Question

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

• A. $61\%$
• B. $71\%$
• C. $81\%$
• D. $91\%$

Answer 1

The correct option is A $61\%$

The pressure at $347{}^{\circ }\text{C}$ can be calculated by using the following formula,
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$ (at constant volume)
$\frac{15}{27+273}=\frac{P_2}{347+273}$

$\Rightarrow P_2=\frac{15\times 620}{300}=31\text{atm}$
Thus, the pressure at $347{}^{\circ }\text{C}$ is $31\text{atm}.$
Now, $\begin{array}{cccccc}2N{H}_3& \rightleftharpoons & N_2& +& 3H_2& \\ 1& & 0& & 0& \text{(Initially)}\\ 1-x& & \frac{x}{2}& & \frac{3x}{2}& \text{(At Equilibrium)}\end{array}$
$\therefore$ Total number of moles after dissociation $=1-x+\frac{x}{2}+\frac{3x}{2}=1+x$
$P_2=31\text{atm}$ (Before dissociation)
$\therefore P_2\propto 1$
($\because P\propto n$ according to the general gas eqn: $PV=nRT$)
$P_3=50\text{atm}$ (After dissociation)
$\therefore P_3\propto 1+x$
Hence, $\frac{1}{1+x}=\frac{31}{50}$ or,
$1+x=\frac{50}{31}$, or
$x=\frac{19}{31}=0.61$
Hence, the percentage of ammonia actually decomposed is $61\%$.