Chapter 14, Problem 4P

Interpretation Introduction

To derive:

The expression of $k_e/k_{\mu }$.

Introduction:

The Arrhenius Law defines the activation energies and the rate constants of the reaction. The reactions can be of two types − catalyzed and uncatalyzed. In catalyzed reaction the presence of catalyst like enzymes reduce the activation energy and hence the reaction can be triggered at a lower energy level. In this question we are deriving an equation for catalytic power, that is, the ratio of catalyzed and uncatalyzed reactions.

Explanation of Solution

The catalyzed reaction is given as:

  $\begin{array}{l}v=k_e\left[ES\right]\\ v=k_e^{\text{'}}\left[E{S}^{\mp }\right]\\ k_e\left[ES\right]=k_e^{\text{'}}\left[E{S}^{\mp }\right]\end{array}$

The equilibrium constant will determine the concentration of EX transition state which is given below:

  $\begin{array}{l}{K}_e^{\mp }=\left[E{X}^{\mp }\right]÷\left[ES\right]\\ \left[E{X}^{\mp }\right]=K_e^{\mp }\left[ES\right]\end{array}$

The relation between free activation energy and equilibrium constant can be given by the following equation:

  $\begin{array}{l}\Delta G_e^{\mp }=-RT\ln K_e^{\mp }\\ K_e^{\mp }=e^{\raisebox{1ex}{$-\Delta G_e^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\end{array}$

After substituting and simplifying the equations we get:

  $\begin{array}{l}{k}_e\left[ES\right]=k_e^{\text{'}}\left[E{X}^{\mp }\right]\\ k_e\left[ES\right]=k_e^{\text{'}}{K}_e^{\mp }\left[ES\right]\\ k_e\left[ES\right]=k_e^{\text{'}}{e}^{\raisebox{1ex}{$-\Delta G_e^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\left[ES\right]\\ k_e=k_e^{\text{'}}{e}^{\raisebox{1ex}{$-\Delta G_e^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\end{array}$

The above reaction was for catalyzed reactions. The reactions for uncatalyzed can be seen below:

  $\begin{array}{l}v=k_{\mu }\left[S\right]\\ v=k_{\mu }^{\text{'}}\left[X^{\mp }\right]\\ k_{\mu }\left[S\right]=k_{\mu }^{\text{'}}\left[X^{\mp }\right]\end{array}$

Similarly, here the equilibrium constant determines the concentration of the X transition state as below:

  $\begin{array}{l}{K}_{\mu }^{\mp }=\left[X^{\mp }\right]/\left[S\right]\\ \left[X^{\mp }\right]=K_{\mu }^{\mp }/\left[S\right]\end{array}$

The relation between free activation energy and equilibrium constant in this case can be given seen in the following equation:

  $\begin{array}{l}\Delta G_{\mu }^{\mp }=-RT\ln K_{\mu }^{\mp }\\ K_{\mu }^{\mp }=e^{\raisebox{1ex}{$-\Delta G_{\mu }^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\end{array}$

Thus, simplifying the equations:

  $\begin{array}{l}{k}_{\mu }\left[S\right]=k_{\mu }^{\text{'}}\left[X^{\mp }\right]\\ k_{\mu }\left[S\right]=k_{\mu }^{\text{'}}{K}_{\mu }^{\mp }/\left[S\right]\\ k_{\mu }\left[S\right]=k_{\mu }^{\text{'}}{e}^{\raisebox{1ex}{$-\Delta G_{\mu }^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\left[S\right]\\ k_{\mu }=k_{\mu }^{\text{'}}{e}^{\raisebox{1ex}{$-\Delta G_{\mu }^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\end{array}$

Hence, it can be considered that:

  $\begin{array}{l}{k}_e^{\text{'}}=k_{\mu }^{\text{'}}\\ k_e/k_{\mu }=e^{\raisebox{1ex}{$-\Delta G_e^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}/e^{\raisebox{1ex}{$-\Delta G_{\mu }^{\mp }$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}\\ k_e/k_{\mu }=e^{\left(\Delta G_{\mu }^{\mp }-\Delta G_e^{\mp }\right)/RT}\end{array}$