Question

the value of Kc = 4.24 at 800K for the reaction

CO(g) + H2O(g) CO2(g) + H2(g)

Calculate the equilibrium concentration of CO2 ,H2O ,CO , H2 at 800K ,if only CO and H2O are pesent initially at concentration of 0.10M each.

Answer 1

For the reaction,

CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g)

Initial concentration:

0.1M

0.1 M

0

0

Let x mole per litre of each of the productbe formed.

At equilibrium:

0.1 - x M

0.1 - x M

x M

x M

where x is the amount of CO2 and H2 at equilibrium. .

Hence, equilibrium constant can be written as,

K_c= x²/(0.1-x)² = 4.24

x² = 4.24(0.01 + x²-0.2x)

x² = 0.0424 + 4.24x2-0.848x

3.24x² – 0.848x + 0.0424 = 0

a = 3.24, b = – 0.848, c = 0.0424

for quadratic equation ax² + bx + c = 0,

Thus solving we get two values of x

x₁= 0.067 x₂= 0.194

Neglecting x₂= 0.194 becuase x could not be more than initial concentration.

Hence the equilibrium concentrations are,

[CO₂] = [H₂] = x = 0.067 M

[CO] = [H₂O] = 0.1 – 0.067 = 0.033 M