For the reaction, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , the degree of dissociation 'x' is related to pressure 'P

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Solution

The correct option is B

\begin{array}{c c c}

\text{} & \text{PCl_5(g)} & \text{ $⇌$ } & \text{ $P C l_{3} (g)$ } & \text{+} & \text{ $P C l_{3} (g)$ }\\

\text{ $I n i . m o l e s$ } & \text{1} & \text{} & \text{0} & \text{} & \text{0} \\

\text{ eq.moles} & \text{(1-x)} & \text{} & \text{x} & \text{} & \text{x=(1+x)} \\

\end{array}

total number of eq.moles = $(1 - x) + x + x = (1 + x)$

eq.Pressure = P

$P_{P C l_{5}} = \frac{(1 - x)}{(1 + x)} \times P; P_{P C l_{5}} = \frac{x}{(1 + x)} \times P; P_{C l_{2}} = \frac{x}{(1 + x)} \times P$

$K = \frac{P_{P C I_{5}} \times P_{C l_{2}}}{P_{P C l_{5}}} = \frac{(\frac{x}{(1 + x)} \times P) (\frac{x}{(1 + x)} \times P)}{(\frac{(1 + x)}{(1 + x)} \times P)} = \frac{x^{2} . P}{(1 - x^{2})}$

$i f x << 1 t h e n (1 - x^{2}) \approx 1); K = x^{2} . P; P α \frac{1}{x^{2}}$

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For the reaction, $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , the degree of dissociation 'x' is related to pressure 'P(x<<1)

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