Bioenergetics and Metabolism Miscellaneous

Bioenergetics and Metabolism Miscellaneous

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Bioenergetics and Metabolism

Bioenergetics and Metabolism Miscellaneous
  1. Oxidation reduction reactions with positive standard redox potential (?E0) have








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    Redox potential E0 can be calculated by the ratio between Gibbs free energy and product of Faradays Number with number of participating moles. It implies that the value of Δ G° will be negative as Faraday’s constant is a positive number.

    Correct Option: B

    Redox potential E0 can be calculated by the ratio between Gibbs free energy and product of Faradays Number with number of participating moles. It implies that the value of Δ G° will be negative as Faraday’s constant is a positive number.

Direction: The standard redox potential values for two half-reactions are given below. The value for Faraday’s constant is 96.48 kJ V–1 mol–1 and Gas constant R is 8.31 JK–1 mol–1.
NAD+ + H+ + 2e-      ? NADH      – 0.315 V
FAD + 2H+ + 2e–      ? FADH2      – 0.219 V

  1. The ?G for the oxidation of NADH by FAD is








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    We have, from Gibbs free energy equation:-
    ΔG = –nFΔE
    Where,
    ΔG = change in Gibbs free energy,
    n = number of electrons per mole product,
    F = Faraday’s constant,
    ΔE = change in electrode potential of the reaction.
    Now,
    ΔGOXIDATION = –nFΔ(EREDUCTION – EOXIDATION)
    Given, from equation:  n = 2 (2e– are exchanged)
    EOXIDATION = –0.315 V [Since NADH is oxidized to NAD+ by loss of 2e–] EREDUCTION = –0.219 V [Since FAD is reduced to FADH2 by accepting 2e–]
    Therefore,
    ΔGOXIDATION = – 2 * 96.48 * {(–0.219) – (–0.315)}
    = – 18.524 kJ/mol
    Therefore ΔG for oxidation of NADH by FAD is:
    –18.524 kJ/mol

    Correct Option: D

    We have, from Gibbs free energy equation:-
    ΔG = –nFΔE
    Where,
    ΔG = change in Gibbs free energy,
    n = number of electrons per mole product,
    F = Faraday’s constant,
    ΔE = change in electrode potential of the reaction.
    Now,
    ΔGOXIDATION = –nFΔ(EREDUCTION – EOXIDATION)
    Given, from equation:  n = 2 (2e– are exchanged)
    EOXIDATION = –0.315 V [Since NADH is oxidized to NAD+ by loss of 2e–] EREDUCTION = –0.219 V [Since FAD is reduced to FADH2 by accepting 2e–]
    Therefore,
    ΔGOXIDATION = – 2 * 96.48 * {(–0.219) – (–0.315)}
    = – 18.524 kJ/mol
    Therefore ΔG for oxidation of NADH by FAD is:
    –18.524 kJ/mol

  1. The value of ?G', given Keq as 1.7, at 23°C will be








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    We know, ΔG’ = ΔG° + RT log Keq
    Where, ΔG’= change in Gibbs free energy
    ΔG° = standard change in Gibbs free energy
    R = Universal Gas constant
    T = absolute temperature
    Keq = equilibrium constant
    Given,
    ΔG° = –18.5 kJ/mol, R= 8.31 J/K/mol
    = 8.31 * 10–3 kJ/K/mol, Keq = 1.7,
    T = 230C = (273 + 23) K = 296K
    Therefore,
    ΔG’= –18.5 + [8.31 * 10–3 *296 * log(1.7)]
    = –17.19 kJ/mol
    Therefore the value of ΔG’ is –17.19 kJ/mol

    Correct Option: A

    We know, ΔG’ = ΔG° + RT log Keq
    Where, ΔG’= change in Gibbs free energy
    ΔG° = standard change in Gibbs free energy
    R = Universal Gas constant
    T = absolute temperature
    Keq = equilibrium constant
    Given,
    ΔG° = –18.5 kJ/mol, R= 8.31 J/K/mol
    = 8.31 * 10–3 kJ/K/mol, Keq = 1.7,
    T = 230C = (273 + 23) K = 296K
    Therefore,
    ΔG’= –18.5 + [8.31 * 10–3 *296 * log(1.7)]
    = –17.19 kJ/mol
    Therefore the value of ΔG’ is –17.19 kJ/mol