JEE Main 2022 Chemistry question
The activation energy of one of the reactions in a biochemical process is 532611 J . When the temperature falls from 310 K to 300 K, the change in rate constant observed is . The value of x is . [Given: ln10 = 2.3, R = 8.3 ]
Answer: 1
Solution
= /2.3R)(1/300 − 1/310) = (532611/(2.3
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