The activation energy of one of the reactions in a biochemical process is 532611 J mol–1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x × 10–3 k310. The value of x is ______.
[Given: ln10 = 2.3, R = 8.3 JK–1 mol–1]
The correct answer is: 1
\(\frac{K_{310}}{K_{300}}=10^3⇒\) \(K_{30}=1×10^{-1}×K_{310}\)







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