Chemistry · Chemical Kinetics

JEE Main 2025 — 2 April, Evening Shift — Question 20

For the reaction $A \rightarrow B$ the following graph was obtained. The time required (in seconds) for the concentration of $A$ to reduce to $2.5 \mathrm{~g} \mathrm{~L}^{-1}$ (if the initial concentration of $A$ was $50 \mathrm{~g} \mathrm{~L}^{-1}$ ) is $\_\_\_\_\_$ . (Nearest integer) Given: $\log 2=0.3010$

Answer: 47

Numerical answer — enter this value.

Step-by-step solution

The graph given is for $1^{\text {st }}$ order reaction

$k=\frac{2.303}{t} \log \frac{\left[A_{0}\right]}{\left[A_{t}\right]}$

$k=\frac{2.303}{25} \log \frac{[50]}{[10]}$

$k=\frac{2.303}{25} \log 5$

$=\frac{2.303 \times 0.7}{25} \mathrm{~s}^{-1}$

$=\frac{2.303 \times 0.7}{25}=\frac{2.303}{t} \log \frac{50}{2.5}$

$\frac{0.699}{25}=\frac{1}{t} \log 20$

$\mathrm{t}=\frac{1.301 \times 25}{0.699}$

$=46.53 \mathrm{~s}$

$\approx 47 \mathrm{~s}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws