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A reaction takes place in 3 steps, the rate constant are $K_{1,} K_{2}$ and $K_{3}$ and energies of activation are 40, 30 and 20 kJ respectively. If overa11 rate constant $K = \frac{K_{1} K_{3}}{K_{2}}$ then overall energy of activation is:

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answer is C.

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Detailed Solution

$K = \frac{K_{1} K_{3}}{K_{2}}$

$∴ lnK = {lnK}_{1} + {lnK}_{3} - {lnK}_{2}$

But $lnK = - \frac{E}{RT} + \ln A$

$∴ - \frac{E}{RT} + lnA = - \frac{E_{1}}{RT} + lnA - \frac{E_{3}}{RT} + lnA + \frac{E_{2}}{RT} - lnA$

$K = A^{'} e^{- \frac{(E_{1} + E_{2} - E_{2})}{RT}}$

$E = E_{1} + E_{3} - E_{2}$

$= 40 + 20 - 30 = 30 kJ$

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