JEE Main 2025 — Chemistry Physical Chemistry

2025mcqeasy

In a reaction $A+B \rightarrow C$ , initial concentrations of $A$ and $B$ are related as $[A]_0=8[B]_0$ . The half lives of $A$ and $B$ are 10 min and 40 min. respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

Official previous-year question

Held on 3 Apr 2025 · Verified 6 Jul 2026.

Options

A
60 min
B
80 min
C
20 min
D
40 min

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Solution

Given : $[\mathrm{A}]_0=8[\mathrm{~B}]_0$

$\left[\mathrm{t}_{1 / 2}\right]_{\mathrm{A}}=10 \mathrm{~min}$ .

$\left[\mathrm{t}_{1 / 2}\right]_{\mathrm{B}}=40 \mathrm{~min}$ .

$\mathrm{I}^{\text {st }}$ order kinetics

$\mathrm{t}=$ ?

$[\mathrm{A}]_{\mathrm{t}}=[\mathrm{B}]_{\mathrm{t}}$

$-\mathrm{k}_{\mathrm{A}} \times \mathrm{t} \quad-\mathrm{k}_{\mathrm{B}} \times \mathrm{t}$

$\Rightarrow \quad[\mathrm{A}]_0 \mathrm{e}=[\mathrm{B}]_0 \mathrm{e}$

$\Rightarrow \quad \frac{[\mathrm{A}]_0}{[\mathrm{~B}]_0}=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \mathrm{t}}$

$\Rightarrow \quad 8=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}}$

$\Rightarrow \quad \ell \mathrm{n} 8=\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}$

$\Rightarrow \quad \ln 8=\ln 2\left(\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{A}}}-\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{B}}}\right) \times \mathrm{t}$

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Solution

Given : $[\mathrm{A}]_0=8[\mathrm{~B}]_0$

$\left[\mathrm{t}_{1 / 2}\right]_{\mathrm{A}}=10 \mathrm{~min}$ .

$\left[\mathrm{t}_{1 / 2}\right]_{\mathrm{B}}=40 \mathrm{~min}$ .

$\mathrm{I}^{\text {st }}$ order kinetics

$\mathrm{t}=$ ?

$[\mathrm{A}]_{\mathrm{t}}=[\mathrm{B}]_{\mathrm{t}}$

$-\mathrm{k}_{\mathrm{A}} \times \mathrm{t} \quad-\mathrm{k}_{\mathrm{B}} \times \mathrm{t}$

$\Rightarrow \quad[\mathrm{A}]_0 \mathrm{e}=[\mathrm{B}]_0 \mathrm{e}$

$\Rightarrow \quad \frac{[\mathrm{A}]_0}{[\mathrm{~B}]_0}=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \mathrm{t}}$

$\Rightarrow \quad 8=\mathrm{e}^{\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}}$

$\Rightarrow \quad \ell \mathrm{n} 8=\left(\mathrm{k}_{\mathrm{A}}-\mathrm{k}_{\mathrm{B}}\right) \times \mathrm{t}$

$\Rightarrow \quad \ln 8=\ln 2\left(\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{A}}}-\frac{1}{\left(\operatorname{ta}_2\right)_{\mathrm{B}}}\right) \times \mathrm{t}$

JEE Main 2025 — Chemistry Physical Chemistry

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JEE Main 2025 — Chemistry Physical Chemistry

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Solution

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