Question

The following concentrations were obtained for the formation of $N{H}_3$ from $N_2$ and $H_2$ at equlibrium at 500K. $\left[N_2\right]$ = 1.5 x ${10}^{-2}M$. $\left[H_2\right]$ = 3.0 x ${10}^{-2}M$ and $\left[N{H}_3\right]$ = 1.2 x ${10}^{-2}M$. Calculate equlibrium constant.

Answer 1

$K_C=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$ $N_2+3H_2\rightleftharpoons 2{NH}_3$

$=\frac{{(1.2\times {10}^{-2})}^2}{(1.5\times {10}^{-2}){(3\times {10}^{-2})}^3}=\frac{1.44\times {10}^{-4}}{1.5\times {10}^{-2}\times 27\times {10}^{-6}}$

$=\frac{1.44}{40.5\times {10}^{-4}}=0.035\times {10}^4$