The activation energy of a reaction is 75.2 kJ mol^–1 in the absence of a catalyst and 50.14 kJ mol^–1 with a catalyst. How many times will the rate of reaction grow in the presence of the catalyst if the reaction proceeds at 25°C?
(R = 8.314 J K^–1 mol^–1)

Solution

By using the Arrhenius equation, we obtained

In k_{1} = In A - \frac{{Ea}_{1}}{RT} In k_{2} = In A - \frac{{Ea}_{2}}{RT} In k_{2} - In k_{1} = \frac{1}{RT} ({Ea}_{1} - {Ea}_{2}) \log \frac{k_{2}}{k_{1}} = \frac{1}{RT} (75.10 kJ - 50.14 kJ) \log \frac{k_{2}}{k_{1}} = \frac{1}{2.303 RT} (25.10 kJ) = \frac{25.10 \times 1000 J {mol}^{- 1}}{2.303 \times 298 k \times 8.314 J K^{- 1} {mol}^{- 1}} \log \frac{k_{2}}{k_{1}} = \frac{25100}{5705.8} = 4.40 \frac{k_{2}}{k_{1}} = Antilog 4.40 = 2.5 \times 10^{4} .

The rate of reaction will grow up by 25000 times in presence of catalyst. The decrease in activation energy takes place by 25.1 kJ mol^–1.

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