Question

Go'=30.5 kJ/mol

Answer 1

Answer:

a) Keq = 4.5x10^-6

b) [oxaloacetate] = 9x10^-9 M

c) 23 oxaloacetate molecules

Explanation:

a) In the standard state we have to:

ΔGo = -R*T*ln(Keq) (eq.1)

ΔGo = 30.5 kJ/moles = 30500 J/moles

R = 8.314 J*K^-1*moles^-1

Clearing Keq:

Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6

b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])

4.5x10^-6 = ([oxaloacetate]/(0.20*10)

Clearing [oxaloacetate]:

[oxaloacetate] = 9x10^-9 M

c) the radius of the mitochondria is equal to:

r = 10^-5 dm

The volume of the mitochondria is:

V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L

1 L of mitochondria contains 9x10^-9 M of oxaloacetate

Thus, 4.18x10^-42 L of mitochondria contains:

molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules