At $227^{o} C$ one mole of $C l_{2}$ and 3 moles of $P C l_{5}$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $P C l_{5}$ and $K_{p}$ for the reaction $P C l_{5}$ (g) $⇌$ $P C l_{3}$ (g) + $C l_{2}$ (g) is:

x = 0.25, $K_{p}$ = 0.12

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x = 0.64, $K_{p}$ = 2.42

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x = 0.33, $K_{p}$ = 0.41

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None of the above

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Solution

The correct option is C
x = 0.33, $K_{p}$ = 0.41

Let x be the degree of dissociation (we usually use α, here it is given as x)

$\begin{matrix} A t s t a r t & 3 & 0 & 1 P C l_{5} (g) & ⇋ & P C l_{3} (g) & + & C l_{2} (g) A t e q u i l i b r i u m & 3 (1 - x) & 3 x & 1 + 3 x \end{matrix}$

$∴ Total number of moles at equilibrium$

$= 3 (1 - x) + 3 x + 1 + 3 x = 3 (1 + x) + 1$
From ideal gas equation,
PV = nRT;
We use P = 2.05 atm, T = 500K,
V = 100 litres and
R = 0.0821 Latm / (Kmol)
We get n = 5 = number of moles at equilibrium = 3(1+x) + 1
Solving we get x = $\frac{1}{3}$ = 0.3333 $\approx$ 0.33
We can't rule out option d here since there is no mention of $K_{p}$
Now let us try and derive the expression of $K_{p}$ in terms of x and equilibrium pressure P.

$K_{p} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$

$K_{p} = \frac{[\frac{3 x P}{3 (1 + x) + 1}] [\frac{(1 + 3 x) P}{3 (1 + x) + 1}]}{[\frac{3 (1 - x)}{3 (1 + x) + 1} P]}$

= $\frac{3 x (3 x + 1)}{4 + 3 x} \times \frac{1 P}{3 (1 - x)}$

= $\frac{(3 x^{2} + x) \times P}{(4 + 3 x) (1 - x)} = \frac{x (3 x + 1) \times P}{(4 + 3 x) (1 - x)}$

Substituting the values of x and P, we obtain $K_{p}$ = 0.42atm

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Similar questions

At $227^{o} C$ one mole of $C l_{2}$ and 3 moles of $P C l_{5}$ are placed in a 100 litre container and allowed to reach equilibrium. It is found that the equilibrium pressure is 2.05 atmospheres. Assuming ideal behaviour, the degree of dissociation 'x' for $P C l_{5}$ and $K_{p}$ for the reaction $P C l_{5}$ (g) $⇌$ $P C l_{3}$ (g) + $C l_{2}$ (g) is:

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