Question

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. Under the conditions, $N{H}_3$ is partially decomposed according to the equation,
$2N{H}_3\leftrightharpoons N_2+3H_2$ .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

The correct option is B $61.3\%$

$2N{H}_3\leftrightharpoons N_2+3H_2$

$\begin{array}{cccc}Initialmole& a& 0& 0\end{array}$

Mole at equilibrium $(a-2x)x3x$

Initial pressure of $N{H}_3$ of a mole = 15 atm ${27}^{\circ }C$

The pressure of 'a' mole of $N{H}_3=patmat{347}^{\circ }C$

$\therefore \frac{15}{300}=\frac{p}{620}$

$\therefore$ p=31 atm

At constant volume and at ${347}^{\circ }C$,mole $\alpha$ pressure

$\therefore \frac{a+2x}{a}=\frac{50}{31}$

$\therefore x=\frac{19}{62}$

$\therefore \%ofN{H}_3$ decomposed = $\frac{2x}{a}\times 100$

$=\frac{2\times 19a}{62\times a}\times 100=61.33\%$