EQUILIBRIUM CONSTANT OF A GAS REACTION IN TERMS OF PARTITION FUNCTION

It's possible to derive an expression for the equilibrium of a reaction from a knowledge of total partition functions of the reactants and products. The equilibrium constant Kp and standard free energy ∆G⁰ are related by

∆G⁰ = − RTln  Kp ........................(1)

For one mole of an ideal gases

A= - KT lnQ

Molar partition function Q = qN  .....(2)

                                                  Nǃ

Applying Stirling's Approximation

ln Nǃ = N ln N−N ; Nǃ = (N )ᴺ

                                         e

A = - KT ln (qe)N.................... (3)

                     N

For one mole of an ideal gas, 

    G= A + PV = A+RT.................... (4)

Substituting the value of A from equation (3) into equation (4)we get,

G = - kT ln (qe)N + RT         ( Nk= R) 

                     N

=-NkT lnq - NkT + NkT ln N + NkT = - 

RT ln (q/N).............. (5)

The expression for free energy of an ideal gas is, 

G = E+ PV - TS = E - TS + RT........... (6)

Substitute G= - RT ln (q/N) and E - TS + RT ........ (6)

S = - RT ln (q/N) - RT² ( dlnq )v + R .....(7)

                                          dT

E is the total energy E₀ is the total energy in the Lowes possible level i.e the zero point energy

E - E₀ = RT² (dlnq)v ; E = E₀ + RT² (dlnq )v−

                       dT.                               dT

                    RT+ RT

G = E₀ − RT ln (q/N)    ...............(8)

E₀ is taken as the energy of the zero level. For calculating the equilibrium constant for a chemical reaction at a given temperature T, we must know ∆G⁰T. The 

∆G⁰T term involves combination of GT⁰ terms for reactants and products. For this reason the difference in zero point energies of the various species must be taken into account.

If the pressure is 1 atm i.e standard conditions,

     G⁰ − E⁰  =− R∂ln (q⁰/N)    .........(9)

         T

For one mole of an ideal gas,

H equal E + PV = E + RT

At standard state, H⁰ = E⁰ + RT

At absolute zero, T = 0. Hence H⁰₀ = E⁰₀ ;

G⁰₀ = H₀⁰ − RT ln (q⁰/N)  or

G⁰₀ − H⁰₀  = − R ln (q⁰/N)       ........(10) or

      T

G⁰ = H⁰₀ − RT ln (q⁰/N)

Consider a reversible reaction, 

aA + bB →cC + dD

∆G⁰ for the reaction is given by 

∆G⁰ ⁼ (cGc⁰ + dɢ⁰D) − ( aɢA⁰ + bɢB⁰)

G⁰ values for the species A, B, C and D are 

G⁰A = H⁰₀A − RT ln (q⁰A) ;

                                   N

Ɠ⁰B = H⁰₀ B − RT ln (qB⁰)

                                    N

ɢ⁰꜀ = H⁰₀ ꜀ - RT ln (q⁰꜀) ; G⁰D = H⁰₀ D − RTln

  (qD⁰)

    N

∆G₀ = ( cH⁰₀c + dH⁰₀D) − ( aH⁰₀ + bH⁰₀B) −

RT ln  [(qc⁰/N)ᶜ(qD⁰/N)d ]

           (q⁰A/N)ᵃ(qB⁰/N)ᵇ

∆G₀ = ∆H⁰₀ RT ln [(qc⁰/N)ᶜ(qD⁰/N)ᵈ]

                               q⁰A/N)ᵃ(qB⁰/N)ᵇ

∆H⁰₀ is enthalpy change at absolute zero, the reactants and products being in their respective standard states. 

∆G⁰ = RT ln Kp, ln Kp = − ∆G⁰/RT

ɢ⁰ = H⁰₀ −RT ln (q⁰/N)