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Question:

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20℃ to 35℃?

342kJmol⁻¹

269kJmol⁻¹

34.7kJmol⁻¹

15.1kJmol⁻¹

Solution:

log(K₂/K₁) = -Ea/(2.303R)[T₁⁻T₂/(T₁.T₂)]
log2 = -Ea/(2.303 × 8.314)[293⁻308/(293 × 308)]
Ea = 0.301 × 2.303 × 8.314 × 293 × 308/15
Ea = 34.67kJmol⁻¹ ≈ 34.7kJmol⁻¹