Metabolism III: Biochemical thermodynamics and use of Gibbs energies to predict respiratory metabolisms

Question 1

Biochemical thermodynamics

Answer 1

• we can predict what respiratory metabolisms are actually possible.
• we have a given electron donor (which can be the C source if a heterotroph) and a given electron acceptor and we know what the product of the latter would be e.g. O2 is reduced to 2H2O.
• we write out a balanced equation for the oxidation of the donor (in any assessment I will provide this for you!).
• we can then determine the change in Gibbs energy under standard conditions (ΔG°) at 25 °C OR the change in Gibbs energy at a given temperature (ΔG).
• two methods for determination.
• if the ΔG for the reaction is positive, it cannot provide energy for ATP biosynthesis and cannot be a valid metabolism.
• if the ΔG for the reaction is negative, it can in theory provide energy for ATP biosynthesis and can be a valid metabolism BUT the value given is a theoretical maximum – we’ll never see that much energy normally!

Question 2

Method I

Answer 2

• determine ΔG based on:
  ΔG = [ΣGf]products– [ΣGf]reactants
• easiest option – works as long as ΔGf values are known for each chemical involved AND that you’re working at 25 °C.

EXAMPLE:
CH4 + 2O2 → CO2 + 2H2O
if: ΔGf
* methane = -50.79 kJ/mol
* oxygen = 0 kJ/mol
* carbon dioxide = -394.38 kJ/mol
* water = -237.18 kJ/mol
* so [-394.38 + 2(-237.18)] – [-50.79 + 2(0)] = -817.95 kJ/mol CH4
oxidised (could also write it as -408.98 kJ/mol O2 reduced etc)

Question 3

Method II

Answer 3

• determine ΔG based on:
  ΔG = ΔH-TΔS
  where:
  ΔH = [ΣHf]products– [ΣHf]reactants =change in standard enthalpy of formation
  ΔS = [ΣS]products– [ΣS]reactants = change in standard entropy
• longwinded option – but the only option if you don’t have ΔGf values for ALL reactants and products OR you need to change temperature.
  COMMON ERRORS!
  1) T is in K not °C.
  2) S values are always in J/mol/K not kJ/mol so you need to convert the units from J to kJ.

EXAMPLE:
CH4 + 2O2 → CO2 + 2H2O

if: ΔH °f
* methane = -74.85 kJ/mol
* oxygen = -22.70 kJ/mol
* carbon dioxide = -393.51 kJ/mol
* water = -285.83 kJ/mol
if: S°f
* methane = +186.2 J/mol/K
* oxygen = 0 J/mol/K
* carbon dioxide = +213.64 J/mol/K
* water = +69.91 J/mol/K
* [-393.51 + 2(-285.83)] – [-74.85 + 2(-22.70)] = ΔH =-844.92 kJ/mol CH4 oxidised
* [+0.21364 + 2(+0.06991)] – [+0.1862 + 2(0)] = ΔS = +0.16726 kJ/mol/K CH4 oxidised

• 25 °C = 298.15 K
• SO: ΔG = ΔH-TΔS = -844.92 – (298.15 * +0.16726) = -894.79 kJ/mol CH4 oxidised
  (not the same as with Method I - the data are all determined in different ways)