Phosphorus pentachloride dissociates as follows in a closed reaction vessel :PCl5( g)⇌PCl3( g)+Cl2( g)If total pressure at equilibrium of the reaction mixture is P and degree of dissociation PC15 is x, the partial pressure of PC13 will be

Phosphorus pentachloride dissociates as follows in a closed reaction vessel : ${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation $PC 1_{5}$ is x, the partial pressure of $PC 1_{3}$ will be

A
$(\frac{x}{1 - x}) P$
B
$(\frac{x}{x + 1}) P$
C
$(\frac{2 x}{1 - x}) P$
D
$(\frac{x}{x - 1}) P$

Solution:

$\begin{array}{l} {PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2} \\ n (1 - x) & n x n x; & n_{total} = n (1 + x) \end{array}$

$p_{{PCl}_{3}} = \{\frac{n x}{n (1 + x)}\} P = \frac{x}{1 + x} P$

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