JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

For a first order reaction $A\rightarrow B$ , the rate constant, $k=5.5\times {10}^{-14}{s}^{-1}$ . The time required for $67%$ completion of reaction is $x\times {10}^{-1}$ times the half life of reaction. The value of $x$ is Nearest integer) (Given : $\mathrm{log}3=0.4771$ )

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

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Solution

For first order reaction, $t=\frac{2.303}{k}\mathrm{log}[\frac{{A}_{0}}{{A}_{t}}]$

Now, ${t}_{67%}=\frac{2.303}{k}\mathrm{log}[\frac{A}{0.33A}]$

& ${t}_{50}=\frac{2.303}{k}\mathrm{log}[\frac{A}{0.5A}]$

${t}_{67%}=\frac{1}{k}\mathrm{ln}(\frac{1}{1-0.67})=\frac{{t}_{1/2}}{\mathrm{ln}2}\times \mathrm{ln}(\frac{1}{1-\frac{2}{3}})$

$\Rightarrow {t}_{67%}=1.585\times {t}_{1/2}$

$X\times {10}^{-1}=1.585$

$\Rightarrow X=15.85$

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JEE Main 2022 — Chemistry Physical Chemistry

2022integermedium

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

For first order reaction, $t=\frac{2.303}{k}\mathrm{log}[\frac{{A}_{0}}{{A}_{t}}]$

Now, ${t}_{67%}=\frac{2.303}{k}\mathrm{log}[\frac{A}{0.33A}]$

& ${t}_{50}=\frac{2.303}{k}\mathrm{log}[\frac{A}{0.5A}]$

${t}_{67%}=\frac{1}{k}\mathrm{ln}(\frac{1}{1-0.67})=\frac{{t}_{1/2}}{\mathrm{ln}2}\times \mathrm{ln}(\frac{1}{1-\frac{2}{3}})$

$\Rightarrow {t}_{67%}=1.585\times {t}_{1/2}$

$X\times {10}^{-1}=1.585$

$\Rightarrow X=15.85$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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

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JEE Main 2022 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2022 — 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