Chapter 20, Problem 10P

Interpretation Introduction

(a)

To calculate:

The change of standard free energy for the given reaction.

Introduction:

Two components that are closely related forms the Oxidative phosphorylation and they are chemiosmosis and electron transport chain. The electrons are transferred from one molecule to another molecules and the energy that is produced in this transfer is used in the formation of gradient that is electrochemical.

Explanation of Solution

The electron acceptor is $O$ and the electron donor is NADH that is determined from the half reactions:

  $NA{D}^{+}+2H^{+}+2e^{-}\to NADH+H^{+}(E_0^{\text{'}}=-0.320V)$

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

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

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

  $\begin{array}{l}=+0.816V-(-0.320V)\\ =+1.136V\end{array}$

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

  $\begin{array}{l}=-2(96.485kJ/V.mol)(1.136V)\\ =-219.2kJ/mol\end{array}$

Interpretation Introduction

(b)

To calculate:

The equilibrium constant of the reaction.

Introduction:

Two components that are closely related forms the Oxidative phosphorylation and they are chemiosmosis and electron transport chain. The electrons are transferred from one molecule to another molecules and the energy that is produced in this transfer is used in the formation of gradient that is electrochemical.

Explanation of Solution

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

  $\begin{array}{l}=e^{\frac{-(-219.2)}{(8.314\times {10}^{-3})(298)}}\\ =2.65\times {10}^{38}\end{array}$

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:

Two components that are closely related forms the Oxidative phosphorylation and they are chemiosmosis and electron transport chain. The electrons are transferred from one molecule to another molecules and the energy that is produced in this transfer is used in the formation of gradient that is electrochemical.

Explanation of Solution

From the given we have,

  $G=-219.2kJ/mol$ for 3 ATP

So with 75 % efficiency,

  $-219.2kJ/mol\times 0.75=-164.4kJ/mol$

For 1 ATP,

  $54.8kJ/mol$

With the equation,

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

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

  $\begin{array}{l}=(2mM)\times e^{\frac{(54.8-30.5)}{(8.314\times {10}^{-3})(298)}}\\ =36.4\end{array}$

The ratio of ATP and ADP maximum for the phosphorylation oxidative occurring at $P_i$ =2 $mM$ is 36.4.