH (g) is in a 2.00 L vessel
which contains 0.0406 mol CH3OH, 0.170 mol CO and 0.302 mol H2 at 500K.
Calculate Kc at this temperature.
[CH3OH] = 0.0406 mol = 0.0203 M
2L
[CO] = 0.17 mol = 0.085 M
2L
[H2] = 0.302 mol = 0.151 M
2L
Kc = [CH3OH] = (0.203) = 10.5
[CO][H2] (0.0850)(0.151)2
6. At 900K Kp = 0.345 for the reaction 2SO2(g) + O2  2 SO3(g) in an
equilibrium mixture the partial pressures of SO2 and O2are 0.165 atm and 0.755
respectively. What is the equilibrium partial pressure of SO3?
Kp = P2SO3 = Pso3 = (Kp x P2SO2 x PO2)1/2 = [(0.345)(0.165)2(0.755)]1/2 = 0.0842 atm
2
P SO2PO2