JEE Main 29 June 2022 Shift 1 Solved Paper

Section: Chemistry
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Question : 85 of 90
 
Marks: +1, -0
The activation energy of one of the reactions in a biochemical process is 532611Jmol−1. When the temperature falls from 310 K to 300K, the change in rate constant observed is k300=x×10−3k310. The value of x is______
[Given .:ln‌10=2.3,R=8.3JK−1mol−1]
[29-Jun-2022-Shift-1]
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