Chemistry·MCQ Practice

Arrhenius Equation — MCQ Practice

NEET UG
Updated 13 Sep 2026

Interactive MCQ Practice

Test your knowledge. Click “Solve” to reveal options, select your answer, then check the result. 6 questions available.

Q1medium

For a first-order reaction, the rate constants at 27∘C27^\circ\text{C} and 37∘C37^\circ\text{C} are 2.0×10−2,s−12.0 \times 10^{-2},\text{s}^{-1} and 4.0×10−2,s−14.0 \times 10^{-2},\text{s}^{-1} respectively. Calculate the activation energy (EaE_a) for this reaction. (Given: R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}, ln⁡2=0.693\ln 2 = 0.693)

Q2medium

For a reaction, the activation energy (EaE_a) is 60 kJ mol−160\,\text{kJ mol}^{-1}. If the rate constant at 300 K300\,\text{K} is 2.0×10−3,s−12.0 \times 10^{-3},\text{s}^{-1}, what will be the approximate rate constant at 310 K310\,\text{K}? (Given: R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}, e0.77≈2.16e^{0.77} \approx 2.16)

Q3easy

Which of the following statements is INCORRECT regarding the Arrhenius equation?

Q4easy

An Arrhenius plot of ln⁡k\ln k versus 1/T1/T for a reaction gives a straight line with a slope of −1.2×104 K-1.2 \times 10^4\,\text{K}. What is the activation energy (EaE_a) for this reaction? (Given: R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1})

Q5hard

A catalyst increases the rate of a reaction by a factor of 10510^5 at 300 K300\,\text{K}. If the activation energy of the uncatalyzed reaction is 100 kJ mol−1100\,\text{kJ mol}^{-1}, what is the activation energy of the catalyzed reaction? (Given: R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}, ln⁡10=2.303\ln 10 = 2.303)

Q6easy

Which of the following plots correctly represents the Arrhenius equation for a typical chemical reaction?