During kinetic study of reaction 2 A + B rightarrow C + D , the following results were obtained : A [ M ] B [ M ] Initial rate of formation of D --- --- --- --- I 0.1 0.1 6.0× 10 -3 II 0.3 0.2 7.2× 10 -2 III 0.3 0.4 2.88× 10 -1 IV 0.4 0.1 2.40× 10 -2 Based on above data, the overall order of the reaction is

Chemistry · Chemical Kinetics

JEE Main 2024 — 5 April, Shift 1 — Question 81

During kinetic study of reaction $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$ , the following results were obtained :

	$\mathbf{A}\left[ \mathbf{M} \right]$	$\mathbf{B}\left[ \mathbf{M} \right]$	Initial rate of formation of $\mathbf{D}$
I	0.1	0.1	$6.0\times {{10}^{-3}}$
II	0.3	0.2	$7.2\times {{10}^{-2}}$
III	0.3	0.4	$2.88\times {{10}^{-1}}$
IV	0.4	0.1	$2.40\times {{10}^{-2}}$

Based on above data, the overall order of the reaction is $\qquad$

Numerical answer — enter this value.

Let the rate law be

\text{Rate} = k[\mathrm{A}]^{m}[\mathrm{B}]^{n}

From experiments II and III, $[\mathrm{A}]$ is constant:

\frac{\text{Rate}_{\mathrm{III}}}{\text{Rate}_{\mathrm{II}}} = \frac{2.88\times10^{-1}}{7.2\times10^{-2}} = 4, \qquad \frac{[\mathrm{B}]_{\mathrm{III}}}{[\mathrm{B}]_{\mathrm{II}}} = \frac{0.4}{0.2} = 2

4 = 2^{n} \Rightarrow n = 2

From experiments I and IV, $[\mathrm{B}]$ is constant:

\frac{\text{Rate}_{\mathrm{IV}}}{\text{Rate}_{\mathrm{I}}} = \frac{2.40\times10^{-2}}{6.0\times10^{-3}} = 4, \qquad \frac{[\mathrm{A}]_{\mathrm{IV}}}{[\mathrm{A}]_{\mathrm{I}}} = \frac{0.4}{0.1} = 4

4 = 4^{m} \Rightarrow m = 1

Thus, the overall order of reaction = m + n = 1 + 2 = 3

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.