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Saturday, 31 October

JEE Main 2024 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 4 April 2024, Shift 1, Chemistry Q85

Question 85 of 90 in this shift, Chemistry question 25 of 30, Section B.

Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below. A+B⇌Step 1Step 3C→Step 2P\mathrm{A + B} \underset{\text{Step 3}}{\overset{\text{Step 1}}{\rightleftharpoons}} \mathrm{C} \xrightarrow{\text{Step 2}} \mathrm{P} Some details of the above reactions are listed below. Step | Rate constant (sec−1\mathrm{sec^{-1}}) | Activation energy (kJ mol−1\mathrm{kJ\ mol^{-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{\mathrm{k_1 k_2}}{\mathrm{k_3}} and the overall activation energy (Ea\mathrm{E_a}) is 400 kJ mol−1\mathrm{kJ\ mol^{-1}}, then the value of Ea3\mathrm{Ea_3} is ________ kJ mol−1\mathrm{kJ\ mol^{-1}} (nearest integer)

Official answer

100

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 29 Jan 2024, Shift 1 · Q87For a reaction taking place in three steps at same temperature, overall rate constant K=K1K2K3K=\frac{K_1K_2}{K_3}. If Ea1Ea_1, Ea2Ea_2 and Ea3Ea_3…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.