Chapter 9, Problem 16RQ

Summary Introduction

Toanalyze:

The free energy changes occurring in the two reactions.

$\text{a}\text{.}\frac{\text{1}}{\text{2}}{\text{ oxygen(O}}_{\text{2}}){\text{+ nicotinamide adenine dinucleotide(NADH) + H}}^{\text{+}}\to {\text{water(H}}_{\text{2}}\text{O)}{\text{+ NAD}}^{\text{+}}$

$\text{b}\text{. Sulphur}({\text{S) + NADH + hydrogen ion(H}}^{\text{+}})\to \text{Hydrogen}\text{sulphide}({\text{H}}_{\text{2}}{\text{S)+ NAD}}^{\text{+}}$

Introduction:

Gibbs free energy is the change in free energy, which is involved in chemical reactions. Gibbs free energy is released when any chemical bond is broken. This energy can be used for carrying out work on a system.

Explanation of Solution

The free energy changes (∆G°) in the given reactions can be calculated using the following procedure:

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

$\Delta {\text{G}}^{°}\text{= -}nF\Delta E^{°}$

$n$= the number of electrons transferred

$F$= the Faraday constant (96,485 Joule/Volt.mole)

$\Delta E^{°}$ = the difference in reduction potential between the electron donor and the electron acceptor.

In this reaction, the electron acceptor is O₂ and the electron donor is NADH. The number of electrons transferred is 2. The reduction potential for O₂ is 0.815V and for NADH it is -0.32V.

$\begin{array}{c}\Delta E^{°} =\Delta E^{°}\left(\text{electron acceptor}\right)-\Delta E^{°}\left(\text{electron donor}\right)\\ \Delta E^{°}=\Delta E^{°}\left({\text{O}}_{\text{2}}\right)-\Delta E^{°}\left(\text{NADH}\right)\\ \Delta E^{°}=\left(0.815V\right)-\left(-0.32V\right)\\ \Delta E^{°}=+1.135V\end{array}$

$\begin{array}{c}\Delta {\text{G}}^{°}=\left(-2\right)\left(96.5kJ/V.mol\right)\left(1.135V\right)\\ \Delta {\text{G}}^{°}=\left(-193\right)\left(1.135V\right)\\ \Delta {\text{G}}^{°}=-219.05kJ/mol.\end{array}$

$\text{b}{\text{. S + NADH + H}}^{\text{+}}\to {\text{H}}_{\text{2}}\text{S}{\text{+ NAD}}^{\text{+}}$

$\Delta {\text{G}}^{°}\text{= -}nF\Delta E^{°}$

Substituting the values of reduction potential for S and NADH.

$\begin{array}{c}\Delta E° =\Delta E° \left(\text{electron acceptor}\right)-\Delta E° \left(\text{electron donor}\right)\\ \Delta E° =\Delta E° \left(S\right)-\Delta E° \left(\text{NADH}\right)\\ \Delta E° =\left(-\text{0}\text{.23 }V\right)-\left(-0.32V\right)\\ \Delta E° =+0.09V\end{array}$

$\begin{array}{c}\Delta {\text{G}}^{°}=\left(-2\right)\left(96.5kJ/V.mol\right)\left(\text{0}\text{.09 }V\right)\\ \Delta {\text{G}}^{°}=\left(-193\right)\left(0.09V\right)\\ \Delta {\text{G}}^{°}=-17.37kJ/mol.\end{array}$

Conclusion

Therefore, it can be concluded that Gibbs free energy for both the reactionsis negative, which implies that these are exergonic as well as spontaneous reactions. In exergonic reactions, energy is released.