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AIIMS 201926 May
Exam
AIIMS
Year
2019
Date
26 May
Subject
Chemistry
Chapter
Chemical Kinetics

AIIMS 2019 · 26 May · Chemistry · Chemical Kinetics

What is the activation energy (kJ / mol) for a reaction if its rate...

What is the activation energy (kJ/mol)(\mathrm{kJ} / \mathrm{mol}) for a reaction if its rate constant doubles when the temperature is raised from 300 Kto400 K300 \mathrm{~K} to 400 \mathrm{~K} ? (R=8.314 J mol−1 K−1)\left(R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)

Options

  1. A69.869.8
  2. B6.926.92
  3. C34.434.4
  4. D3.443.44

Correct answer

B. 6.926.92

Step-by-step solution

 log⁡k2k1=Ea2.303R(1T1−1T2)log⁡21=Ea2.303×8.314(1300−1400)0.3010=Ea2.303×8.314(100300×400)⇒Ea=0.3010×2.303×8.314×3×400=6.916kJmol−1\begin{aligned} & \text{ } \log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ & \log \frac{2}{1}=\frac{E_a}{2.303 \times 8.314}\left(\frac{1}{300}-\frac{1}{400}\right) \\ & 0.3010=\frac{E_a}{2.303 \times 8.314}\left(\frac{100}{300 \times 400}\right) \\ & \Rightarrow E_a=0.3010 \times 2.303 \times 8.314 \times 3 \times 400 \\ & =6.916 \mathrm{ kJ} \mathrm{ mol}^{-1}\end{aligned}
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