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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:

Option: 1

-6RT


Option: 2

-6(2.303)RT


Option: 3

+6RT


Option: 4

+6(2.303)RT


Answers (1)

best_answer

The equation for activation energy is given as follows:

\mathrm{logk\: =\: \frac{-E_{a}}{2.303RT}\: +\: log\, A_{o}}

Now,

\\\mathrm{ln\left (\frac{k_2}{k_1} \right )= -\frac{E_{a_2}-E_{a_1}}{RT}}

\\\mathrm{log\frac{10^6}{1}\: =\: \frac{-1}{2.303RT}}\mathrm{\left ( E_{a_2}-E_{a_1} \right )} \\\\\mathrm{\left ( E_{a_2}-E_{a_1} \right )=\mathrm{ -6(2.303)RT}}

Catalyst decrease the required activation energy for a reaction to proceed and thus increases the reaction rate.

Therefore, Option(2) is correct.

Posted by

Divya Prakash Singh

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