Question

At 25°C, the equilibrium partial pressures of NO2 and N2O4 are 0.150 atm and 0.200 atm, respectively. If the volume is increased by 1.60 fold at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

PNO2 = atm

PN2O4 = atm

Answer 1

From the initial conditions, you can calculate Kp as:

Kp = P(N2O4) / P(NO2)^2 = 0.200 / (0.150)^2 = 8.89

After the volume expansion by 1.6 fold, the reaction quotient Q, is
Q = (0.125)/0.0938)^2 = 14.2

Since Q>K, the reaction will shift to the left in order to reach equilibrium. So, at equilibrium,
P(N2O4) = 0.125 - x and P(NO2) = 0.0938 + 2x

At the new equilibrium, then,

Kp = 8.89 = (0.125 - x) / (0.0938 + 2x)^2

x = 0.00997 atm

PN2O4 = 0.125 - 0.00997 = 0.115 atm
PNO2 = 0.00938 + (2*0.00997) = 0.029 atm