Two moles of PCl5 were heated in a closed vessel of 2 L. At equilibrium 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is

PCl5a(1-x) ⇌PCl3ax+Cl2axa=2,x=0.4,V=2L∴PCl5=2(1−0.4)2=0.6molL−1PCl3=Cl2=2×0.42=0.4molL−1∴Kc=0.4×0.406=0.267

Two moles of PCl5 were heated in a closed vessel of 2 L. At equilibrium 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is

PCl5a(1-x) ⇌PCl3ax+Cl2axa=2,x=0.4,V=2L∴PCl5=2(1−0.4)2=0.6molL−1PCl3=Cl2=2×0.42=0.4molL−1∴Kc=0.4×0.406=0.267

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Two moles of PCl₅ were heated in a closed vessel of 2 L. At equilibrium 40% of PCl₅ is dissociated into PCl₃ and Cl₂. The value of equilibrium constant is

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0.53

0.267

2.63

5.3

answer is B.

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Detailed Solution

$\underset{a (1 - x)}{{PCl}_{5}} ⇌ \underset{ax}{{PCl}_{3}} + \underset{ax}{{Cl}_{2}}$
$\begin{array}{l} a = 2, x = 0 .4, V = 2 L \\ ∴ [{PCl}_{5}] = \frac{2 (1 - 0 .4)}{2} = 0 .6 {molL}^{- 1} \\ [{PCl}_{3}] = [{Cl}_{2}] = \frac{2 \times 0 .4}{2} = 0 .4 {molL}^{- 1} \\ ∴ K_{c} = \frac{0 .4 \times 0 .4}{06} = 0 .267 \end{array}$