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Chemical Kinetics: JEE Main Chemistry Question with Solution

Q: Can we directly use mass instead of molarity in the first-order rate law?

Yes, because molar mass and volume cancel out in the ratio .

If compound A reacts with B following first order kinetics with rate constant

2.011 \times 10^{-3}\mathrm{~s}^{-1}

. The time taken by A (in seconds) to reduce from

7\mathrm{~g}

to

2\mathrm{~g}

will be (Nearest Integer)

[\log 5 = 0.698, \log 7 = 0.845, \log 2 = 0.301]

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Step-by-step solutionView

Correct answer

The time taken for compound A to reduce from 7 g to 2 g is approximately 623 seconds.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Rate constant $k = 2.011 \times 10^{-3}\text{ s}^{-1}$ , initial amount $[A]_0 = 7\text{ g}$ , final amount $[A]_t = 2\text{ g}$ , $\log 7 = 0.845$ , $\log 2 = 0.301$ .

02find

Time $t$ in seconds required for the reactant amount to decrease from $7\text{ g}$ to $2\text{ g}$ .

03visualise

In first-order kinetics, the ratio of masses equals the ratio of molar concentrations when volume is constant: $\frac{[A]_0}{[A]_t} = \frac{7}{2}$ .

04strategise

Use the integrated rate equation for a first-order reaction: $t = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{[A]_t}\right) = \frac{2.303}{k} (\log 7 - \log 2)$ .

05execute

$t = \frac{2.303 \times (0.845 - 0.301)}{2.011 \times 10^{-3}} = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}} = \frac{1.252832}{2.011 \times 10^{-3}} \approx 622.99\text{ s} \approx 623\text{ s}$

✓verify

Check order of magnitude: $t_{1/2} = 0.693 / (2.011 \times 10^{-3}) \approx 345\text{ s}$ . The decrease from $7\text{ g}$ to $2\text{ g}$ represents a factor of $3.5$ , which is between 1 and 2 half-lives ( $345\text{ s} < t < 690\text{ s}$ ). $623\text{ s}$ lies neatly in this range.

Hints that build this answer step by step

Which integrated rate equation correctly relates time $t$ to initial mass $[A]_0$ and final mass $[A]_t$ for a first-order reaction?
$t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]_t}\right)$
What is the value of $\log\left(\frac{7}{2}\right)$ using the provided logarithmic values?
$0.544$
Substituting $k = 2.011 \times 10^{-3}\text{ s}^{-1}$ and $\log(7/2) = 0.544$ into $t = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}}$ , what is the value of $t$ to the nearest integer?
623

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we directly use mass instead of molarity in the first-order rate law?

Yes, because molar mass and volume cancel out in the ratio $[A]_0/[A]_t$ .

Answer

The time taken for compound A to reduce from 7 g to 2 g is approximately 623 seconds.

Why each option works or fails

Step-by-step solution

given: Rate constant $k = 2.011 \times 10^{-3}\text{ s}^{-1}$ , initial amount $[A]_0 = 7\text{ g}$ , final amount $[A]_t = 2\text{ g}$ , $\log 7 = 0.845$ , $\log 2 = 0.301$ .
find: Time $t$ in seconds required for the reactant amount to decrease from $7\text{ g}$ to $2\text{ g}$ .
visualise: In first-order kinetics, the ratio of masses equals the ratio of molar concentrations when volume is constant: $\frac{[A]_0}{[A]_t} = \frac{7}{2}$ .
strategise: Use the integrated rate equation for a first-order reaction: $t = \frac{2.303}{k} \log_{10}\left(\frac{[A]_0}{[A]_t}\right) = \frac{2.303}{k} (\log 7 - \log 2)$ .
execute: $t = \frac{2.303 \times (0.845 - 0.301)}{2.011 \times 10^{-3}} = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}} = \frac{1.252832}{2.011 \times 10^{-3}} \approx 622.99\text{ s} \approx 623\text{ s}$
verify: Check order of magnitude: $t_{1/2} = 0.693 / (2.011 \times 10^{-3}) \approx 345\text{ s}$ . The decrease from $7\text{ g}$ to $2\text{ g}$ represents a factor of $3.5$ , which is between 1 and 2 half-lives ( $345\text{ s} < t < 690\text{ s}$ ). $623\text{ s}$ lies neatly in this range.

Chemical Kinetics: JEE Main Chemistry Question with Solution

What feels right?

The time taken for compound A to reduce from 7 g to 2 g is approximately 623 seconds.

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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