Question

The following concentrations were obtained for the formation of $N{H}_3$ and $N_2$ and $H_2$ at equilibrium at $500K$.

$N_2=1.5\times {10}^{-2}M$, $H_2=3.0\times {10}^{-2}M$ and $N{H}_3=1.2\times {10}^{-2}M$.

Calculate equilibrium constant.

Answer 1

$N_2+3H_2\to 2N{H}_3$

$K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$

Given;

$\left[N_2\right]=1.5\times {10}^{-2}$ M

$\left[H_2\right]=3.0\times {10}^{-2}$ M

$[N{H}_3=1.2\times {10}^{-2}]$ M

$K_c=\frac{(1.2\times {10}^{-2}{)}^2}{1.5\times {10}^{-2}\times (3\times {10}^{-2}{)}^3}=3.55\times {10}^{-2}$