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

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Solution

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

$A + B ⇌_{Step 3}^{Step 1}$ $C \overset{Step 2}{\to}$ $P$

Some details of the above reaction are listed below:

Step Rate constant ( $\sec^{- 1}$ ) Activation energy (kJ ${mol}^{- 1}$ )

1 $k_{1}$ $300$

2 $k_{2}$ $200$

3 $k_{3}$ ${Ea}_{3}$

If the overall rate constant of the above transformation (k) is given as $k = \frac{k_{1} k_{2}}{k_{3}}$ and the overall activation energy (Ea) is 400 kJ ${mol}^{- 1}$ , then the value of ${Ea}_{3}$ is ______ kJ ${mol}^{- 1}$ (nearest integer). [2024]

Ans.
(100)

$k = \frac{k_{1} \times k_{2}}{k_{3}}$

$A e^{- \frac{E_{a}}{R T}} = \frac{A_{1} e^{- \frac{E_{a 1}}{R T}} \times A_{2} e^{- \frac{E_{a 2}}{R T}}}{A_{3} e^{- \frac{E_{a 3}}{R T}}}$

$A e^{- E_{a} / R T} = \frac{A_{1} \times A_{2}}{A_{3}} e^{(- E_{a 1} - E_{a 2} + E_{a 3}) / RT}$

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

$400 = 300 + 200 - E_{a 3}$

$E_{a 3} = 100 {kJ mol}^{- 1}$

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