Go\'=30.5 kJ/mol a) Calculate the equilibrium constant for this reaction at 25C.

Question

Go'=30.5 kJ/mol

a) Calculate the equilibrium constant for this reaction at 25C.

b) Because deltaGo' assumes standard pH of 7, the equilibrium constant calculated in (a) corresponds to Keq'= [oxaloacetate] [NADH]/ [L-malate] [NAD+]

The measured concentration of L-malate in rat liver mitochondria is about 0.20 mM when [NAD+]/[NADH] is 10. Calculate the concentration of oxaloaceteate at pH 7 in mitochondria.

c) To appreciate the magnitude of the mitochondrial oxaloacetate concentration, calculate the number of oxaloacetate molecules in a single rat mitochodrion. Assume the mitochondrion is a sphere of diameter 2.0 micro meters.

Explanation / Answer

a. At standard state, dGo = -RT ln Keq

or, 30.5 kJ/mole = - 8.314 J.K-1.mole-1 * 298 K * ln Keq

or, Keq = 4.5 * 10-6

b. Keq'= [oxaloacetate] [NADH]/ [L-malate] [NAD+]

or, 4.5 * 10-6 = [oxaloacetate]/(0.20 mM)*10

or, [oxaloacetate] = 9*10-6 mM = 9*10-9 M

c. Radius of mitochondrion = 1 micrometer = 10-6 m = 10-5 dm

volume of mitochondrion = (4/3) r3 = (4/3) (10-5 dm)3 = (4/3)(10-15) dm3 = (4/3)(10-15) liter

According to the definition of molarity we can say:

1 liter of mitochondrion plasma contains 9*10-9 M oxaloacetate

So, (4/3)(10-15) liter of mitochondrion plasma contains [(4/3)(10-15) * 9*10-9 M] oxaloacetate

or, (4/3)(10-15) liter of mitochondrion plasma contains [(4/3)(10-15) * 9*10-9 * 6.023*1023 ] oxaloacetate molecules or 23 oxaloacetate molecules.

Ans: 23 molecules.

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.