Question

Consider the reaction:

CO(g)+H2O(g)⇌CO2(g)+H2(g)

K=0.118 at 4000 K

A reaction mixture initially contains a CO partial pressure of 1361 mbar and a H2O partial pressure of 1781 mbar at 4000 K.

Part A: Calculate the equilibrium partial pressure of CO in bar using three decimal places

Part B: Calculate the equilibrium partial pressure of H2O in bar using three decimal places.

Part C: Calculate the equilibrium partial pressure of CO2 in bar using three decimal places

Part D: Calculate the equilibrium partial pressure of H2 in bar using three decimal places

Answer 1

Initial pressure of CO = 1361 mbar = 1.361 bar

Initial pressure of H2O = 1781 mbar = 1.781 bar

ICE Table:

Equilibrium constant expression is

Kp = p(CO2)*p(H2)/p(CO)*p(H2O)

0.118 = (1*x)(1*x)/((1.361-1*x)(1.781-1*x))

0.118 = (1*x^2)/(2.424-3.142*x + 1*x^2)

0.286-0.3708*x + 0.118*x^2 = 1*x^2

0.286-0.3708*x-0.882*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -0.882

b = -0.3708

c = 0.286

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.147

roots are :

x = -0.8172 and x = 0.3968

since x can't be negative, the possible value of x is

x = 0.3968

At equilibrium:

p(CO) = 1.361 - x = 1.361 - 0.3968 = 0.964 bar

p(H2O) = 1.781 - x = 1.781 - 0.3968 = 1.38 bar

p(CO2) = x = 0.3968 bar

p(H2) = x = 0.3968 bar

A)

0.964 bar

B)

1.38 bar

C)

0.397 bar

D)

0.397 bar