Verified Solution • 2025–26 Session moderate

QJEEM24Chff95: JEE Main 2024 Chemistry - Class 12 JEEM24Chff95 Solution: 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.

JEE Main 2024 (Online) 4th April Morning Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 21 English

Some details of the above reactions are listed below.

Step Rate constant (sec −1^{-1} ) Activation energy (kJ mol −1^{-1} )
1 k1\mathrm{k_1} 300
2 k2\mathrm{k_2} 200
3 k3\mathrm{k_3} Ea3\mathrm{Ea_3}

If the overall rate constant of the above transformation (k) is given as k=k1k2k3\mathrm{k=\frac{k_1 k_2}{k_3}} and the overall activation energy (Ea)(\mathrm{E}_{\mathrm{a}}) is 400 kJ mol m−1400 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1} , then the value of Ea3\mathrm{Ea}_3 is ________ integer)

Step-by-Step Verified Solution

amp;k=k1k2k3amp;Eaeff=Ea1+Ea2−Ea3\begin{aligned}& k=\frac{k_1 k_2}{k_3} \\& E_{a_{e f f}}=E_{a_1}+E_{a_2}-E_{a_3}\end{aligned}

amp;400=300+200−Ea3amp;400=500−Ea3amp;Ea3=100 kJ mole−1\begin{aligned}& 400=300+200-E_{\mathrm{a}_3} \\& 400=500-E_{\mathrm{a}_3} \\& E_{\mathrm{a}_3}=100 \mathrm{~kJ} \mathrm{~mole}^{-1}\end{aligned}

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