I already know the answer to this entire problem, which is -113.5 kj/mol. Howeve

Question

I already know the answer to this entire problem, which is -113.5 kj/mol. However, my setup is obviously wrong because my calculations don't add up to that value.

Balanced equation for Glucose to lactate: Glucose + 2 ADP + 2Pi -> 2 Lactate + 2 ATP + 2 H2O. Standard delta G = -123.1 kj/mol

Calculate the standard free-energy change of this reaction the fact that the standard delta G is -25 kJ/mol for the following reaction:

Pyruvate + NADH + H+ -> Lactate + NAD+

What is the free-energy change (delta G not standard delta G) of this reaction when the concentrations of reactants are: glucose, 5 mM; lactate, 0.05 mM; ATP, 2 mM; ADP, 0.2 mM; and Pi, 1 mM? T = 298 K; R = 0.008315 kj/mol

The answer is -113.5 kj/mol, but I have no idea how. Appreciate the help.

Explanation / Answer

consider the given reaction

Glucose + 2 ADP + 2Pi -> 2 Lactate + 2 ATP + 2 H2O

we know that

dG = dGo + RT lnK

in this case

K = [ lactate]^2 [ATP]^2 / [glucose] [ADP]^2 [Pi]^2

now

using the given values

we get

K = [ 0.05 x 10-3]^2 [ 2 x 10-3]^2 / [5 x 10-3 ] [ 0.2 x 10-3]^2 [1 x 10-3]^2

K = 50

now

dG = dGo + RT lnK

given

dGo = -123.1 x 1000 J/mol

so

dG = (-123.1 x 1000) + ( 8.315 x 298 x ln 50)

dG = (-123.1 x 1000) + ( 9.69348 x 1000)

dG = -113.4065 x 1000

dG = -113.4065 kJ/mol

soo

the free energy change is -113.4065 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

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