Chemistry · Chemical Kinetics

JEE Main 2024 — 8 April, Shift 1 — Question 78

Consider the following reaction $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ The time taken for A to become $1 / 4^{\text {th }}$ of its initial concentration is twice the time taken to become $1 / 2$ of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is $\qquad$

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

For $1^{\text {st }}$ order reaction

$75 \%$ life $=2 \times 50 \%$ life So order with respect to $A$ will be first order.

So order with respect to B will be zero.

Overall order of reaction $=1+0=1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Rate Laws and Rate Constant