MCQ
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.

$\mathrm{A}+\mathrm{B} \underset{\text { Step } 3}{\text { Step } 1} \mathrm{C} \xrightarrow{\text { Step } 2} \mathrm{P}$

Some details of the above reaction are listed below.

Step

Rate constant $\left(\sec ^{-1}\right)$

Activation energy 

$\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$

$1$ ${k}_1$ $300$
$2$ ${k}_2$ $200$
$3$ ${k}_3$ $\mathrm{Ea}_3$

If the overall rate constant of the above transformation (k) is given as $\mathrm{k}=\frac{\mathrm{k}_1 \mathrm{k}_2}{\mathrm{k}_3}$ and the overall activation energy $\left(E_2\right)$ is $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then the value of $\mathrm{Ea}_3$ is $\qquad$ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer)

  • A
    $70$
  • B
    $98$
  • $100$
  • D
    $90$

Answer

Correct option: C.
$100$
c
$\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$

$A \mathrm{e}^{\frac{-\mathrm{E}_2}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{a}_1}}{\mathrm{RT}}} \mathrm{A}_2 \mathrm{e}^{\frac{-\mathrm{E}_{2_2}}{\mathrm{RT}}}}{\mathrm{A}_3 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{a}_1}}{\mathrm{RT}}}}$

$A \mathrm{e}^{\frac{-\mathrm{E}_2}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \mathrm{e}^{\frac{-\left(\mathrm{E}_{\mathrm{a}_2}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{E}_3}\right)}{\mathrm{RT}}}$

$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}$

$400=300+200-\mathrm{E}_{\mathrm{a}_3}$

$\mathrm{E}_{\mathrm{a}_3}=100 \mathrm{~kJ} / \mathrm{mole}$

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