Question

The value of Kc = 4.24 at 800 Kfor the reaction CO(g) + H2O (g) gives rise to CO2(g) + H2(g) Calculate equilibrium concentrations of CO2, H2, CO and H2O at 800 K , if only CO and H2O are present initially at concentration of 0.10 M 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,

Kc= 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