JEE Main 2018 · Chemistry · Chemical Equilibrium · medium

PCl5(g)rightleftharpoons PCl3(g)+Cl2(g) is 40% dissociated at an equilibrium total pressure of 1 atm. If the total pressure of an equilibrium mixture…

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

40\%

dissociated at an equilibrium total pressure of

1\ \mathrm{atm}

. If the total pressure of an equilibrium mixture starting from pure

\mathrm{PCl_5}

at the same temperature is raised to

4\ \mathrm{atm}

, the degree of dissociation becomes approximately

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Answer: (C) $0.21$

For initial

1

mol

\mathrm{PCl_5}

and degree

\alpha

K_p=\dfrac{\alpha^2P}{1-\alpha^2}

. At

P=1

atm,

\alpha=0.4

K_p=0.16/0.84=0.1905

atm. At

P=4

atm:

\dfrac{\alpha^2}{1-\alpha^2}=\dfrac{0.1905}{4}=0.0476

, so

\alpha^2=0.0476/1.0476=0.04545

and

\alpha=0.213

. Note

\alpha\propto P^{-1/2}

only holds when

\alpha\ll1

(which would give

0.28

); here the

1-\alpha^2

term cannot be dropped.

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