Question

Calculate the equilibrium constant for the following reaction: 2Cl2(g) + 2H2O(g) ⇌ 4HCl(g) + O2(g) Given: 2MgCl2(s) + O2(g) ⇌ 2MgO(s) + 2Cl2(g) K = 8.59 MgCl2(s) + H2O(g) ⇌ MgO(s) + 2HCl(g) K = 8.29

Answer 1

Answer : The equilibrium constant for the final reaction will be, 8.00

Explanation :

The following equilibrium reactions are :

(1) $2MgCl_2(s)+O_2(g)\rightleftharpoons 2MgO(g)+2Cl_2(g)$       $K_1=8.59$

(2) $MgCl_2(s)+H_2O(g)\rightleftharpoons MgO(s)+2HCl(g)$       $K_2=8.29$

The final equilibrium reaction is :

$2Cl_2(g)+2H_2O(l)\rightleftharpoons 4HCl(g)+O_2(g)$         $K=?$

Now we have to calculate the value of $K$ for the final reaction.

First reverse the equation 1 that means we are writing the equilibrium constant as $K_1^{-1}\text{ or }\frac{1}{K_1}$ and multiplying the equation 2 by 2 that means we are writing the equilibrium constant as $(K_2)^2$, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K=\frac{(K_2)^2}{K_1}$

Now put all the given values in this expression, we get :

$K=\frac{(8.29)^2}{8.59}=8.00$

Therefore, the equilibrium constant for the final reaction will be, 8.00