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 $k_{300} = x \times 10^{- 3} k_{310 °}$ . The value of $x$ is
[Given: $\ln 10 = 2.3 R = 8.3 J K^{- 1} {mol}^{- 1}$ ]

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ln K 2 K 1 = E a R 1 T 1 - 1 T 2 ln K 2 K 1 = 532611 8 . 3 × 10 310 × 300 where K 2 is at 310 K & K 1 is at 300 K ln K 2 K 1 = 6 . 9 = 3 × ln 10 ln K 2 K 1 = ln 10 3 K 2 = K 1 × 10 3 K 1 = K 2 × 10 3 So

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