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ChemistryNEET 2024Arrhenius equation and activation energy

Chemical Kinetics

The rate of a reaction quadruples when the temperature changes from 27∘27^\circC to 57∘57^\circC. Calculate the energy of activation. (Given R=8.314 J K−1 mol−1R = 8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}, log⁡4=0.6021\log 4 = 0.6021.)

  1. A.380.4 kJ/mol
  2. B.3.80 kJ/mol
  3. C.3804 kJ/mol
  4. D.38.04 kJ/mol✓

Solution

log⁡k2k1=Ea2.303R(1T1−1T2)\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) with k2/k1=4k_2/k_1 = 4, T1=300T_1 = 300 K, T2=330T_2 = 330 K: 0.6021=Ea2.303×8.314×30300×330⇒Ea≈38.040.6021 = \frac{E_a}{2.303 \times 8.314} \times \frac{30}{300 \times 330} \Rightarrow E_a \approx 38.04 kJ/mol.

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