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 1390 torr and a H2O partial pressure of 1710 torr at 2000 K.

Calculate the equilibrium partial pressure of CO2.

Answer 1

ICE Table:

                    p(CO)               p(H2O)              p(CO2)              p(H2)             

initial             1390.0              1710.0              0                   0                 

change              -1x                 -1x                 +1x                 +1x               

equilibrium         1390.0-1x           1710.0-1x           +1x                 +1x               

Equilibrium constant expression is
Kp = p(CO2)*p(H2)/p(CO)*p(H2O)
0.0611 = (1*x)(1*x)/((1390-1*x)(1710-1*x))
0.0611 = (1*x^2)/(2376900-3100*x + 1*x^2)
145228.59-189.41*x + 0.0611*x^2 = 1*x^2
145228.59-189.41*x-0.9389*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -0.9389
b = -1.894*10^2
c = 1.452*10^5

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 = 5.813*10^5

roots are :
x = -5.069*10^2 and x = 3.052*10^2

since x can't be negative, the possible value of x is
x = 3.052*10^2

At equilibrium:
p(CO2) = x = 305.2 torr

Answer: 305 torr