Question

g 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 ∘C (310 K). Δ????∘′ for the reaction is +29.7 kJ/mol . Assume that the reaction occurs at pH 7. [malate]=1.37 mM [oxaloacetate]=0.130 mM [NAD+]=460 mM [NADH]=180 mM Δ????:

Answer 1

The free energy change for malate dehydrogenase reaction is -50kJ/mol

Step-by-step explanation:

The malate dehydrogenase reaction is:

Malate + NAD⁺ ⇄ Oxaloacetate + NADH + H⁺

Where equilibrium constant, k, could be expressed as:

`K = ([Oxaloacetate][NADH][H^+])/([NAD^+][Malate])`

Replacing with the listed concentrations:

`K = ([0,130mM][180mM][10^(-7)M])/([460mM][1,37mM])`

K = 3,713x10⁻⁹

ΔG° is defined as:

ΔG° = RT ln K (1)

Where:

R is gas constant 8,314J/molK

T is temperature 310K

And K = 3,713x10⁻⁹

ΔG° = -50000 J/mol ≡ -50kJ/mol

I hope it helps!