JEE Main 2022 — Chemistry Physical Chemistry

2022mcqhard

For a reaction at equilibrium

$A(g)\rightleftharpoons B(g)+\frac{1}{2}C(g)$

the relation between dissociation constant $(K)$ , degree of dissociation $(\alpha )$ and equilibrium pressure $(p)$ is given by :

Official previous-year question

Held on 24 Jun 2022 · Verified 6 Jul 2026.

Options

A
$K=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{{(2+\alpha )}^{\frac{1}{2}}(1-\alpha )}$
B
$K=\frac{{\alpha }^{\frac{1}{2}}{p}^{\frac{3}{2}}}{{(1+\frac{3}{2}\alpha )}^{\frac{1}{2}}(1-\alpha )}$
C
$K=\frac{{(\alpha p)}^{\frac{3}{2}}}{{(1+\frac{3}{2}\alpha )}^{\frac{1}{2}}(1-\alpha )}$
D
$K=\frac{{(\alpha p)}^{\frac{3}{2}}}{(1+\alpha ){(1-\alpha )}^{\frac{1}{2}}}$

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Solution

$\begin{matrix} & {A}_{(g)} & \rightleftharpoons & {B}_{(g)} & + & \frac{1}{2}{C}_{(g)} \\ \mathrm{at}t=0 & 1 & & 0 & & 0 \\ \mathrm{at}\mathrm{eq}. & 1-\alpha & & \alpha & & \frac{\alpha }{2}\end{matrix}$

total moles at eq. $=1-\alpha +\alpha +\frac{\alpha }{2}=1+\frac{\alpha }{2}$

${P}_{B}=\frac{\alpha }{1+\frac{\alpha }{2}}p{P}_{c}=\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}p{P}_{A}=\frac{1-\alpha }{1+\frac{\alpha }{2}}p$

${K}_{p}=\frac{{P}_{B}\times {P}_{C}^{\frac{1}{2}}}{{P}_{A}}=\frac{(\frac{\alpha }{1+\frac{\alpha }{2}}p){(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}p)}^{\frac{1}{2}}}{(\frac{1-\alpha }{1+\frac{\alpha }{2}}p)}$ ${K}_{p}=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{{2}^{\frac{1}{2}}(1-\alpha ){(1+\frac{\alpha }{2})}^{\frac{1}{2}}} {k}_{p}=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{(1-\alpha ){(2+\alpha )}^{\frac{1}{2}}}$

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Options

$K=\frac{{\alpha }^{\frac{3}{2}}{p}^{\frac{1}{2}}}{{(2+\alpha )}^{\frac{1}{2}}(1-\alpha )}$

$K=\frac{{\alpha }^{\frac{1}{2}}{p}^{\frac{3}{2}}}{{(1+\frac{3}{2}\alpha )}^{\frac{1}{2}}(1-\alpha )}$

$K=\frac{{(\alpha p)}^{\frac{3}{2}}}{{(1+\frac{3}{2}\alpha )}^{\frac{1}{2}}(1-\alpha )}$

$K=\frac{{(\alpha p)}^{\frac{3}{2}}}{(1+\alpha ){(1-\alpha )}^{\frac{1}{2}}}$

Solution

total moles at eq. $=1-\alpha +\alpha +\frac{\alpha }{2}=1+\frac{\alpha }{2}$

${P}_{B}=\frac{\alpha }{1+\frac{\alpha }{2}}p{P}_{c}=\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}p{P}_{A}=\frac{1-\alpha }{1+\frac{\alpha }{2}}p$

JEE Main 2022 — Chemistry Physical Chemistry

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JEE Main 2022 — Chemistry Physical Chemistry

Options

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Solution

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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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