Question

$2$ mol of $N_2$ is mixed with $6$ mol of $H_2$ in a closed vessel of one litre capacity. If $50\%$ $N_2$ is converted into $N{H}_3$ at equilibrium, the value of $K_c$ for the reaction.
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

• A. $4/27$
• B. $27/4$
• C. $2/27$
• D. $20$

Answer 1

Answer: (A)

$4/27$

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
at $t=02$ mol $6$ mol
at equilibrium $50\%N_2$ is converted into $N{H}_3$
at equation $(2-1)$
$K_C=\frac{[N{H}_3{]}^2}{\left[H_2{]}^3\right[N_2]}=\frac{(2{)}^2}{(3{)}^3\times 1}=\frac{4}{27}$