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Boyle's Law

In an evacuat...

Question

In an evacuated vessel of capacity $49.2 L$ , $4$ moles of $A r (g)$ and $5$ moles of $P C l_{5} (g)$ were introduced.
The temperature is maintained at $240 K$ (assume that at this temperature decomposition of $P C l_{5}$ takes place). At equilibrium, the total pressure of the mixture was found to be $4.8 atm$ .
(Given $R = 0.082 {atm L mol}^{- 1} K^{- 1}$ ) Correct statement (s) is/are:

The degree of dissociation of

P C l_{5} (g)

into

P C l_{3} (g)

and

C l_{2} (g)

0.4

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K_{p}

for the reaction

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

2.7 atm

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At equilibrium total moles of gases are

12

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On removing

A r (g)

from the equilibrium mixture at constant pressure and temperature, the extent of dissociation of

P C l_{5}

will increase

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Solution

The correct option is C At equilibrium total moles of gases are $12$
$P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g) t =0 5 t = t_{eq} 5 - x x x n_{T} = 4 + 5 - x + x + x = 9 + x We know, P V = n R T$
Here, $P = 4.8 atm, V = 49.2 L, T = 240 K$
$\Rightarrow n = \frac{4.8 \times 49.2}{0.082 \times 240} ∴ 9 + x = 12 \Rightarrow x = 12 - 9 \Rightarrow x = 3 degree of dissociation, α = \frac{3}{5} = 0.6$

$∴ K_{P} = \frac{\frac{3}{12} \times 4.8 \times \frac{3}{12} \times 4.8}{\frac{2}{12} \times 4.8} = \frac{3}{12} \times \frac{4.8 \times 3}{2} = 1.8$
On removing $A r (g)$ at constant volume will have no effect on equilibrium.

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