The rate of a reaction quadruples when the temperature changes from $300$ to $310 K$ . The activation energy of this reaction is:

(Assume Activation energy and pre-exponential factor are independent of temperature; $\ln (2) = 0.693; R = 8.314 J {mol}^{- 1} K^{- 1}$ )

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TestHub · Accepted Answer

The activation energy, rate constant and temperature can be related using Arrhenius equation.
$k = {Ae}^{- \frac{E_{a}}{RT}}$
$\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$\ln (4) = \frac{E_{a}}{R} (\frac{1}{300} - \frac{1}{310})$

$2 \times 0.693 = \frac{E_{a}}{R} (\frac{10}{300 \times 310})$

$E_{a} = \frac{2 \times 0.693 \times 300 \times 310 \times 8.314}{10} = 107.2 kJ/mol$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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