Question

value of kc =4.24 at 800k for reaction CO(g)+H2O(g) CO2(g) +H2(g).calculate equili conc of CO2 CO and H2O if initial conc of CO and H2O are 0.01M each.

Answer 1

using the general formula for kc and comparing with given value

Answer 2

Answer:

Equilibrium concentration of CO = [CO] = x = 0.00675 M

Equilibrium concentration of $H_2O= [H_2O]$ = x = 0.00675 M

Explanation:

Equilibrium constant of given reaction $K_c= 4.24$

 $CO(g)+H_2O(g)\rightleftharpoons CO_2(g) +H_2(g)$

Initial

 0.01 M       0.01 M

At equilibrium:

(0.01 M -x)   (0.01 M -x)       x      x

An equilibrium expression of given reaction is given as:

$K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_c=\frac{x\times x}{(0.01 M -x)\times (0.01 M -x) }$

$4.24=(\frac{x^2}{(0.01 M -x) })^2$

$\sqrt{4.24}=\frac{x^2}{(0.01 M -x)}$

$2.06=frac{x}{0.01 M-x}$

On solving we get:

x = 0.00675 M

Equilibrium concentration of CO = [CO] = x = 0.00675 M

Equilibrium concentration of $H_2O= [H_2O]$ = x = 0.00675 M