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Chemical Kinetics and Nuclear Chemistry question

2024 · 4 Apr · Shift 1 · Q23
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Chemical Kinetics and Nuclear Chemistry question

2024 · 4 Apr · Shift 1 · Q23

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

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 34 English

Some details of the above reactions are listed below.

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

If the overall rate constant of the above transformation (k) is given as k=k1k2k3\mathrm{k=\frac{k_1 k_2}{k_3}}k=k3​k1​k2​​ and the overall activation energy (Ea)(\mathrm{E}_{\mathrm{a}})(Ea​) is 400 kJ mol m−1400 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1}400 kJ mol m−1, then the value of Ea3\mathrm{Ea}_3Ea3​ is ‾\underline{\hspace{2cm}}​ integer)

Numerical answer
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