Question:

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]

Updated On: Sep 24, 2024

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Correct Answer: 1

The correct answer is: 1

\(\frac{K_{310}}{K_{300}}=10^3⇒\) \(K_{30}=1×10^{-1}×K_{310}\)

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