Chemistry-Activation Energy

NEET Chemistry Activation Energy MCQ Question

Type: MCQ-numerical-Easy-Class 12

Given the rate constants of a reaction at 500K and 700K are 0.02 s⁻¹ and 0.07 s⁻¹ respectively, what is the activation energy (Eₐ) in J/mol for this reaction?

A

34,710 J/mol

B

41,840 J/mol

C

52,360 J/mol

D

60,420 J/mol

Correct Answer

Option B

Detailed Explanation

Using the formula log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2), where k1 = 0.02 s⁻¹, k2 = 0.07 s⁻¹, R = 8.314 J/mol·K, T1 = 500K, T2 = 700K, we can solve for Ea.

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