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JEE MAINS 2024
09.04.24 S1
Question
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.
$\mathrm{A}+\mathrm{B} \underset{\text { Step 3 }}{\stackrel{\text { Step 1 }}{\rightleftharpoons}} \mathrm{C} \xrightarrow{\text { Step 2 }} \mathrm{P}$
Some details of the above reaction are listed below. 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 $\left(E_a\right)$ is $400 \mathrm{~kJ} \mathrm{~mol}^{-1}$, then the value of $\mathrm{Ea}_3$ is $\$$ iqquad $\$ \mathrm{kJmol}^{-1}$ (nearest integer)
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Solution
$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3} \\ & A e^{\frac{-E_a}{R T}}=\frac{A_1 e^{\frac{-E_{a_1}}{R T}} A_2 e^{\frac{-E_{a_2}}{R T}}}{A_3 e^{\frac{-E_{a_3}}{R T}}} \\ & A e^{\frac{-E_a}{R T}}=\frac{A_1 A_2}{A_3} e^{\frac{-\left(E_{a_1}+E_{a_2}-E_{a_3}\right)}{R T}} \\ & \mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3} \\ & 400=300+200-\mathrm{E}_{\mathrm{a}_3} \\ & \mathrm{E}_{\mathrm{a}_3}=100 \mathrm{~kJ} / \mathrm{mole}\end{aligned}$
Question Tags
JEE Main
Chemistry
Hard
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