JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

The first order rate constant for the decomposition of ${\mathrm{CaCO}}_{3}$ at $700K$ is $6.36\times {10}^{-3}{s}^{-1}$ and activation energy is $209\mathrm{kJ}{\mathrm{mol}}^{-1}$ . Its rate constant (in ${s}^{-1})$ at $600K$ is $x\times {10}^{-6}$ . The value of $x$ is (Nearest integer)

[Given $R=8.31J{K}^{-1}{\mathrm{mol}}^{-1};\mathrm{log}6.36\times {10}^{-3}=-2.19,{10}^{-4.79}=1.62\times {10}^{-5}]$

Official previous-year question

Held on 27 Aug 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

${K}_{700}=6.36\times {10}^{-3}{s}^{-1}$

${K}_{600}=x\times {10}^{-6}{s}^{-1}$

${E}_{a}=209\mathrm{kJ}/\mathrm{mol}$

Applying :

$\mathrm{log}(\frac{{K}_{{T}_{2}}}{{K}_{{T}_{1}}})=\frac{-{E}_{a}}{2.303R}(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}})$

$\mathrm{log}(\frac{{K}_{700}}{{K}_{600}})=\frac{-{E}_{a}}{2.303R}(\frac{1}{700}-\frac{1}{600})$

$\mathrm{log}(\frac{6.36\times {10}^{-3}}{{K}_{600}})=\frac{+209\times 1000}{2.303\times 8.31}(\frac{100}{700\times 600})$

$\mathrm{log}(6.36\times {10}^{-3})-{\mathrm{logK}}_{600}=2.6$

$\mathrm{log}(6.36\times {10}^{-3})-\mathrm{log}{K}_{600}=2.6$

$\Rightarrow {\mathrm{logK}}_{600}=-2.19-2.6=-4.79$

$\Rightarrow {K}_{600}={10}^{-4.79}=1.62\times {10}^{-5}$

$=16.2\times {10}^{-6}$

$=x\times {10}^{-6}$

$\Rightarrow x=16$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2022 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

[Given $R=8.31J{K}^{-1}{\mathrm{mol}}^{-1};\mathrm{log}6.36\times {10}^{-3}=-2.19,{10}^{-4.79}=1.62\times {10}^{-5}]$

Official previous-year question

Held on 27 Aug 2021 · Verified 6 Jul 2026.

Did you get this right?

Solution

${K}_{700}=6.36\times {10}^{-3}{s}^{-1}$

${K}_{600}=x\times {10}^{-6}{s}^{-1}$

${E}_{a}=209\mathrm{kJ}/\mathrm{mol}$

Applying :

$\mathrm{log}(\frac{{K}_{{T}_{2}}}{{K}_{{T}_{1}}})=\frac{-{E}_{a}}{2.303R}(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}})$

$\mathrm{log}(\frac{{K}_{700}}{{K}_{600}})=\frac{-{E}_{a}}{2.303R}(\frac{1}{700}-\frac{1}{600})$

$\mathrm{log}(\frac{6.36\times {10}^{-3}}{{K}_{600}})=\frac{+209\times 1000}{2.303\times 8.31}(\frac{100}{700\times 600})$

$\mathrm{log}(6.36\times {10}^{-3})-{\mathrm{logK}}_{600}=2.6$

$\mathrm{log}(6.36\times {10}^{-3})-\mathrm{log}{K}_{600}=2.6$

$\Rightarrow {\mathrm{logK}}_{600}=-2.19-2.6=-4.79$

$\Rightarrow {K}_{600}={10}^{-4.79}=1.62\times {10}^{-5}$

$=16.2\times {10}^{-6}$

$=x\times {10}^{-6}$

$\Rightarrow x=16$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2021 — 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

Keep practicing Physical Chemistry

JEE Main 2021 — 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

Keep practicing Physical Chemistry