Answer:
[tex]\Delta G_{rxn}=23.7\ kJ/mol[/tex]
Explanation:
According to the reaction,
[tex]Malate + NAD^+\rightleftharpoons Oxaloacetate + NADH[/tex]
The expression for the equilibrium constant is:-
[tex]K =\frac {[oxaloacetate][NADH]}{[malate][NAD^+]}[/tex]
Given:-
[malate] = 1.13 mM
[oxaloacetate] = 0.270 mM
[tex][NAD^+][/tex] = 390 mM
[NADH] = 160 mM
Thus,
[tex]K = \frac{(0.270)(160)}{(1.13)(390)} = 0.0980[/tex]
The expression for calculation of free energy change is shown below as:-
[tex]\Delta G_{rxn}= \Delta G^0_{rxn} + 2.303RT\times log\ K[/tex]
Where, R is gas constant, 8.314 J/K.mol
T is the temperature in Kelvins
So, Given:- [tex]\Delta G^0_{rxn}=29.7\ kJ/mol[/tex]
T = 310 K
Thus,
[tex]\Delta G_{rxn}=29.7 + (2.303\times 0.008314\times 310\times log 0.0980)[/tex]
[tex]\Delta G_{rxn}=23.7\ kJ/mol[/tex]
