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Saturday, 31 October

JEE Main 2022 · ChemistryNumerical answerNumerical valueEasyData or graph

JEE Main 28 July 2022, Shift 2, Chemistry Q58

Question 58 of 90 in this shift, Chemistry question 28 of 30, Section B.

For a reaction, given below is the graph of ln k vs 1T\frac{1}{T}. The activation energy for the reaction is equal to ______ cal mol−1\mathrm{mol^{-1}}. (nearest integer) (Given : R = 2 cal K−1 mol−1\mathrm{K^{-1}\ mol^{-1}})

The figure, in words

Graph of ln k (y-axis, marks at 10 and 20, origin 0.0) versus 1/T (K^-1, x-axis, marks 1 to 5): a straight line falling from ln k = 20 at 1/T = 0 to ln k = 0 at 1/T = 5.

Official answer

8

NTA final key.