Question

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

• A. $4/27$
• B. $27/4$
• C. $1/27$
• D. $27$
• E. $9$

Answer 1

The correct option is A $4/27$

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

Since total volume is 1 L, the number of moles is equal to the molar concentration.

$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in closed vessel of one litre capacity.
If $50$% $N_2$ is converted into ${NH}_3$ at equilibrium.

$2\times \frac{50}{100}=1$ mole of $N_2$ will react with
$3\times 1=3$ moles of $H_2$ to form $2\times 1=2$ moles of ${NH}_3$

$2-1=1$ mole of $N_2$ and $6-3=3$ moles of $H_2$ will remain at equilibrium.

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

$K_c=\frac{2^2}{1\times 3^3}$

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