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Arrhenius Equation

Hardchemistry

When the temperature of a reaction is raised from 300 K to 310 K, the rate roughly doubles; what does this reveal about the activation energy involved?

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About This Question

Subject
chemistry
Chapter
chemical kinetics
Topic
arrhenius equation
Difficulty
Hard
Year
2025
Tags
Arrhenius equationactivation energytemperature coefficientrate constanttwo-temperature form

Solution

Correct Answer:

The Arrhenius equation, k = A e^(-Ea/RT), shows that the rate constant rises with temperature because more molecules acquire energy exceeding the activation barrier. Using the two-temperature form, log(k_2/k_1) = (Ea/2.303R)(1/T_1 - 1/T_2), a doubling of rate (k_2/k_1 = 2) over 300 K to 310 K can be solved for Ea. Substituting log 2 = 0.301, R = 8.314 J/mol·K, and the temperature term (1/300 - 1/310) = 1.075 × 10^-4 gives Ea = (0.301 × 2.303 × 8.314)/(1.075 × 10^-4) ≈ 53000 J/mol, about 50 kJ/mol. This is why the common rule of thumb that rate doubles for a 10 K rise corresponds to typical activation energies near 50 kJ/mol. Zero activation energy would make rate nearly temperature-independent. A negative value is physically meaningless for an elementary step. A value above 500 kJ/mol would make the rate enormously temperature-sensitive. Examiners frequently test whether a student can connect activation energy with the underlying principle rather than merely recalling an isolated fact. Understanding arrhenius equation in this way ties directly into the wider study of chemical kinetics, where the same reasoning recurs across many problems. Plausibility check: the familiar doubling rule matching a moderate barrier confirms roughly 50 kJ/mol.

This hard difficulty chemistry question is from the chapter chemical kinetics, covering the topic of arrhenius equation. It appeared in the 2025 exam.

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