Question

16.60

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 1772 torr at 2000 K .

Part A: Calculate the equilibrium partial pressure of CO2 .

Part B: Calculate the equilibrium partial pressure of H2 .

Answer 1

initially,
p(CO) = 1332 torr
= 1332/760 atm
= 1.753 atm

p(H2O) = 1772 torr
= 1772/760 atm
= 2.332 atm

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

initial             1.753               2.332               0                   0                 

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

equilibrium         1.753-1x            2.332-1x            +1x                 +1x               

Equilibrium constant expression is
Kp = p(CO2)*p(H2)/p(CO)*p(H2O)
0.0611 = (1*x)(1*x)/((1.753-1*x)(2.332-1*x))
0.0611 = (1*x^2)/(4.088-4.085*x1*x^2)
0.24978-0.24959*x0.0611*x^2 = 1*x^2
0.24978-0.24959*x-0.9389*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -0.9389
b = -0.2496
c = 0.2498

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

roots are :
x = -0.6656 and x = 0.3997

since x can't be negative, the possible value of x is
x = 0.3997

At equilibrium:
p(CO2) = 0+1x = 0+1*0.3997 = 0.3997 atm
p(H2) = 0+1x = 0+1*0.3997 = 0.3997 atm

Answer:
A)
0.3997 atm

B)
0.3997 atm