Answer : The equilibrium constant for the final reaction will be, 8.00
Explanation :
The following equilibrium reactions are :
(1) [tex]2MgCl_2(s)+O_2(g)\rightleftharpoons 2MgO(g)+2Cl_2(g)[/tex] [tex]K_1=8.59[/tex]
(2) [tex]MgCl_2(s)+H_2O(g)\rightleftharpoons MgO(s)+2HCl(g)[/tex] [tex]K_2=8.29[/tex]
The final equilibrium reaction is :
[tex]2Cl_2(g)+2H_2O(l)\rightleftharpoons 4HCl(g)+O_2(g)[/tex] [tex]K=?[/tex]
Now we have to calculate the value of [tex]K[/tex] for the final reaction.
First reverse the equation 1 that means we are writing the equilibrium constant as [tex]K_1^{-1}\text{ or }\frac{1}{K_1}[/tex] and multiplying the equation 2 by 2 that means we are writing the equilibrium constant as [tex](K_2)^2[/tex], we get the final equilibrium reaction and the expression of final equilibrium constant is:
[tex]K=\frac{(K_2)^2}{K_1}[/tex]
Now put all the given values in this expression, we get :
[tex]K=\frac{(8.29)^2}{8.59}=8.00[/tex]
Therefore, the equilibrium constant for the final reaction will be, 8.00
