JEE Main 2020 — Chemistry Physical Chemistry

2020mcqeasy

A flask contains a mixture of compounds $A$ and $B$ . Both compounds decompose by first-order kinetics. The half-lives for $A$ and $B$ are $300s$ and $180s$ , respectively. If the concentrations of $A$ and $B$ are equal initially, the time required for the concentration of A to be four times that of $B$ (in s) is : $(\mathrm{Use}\mathrm{ln}2=0.693)$

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

A
$180$
B
$900$
C
$300$
D
$120$

Did you get this right?

Solution

${C}_{t}={C}_{0}{e}^{-\mathrm{kt}}$

${k}_{A}=\frac{\mathrm{ln}2}{300}$ $(k=\frac{\mathrm{ln}2}{{T}_{1/2}})$

${k}_{B}=\frac{\mathrm{ln}2}{180}$

${({C}_{t})}_{A}={({C}_{0})}_{A}{e}^{-{k}_{A}t}$

${({C}_{t})}_{B}={({C}_{0})}_{B}{e}^{-{k}_{B}t}$

$\frac{{({C}_{t})}_{,B}}{({C}_{t}),A}=\frac{{({C}_{0})}_{,B}}{({C}_{0}),A}\times {e}^{({k}_{A}-{k}_{B})t}$

$\frac{1}{4}={e}^{({k}_{A}-{k}_{B})t}$

$\mathrm{ln}\frac{1}{4}=[\frac{\mathrm{ln}2}{300}-\frac{\mathrm{ln}2}{180}]t$

$-2\mathrm{ln}2=[\frac{\mathrm{ln}2}{300}-\frac{\mathrm{ln}2}{180}]t$

$2=(\frac{120}{180\times 300})t$

$t=\frac{2\times 180\times 300}{120}=900\mathrm{sec}$

Your note

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JEE Main 2020 — Chemistry Physical Chemistry

2020mcqeasy

Official previous-year question

Held on 5 Sept 2020 · Verified 6 Jul 2026.

Options

A
$180$
B
$900$
C
$300$
D
$120$

Did you get this right?

Solution

${C}_{t}={C}_{0}{e}^{-\mathrm{kt}}$

${k}_{A}=\frac{\mathrm{ln}2}{300}$ $(k=\frac{\mathrm{ln}2}{{T}_{1/2}})$

${k}_{B}=\frac{\mathrm{ln}2}{180}$

${({C}_{t})}_{A}={({C}_{0})}_{A}{e}^{-{k}_{A}t}$

${({C}_{t})}_{B}={({C}_{0})}_{B}{e}^{-{k}_{B}t}$

$\frac{{({C}_{t})}_{,B}}{({C}_{t}),A}=\frac{{({C}_{0})}_{,B}}{({C}_{0}),A}\times {e}^{({k}_{A}-{k}_{B})t}$

$\frac{1}{4}={e}^{({k}_{A}-{k}_{B})t}$

$\mathrm{ln}\frac{1}{4}=[\frac{\mathrm{ln}2}{300}-\frac{\mathrm{ln}2}{180}]t$

$-2\mathrm{ln}2=[\frac{\mathrm{ln}2}{300}-\frac{\mathrm{ln}2}{180}]t$

$2=(\frac{120}{180\times 300})t$

$t=\frac{2\times 180\times 300}{120}=900\mathrm{sec}$

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 2020 — Chemistry Physical Chemistry

Options

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2020 — Chemistry Physical Chemistry

Options

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