Chapter 9, Problem 17RQ

Summary Introduction

To analyze:

The difference in energy between the oxidation of nicotinamide adenine dinucleotide (NADH) by oxygen (O₂)and sulfur(S).

Introduction:

The change in free energy is Gibbs free energy (∆G°)that is released or utilized in every chemical reaction occurring in nature. This energy is needed to do any work where it is required.

Explanation of Solution

In this reaction, the energy produced for the oxidation of NADH by O₂ is as follows:-

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

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

$n$= the number of electrons transferred

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

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

In this reaction, twoelectrons are being transferred, therefore, $n$= 2. In this reaction, the electron acceptor is O₂ and the electron donor is NADH. The number of electrons transferred is two. The reduction potential for O₂ is 0.815V and for NADH it is -0.32V.

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

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

In the other reaction, the energy produced for the oxidation of NADH by S is as follows:-

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

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

Substituting the values of reduction potential for S and NADH.

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

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

The difference between the energy produced during the oxidation of NADH by O₂ and S is:

$\Delta G^{°}\text{=}\Delta G^{°}\left({\text{O}}_{\text{2}}\right)-\Delta G^{°}\left(\text{S}\right)$

$\Delta G^{°}=\left(-219.05kJ/mol\right)\text{- }\left(-17.37kJ/mol\right)$

$\Delta G^{°}\text{=}\left(\text{-219}\text{.05kJ/mol}\text{+ 17}\text{.37kJ/mol }\right)$

$\Delta G^{°}\text{=}\text{}\left(\text{-201}\text{.68kJ/mol}\right)$

Conclusion

Therefore, it can be concluded that the difference between the energy produced during oxidation of NADH by O₂ and S is -201.68 kJ/mol.