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

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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