Chemistry · Chemical Kinetics

JEE Main 2024 — 4 April, Shift 2 — Question 79

Consider the following reaction, the rate expression of which is given below A+B→C\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} rate =k[A]1/2[ B]1/2=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2}

The reaction is initiated by taking 1 M concentration AA and BB each. If the rate constant ( k ) is 4.6×10−2 s−14.6 \times 10^{-2} \mathrm{~s}^{-1}, then the time taken for A to become 0.1 M is \qquad sec. (nearest integer)

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

K=2.303tlog⁡10.1\quad \mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{1}{0.1}

4.6×10−2=2.303t4.6 \times 10^{-2}=\frac{2.303}{t}

t=50sec\mathrm{t}=50 \mathrm{sec}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Integrated Rate Laws
Consider the following reaction, the rate expression of which is… | JEE Main 2024 PYQ with Solution · DhiX AI