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Chemical Kinetics

Question

The rate of a particular reaction triples when temperature changes from 50°C to 100°C. Calculate the activation energy of the reaction. [log 3 = 0.4771 (R = 8.314 ] K^–1mol^–1)

Solution

By using the Arrhenius equation we get

\log \frac{k_{2}}{k_{1}} = \frac{Ea}{2.303 RT} (\frac{1}{T_{1}} - \frac{1}{T_{2}}) \log 3 = \frac{Ea}{2.303 \times 8.314} (\frac{1}{323} - \frac{1}{373}) Ea = \frac{0.4771 \times 19.147 \times 323 \times 373}{50} = \frac{1100579.7}{50} = 22011.59 J = 220.12 kJ {mol}^{- 1} .