Question

One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If $0.25$ percent of nitrogen is converted to ammonia by the following reaction
$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$
then calculate the equilibrium constant, $K_c$ in concentration units. What will be the value of $K_c$ for the following equilibrium
$\frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g)\rightleftharpoons 2N{H}_3(g)$

Answer 1

$\left[N_2\right]=\frac{0.75}{4},\left[H_2\right]=\frac{2.25}{4},\left[N{H}_3\right]=\frac{0.50}{4}$
$K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}=\frac{(0.50{)}^2}{(0.75)(2.25{)}^3}\times 16$
$=0.468L^{2}mo{l}^{-2}$
Also for: $\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$
$K_c^{\prime }=\sqrt{K_c}=0.68$