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  4. The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is

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Q. The activation energy for a reaction which doubles the rate when the temperature is raised from 298 K to 308 K is

AIEEEAIEEE 2012Chemical Kinetics

Solution:

Activation energy can be calculated from the equation.
$\frac{log\,K_{2}}{log\,K_{1}} = \frac{-E_{a}}{2.303\,R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$
Given $\frac{log\,K_{2}}{log\,K_{1}} = 2\, T_{2} = 308;\, T_{1} = 298$
$\therefore \quad log\,2 = \frac{-E_{a}}{2.303 \times8.314}\left(\frac{1}{308}-\frac{1}{298}\right)$
$E_{a} = 52.9 \,kJ \,mol^{-1}$