A substance undergoes first order parallel reaction as shown. Calculate net activation energy if it is a known that B, C and Dare formed in a molar ratio of 2 : 3 : 4 respectively.

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answer is 4.

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Detailed Solution

It is know that,

$\frac{3 K_{1}}{2 K_{2}} = \frac{2}{3} \frac{2 K_{2}}{K_{3}} = \frac{3}{4}$

$\frac{K_{1}}{K_{2}} = \frac{4}{9} \frac{K_{2}}{K_{3}} = \frac{3}{8} = \frac{9}{24}$

$\begin{array}{l} K_{1} : K_{2} : K_{3} \\ 4 : 9 : 24 \end{array}$

$\begin{array}{l} E = \frac{K_{1} E_{1} + K_{2} E_{2} + K_{3} E_{9}}{K_{1} + K_{2} + K_{3}} \\ = \frac{4 x \times 8 + 9 x \times 6 + 24 x \times 3}{37 x} \\ = \frac{32 + 54 + 72}{37} = 4 \end{array}$

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