JEE Main 2020 · Chemistry · Chemical Equilibrium · medium

For PCl5(g)rightleftharpoons PCl3(g)+Cl2(g), the degree of dissociation is 0.40 at an equilibrium total pressure of 2 atm (temperature constant). The…

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For

\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}

, the degree of dissociation is

0.40

at an equilibrium total pressure of

2\ \mathrm{atm}

(temperature constant). The degree of dissociation when the equilibrium total pressure is raised to

8\ \mathrm{atm}

is nearest to

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Answer: (C) $0.21$

K_p=\dfrac{\alpha^2P}{1-\alpha^2}=\dfrac{0.16\times2}{0.84}=0.381\ \mathrm{atm}

. At

8

atm:

\dfrac{\alpha^2\cdot8}{1-\alpha^2}=0.381\Rightarrow \alpha^2=\dfrac{0.381}{8.381}=0.04545

, so

\alpha=0.213\approx0.21

. Note

\alpha

does not fall by the naive factor

\sqrt{4}=2

(which would give 0.20) because of the

1-\alpha^2

term.

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