Question

If $\Delta G$ for a reaction is $-8290.8cal$. The $K_c$ for the reaction $300K$ is ${10}^n$. The value of $n$ is :

Answer 1

The relationship between the standard Gibbs free energy change and the equilibrium constant K is $-\Delta G^0=RTlnK$.

Substitute values in the above expression.

$8290.8=2\times 300\times lnK$

$lnK=13.818$

$K={10}^6={10}^n$

Hence, $n=6$