Ammonia under a pressure of 15 atm at 27∘C is heated to 347∘C in

Discussion Forum : Chemical Equilibrium

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,
$2 N H_{3} ⇋ N_{2} + 3 H_{2}$ .The vessel is such that the volume remains effectively constant whereas pressure increases to
$50 a t m$ . Calculate the percentage of $N H_{3}$ actually decomposed.

Options:

A .

65 %

B .

61.3 %

C .

62.5 %

D .

64 %

Answer: Option B
:
B

$2 N H_{3} ⇋ N_{2} + 3 H_{2}$
$\begin{matrix} I n i t i a l m o l e & a & 0 & 0 \end{matrix}$
Mole at equilibrium $(a - 2 x) x 3 x$
Initial pressure of $N H_{3}$ of a mole = 15 atm $27^{\circ} C$
The pressure of 'a' mole of $N H_{3} = p a t m a t 347^{\circ} C$
$∴ \frac{15}{300} = \frac{p}{620}$
$∴$ p=31 atm
At constant volume and at $347^{\circ} C$ ,mole $α$ pressure
$∴ \frac{a + 2 x}{a} = \frac{50}{31}$
$∴ x = \frac{19}{62}$
$∴ % o f N H_{3}$ decomposed = $\frac{2 x}{a} \times 100$
$= \frac{2 \times 19 a}{62 \times a} \times 100 = 61.33 %$

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