Question

#14.60 Consider the following reaction: CO(g)+H2O(g)⇌CO2(g)+H2(g) Kp=0.0611 at 2000 K A reaction mixture initially contains a CO partial pressure of 1332 torr and a H2O partial pressure of 1756 torr at 2000 K.

1) Calculate the equilibrium partial pressure of CO2.

2) Calculate the equilibrium partial pressure of H2.

Answer 1

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

Kp=0.0611 at 2000 K

But,

Kp = [P(products)] / [P(reactants)]

      = [P_CO2 x P_H2] / [P_CO x P_H2O]

Set up an ICE table:

Cmpd...Initial...Change...Equil

CO.......1332......-x........(1332-x)

H₂O.......1756....-x........(1756-x

CO₂.......0..........+x...........x

H₂..........0..........+x..........x

Substitute these Equil values into the Kp expression.

Kp = ((x)*(x))/((1332-x)*(1756-x)) = 0.0611

Multiply this out to obtain a quadratic equation

((x)*(x))/((1332-x)*(1756-x)) = 0.0611

x² = 0.0611 * ((1332-x) * (1756-x))

x² = 0.0611 x (2338992 - 3088x + x²)

x² = 0.0611x² – 188.7x + 142912.4

0.9389x² + 188.7x – 142912.4 = 0

Using the quadratic formula, we solve this for x: The negative root is a negative number which is not realistic and is discarded. The positive root gives

x = 302.4 Torr

From the ICE table, we see that this is the partial pressure for CO₂ at equilibrium:

P(CO₂) = x = 302.4 Torr

P(H₂) = x = 302.4 Torr