Thermodynamics of Bioreactions
99
For the corresponding reaction where the produced water is in equilibrium with liquid water at 25°C
fO;+H,
(HjO)^ ; AC,
(H20)e3s <->(H20),
; AG, =0
(2)
The total reaction to form water in the liquid phase has A G = A G , + A
G2
= A G j
AG = A G °+ JR7,ln—
= -5 4 6 4 0 -2 0 5 3 = -56693 cal mole1
(3)
760
since the partial pressure of H30 in equilibrium with liquid water at 25°C is 23.7 mm Hg. Apart from round-
off errors this is the same result as was obtained in Example 4.1 where a standard free energy AG° was
defined in a reference frame of liquid water at 25 C rather than gaseous water. Similarly, if in a reaction
with no liquid water present we can remove the produced water vapor to a partial pressure of 23.7 mm Hg
while the partial pressures of 0 2 and H3 are kept at 1 atm we can shift the free energy of the gas phase
reaction from -54.64 to -56.69 kcal mole'1
H3Q formed._________________________________
The two simple examples 4.1 and 4.2 are given to show how thermodynamic properties can be
calculated using eqs. (4. l)-(4.4). It is demonstrated how the value of A G can be pushed in a desired
direction by fixing the concentrations (or for gas phase reactions the partial pressures) o f reactants
at levels different from those at standard conditions. Since the reaction only proceeds spontaneously
when A G is negative we may force it thermodynamically by decreasing the product concentrations
and increase the substrate concentrations. For reactions to run inside a living organism it is
therefore important that the levels of substrates and products are allowed to vary without affecting
the overall thermodynamic feasibility of the reaction. For this reason the first reaction in a pathway
is typically designed to have a large and negative A G as it may hereby be thermodynamically
feasible even when the substrate concentration gets very low. Similarly the last reaction in a
pathway also typically has a large and negative A G to make it thermodynamic feasible even when
the product concentration gets very high. This is illustrated for the EMP pathway in Example 4.3.
The typical way in which a pathway reaction is forced to proceed in a thermodynamically
unfavored direction is by coupling it to a thermodynamically favoured reaction. This is where ATP
plays a crucial role as a co-factor in pathway reactions. Thus, phosphorylation of glucose to
glucose-6 phosphate (G6P) is coupled to the hydrolysis of one ATP molecule to ADP
ATP + ADP + Pj = 0 ; A G ° = —30.5 kJ mole’1
(4.6)
- glucose - P; + G6P = 0 ; A G °=13.8
kJ mole'1
(4.7)
Addition of (4.6) and (4.7) yields
-glucose-A TP + ADP + G6P = 0
;
AG° = -30.5 + 13.8 = -16.7 kJ m ole1
(4.8)