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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 22 January · Evening Shift · Chemistry · Chemical Kinetics

Correct statements regarding Arrhenius equation among the following...

Correct statements regarding Arrhenius equation among the following are :
A. Factor e−Ea/RTe^{-E a / R T} corresponds to fraction of molecules having kinetic energy less than Ea.
B. At a given temperature, lower the Ea, faster is the reaction.
C. Increase in temperature by about 10∘C10^{\circ} \mathrm{C} doubles the rate of reaction.
D. Plot of log⁡k\log \mathrm{k} vs 1 T\frac{1}{\mathrm{~T}} gives a straight line with slope =−EaR=-\frac{E a}{R}.
Choose the correct answer from the options given below :

Options

  1. AB and D Only
  2. BA and B Only
  3. CB and C Only
  4. DA and C Only

Correct answer

C. B and C Only

Step-by-step solution

The Arrhenius equation is given by k=Ae−Ea/RTk = A e^{-E_a/RT}.
Statement A: The factor e−Ea/RTe^{-E_a/RT} represents the fraction of molecules having kinetic energy equal to or greater than the activation energy EaE_a. Thus, statement A is incorrect.
Statement B: From the equation, as EaE_a decreases, the value of kk increases, leading to a faster reaction rate. Thus, statement B is correct.
Statement C: For many reactions, the temperature coefficient is approximately 2, meaning the rate of reaction doubles for every 10∘C10^{\circ}C rise in temperature. Thus, statement C is correct.
Statement D: Taking the natural logarithm of the Arrhenius equation, ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}. Converting to base 10, log⁡k=log⁡A−Ea2.303RT\log k = \log A - \frac{E_a}{2.303RT}. A plot of log⁡k\log k vs 1T\frac{1}{T} gives a straight line with slope =−Ea2.303R= -\frac{E_a}{2.303R}. The statement says slope =−EaR= -\frac{E_a}{R}, which is only true for a ln⁡k\ln k vs 1T\frac{1}{T} plot. Thus, statement D is incorrect.
Therefore, statements B and C are correct.
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