NEET UG 2022 — Chemistry Physical Chemistry

2022mcqhard

$3{O}_{2}(g)\rightleftharpoons 2{O}_{3}(g)$

for the above reaction at $298K,{K}_{c}$ is found to be $3.0\times {10}^{-59}$ . If the concentration of ${O}_{2}$ at equilibrium is $0.040M$ then concentration of ${O}_{3}$ in $M$ is

Official previous-year question

Held on 30 Apr 2022 · Verified 9 Jul 2026.

Options

A
$1.9\times {10}^{-63}$
B
$2.4\times {10}^{31}$
C
$1.2\times {10}^{21}$
D
$4.38\times {10}^{-32}$

Did you get this right?

Solution

${K}_{c}=3.0\times {10}^{-59}$

$[{O}_{2}(g)]=4\times {10}^{-2}.$

The given reaction $3{O}_{2}(g)\rightleftharpoons 2{O}_{3}(g)$

${K}_{c}=\frac{productofmolarconcentrati\mathrm{on}ofproduc\mathrm{ts}}{productofmolarconcentrationofreactan\mathrm{ts}}=\frac{{[{O}_{3}(g)]}^{2}}{{[{O}_{2}(g)]}^{3}}$

Put all the given values in the expression of $\mathrm{Kc}$

$3.0\times {10}^{-59}=\frac{{[{O}_{3}(g)]}^{2}}{{[4\times {10}^{-2}]}^{3}}$

$\Rightarrow {[{O}_{3}(g)]}^{2}=3.0\times {10}^{-59}\times {(4\times {10}^{-2})}^{3}$

$\Rightarrow {[{O}_{3}(g)]}^{2}=192\times {10}^{-65}$

Hence, the concentration of ${O}_{3}$ = $4.38\times {10}^{-32}M$

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NEET UG 2022 — Chemistry Physical Chemistry

2022mcqhard

$3{O}_{2}(g)\rightleftharpoons 2{O}_{3}(g)$

for the above reaction at $298K,{K}_{c}$ is found to be $3.0\times {10}^{-59}$ . If the concentration of ${O}_{2}$ at equilibrium is $0.040M$ then concentration of ${O}_{3}$ in $M$ is

Official previous-year question

Held on 30 Apr 2022 · Verified 9 Jul 2026.

Options

A
$1.9\times {10}^{-63}$
B
$2.4\times {10}^{31}$
C
$1.2\times {10}^{21}$
D
$4.38\times {10}^{-32}$

Did you get this right?

Solution

${K}_{c}=3.0\times {10}^{-59}$

$[{O}_{2}(g)]=4\times {10}^{-2}.$

The given reaction $3{O}_{2}(g)\rightleftharpoons 2{O}_{3}(g)$

${K}_{c}=\frac{productofmolarconcentrati\mathrm{on}ofproduc\mathrm{ts}}{productofmolarconcentrationofreactan\mathrm{ts}}=\frac{{[{O}_{3}(g)]}^{2}}{{[{O}_{2}(g)]}^{3}}$

Put all the given values in the expression of $\mathrm{Kc}$

$3.0\times {10}^{-59}=\frac{{[{O}_{3}(g)]}^{2}}{{[4\times {10}^{-2}]}^{3}}$

$\Rightarrow {[{O}_{3}(g)]}^{2}=3.0\times {10}^{-59}\times {(4\times {10}^{-2})}^{3}$

$\Rightarrow {[{O}_{3}(g)]}^{2}=192\times {10}^{-65}$

Hence, the concentration of ${O}_{3}$ = $4.38\times {10}^{-32}M$

Your note

NEET UG Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every NEET UG Physical Chemistry PYQ, year by year.

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NEET UG 2022 — Chemistry Physical Chemistry

Options

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Solution

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NEET UG Physical Chemistry in other years

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NEET UG 2022 — Chemistry Physical Chemistry

Options

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Solution

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NEET UG Physical Chemistry in other years

More NEET UG Physical Chemistry questions

Other Chemistry topics for NEET UG

Organic Chemistry

Inorganic Chemistry

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