Question

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 1344 torr and a H2O partial pressure of 1780 torr at 2000 K. calculate the equilibrium partial pressure of CO2. calculate the same for H2.

Answer 1

Apply Equilibrium law.

Kp = P-CO2 * P-H2 /(P-CO * P-H2O)

Kp = 0.0611

P-CO = 1344

P- H2O = 1780

P-CO2 = 0

P-H2 = 0

find equilibrium pressures

P-CO = 1344 - x

P- H2O = 1780 - x

P-CO2 = 0+x

P-H2 = 0 + x

substitute in K

Kp = P-CO2 * P-H2 /(P-CO * P-H2O)

0.0611 = x*x /(1344 - x)(1780 - x)

solve for x

0.0611 *(1344 *1780 - (1344 +1780 )x + x^2) = x^2

146170.752 - 190.8764x + 0.0611 x^2 = x^2

(1-0.0611 )x^2 +190.8764x - 146170.752 = 0

0.9389x^2 + 190.8764x - 146170.752 = 0

x = 305.8

P-CO = 1344 - 305.8 = 1038.2

P- H2O = 1780 - 305.8 = 1474.2

P-CO2 = 0+x = 305.8

P-H2 = 0 + x = 305.8

Proof

Q = 305.8*305.8/(1038.2*1474.2) = 0.0611

so this is correct