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Friday, 9 October

JEE Main 2022 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 29 June 2022, Shift 1, Chemistry Q85

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

The activation energy of one of the reactions in a biochemical process is 532611 J mol−1532611\ \mathrm{J\ mol^{-1}}. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×10−3 k310k_{300} = x \times 10^{-3}\ k_{310}. The value of xx is ____________. [Given : ln⁡10=2.3\ln 10 = 2.3 R=8.3 J K−1 mol−1R = 8.3\ \mathrm{J\ K^{-1}\ mol^{-1}}]

Official answer

1

NTA final key (2022 Session 1).