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 = F^o + RT lnP
or, F = F^o + 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 a_A and a_B are the activities of A and B at the start of the reaction and a_C and a_D are the activities of C and D respectively at temperature T then the free energies per mole of A, B, C and D are-
F_A = F^o_A + RTlna_A
F_B = F^o_B + RTlna_B
F_C = F^o_C + RTlna_C
F_D = F^o_D + RTlna_D
When F^o_A, F^o_B, F^o_C and F^o_D are the standard free energies at unit activity. Hence, the free energy change for the reaction becomes-
ΔF = ΣΔF_Product − ΣΔF_Reactant
or, ΔF = (cF_C + dF_D) − (aF_A + bF_B)
or, ΔF = (cF^o_C + RTlna_C + dF^o_D + RTlna_D) − (aF^o_A + RTlna_A + bF^o_B + RTlna_B)
or, ΔF = [(cF^o_C + dF^o_D) − (aF^o_A + bF^o_B)] + [(RTlna_C + RTlna_D) − (RTlna_A + RTlna_B)]

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