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Chemical Kinetics: JEE Main Chemistry Question with Solution

The number of correct statement/s from the following is A. Larger the activation energy, smaller is the value of the rate constant. B. The higher is the activation energy, higher is the value of the temperature coefficient. C. At lower temperatures, increase in temperature causes more change in the value of

\mathrm{k}

than at higher temperature D. A plot of

\ln \mathrm{k} v v \frac{1}{T}

is a straight line with slope equal to

-\frac{\mathrm{E}_\mathrm{a}}{\mathrm{R}}

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Step-by-step solutionView

Correct answer

All four statements (A, B, C, and D) are correct, so the answer is 4.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Four statements regarding the Arrhenius equation $k = A e^{-E_a/RT}$ : A. Larger activation energy implies smaller rate constant. B. Higher activation energy implies higher temperature coefficient. C. At lower temperatures, an increase in temperature causes a larger relative change in $k$ than at higher temperatures. D. A plot of $\ln k$ vs $1/T$ is a straight line with slope equal to $-E_a/R$ .

02find

The total count of correct statements among A, B, C, and D.

03strategise

Use the Arrhenius relation $k = A e^{-E_a/(RT)}$ and its logarithmic form $\ln k = \ln A - \frac{E_a}{RT}$ to test each statement individually.

04execute

Statement A: Since $k = A e^{-E_a/RT}$ , as $E_a$ increases, the exponent $-E_a/RT$ becomes more negative, hence $k$ decreases. Thus, A is correct. Statement B: The temperature coefficient is $\frac{k_{T+10}}{k_T} = \exp\left[\frac{E_a}{R}\left(\frac{10}{T(T+10)}\right)\right]$ . As $E_a$ increases, this ratio increases. Thus, B is correct. Statement C: From $\frac{d\ln k}{dT} = \frac{E_a}{RT^2}$ , the fractional change in $k$ per degree, $\frac{1}{k}\frac{dk}{dT}$ , is inversely proportional to $T^2$ . Hence, at lower $T$ , temperature increase causes a greater change/sensitivity in $k$ than at higher $T$ . Thus, C is correct. Statement D: Plotting $\ln k$ vs $\frac{1}{T}$ gives $\ln k = \ln A - \left(\frac{E_a}{R}\right)\frac{1}{T}$ , which is a straight line $y = c + mx$ with slope $m = -\frac{E_a}{R}$ . Thus, D is correct. Total correct statements = 1 + 1 + 1 + 1 = 4.

✓verify

All four statements are well-established theoretical properties of the Arrhenius rate law in Chemical Kinetics. Total count is 4.

Hints that build this answer step by step

According to the Arrhenius equation $k = A e^{-E_a/(RT)}$ , how does the rate constant $k$ depend on the activation energy $E_a$ and what is the slope of $\ln k$ versus $1/T$ ?
Larger $E_a$ yields a smaller $k$ , and the slope of $\ln k$ vs $1/T$ is $-E_a/R$ .
From the differential Arrhenius form $\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$ , how do the temperature coefficient (fractional change in $k$ with temperature) and temperature level influence the sensitivity of $k$ ?
Higher $E_a$ gives a higher temperature coefficient, and the sensitivity is greater at lower temperatures (due to $1/T^2$ ).
How many of the statements A, B, C, and D are correct?
4

✓ 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

Why does a temperature increase cause a greater effect at lower temperatures in statement C?

Because the derivative $\frac{d\ln k}{dT} = \frac{E_a}{RT^2}$ has $T^2$ in the denominator. When $T$ is small, $\frac{1}{T^2}$ is much larger, meaning the rate constant is much more sensitive to changes in temperature at low temperatures.

Answer

All four statements (A, B, C, and D) are correct, so the answer is 4.

Why each option works or fails

Step-by-step solution

given: Four statements regarding the Arrhenius equation $k = A e^{-E_a/RT}$ : A. Larger activation energy implies smaller rate constant. B. Higher activation energy implies higher temperature coefficient. C. At lower temperatures, an increase in temperature causes a larger relative change in $k$ than at higher temperatures. D. A plot of $\ln k$ vs $1/T$ is a straight line with slope equal to $-E_a/R$ .
find: The total count of correct statements among A, B, C, and D.
strategise: Use the Arrhenius relation $k = A e^{-E_a/(RT)}$ and its logarithmic form $\ln k = \ln A - \frac{E_a}{RT}$ to test each statement individually.
execute: Statement A: Since $E_a$ , as $E_a$ increases, the exponent $-E_a/RT$ becomes more negative, hence $k$ decreases. Thus, A is correct. Statement B: The temperature coefficient is $\frac{k_{T+10}}{k_T} = \exp\left[\frac{E_a}{R}\left(\frac{10}{T(T+10)}\right)\right]$ . As $E_a$ increases, this ratio increases. Thus, B is correct. Statement C: From $\frac{d\ln k}{dT} = \frac{E_a}{RT^2}$ , the fractional change in $k$ per degree, $\frac{1}{k}\frac{dk}{dT}$ , is inversely proportional to $T^2$ . Hence, at lower $T$ , temperature increase causes a greater change/sensitivity in $k$ than at higher $T$ . Thus, C is correct. Statement D: Plotting $\ln k$ vs $\frac{1}{T}$ gives $\ln k = \ln A - \left(\frac{E_a}{R}\right)\frac{1}{T}$ , which is a straight line $y = c + mx$ with slope $m = -\frac{E_a}{R}$ . Thus, D is correct. Total correct statements = 1 + 1 + 1 + 1 = 4.
verify: All four statements are well-established theoretical properties of the Arrhenius rate law in Chemical Kinetics. Total count is 4.