Q22 / 180 · COMEDK 2026 (9 May Shift 2) · Chemistry

A first order reaction is 50%complete in 30 minutes at 300 K and in 10 minutes at 320 K. The activation energy of the reaction (E$_a$) is:
[R=8.314 J K$^{-1}$ mol$^{-1}$; log 2 = 0.3010; log 3=0.4771]

  • A
    23.7 kJ mol$^{-1}$
  • Correct answer
    B
    43.8 kJ mol$^{-1}$
  • C
    52.5 kJ mol$^{-1}$
  • D
    75.2 kJ mol$^{-1}$

Solution

For a first-order reaction, the rate constant is given by $k = \dfrac{0.693}{t_{1/2}}$.

At $T_1 = 300 \text{ K}$, $k_1 = \dfrac{0.693}{30} \text{ min}^{-1}$

At $T_2 = 320 \text{ K}$, $k_2 = \dfrac{0.693}{10} \text{ min}^{-1}$

$\dfrac{k_2}{k_1} = \dfrac{30}{10} = 3$

Using the Arrhenius equation:

$\log \left( \dfrac{k_2}{k_1} \right) = \dfrac{E_a}{2.303 R} \left( \dfrac{T_2 - T_1}{T_1 T_2} \right)$

$\log 3 = \dfrac{E_a}{2.303 \times 8.314} \left( \dfrac{320 - 300}{300 \times 320} \right)$

$0.4771 = \dfrac{E_a}{19.147} \left( \dfrac{20}{96000} \right)$

$E_a = \dfrac{0.4771 \times 19.147 \times 96000}{20}$

$E_a = 43848 \text{ J mol}^{-1} = 43.8 \text{ kJ mol}^{-1}$

Answer: 43.8 kJ mol$^{-1}$

Hence, the correct answer is option: 'B'.

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