Chemistry - Chemical Equilibrium Question with Solution | TestHub

The relationship between $K_P$ and $K_C$ is given by $K_P=K_C(RT{)}^{\Delta n_g}$.

Given $K_P=5.76\times 10^{4}at{m}^2$, $R=0.08L-atm/K-mol$,

 

.From the units of $K_P$ (atm²), $\Delta n_g=2$.

$5.76\times 10^4=K_C(0.08\times 300{)}^2=K_C(24{)}^2=K_C(576)$.

$K_C=\frac{5.76\times 10^4}{576}=100M^2$.