Question

2mol 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.
$N_2\left(g\right)3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$ is:

• A. 4/27
• B. 27/4
• C. 2/27
• D. 20

Answer 1

The correct option is A 4/27

$N_2+3H_2\to 2N{H}_2$

$2$ moles of $N_2$ will react with $6$ moles of $H_2$ and $2$ moles of $N_2$ gives $4$ moles of $N{H}_3$ But ($50$ of $N_2$ only reacts)

Hence $1$ mole of $N_2$ gives $2$ moles of $N{H}_3$

$N_2+3H_2\to 2N{H}_2$

Initial $2$ $6$ $0$

Eq $1$ $3$ $2$

$K_c$ $=\frac{[N{H}_3{]}_2}{\left[N_2\right][H_2{]}^3}=\frac{2^2}{3^3}=\frac{4}{27}$