Question

Consider the following reaction: CO(g) + H2O(g) = CO2(g) + H2(g) K = 0.0611 at 2000 K A reaction mixture initially contains a CO partial pressure of 1360 torr and a H2O partial pressure of 1770 torr at 2000 K.

1. Calculate the equilibrium partial pressure of CO2.

2. Calculate the equilibrium partial pressure of H2.

Answer 1

CO +     H2O ------> CO2 + H2 , Kp = 0.0611

1360 1770 0 0 (at t=0)

1360-x 1770-x x x (at equilibrium)  

Kp = x * x/ (1360-x)(1770-x)

0.0611 * (1360-x) * (1770-x) = x^2

0.0611 * (2407200 - 1360x - 1770x + x^2) = x^2

0.0611 * (2407200 - 3130x + x^2) = x^2

147079.92 - 191.243x + 0.0611x^2 = x^2

0.9389x^2 + 191.243x - 147079.92 = 0

After solving above quadratic equation;

x = 306.731

Hence, at equilibrium;

P(CO2) = P(H2) = x = 306.731 torr

P(CO) = 1360 - x = 1360 - 306.731 = 1053.269 torr

P(H2O) = 1770 - x = 1770 - 306.71 = 1463.269 torr

Let me know if any doubts.