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 convened into $N{H}_3$ at equilibriumm, the value $K_c$ for the reaction is:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

• A. 4/27
• B. 27/4
• C. 2/27
• D. 24

Answer 1

The correct option is A 4/27

Initial $2$ mole $N_2$ $6$ mol $H_2$

At $equatio{n}^m$

$2-\left(0.5\right)\times \frac{1}{100}\to N_2$

$6-\left(0.5\right)\times \frac{1}{100}\to H_2$

$N{H}_3=\frac{2(0.5\times 1)}{100}$

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

$K_c=\frac{4}{27}$