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Friday, 9 October

JEE Main 2022 · ChemistryNumerical answerNumerical valueEasyData or graph

JEE Main 24 June 2022, Shift 1, Chemistry Q85

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

The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700−1000700-1000 K. The data has been analysed by plotting ln⁡k\ln k vs 103T\frac{10^3}{T} graph. The value of activation energy for the reaction is ______ kJ mol−1\mathrm{kJ\ mol^{-1}}. (Nearest integer) (Given : R=8.31 J K−1 mol−1R=8.31\ \mathrm{J\ K^{-1}\ mol^{-1}})

The figure, in words

Graph of ln k (y-axis) vs 10^3/T (x-axis): straight line with negative slope, labelled Slope = -18.5.

Official answer

154

NTA final key (2022 Session 1).

Same idea in other shifts

Asked 2× in all
  1. 6 Apr 2023, Shift 1 · Q87For the adsorption of hydrogen on platinum, the activation energy is 30 kJ mol−1^{-1} and for the adsorption of hydrogen on nickel, the…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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.