Question

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:
A.-6RT
B.+6RT
C.$- 6\left( {2.303} \right){\text{RT}}$
D. $+ 6\left( {2.303} \right){\text{RT}}$

Answer 1

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant ‘k’ is given by the Arrhenius equation which is written as:
${\text{k = A}}{{\text{e}}^{\dfrac{{{\text{ - }}{{\text{E}}_{\text{a}}}}}{{{\text{RT}}}}}}$
Here, the pre – exponential factor A represents a constant called the frequency factor, ${{\text{E}}_{\text{a}}}$ represents the energy of activation, R represents the gas constant and T is the absolute temperature.
The addition of a catalyst lowers the activation energy of the reaction.

Complete step by step answer:
Given that the rate of a certain biochemical reaction at physiological temperature (T) occurs ${10^6}$ times faster with enzyme than without the enzyme.
We need to find out the change in activation energy when the addition of the enzyme takes place.
The energy of activation is an important quantity as it is characteristic of the reaction.
The Arrhenius equation can also be written in another form by taking log on both sides of the equation:
$\log {\text{k}} = {\text{logA}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303RT}}}}$
Let the rate constant in the absence of a catalyst be named as k. Then, the Arrhenius equation can be written as
$\log {\text{k}} = {\text{logA}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303RT}}}}$
Let the rate constant in the presence of catalyst be ${\text{k'}}$ and the activation energy be ${{\text{E}}_{\text{a}}}'$. Then, the Arrhenius equation can be written as
$\log {\text{k'}} = {\text{logA}} - \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303RT}}}}$
Since the rate of the reaction increases by ${10^6}$ times faster with enzyme than without the enzyme, so the rate constant ${\text{k'}}$ will be equal to ${10^6}$ k.
Now, subtract the Arrhenius equation in absence of enzyme catalyst from that in presence of enzyme catalyst.

$$\log {\text{k}}' - \log {\text{k}} = - \dfrac{1}{{{\text{2}}{\text{.303RT}}}}\left( {{{\text{E}}_{\text{a}}}{\text{' - }}{{\text{E}}_{\text{a}}}} \right) \\\ \Rightarrow {\text{log}}\dfrac{{{\text{k'}}}}{{\text{k}}} = - \dfrac{1}{{{\text{2}}{\text{.303RT}}}}\left( {{{\text{E}}_{\text{a}}}{\text{' - }}{{\text{E}}_{\text{a}}}} \right) \\\ \Rightarrow {\text{log}}{10^6} = - \dfrac{1}{{{\text{2}}{\text{.303RT}}}}\left( {{{\text{E}}_{\text{a}}}{\text{' - }}{{\text{E}}_{\text{a}}}} \right) \\\ \Rightarrow 6 = - \dfrac{1}{{{\text{2}}{\text{.303RT}}}}\left( {{{\text{E}}_{\text{a}}}{\text{' - }}{{\text{E}}_{\text{a}}}} \right) \\\ \Rightarrow {{\text{E}}_{\text{a}}}{\text{' - }}{{\text{E}}_{\text{a}}} = - 6\left( {{\text{2}}{\text{.303RT}}} \right) \\\$$

So, the correct option is C.

Note:
The Arrhenius equation can also be written in another form:
$\log \dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left( {\dfrac{{{{\text{T}}_{\text{2}}}{\text{ - }}{{\text{T}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}{{\text{T}}_{\text{2}}}}}} \right)$
Or $\log \dfrac{{{{\text{k}}_{\text{2}}}}}{{{{\text{k}}_{\text{1}}}}} = \dfrac{{{{\text{E}}_{\text{a}}}}}{{{\text{2}}{\text{.303R}}}}\left( {\dfrac{1}{{{{\text{T}}_{\text{1}}}}} - \dfrac{1}{{{{\text{T}}_2}}}} \right)$
Here, ${{\text{k}}_{\text{1}}}$ represents the value of the rate constant at the temperature ${{\text{T}}_{\text{1}}}$ and ${{\text{k}}_2}$ represents the value of the rate constant at the temperature ${{\text{T}}_2}$ .
The increase in rate of reaction with temperature is not due to the increase in the total number of collisions but due to the increase in the total number of effective collisions.