Answer:
The concentration of [ADP] = 21.896*10^-6 μM
Explanation:
Given Data:
creatine + ATP -----------> ADP + creatine phosphate
ΔG∘ = -12.6 KJ/mole = -12600 J/mole
ΔG = -0.1 KJ/mole = -100 J/mole
[Creatine phosphate] = 25 mM = 25*10^-3 M
[Creatine] = 17 mM = 17*10^-3 M
[ATP] =5mM = 5*10^-3M
Calculating the concentration of [ADP] using the formula;
ΔG = ΔG∘ + RTlnQc
Substituting, we have
-12600 = -100 + 8.314*298lnQc
-12600+100 = 8.314*298lnQc
-12500 = 2477.57lnQc
lnQc = -12500/2477.57
lnQc = -5.045
Qc = e^ -5.045
Qc = 6.44*10^-3
But,
Qc = [Creatine phosphate]*[ADP]/[creatine]*[ATP]
6.44*10^-3 = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)
6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5
6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]
5.474*10^-7 = 25*10^-3[ADP]
[ADP] = 5.474*10^-7 /25*10^-3
= 2.1896 *10^-5 M
= 21.896*10^-6 μM
Therefore, the concentration of [ADP] = 21.896*10^-6 μM
