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Chemical Kinetics: JEE Main Chemistry Question with Solution

\mathrm{A} \rightarrow \mathrm{B}

The rate constants of the above reaction at

200\text{ K}

and

300\text{ K}

are

0.03\text{ min}^{-1}

and

0.05\text{ min}^{-1}

respectively. The activation energy for the reaction is

\mathrm{J}

(Nearest integer)

(\text{Given : } \ln 10 = 2.3

\mathrm{R} = 8.3\text{ J K}^{-1}\text{ mol}^{-1}

\log 5 = 0.70

\log 3 = 0.48

\log 2 = 0.30)

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Step-by-step solutionView

Correct answer

The activation energy of the reaction is approximately $2521\text{ J}$ (or $2522\text{ J}$ depending on rounding, specifically $2521\text{ J}$ ).

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

$T_1 = 200\text{ K}$ , $k_1 = 0.03\text{ min}^{-1}$ , $T_2 = 300\text{ K}$ , $k_2 = 0.05\text{ min}^{-1}$ , $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ , $\ln 10 = 2.3$ , $\log 5 = 0.70$ , $\log 3 = 0.48$ .

02find

Activation energy $E_a$ in $\text{J}\text{ mol}^{-1}$ to the nearest integer.

03visualise

The Arrhenius relation gives the temperature dependence of the rate constant: $\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ .

04strategise

Substitute the ratio $\frac{k_2}{k_1} = \frac{5}{3}$ into $\log(5/3) = \log 5 - \log 3 = 0.70 - 0.48 = 0.22$ . Then solve for $E_a = 0.22 \times 2.3 \times 8.3 \times \left(\frac{200 \times 300}{300 - 200}\right)$ .

05execute

$E_a = 0.22 \times 2.3 \times 8.3 \times 600 = 2519.88 \approx 2520\text{ J}$ .

✓verify

Round $2519.88$ to the nearest integer, giving $2520\text{ J}$ . The units of rate constants are in $\text{min}^{-1}$ , which cancel out in the ratio $k_2/k_1$ , so converting time units is not necessary.

Hints that build this answer step by step

Which form of the Arrhenius equation relates the rate constants $k_1$ and $k_2$ at temperatures $T_1$ and $T_2$ ?
$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$
Using the given values, what is the value of $\ln\left(\frac{k_2}{k_1}\right) = \ln\left(\frac{0.05}{0.03}\right)$ ?
$0.506$
What is the value of $\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ for $T_1 = 200\text{ K}$ and $T_2 = 300\text{ K}$ ?
$\frac{1}{600}\text{ K}^{-1}$
Substituting $\ln(k_2/k_1) = 0.506$ , $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ , and $\Delta(1/T) = \frac{1}{600}\text{ K}^{-1}$ , what is $E_a$ to the nearest integer?
$2520\text{ J}$ (or $2521\text{ J}$ )

✓ 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

Do we need to convert rate constants from min^-1 to s^-1?

No, because the formula uses the ratio k2/k1. The unit conversion factor (1/60) cancels out completely.

Why use 2.3 instead of 2.303 for ln 10?

The problem explicitly specifies the value 'ln 10 = 2.3'. Always adhere strictly to the constants provided in the question stem.

Answer

The activation energy of the reaction is approximately $2521\text{ J}$ (or $2522\text{ J}$ depending on rounding, specifically $2521\text{ J}$ ).

Why each option works or fails

Step-by-step solution

given: $T_1 = 200\text{ K}$ , $k_1 = 0.03\text{ min}^{-1}$ , $T_2 = 300\text{ K}$ , $k_2 = 0.05\text{ min}^{-1}$ , $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ , $\ln 10 = 2.3$ , $\log 5 = 0.70$ , $\log 3 = 0.48$ .
find: Activation energy $E_a$ in $\text{J}\text{ mol}^{-1}$ to the nearest integer.
visualise: The Arrhenius relation gives the temperature dependence of the rate constant: $\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ .
strategise: Substitute the ratio $\frac{k_2}{k_1} = \frac{5}{3}$ into $\log(5/3) = \log 5 - \log 3 = 0.70 - 0.48 = 0.22$ . Then solve for $E_a = 0.22 \times 2.3 \times 8.3 \times \left(\frac{200 \times 300}{300 - 200}\right)$ .
execute: $E_a = 0.22 \times 2.3 \times 8.3 \times 600 = 2519.88 \approx 2520\text{ J}$ .
verify: Round $2519.88$ to the nearest integer, giving $2520\text{ J}$ . The units of rate constants are in $\text{min}^{-1}$ , which cancel out in the ratio $k_2/k_1$ , so converting time units is not necessary.

Chemical Kinetics: JEE Main Chemistry Question with Solution

What feels right?

The activation energy of the reaction is approximately $2521\text{ J}$ (or $2522\text{ J}$ depending on rounding, specifically $2521\text{ J}$ ).

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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