JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

The equilibrium composition for the reaction

${\mathrm{PCl}}_{3}+{\mathrm{Cl}}_{2}\underset{}{\rightleftharpoons }{\mathrm{PCl}}_{5}$ at $298K$ is given below:

${[{\mathrm{PCl}}_{3}]}_{\mathrm{eq}}=0.2\mathrm{mol}{L}^{-1},{[{\mathrm{Cl}}_{2}]}_{\mathrm{eq}}=0.1\mathrm{mol}{L}^{-1},{[{\mathrm{PCl}}_{5}]}_{\mathrm{eq}}=0.40\mathrm{mol}{L}^{-1}$

If $0.2\mathrm{mol}$ of ${\mathrm{Cl}}_{2}$ is added at the same temperature, the equilibrium concentrations of ${\mathrm{PCl}}_{5}$ is _____ $\times {10}^{–2}\mathrm{mol}{L}^{–1}$

Given: ${K}_{c}$ for the reaction at $298K$ is $20$

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

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Solution

$\underset{0.2}{{\mathrm{PCl}}_{3}(g)}+\underset{0.1}{{\mathrm{Cl}}_{2}(g)}\underset{}{\rightleftharpoons }\underset{0.4}{{\mathrm{PCl}}_{5}(g)}$

${K}_{c}=\frac{0.4}{0.2\times 0.1}=20$

If $0.2$ moles of ${\mathrm{Cl}}_{2}$ is added. When reactant is added the reaction takes place in forward direction.

$20={K}_{c}=\frac{0.4+X}{(0.3-X)(0.2-X)}$

$\Rightarrow (0.4+X)=20(0.3-X)(0.2-X)$

$\therefore 0.4+X=20(0.06+{X}^{2}-0.5X)$

$0.4+X=1.2+20{X}^{2}-10X$

$20{X}^{2}-11X+0.8=0$

$X=\frac{11\pm \sqrt{121-64}}{40}$

$=\frac{11-7.55}{40}\simeq 0.08625$

$\therefore ({\mathrm{PCl}}_{5})=0.48625\simeq 48.625\times {10}^{-2}$ or $49\times {10}^{-2}$

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JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

The equilibrium composition for the reaction

${\mathrm{PCl}}_{3}+{\mathrm{Cl}}_{2}\underset{}{\rightleftharpoons }{\mathrm{PCl}}_{5}$ at $298K$ is given below:

${[{\mathrm{PCl}}_{3}]}_{\mathrm{eq}}=0.2\mathrm{mol}{L}^{-1},{[{\mathrm{Cl}}_{2}]}_{\mathrm{eq}}=0.1\mathrm{mol}{L}^{-1},{[{\mathrm{PCl}}_{5}]}_{\mathrm{eq}}=0.40\mathrm{mol}{L}^{-1}$

If $0.2\mathrm{mol}$ of ${\mathrm{Cl}}_{2}$ is added at the same temperature, the equilibrium concentrations of ${\mathrm{PCl}}_{5}$ is _____ $\times {10}^{–2}\mathrm{mol}{L}^{–1}$

Given: ${K}_{c}$ for the reaction at $298K$ is $20$

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\underset{0.2}{{\mathrm{PCl}}_{3}(g)}+\underset{0.1}{{\mathrm{Cl}}_{2}(g)}\underset{}{\rightleftharpoons }\underset{0.4}{{\mathrm{PCl}}_{5}(g)}$

${K}_{c}=\frac{0.4}{0.2\times 0.1}=20$

If $0.2$ moles of ${\mathrm{Cl}}_{2}$ is added. When reactant is added the reaction takes place in forward direction.

$20={K}_{c}=\frac{0.4+X}{(0.3-X)(0.2-X)}$

$\Rightarrow (0.4+X)=20(0.3-X)(0.2-X)$

$\therefore 0.4+X=20(0.06+{X}^{2}-0.5X)$

$0.4+X=1.2+20{X}^{2}-10X$

$20{X}^{2}-11X+0.8=0$

$X=\frac{11\pm \sqrt{121-64}}{40}$

$=\frac{11-7.55}{40}\simeq 0.08625$

$\therefore ({\mathrm{PCl}}_{5})=0.48625\simeq 48.625\times {10}^{-2}$ or $49\times {10}^{-2}$

JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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