The rate constant of a reaction is 0.002 per second at 300 Kelvin and 0.02 per second at 310 Kelvin. Calculate the activation energy in Joules per mole.

A. 107.5 kJ mol^-1 — Using the Arrhenius equation, the activation energy is calculated to be approximately 107.5 kJ/mol.

JEE Advanced · ChemistryMedium

$The rate constant of a reaction is k_{1} = 0.002 s^{- 1} at T_{1} = 300 K and k_{2} = 0.02 s^{- 1} at T_{2} = 310 K . Calculate the activation energy E_{a} in J mol^{- 1} . (Use R = 8.314 J K^{- 1} mol^{- 1})$

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Correct answer: A — $107.5 kJ mol^{- 1}$

$Use the Arrhenius equation: ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}}) .$
$Substitute values: ln (10) = \frac{E _{a}}{8.314} (\frac{10}{300 \times 310}) .$
$2.303 = \frac{E _{a}}{8.314} (\frac{10}{93000}) .$
$E_{a} = \frac{2.303 \times 8.314 \times 9300}{1} \approx 107500 J mol^{- 1} .$
$Hence the answer is (A).$