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Chemical Kinetics

Question

The activation energy of a reaction is 94.14 kJ mol^–1 and the value of rate constant at 313 K is 1.8 x 10^–5 sec^–1. Calculate the frequency factor A.

Solution

We have given

E_{a} = 94.14 kJ {mol}^{- 1} = 94140 J {mol}^{- 1} T = 313, k = 1.8 \times 10^{- 5} \sec^{- 1}

thus by usingv Arrhenius equation,

Now,

\log k = \frac{- E_{a}}{2.303 RT} + \log A

We get,

logA = \log (1.8 \times 10^{- 5}) + \frac{94140}{2.303 \times 8.314 \times 313} = (\log 1.8) - 5 + 15.7082 = 0.2553 - 5 + 15.7082 = 10.9635

Therefore,

A = anti \log (10.9635) = 9.194 \times 10^{10} collisions / \sec .