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Chemistry · Chemical Kinetics

JEE Main 2026 — 4 April, Evening Shift — Question 65

For a first order reaction A→B\mathrm{A} \rightarrow \mathrm{B}

t/ min[A]/M
00.6500
x0.0650
200.00065

x=\mathrm{x}= ____\_\_\_\_ min. (Nearest integer)

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

K=1tln⁡[A0At]K=\frac{1}{t} \ln \left[\frac{A_{0}}{A_{t}}\right] K=120ln⁡[0.65000.00065]⇒3ln⁡1020\mathrm{K}=\frac{1}{20} \ln \left[\frac{0.6500}{0.00065}\right] \Rightarrow \frac{3 \ln 10}{20} min −1^{-1} Also, x=1Kln⁡[A0Ax]⇒203ln⁡10ln⁡(0.65000.0650)x=\frac{1}{K} \ln \left[\frac{A_{0}}{A_{x}}\right] \Rightarrow \frac{20}{3 \ln 10} \ln \left(\frac{0.6500}{0.0650}\right)

=203 min⇒6.7 min=\frac{20}{3} \mathrm{~min} \Rightarrow 6.7 \mathrm{~min}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws

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