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5.60 Thermodynamics & Kinetics
Spring 2008
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5.60 Spring 2008
Lecture #14
page 1
Multicomponent Systems, Partial Molar Quantities,
and the Chemical Potential
So far we’ve worked with fundamental equations for a closed (no
mass change) system with no composition change.
dU =TdS − pdV
dA = −SdT − pdV
dH =TdS +Vdp
dG = −SdT +Vdp
How does this change if we allow the composition of the system to
change? Like in a chemical reaction or a biochemical process?
• Consider Gibbs free energy of a 2-component system
⇒
G (T , p , n1 , n2 )
⎛ ∂G ⎞
⎛ ∂G ⎞
⎛ ∂G ⎞
⎛ ∂G ⎞
dG = ⎜
dT + ⎜
dp + ⎜
dn1 + ⎜
dn2
⎟
⎟
⎟
⎟
∂p ⎠T ,n ,n
∂n1 ⎠T , p ,n
∂n2 ⎠T , p ,n
⎝ ∂T ⎠ p ,n ,n
⎝
⎝
⎝
1
2
−S
We define
2
1
2
µ1
V
⎛ ∂G ⎞
⎟
⎝ ∂ni ⎠T , p ,nj ≠i
µi ≡ ⎜
1
µ2
as the chemical potential of species “i”
µi (T , p , nj ) is an intensive variable
This gives a new set of fundamental equations for open systems
(mass can flow in and out, composition can change)
dG = −SdT +Vdp + ∑ µi dni
i
dH =TdS +Vdp + ∑ µi dni
i
dU =TdS − pdV + ∑ µi dni
i
dA = −SdT − pdV + ∑ µi dni
i
5.60 Spring 2008
Lecture #14
page 2
⎛ ∂G ⎞
⎛ ∂H ⎞
⎛ ∂U ⎞
⎛ ∂A ⎞
=⎜
=⎜
=⎜
⎟
⎟
⎟
⎟
⎝ ∂ni ⎠T , p ,nj ≠i ⎝ ∂ni ⎠S , p ,nj ≠i ⎝ ∂ni ⎠S ,V ,nj ≠i ⎝ ∂ni ⎠T ,V ,nj ≠i
µi = ⎜
• At equilibrium, the chemical potential of a species is the same
everywhere in the system
Let’s show this in a system that has one component and two parts,
(for example a solid and a liquid phase, or for the case of a cell
placed in salt water, the water in the cell versus the water out of
the cell in the salt water)
Consider moving an infinitesimal amount dn1 of component #1 from
phase a to phase b at constant T,p. Let’s write the change in state.
dn1 (T , p ,phase a ) = −dn1 (T , p ,phase b )
dG = ⎡⎣ µ1(b ) − µ1(a ) ⎤⎦ dn1
µ1(b ) < µ1(a ) ⇒ dG < 0
⇒
spontaneous conversion from (a) to (b)
µ1(a ) < µ1(b ) ⇒ dG > 0
⇒
spontaneous conversion from (b) to (a)
At equilibrium there cannot be any spontaneous processes, so
µ1(a ) = µ1(b ) at equilibrium
e.g. liquid water and ice in equilibrium
ice
water
µ ice (T , p ) = µ water (T , p )
at coexistence equilibrium
5.60 Spring 2008
Lecture #14
page 3
For the cell in a salt water solution, µ water (cell ) (T , p ) > µ water (solution ) (T , p ) and
the cell dies as the water flows from the cell to the solution (this is
what we call osmotic pressure)
The chemical potential and its downhill drive to equilibrium will be
the guiding principle for our study of phase transitions, chemical
reactions, and biochemical processes
• Partial molar quantities
µi is the Gibbs free energy per mole of component “i”, i.e. the
partial molar Gibbs free energy
⎛ ∂G
= µi = Gi
⎜
⎟
⎝ ∂ni ⎠
T , p ,nj ≠i
G = n1 µ1 + n2 µ2 + " + ni µi = ∑ ni µi = ∑ ni Gi
i
i
Let’s prove this, using the fact that G is extensive.
G (T , p , λn1 , λn2 ) = λG (T , p , n1 , n2 )
dG
(T , p , λn1 , λn2 ) = G (T , p,n1 ,n2 )
dλ
⎛ ∂G ⎞
⎛ ∂G ⎞
⎛ ∂ ( λn1 ) ⎞
⎛ ∂ ( λn2 ) ⎞
+⎜
=G
⎜⎜
⎟⎟
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝ ∂ ( λn1 ) ⎠T , p ,λn2 ⎝ ∂λ ⎠T , p ,λn2 ⎝ ∂ ( λn2 ) ⎠T , p ,λn1 ⎝ ∂λ ⎠T , p ,λn1
n1
⇒
n1 µ1 + n2 µ2 = G
n2
5.60 Spring 2008
Lecture #14
page 4
We can define other partial molar quantities similarly.
⎛ ∂A ⎞
= Ai
⎜
⎟
⎝ ∂ni ⎠T , p ,nj ≠i
⇒
A = n1A1 + n2A2 + " + ni Ai = ∑ ni Ai
i
partial molar Helmholtz free energy
⎛ ∂A ⎞
note what is kept constant
⎛ ∂H ⎞
= Hi
⎜
⎟
⎝ ∂ni ⎠T , p ,n j ≠i
⇒
⇒ not to be confused with ⎜ ⎟
⎝ ∂ni ⎠T ,V ,n
j ≠i
H = n1H1 + n2H2 + " + ni Hi = ∑ ni Hi
i
partial molar enthalpy
⎛ ∂U ⎞
= Ui
⎜
⎟
⎝ ∂ni ⎠T , p ,n j ≠i
= µi
⇒
U = n1U1 + n2U2 + " + niUi = ∑ niUi
i
partial molar energy
__________________________________________________
Let’s compare µ of a pure ideal gas to µ in a mixture of ideal gases.
• Chemical potential in a pure (1-component) ideal gas
From
G (T , p ) = G o (T ) + RT ln
p
p0
⇒
µ (T , p ) = µ o (T ) + RT ln
• Chemical potential in a mixture of ideal gases
Consider the equilibrium
pA′ + pB′ = ptot
At equilibrium
mixed pure
A
A,B
p'A, p'B pA
\
µA ( mix ,T , ptot ) = µA ( pure ,T , pA )
Fixed Partition
p
p0
5.60 Spring 2008
Lecture #14
Also
page 5
pA ( pure ) = pA′ ( mix ) = ptot XA
Dalton’s Law
So
µA (mix ,T , ptot ) = µA ( pure ,T , ptot XA )
⎛p X ⎞
= µAo (T ) + RT ln ⎜ tot A ⎟
⎝ p0 ⎠
= µAo (T ) + RT ln
ptot
+ RT ln X A
p0
µA ( pure ,T , ptot )
∴
Note
µA ( mix ,T , ptot ) = µA ( pure ,T , ptot ) + RT ln XA
XA < 1
⇒
µA (mix ,T , ptot ) < µA ( pure ,T , ptot )
The chemical potential of A in the mixture is always less than the
chemical potential of A when pure, at the same total pressure. This
is at heart a reflection about entropy, the chemical potential of “A”
in the mixture is less than if it were pure, under the same (T,p)
conditions, because of the underlying (but hidden in this case)
entropy change!