The rate of a reaction quadruples when temperature changes from 27^°C to 57^°C. Calculate the energy of activation.Given R = 8.314 J K^-1 mol^-1, log 4 = 0.6021

1. 3804 kJ/mol

2. 38.04 kJ/mol

3. 380.4 kJ/mol

4. 3.80 kJ/mol

Doubtly Asked 30 Oct 2026 NEET UG 2024 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 2. 38.04 kJ/mol

Key idea: $\log\dfrac{k_2}{k_1}=\dfrac{E_a}{2.303R}\dfrac{T_2-T_1}{T_1T_2}$.

$T_1=300$ K, $T_2=330$ K, $k_2/k_1=4$.

$$E_a=\frac{2.303\times8.314\times0.6021\times300\times330}{30}\approx38046\ \mathrm{J/mol}=38.04\ \mathrm{kJ/mol}$$

Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?