Answer : The value of [tex]K'_c[/tex] for the reaction is, [tex]1.9\times 10^{19}[/tex]
Explanation :
The given equilibrium reactions are:
(1) [tex]CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g)[/tex] [tex]K_c=4.4\times 10^9[/tex]
(2) [tex]2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)[/tex] [tex]K'_c=?[/tex]
As we know that, when we are multiplying the equation (1) by 2 then we get the equation (2). Thus, the relation between the two equilibrium constant will be,
[tex]K'_c=(K_c)^2[/tex]
So,
[tex]K'_c=(4.4\times 10^9)^2[/tex]
[tex]K'_c=1.9\times 10^{19}[/tex]
Therefore, the value of [tex]K'_c[/tex] for the reaction is, [tex]1.9\times 10^{19}[/tex]
