Question

If a mixture of $N_2$ and $H_2$ in the ratio 1 : 3 at 50 atmosphere and ${650}^{o}C$ is allowed to react till equilibrium is reached. Ammonia present at equilibrium was at 25 atm pressure. Calculate the equilibrium constant ($K_P$) for the reaction.
$N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$

• A. $1.677\times {10}^{-3}at{m}^{-2}$
• B. $2.677\times {10}^{-3}at{m}^{-2}$
• C. $0.677\times {10}^{-3}at{m}^{-2}$
• D. $3.677\times {10}^{-3}at{m}^{-2}$

Answer 1

The correct option is A $1.677\times {10}^{-3}at{m}^{-2}$

Equilibrium of formation of $N{H}_3$ may be given as,
$\begin{array}{cccccc}& N_2& +& 3H_2\rightleftharpoons 2N{H}_3& & \\ Att=0& 1& & 3& & 2\\ Atequilibrium& (1-x)& & 3(1-x)& & 2x\end{array}$
Mole fraction of $N{H}_3=\frac{2x}{(4-2x)}$
Partial pressure of $N{H}_3=\frac{2x}{(4-2x)}\times p=\frac{2x}{(4-2x)}\times 50$
$25=\frac{2x\times 50}{(4-2x)}or=0.666$
$K_p=\frac{16x^2(2-x{)}^2}{27p^2(1-x{)}^4}=\frac{16\times \left(0.666{)}^2\right(2-0.666{)}^2}{27\times (50{)}^2\times (0.334{)}^2}=1.677\times {10}^{-3}at{m}^{-2}$