Question

Ammonia at $15$ atm and ${27}^{\circ }C$ is heated to ${347}^{\circ }C$ in a closed vessel in the presence of a catalyst. Under the condition, $N{H}_3$ is partially decomposed according to the equation, $2N{H}_3\iff N_2+3H_2$. The vessel is such that the volume remains effectively constant whereas pressure increases to $50$ atm. Calculate the percentage of $N{H}_3$ actually decomposed.

• A. $81.4\%$
• B. $69\%$
• C. $61.3\%$
• D. $71.1\%$

Answer 1

Answer: (C)

$61.3\%$

Given that ammonia under a pressure of $15$ atm at ${27}^{\circ }C$ is heated to ${347}^{\circ }C$ in a closed vessel in the presence of a catalyst.
$\frac{P_1}{P_2}=\frac{T_1}{T_2}$
Therefore, $P_2=\frac{T_2}{T_1}\times P_1$
Therefore, $P_2=31$ atm.

The reaction is as follows:
$2N{H}_3\iff N_2+3H_2$
Initial partial $31$ $0$ $0$
pressures:
At equilibrium: $31(1-\alpha )\frac{31\alpha }{2}\frac{31\times 3\alpha }{2}$
Total pressure at equilibrium $=31$ ($1+$ $\alpha$) $=50$ atm.
Therefore, $\alpha$ $=0.613$ or $61.3\%$.