Question

$2$ moles of $N_2$ is mixed with $6$ moles of $H_2$ in a closed vessel of $1$ liter capacity.

If $50\%$ $N_2$ is converted to ${NH}_3$ at equilibrium, then what is the value of $K_c$ for the below given reaction?

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

• A. $\frac{4}{27}$
• B. $\frac{27}{4}$
• C. $\frac{1}{27}$
• D. $27$

Answer 1

The correct option is A $\frac{4}{27}$

The given reaction is:-

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

Initial moles : $2$ $6$ $0$

At eqm : $2-1=1$ $6-3=3$ $2$ ($\because 50$% of ${NO}_2$ is dissociated)

Now, $K_C=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

$=\frac{2\times 2}{1\times 3\times 3\times 3}=\frac{4}{27}$