Question

Consider the malate dehydrogenase reaction from the citric acid cycle. Given the listed concentrations, calculate the free energy change for this reaction at energy change for this reaction at 37.0 ∘C37.0 ∘C (310 K). ΔG∘′ΔG∘′ for the reaction is +29.7 kJ/mol+29.7 kJ/mol . Assume that the reaction occurs at pH 7.

[malate]=1.27 mM

[oxaloacetate]=0.150 mM

[NAD+]=150 mM

[NADH]=6.0×101 mM

ΔG:_______________kJ⋅mol−1

Answer 1

The malate mdehydrogenase reaction is
Malate + NAD^+ -------> Oxaloacetate + NADH + H^+
equilibrium constant
K   = [oxaloaceatate][NADH][H^+]/[NAD^+][Malate]
[malate]=1.27mM   = 1.27*10^-3M

[oxaloacetate]=0.150 mM   = 0.15*10^-3M

[NAD+]=150 mM   = 150*10^-3M

[NADH]=6.0×101 mM   = 6*10^1*10^-3M
PH = 7
-log[H^+] = 7
[H^+] = 10^-7M
K = 0.15*10^-3 *6*10^1*10^-3*10^-7/150*10^-3*1.27*10^-3
   = 9*10^-13/0.0001905
   =4.72*10^-9

G⁰    = -RTlnk

             = 8.314*298ln(4.72*10^-9)

             = 8.314*298*-19.1714

            = -47498J/mole

            = 47.498KJ/mole >>>>answer