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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2024 · 4 April · Morning Shift · Chemistry · Chemical Kinetics

Consider the following transformation involving first order...

Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. (A+B \underset{\text{ Step 3 }}{\text{ Step } 1} C \xrightarrow{\text{ Step 2 }} P) Some details of the above reactions are listed below. ( Step  Rate constant (sec−1) Activation energy (kJ mol−1)1k13002k22003k3Ea3\begin{array}{ccc} \text{ Step } & \text{ Rate constant }\left(\mathbf{s e c}^{-\mathbf{1}}\right) & \text{ Activation energy }\left(\mathbf{k J ~ m o l}^{-\mathbf{1}}\right) \\ 1 & \mathrm{k}_1 & 300 \\ 2 & \mathrm{k}_2 & 200 \\ 3 & \mathrm{k}_3 & \mathrm{Ea}_3 \end{array} If the overall rate constant of the above transformation (k)(k) is given as (k=\frac{k_1 k_2}{k_3}) and the overall activation energy (\left(\mathrm{E}_{\mathrm{a}}\right)) is (400 \mathrm{~kJ} \mathrm{~mol}^{-1}), then the value of (\mathrm{Ea}_3) is (\qquad) (\mathrm{kJ} \mathrm{mol}^{-1}) (nearest integer)

Correct answer

100

Step-by-step solution

K=K1 K2 K3Ae−E3RT=A1e−Ea1RTA2e−E32RTA3e−Ek3RTAe−E3RT=A1 A2 A3e−(Ea1−Ea2−E33)RTEa=Ea1+Ea2−Ea3400=300+200−Ea3Ea3=100 kJ/mole\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3} \\ & \mathrm{Ae}^{\frac{-\mathrm{E}_3}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{a}_1}}{\mathrm{RT}}} \mathrm{A}_2 \mathrm{e}^{\frac{-\mathrm{E}_{3_2}}{\mathrm{RT}}}}{\mathrm{A}_3 \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{k}_3}}{\mathrm{RT}}}} \\ & \mathrm{Ae}^{\frac{-\mathrm{E}_3}{\mathrm{RT}}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \mathrm{e}^{\frac{-\left(\mathrm{E}_{\mathrm{a}_1}-\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{3_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}\end{aligned}
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