Question

Ammonia under a pressure of $15$ atm at $27^{o}C$ is heated to $347^{o}C$ in a closed vessel in the presence of a catalyst. Under the 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 $50atm$. Calculate the percentage of $N{H}_3$ actually decomposed.

• A. $65\%$
• B. $61.3\%$
• C. $62.5\%$
• D. $64\%$

Answer 1

Answer: (B)

Initial pressure of $N{H}_3$ of a mole $=15$ atm at $27°C$
The pressure of ‘a ’ mole of $N{H}_3-p$ atm at $347°C$
$\therefore \frac{15}{300}=\frac{P}{620}$
$\therefore P=3$atm
At constant volume and at $347^{o}C$, mole $\propto$ pressure a $\propto$ 31 (before equilibrium)
$\therefore a+2x\propto 50\left(afterequilibrium\right)$
$\therefore \frac{a+2x}{a}=\frac{50}{31}$
$\therefore x=\frac{19}{62}a$
$\therefore \%ofN{H}_3$ decomposed $=\frac{2x}{a}\times 100$
$=\frac{2\times 19a}{62\times a}\times 100=61.33\%$