Engineering Chemistry S Datta

THERMODYNAMICS 91
Therefore,
dn 1
= n 1
∆x, dn 2
= n 2
∆x, dn i
= n i
∆x
Since, G is an extensive property, therefore it will increase by the amount G∆x
hence,
So, we have,
dG = G∆x
dG = µ 1
n 1
∆x + µ 2
n 2
∆x + ... + µ i
n i
∆x + ...
or, G∆x = µ 1
n 1
∆x + µ 2
n 2
∆x + .. + µ i
n i
∆x + ...
Hence, G = µ 1
n 1
+ µ 2
n 2
+ ... + µ i
n i
+ ... = ∑µ i
n i
Now, complete differentiation of this equation gives us,
Subtracting equation (a) from (b)
dG = (µ 1
dn 1
+ µ 2
dn 2
+ µ 3
dn 3
+ ...) + (n 1
dµ 1
+ n 2
dµ 2
+ n 3
dµ 3
+ ...) ...(b)
n 1
dµ 1
+ n 2
dµ 2
+ n 3
dµ 3
+ ... = 0
We have thus three useful relations for chemical potentials with the composition of the
system,
(i) G = µ 1
n 1
+ µ 2
n 2
+ µ 3
n 3
+ ... = ∑µn
(ii) dG = µ 1
dn 1
+ µ 2
dn 2
+ µ 3
dn 3
+ ... = ∑µdn
(iii) 0 = n 1
dµ 1
+ n 2
dµ 2
+ n 3
dµ 3
+ ... = ∑ndµ
These relations are commonly known as Gibbs-Duhem relations.
Deduction of Van’t Hoff’s reaction isotherm (Application of thermodynamics to
homogeneous equilibrium or thermodynamics of chemical equilibrium):
Let us consider a general reaction
aA + bB → cC + dD ...(i)
We, know, G = H – TS = E + PV – TS [Q H = E + PV]
So,
dG = dE + PdV + VdP – TdS – S dT.
Again,
dq = dE + PdV and dS = dq/T
∴
dG = V.dP – S.dT
At constant temperature, (dG) T
= V.dP.
Free energy change for one mole of any gas at a constant temperature is given by:
Integrating both sides,
dG = V.dP = R.T dP P
z
L
NM Q
RT
PV = RT or V =
P
dP
∫ dG = RT
P
So, G = G 0 + RT ln P ...(ii)
where G 0 = integration constant known as standard free energy i.e.,
G 0 = G when P = 1 atm.
Let the energy/mole of A, B, C and D at their respective pressures P A
, P B
, P C
and P D
are
G A
, G B
, G C
and G D
, respectively. Then, free energy change for the reaction (i) is given by
∆G = G products
– G reactants
= (cG C
+ dG D
) – (aG A
+ bG B
) ...(iii)
O
QP