Thermodynamic Derivation of Law of Mass Action
Thermodynamic Derivation of Law of Mass Action
We know that-dF = VdP − SdT
At constant T-
dF = VdP
or, dF = (RT/P)dP (As PV = RT for one mole)
On integrating the above equation-
∫dF = RT∫dP/P
F = Fo + RT lnP
or, F = Fo + RT lna
Since, PV = nRT
So, P = (n/V)RT = CRT
P ∝ c and also a ∝ c
so, P ∝ a
So, we can use a in place of P
Let us consider a general reversible reaction-
aA + bB ⇌ cC + dD
If aA and aB are the activities of A and B at the start of the reaction and aC and aD are the activities of C and D respectively at temperature T then the free energies per mole of A, B, C and D are-
FA = FoA + RTlnaA
FB = FoB + RTlnaB
FC = FoC + RTlnaC
FD = FoD + RTlnaD
When FoA, FoB, FoC and FoD are the standard free energies at unit activity. Hence, the free energy change for the reaction becomes-
ΔF = ΣΔFProduct − ΣΔFReactant
or, ΔF = (cFC + dFD) − (aFA + bFB)
or, ΔF = (cFoC + RTlnaC + dFoD + RTlnaD) − (aFoA + RTlnaA + bFoB + RTlnaB)
or, ΔF = [(cFoC + dFoD) − (aFoA + bFoB)] + [(RTlnaC + RTlnaD) − (RTlnaA + RTlnaB)]
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