Question

Ammonia kept under pressure of 15 atm at 27⁰C is heated to 347⁰C in a closed vessel in the presence of catalyst. Under these conditions, NH₃ 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 50 atm. Calculate the percentage of NH₃ actually decomposed.

• A. 61.3%
• B. 63.5%
• C. 65.3%
• D. 66.6%

Answer 1

The correct option is A 61.3%

$\begin{array}{cccccc}{\text{2NH}}_3& \rightleftharpoons & N_2+3H_2& & & \\ \text{Initial moles}& a& & 0& 0& \\ \text{At equilibrium}& (a-2x)& x& x& & \end{array}$

Initial pressure of ${NH}_3$ if ${}^{\prime }{a}^{\prime }$ mole $=15$ atm at ${27}^{\circ }C$. The pressure of $a^{\prime }$ mole of ${NH}_3=p$ atm at ${347}^{\circ }C$

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

$\therefore p=31atm$

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

$a\propto 31$
$\therefore (a-2x)\propto 50$
$\therefore (a-2x)50$
$a31$

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

$\therefore \%$ of ${NH}_3$ decomposed $=\frac{2x}{x}\times 100$
$=\frac{2\times 19a}{62\times a}\times 100=61.29\%$