Chapter 9, Problem 47TQ

Summary Introduction

To review:

The standard free energy for the following reactions:

$\begin{array}{l}\text{a}\text{. NADH}+{\text{H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}\text{O}\hfill \\ \text{b}\text{. Cytochrome c }\left({\text{Fe}}^{\text{2+}}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\to \text{Cytochrome c }\left({\text{Fe}}^{\text{3+}}\right)+{\text{H}}_{\text{2}}\text{O}\hfill \\ \hfill \end{array}$

Introduction:

Gibbs free energy is the potential that can be used for the calculation of maximum reversible work that is performed at constant temperature and pressure. The phosphoryl group transfer potential of a compound can be defined as ameasure of the strength of attachment of a group to amolecule. It usually refers to the differences in the standard free energies of the molecule with and without the group.

Explanation of Solution

According to the standard reduction potentials (E°’) given in table 9.1, E°’ of some redox half-reactions are given below:

Redox half-reaction	E°’ (in Volt or V)
$\frac{1}{2}{\text{O}}_{\text{2}}+2{\text{H}}^{\text{+}}+2e^{-}\to {\text{H}}_{\text{2}}\text{O}$	$+\text{0}\text{.82}$
${\text{NAD}}^{\text{+}}+{\text{H}}^{\text{+}}+{\text{2e}}^{\text{-}}\to \text{NADH}$	-0.32
$\text{Cytochrome c }\left({\text{Fe}}^{\text{3+}}\right)+{\text{e}}^{-}+{\text{H}}_{\text{2}}\text{O}\to {\text{cytochrome c (Fe}}^{\text{2+}}\text{)}$	+0.235

The calculation of standard free energy for the following reactions is described below:

$\text{a}\text{. NADH}+{\text{H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}\text{O}$

For calculating the standard free energy, the number of electrons transferred needs to be balanced. For a reaction, the standard free energy can be calculated by using the Nernst equation $\Delta \text{G}=\text{-nF}\Delta \text{E}°\text{'}$ .

Where,

n is the number of electrons transferred,

F is Faraday’s constant, which is 96.15 kJ/V.mol (kilojoule per Volt. mole) and

∆E°’ is overall cell potential.

∆E^°’ can be calculated by the following formula:

$\Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\text{electron acceptor}\right)-{\text{E}}^{°\text{'}}\left(\text{electron donor}\right)\text{V}$

In the given case, E°’ of electron acceptor is +0.82 V and that of electron donor is -0.32 V.

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\text{electron acceptor}\right)-{\text{E}}^{°\text{'}}\left(\text{electron donor}\right)\text{V}\left(\text{volt}\right)\\ \Delta {\text{E}}^{°\text{'}}=\text{0}\text{.82-}\left(\text{-0}\text{.32}\right)\text{V}\\ \Delta {\text{E}}^{°\text{'}}=\text{1}\text{.14V}\end{array}$

Putting the values of n, F, and ∆E^°’ in the Nernst equation:

$\begin{array}{l}{\text{ΔG}}^{o\text{'}}=\text{-nΔFE}°\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-2}\times \left(\text{96}\text{.15 kJ/V}\text{.mol}\right)\times \text{1}\text{.14V}\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-219}\text{.222kJ/mol}\hfill \end{array}$

Thus, the standard free energy of the reaction is $\text{-219}\text{.222 kJ/mol}$

$\text{b}\text{. Cytochrome c }\left({\text{Fe}}^{\text{2+}}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\to \text{Cytochrome c }\left({\text{Fe}}^{\text{3+}}\right)+{\text{H}}_{\text{2}}\text{O}$

In the given case, E°’ of the electron acceptor is +0.82 V and that of the electron donor is -0.235 V.

For a reaction, the standard free energy can be calculated by using the Nernst equation.

$\begin{array}{l}\Delta {\text{E}}^{°\text{'}}={\text{E}}^{°\text{'}}\left(\text{electron acceptor}\right)-{\text{E}}^{°\text{'}}\left(\text{electron donor}\right)\text{V }\left(\text{volt}\right)\\ \Delta {\text{E}}^{°\text{'}}=\text{0}\text{.82-}\left(\text{0}\text{.235}\right)\text{V}\\ \Delta {\text{E}}^{°\text{'}}=0.585\mathrm{V}\end{array}$

Putting the value of n, F, and ∆ E°’ in the Nernst equation:

$\begin{array}{l}{\text{ΔG}}^{o\text{'}}=\text{-nΔFE}°\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-2}\times \left(\text{96}\text{.15 kJ/V}\text{.mol}\right)\times \text{0}\text{.585V}\hfill \\ \Delta {\text{G}}^{\text{o’}}=\text{-112}\text{.4955kJ/mol}\hfill \end{array}$

Thus, the standard free energy of the reaction is $\text{-112}\text{.4955 kJ/mol}$ .

Conclusion

Thus, it can be concluded that the standard free energy of $\text{NADH}+{\text{H}}^{\text{+}}+\frac{1}{2}{\text{O}}_{\text{2}}\to {\text{NAD}}^{\text{+}}+{\text{H}}_{\text{2}}\text{O}$ is $\text{-219}\text{.222 kJ/mol}$ and the standard free energy of $\text{Cytochrome c }\left({\text{Fe}}^{\text{2+}}\right)+\frac{1}{2}{\text{O}}_{\text{2}}\to \text{Cytochrome c }\left({\text{Fe}}^{\text{3+}}\right)+{\text{H}}_{\text{2}}\text{O}$ is $\text{-112}\text{.4955 kJ/mol}$ .