The following results were obtained during kinetic studies of the reaction. $2A+B\rightarrow$ product Experiment $[A]$ $(\mathrm{in} \mathrm{mol} {L}^{-1})$ $[B]$ $(in mol {L}^{-1})$ Initial rate of reaction $(\mathrm{in} \mathrm{mol} {L}^{-1} \mathrm{min}{ }^{-1})$ $I$ $0.10$ $0.20$ $6.93\times {10}^{-3}$ $\mathrm{II}$ $0.10$ $0.25$ $6.93\times {10}^{-3}$ $\mathrm{III}$ $0.20$ $0.30$ $1.386\times {10}^{-2}$ The time (in minutes) required to consume half of $\text{A}$ is

JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

The following results were obtained during kinetic studies of the reaction.

$2A+B\rightarrow$ product

Experiment	$[A]$ $(\mathrm{in} \mathrm{mol} {L}^{-1})$	$[B]$ $(in mol {L}^{-1})$	Initial rate of reaction $(\mathrm{in} \mathrm{mol} {L}^{-1} \mathrm{min}{ }^{-1})$
$I$	$0.10$	$0.20$	$6.93\times {10}^{-3}$
$\mathrm{II}$	$0.10$	$0.25$	$6.93\times {10}^{-3}$
$\mathrm{III}$	$0.20$	$0.30$	$1.386\times {10}^{-2}$

The time (in minutes) required to consume half of $\text{A}$ is

Official previous-year question

Held on 9 Jan 2019 · Verified 6 Jul 2026.

Options

A
$100$
B
$10$
C
$5$
D
$1$

Did you get this right?

Solution

Let's assume $x$ and $y$ are the order of the reaction with respect to $A$ and $B$ , respectively.

From equation $(1)$ and $(2)$ ,

${(\frac{0.1}{0.1})}^{x}{(\frac{0.2}{0.25})}^{y}=\frac{6.93\times {10}^{-3}}{6.93\times {10}^{-3}}\Rightarrow y=0$

From equation $(1)$ and $(3)$ and put the value $y=0$ ,

${(\frac{0.1}{0.2})}^{x}{(\frac{0.2}{0.3})}^{0}=\frac{6.93\times {10}^{-3}}{1.386\times {10}^{-2}}$ $\therefore y=0$

${(\frac{1}{2})}^{x}=(\frac{1}{2})\Rightarrow x=1$

The rate law will be $R=K{[A]}^{1}{[B]}^{0}$

Hence, the reaction is of the first order.

$K=\frac{R}{[A]}=\frac{6.93\times {10}^{-3}}{0.1}=6.93\times {10}^{-2}$

Here ${t}_{\frac{1}{2}}$ for the first-order reaction is given by, ${t}_{\frac{1}{2}}=\frac{0.693}{K}=\frac{0.693}{(6.93\times {10}^{-2})}=10$

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Solution

Let's assume $x$ and $y$ are the order of the reaction with respect to $A$ and $B$ , respectively.

From equation $(1)$ and $(2)$ ,

${(\frac{0.1}{0.1})}^{x}{(\frac{0.2}{0.25})}^{y}=\frac{6.93\times {10}^{-3}}{6.93\times {10}^{-3}}\Rightarrow y=0$

From equation $(1)$ and $(3)$ and put the value $y=0$ ,

${(\frac{0.1}{0.2})}^{x}{(\frac{0.2}{0.3})}^{0}=\frac{6.93\times {10}^{-3}}{1.386\times {10}^{-2}}$ $\therefore y=0$

${(\frac{1}{2})}^{x}=(\frac{1}{2})\Rightarrow x=1$

The rate law will be $R=K{[A]}^{1}{[B]}^{0}$

Hence, the reaction is of the first order.

$K=\frac{R}{[A]}=\frac{6.93\times {10}^{-3}}{0.1}=6.93\times {10}^{-2}$

Here ${t}_{\frac{1}{2}}$ for the first-order reaction is given by, ${t}_{\frac{1}{2}}=\frac{0.693}{K}=\frac{0.693}{(6.93\times {10}^{-2})}=10$

JEE Main 2019 — Chemistry Physical Chemistry

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JEE Main 2019 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry