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:

Step-by-step solution

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