The rate of a certain biochemical reaction at physiological temperature T occurs 10 6 times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

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K = Ae - E / RT . . . . . . ( 1 ) 10 6 K = Ae − E c RT … … . ( 2 ) On dividing eq, 2 by eq, 1 10 6 = e E − E c / RT on taking logarithm both sides 6 ln 10 = E − E c RT 2 .303 × 6 = E − E c RT E − E c = 2 .303 × ( 6 RT ) ΔE a = E c − E = − 2 .303 × 6 RT

Chemistry - Chemical Kinetics Question with Solution | TestHub

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