Chapter 20, Problem 9P

Interpretation Introduction

(a)

To calculate:

The change of standard free energy for the given reaction.

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Explanation of Solution

The electron acceptor is $O$ and the electron donor is succinate which is determined from the half reactions:

  $Fumarate+2H^{+}+2e^{-}\to Succinate+(E_0^{\text{'}}=+0.031V)$

  $\frac{1}{2}{O}_2+2H^{+}+2e^{-}\to H_{2}O,(E_0^{\text{'}}=+0.816V)$

The free change of energy for the reaction is calculated by determining the change of the standard potential:

  $\Delta E_0^{\text{'}}=acceptor\Delta E_0^{\text{'}}-donor\Delta E_0^{\text{'}}$

  $\begin{array}{l}=+0.816V-(0.031V)\\ =+0.785V\end{array}$

  $\begin{array}{l}\Delta G^{0\text{'}}=-nF\Delta E_0^{\text{'}}\\ n=2\\ F=96,485kJ/Vmol\end{array}$

  $\begin{array}{l}=-2(96.485kJ/V.mol)(0.785V)\\ =-151.5kJ/mol\end{array}$

Interpretation Introduction

(b)

To calculate:

The equilibrium constant of the reaction.

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Explanation of Solution

  $\begin{array}{l}\Delta G^{0\text{'}}=-RT\ln K_{eq}\\ K_{eq}=e^{\frac{-\Delta G}{RT}}\end{array}$

  $\begin{array}{l}=e^{\frac{-(-151.5)}{(8.314\times {10}^{-3})(298)}}\\ =3.60\times {10}^{26}\end{array}$

Interpretation Introduction

(c)

To determine:

The release of the actual free energy accompanying the coenzyme Q reductase of NADH is equal to releasing amount under the normal condition

Introduction:

Succinate dehydrogenase is also known as SQR succinate Q reductase is seen in the cells of bacteria as a complex enzyme. It is also found in the eukaryotes in the membrane of inner mitochondria. Thus enzyme takes part in the electron transport chain and citric acid cycle.

Explanation of Solution

From the given we have,

  $G=-151.3kJ/mol$ for 2 ATP

So with 70% efficiency,

  $-151.3kJ/mol\times 0.70=-106.1kJ/mol$

For 1 ATP,

  $53.1kJ/mol$

With the equation,

  $\Delta G=\Delta G^0+RT\ln \frac{(ATP)}{(ADP)(P_i)}$

  $\frac{(ATP)}{(ADP)}=(P_i)\times e^{\frac{\Delta G-\Delta G^{0\text{'}}}{RT}}$

  $\begin{array}{l}=(1mM)\times e^{\frac{(53.05-30.5)}{(8.314\times {10}^{-3})(298)}}\\ =8.89\end{array}$

The ratio of ATP and ADP maximum for the phosphorylation oxidative occurring at $P_i$ =1 $mM$ is 8.89.