Equilibrium
EQUILIBRIUM
• Stable
• Metastable
• Unstable
Equilibrium Model
6th corollary of 2nd law of thermodynamic:
For an isolated system the entropy either
increases or in the limit remains constant.
Surrounding
System
Q
T
Equilibrium Model (cont.)
A system plus surroundings form an isolated system:
dS)syst + dS)sur ≥ 0 ……..........(1)
But:
dS)sur = – δQ / T…………...........(2)
.·. ds)sys – δQ / T ≥ 0……….......(3)
Equilibrium Model (cont.)
Integration:
T(S2 –S1) ≥ Q…………….(4)
1st Law of thermodynamics gives :
δQ – ΔW = du…………….(5)
.·.Q – W = U2 –U1………….(6)
.·.W = Q – (U2 –U1)………..(7)
Substituting for Q from eqn. (4)
.·.W ≤ T (S2 –S1) – (U2 –U1)............(8)
Equilibrium Model (cont.)
Let A = U – TS ……………(9)
W ≤ A1 – A2 ………………(10)
Where:
A: Helmholtz free energy Function.
Equilibrium Model (cont.)
Since the decrease in A of the system is associated
with an increase in entropy of the system and
surroundings then the minimum value of A
corresponds to equilibrium condition.
A given system of prescribed V&T moves to
equilibrium state when it produces work, and the
criterion therefore for spontaneous change for a
closed system is dA<0
Equilibrium Model (cont.)
For a system of constant volume
W=0 .·.A1 > A2
(dA)T,V ≤ 0 ……………………(11)
For a reversible process A1 = A2, for all
other processes there is a decrease in A,
and equilibrium is reached at the
minimum value of A.
Equilibrium Model (cont.)
For the case of the system of variable
volume and constant pressure P, and
temperature T. There will be work done
by the system to maintain the constant
pressure
W (total work) = W′ +P (V2 – V 1)
………….....(12)
W′ = shaft work
P (V2 – V1) = flow work
Equilibrium Model (cont.)
Substituting In equation. (7)
W′ + P (V2 – V1) = Q – (U2 – U1)……...(13)
W′ + P (V2 – V1) = T(S2 – S1) – (U2 – U1)
…….…….(14)
Hence
W′ ≤ T(S2 – S1) – [(U2 + P2V2) – (U1 + P1V1)]
………………(15)
W′ ≤ G1 – G2 …………………………..(16)
Equilibrium Model (cont.)
G = Gibbs free energy function
G = H – TS ………………………..(17)
The maximum useful work is attained
when G reaches its minimum value
.·. At equilibrium
dA)T,V = 0 …………............(18) &
dG)T,P = 0 ………………………..(19)
Chemical Equilibrium of Ideal Gases:
For a multi-component system, the equilibrium
condition may be represented by the equation :
dG)P,T = 0
d Σ (ni gi )T,P = 0 ………….(20)
ni = No of mole of species i
Gi = Gibbs function of species i
Chemical Equilibrium of I. G. (cont.)
For the reaction:
aA+bB cC+dD
Considering A, B, C, and D are ideal gases
For an ideal gas:
g = u + Pv – Ts
dg = du + Pdv + vdP – Tds – sdT
(P is he partial pressure)
dg = vdP – sdT
Chemical Equilibrium of I. G. (cont.)
For an ideal gas (Pv = RT) at constant
temperature.
dg)P,T = RTdP / P
Integration at constant (T) from gO to g
g = gO + RT Ln P
gO is the Gibbs function at reference
temperature.
P in atmospheres (as the reference pressure is
assumed to be 1 atm)
Chemical Equilibrium of I. G. (cont.)
Applying equation (20) for the reaction
c(goC + RT Ln PC) + d(goD + RT Ln PD) –
a(goA + RT Ln PA) – b(goB + RT Ln PB) = 0
Collecting terms and rearranging
cgoC + dgoD – agoA – bgoB = – RT Ln [(PCc
PDd)/(PAa PBb)]
L.H.S. =∆Go
Chemical Equilibrium of I. G. (cont.)
KP = (PCc PDd)/(PAa PBb) ,
KP = equilibrium constant
.·.RT Ln KP = – ∆Go,
.·. Ln KP = – ∆Go / RT
Degree of reaction ε:
ε=[N ) 1 max – N1]/[N1)max – N1)min]
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