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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2024 · Paper 2 · Chemistry · Chemical Kinetics

Calculate the activation energy of a reaction, whose rate constant...

Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K.

Options

  1. A3.45kJ/mol3.45 \mathrm{ kJ} / \mathrm{mol}
  2. B6.90kJ/mol6.90 \mathrm{ kJ} / \mathrm{mol}
  3. C9.68kJ/mol9.68 \mathrm{ kJ} / \mathrm{mol}
  4. D19.6kJ/mol19.6 \mathrm{ kJ} / \mathrm{mol}

Correct answer

A. 3.45kJ/mol3.45 \mathrm{ kJ} / \mathrm{mol}

Step-by-step solution

log⁡(k2k1)=Ea2.303R[1T1−1T2]or,log⁡(2kk)=Ea2.303×8.3×[600−300]300×600\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right] or, \log \left(\frac{2 k}{k}\right)=\frac{E_a}{2.303 \times 8.3} \times \frac{[600-300]}{300 \times 600} E_a ≈3.45kJ\approx3.45 \mathrm{ kJ}/ mol\mathrm{mol}
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