Consider the reaction

The rate constant for two parallel reactions were found to be $10^{- 2} {dm}^{3} {mol}^{- 1} s^{- 1} and 4 \times 10^{- 2} {dm}^{3} {mol}^{- 1} s^{- 1}$ .
if the corresponding energies of activation of the parallel reaction are 100 and 120 kJ/mol respectively, what is the net energy of activation (E_a)of A?

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120 kJ/ mol

116 kJ/ mol

220 kJ/ mol

100 kJ / mol

answer is C.

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Detailed Solution

$\begin{array}{l} K_{A V} = K_{1} + K_{2} \\ {Ae}^{- E_{a 1 / Rτ}} = {Ae}^{- E_{a 1 / Rτ}} + {Ae}^{- E_{a_{2} / Rτ}} \\ {Ae}^{- E_{a / Rτ}} (\frac{- E_{a}}{R}) = {Ae}^{- E_{a_{1} / R T}} (\frac{- E_{a_{1}}}{R}) + {Ae}^{- E_{a_{2} / Rτ}} (\frac{- E_{a_{2}}}{R}) \\ K_{A V} (E_{aτ}) = K_{1} E_{a_{1}} + K_{2} E_{a_{2}} \\ E_{a_{1} τ} = \frac{K_{1} E_{a_{1}} + K_{2} E_{a_{2}}}{K_{1} + K_{2}} = \frac{10^{- 2} \times 100 + 4 \times 10^{- 2} \times 120}{5 \times 10^{- 2}} = \frac{580}{5} = 116 \end{array}$