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In a $500 ml$ flask, the degree of dissociation of ${PCl}_{5}$ at equilibrium is $40 %$ and the initial amount is 5 moles. The value of equilibrium constant in mol lit $^{- 1}$ for the decomposition of ${PCl}_{5}$ is

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4.66

2.66

3.33

5.32

answer is B.

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Detailed Solution

${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2}$

At a $o o$

At equilibrium $a (1 - α) a α a α$

$α = \frac{40}{100} = 0.4$

$a = \frac{5}{0.5} = 10 {molL}^{- 1}$

$= \frac{(10 \times 0.4) x (10 x 0.4)}{10 (1 - 0.4)}$

$∴$ Equilibrium constant( $(k) = 2.66 mol L^{- 1}$

Hence the correct answer (B) 2.66

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