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Q.

The activation energy of one of the reactions in a biochemical process is 532611Jmol1 When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300=x×103k310. The value of x is  ________.

[Given :  ln10=2.3 R=8.3JK1mol1]

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answer is 1.

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Detailed Solution

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ℓnK2K1=EaR1T11T2ℓnK2K1=5326118.3×10310×300
where K2 is at 310 K & K1 is at 300 K
lnK2K1=6.9
=3×ℓn10ℓnK2K1=ln103K2=K1×103K1=K2×10-3
So K = 1

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