CHEMICAL POTENTIAL AND GIBBS DISTRIBUTION

Chemical Equilibrium

• Remember that T was the property, two system share when they

are in thermal equilibrium (flow of energy).

• The Chemical Potential governs the flow of particles

(diffusive contact) between the systems, just as the temperature

governs the flow of energy.

We will show that if of the two system is different, particles will flow

from the system with higher to the system with lower , until

µ1µ 2µ

21 µµ =

),,( NVτµ

Consider 1 + 2 as shown in the figure:

dN2

positive

dN1

negative

System 1 System 2

Reservoir Energy exchange

, V1, N1 , V2, N2

We have shown earlier that for a system S in thermal equilibrium with

reservoir , the Helmholtz free energy F is minimum compatible with

and other constraints.

)( 212121 σστ +−+=+= UUFFF

21

210

dNdN

dNdNdN

−=⇒+==

00 22

21

1

1 =

∂∂+

∂∂== dN

N

FdN

N

FdF

ττ

This is also true for S1 + S2 (here N = N1 + N2 = constant).

is a minimum.

Since N is constant,

At the minimum,

Also hold constant.21,VV

ℜ

NV ,,τ

m).equilibriufor (condition ),,(

)( 0

2

2

1

1

12

2

1

1

jjj NVN

F

N

F

dNN

F

N

F

τµττ

ττ

=

∂∂=

∂∂⇒

∗=

∂∂−

∂∂⇒

),,(:,

NVN

F

V

τµτ

=

∂∂

21 µµ >

The chemical potential is defined as

)(∗ 0<dF ⇒< 0dN

µIf we see from that when Particles flow

from the system with larger to the system with lower µ

Since Particles are not divisible, the strict definition of is in terms of

a difference (as opposed to derivative)

.,...,, 2,1 NNVj

j N

F

τ

µ

∂∂=

)1,,(),,(),,( −−≡ NVFNVFNV τττµ

If several chemical species are present , each has its own chemical

potential

µ

Example: Chemical Potential of the Ideal

Gas Remember: The free energy of the monatomic ideal gas is

[ ]

τπλ

λλ

ττ

τ

M

nVnV

Z

N

ZNZNF

or

ZF

DB

DBQQ

DB

N

2

331

11

2 and

ion)concentrat (Quantum 1

, where

!log!loglog

ln

=

===

−=−−=

−=

Use

[ ]

. where,logor

)log(loglog

)!1(

!loglog)1()1()(

11

1

VNnn

n

ZNNZ

N

NZNNNFNF

Q==

=−−=⇒

−

−+−−=−−

τµ

ττµ

τ

).log( Qnp ττµ =

τnp =

µ depends on concentration, not on N or V separately.

Increases as n increases! This is what one would expect intuitively.

Using Ideal gas law we can write

µ

Internal and Total Chemical Potential

Consider the two systems of charged particles. A potential step

between them can be established by applying a voltage.

V∆

S1

+

τµ ,1

S2

-τµ ,2

Initially yielding 1212 (initial) , µµµµµ −=∆>

Now apply to S1 such that

If F2 is fixed, the potential step rises F1 by

For diffusive equilibrium we need

The barrier brings S1 and S2 into diffusive equilibrium.

)initial()( 12 µ∆=−=∆ VVqVq

energy) potentialin (increase )initial(1 µ∆N

V∆

[ ] )final()initial()initial()initial()initial()final( 22

)initial(

1211 µµµµµµµ

==−+=∆

Vq∆

Note: The chemical potential potential energy.

The difference in is equal to a potential barrier that will bring two

systems into equilibrium.

=µ

When external potentials are present

intexttot µµµµ +==

.

0

intext

tot

µµµ

∆−=∆⇒=∆

12 µµ =

Internal chemical potential, defined as if no external

potential is present.

The equilibrium condition can than be written as

:intµ µ

Example: Variation of barometric pressure with altitude.

Place the zero of the potential energy

at ground level, then the potential

energy per molecule at height h is

Mgh, where M is the particle mass

and g the gravitational acceleration.

The internal chemical potential of

the particles is given by

.

,log

ext

int

Mgh

nn

Q

=

=

µ

τµSystem (1)

System (2)

h

A model of the variation of atmospheric pressure with altitude.

The total chemical potential is

In equilibrium, the chemical potential must be every where identical:

Mghnhnh

Q+

= )(log)( τµ

−=⇒

=+

τ

ττ

Mghnhn

nnMghn

hnQQ

exp)0()(

)0(log)(log

.

,exp)0(exp)0()(

Mghwhere

hhpMghphp

c

c

ττ

=

−=

−=

For an ideal gas τnp =

µ

This is barometric pressure equation.

At the characteristic height the atmospheric pressure

decreases by the fraction .To estimate the characteristic

height, consider an isothermal atmosphere composed of nitrogen

molecules with a molecular weight of 28.

Mass of an N2 molecule is .

At temperature 290K the value of

With g = 980cms-2, the characteristic height is then 8.5km.

Mghcτ=

Barometric pressure equation gives us dependence of the pressure on

altitude in an isothermal atmosphere of a single chemical species.

37.01 ≈−e

TkB≡τ

gm.1048 24−×

erg.100.4 is 14−×

Batteries

One of the most vivid examples of chemical potentials and potential steps

is the electrochemical battery. In the lead-acid battery the negative

electrode consists of metallic lead, Pb, and the positive electrode is a

layer of reddish-brown lead oxide, PbO2, on a Pb substrate. The

electrodes are immersed in diluted sulfuric acid, H2SO4, which is partially

ionized into H+ ions SO4-- ions. It is the ion that matter.

PbO2 Pb

Convert to PbSO42e-

SO4-- 2H+ + H2SO4

2H2O

2e-

Pb

(-) electrode (+) electrode

The lead-acid battery consists of a Pb and a PbO2 electrode immersed in

partially ionized H2SO4. One SO4 ion converts one Pb atom into PbSO4 + 2e-;

two H+ ions plus one un-ionized H2SO4 molecule convert one PbO2

molecule into PbSO4 + 2H2O, consuming two electrons.

In the discharge process both the metallic Pb of the negative electrode

and the PbO2 of the positive electrode are converted to lead sulfate,

PbSO4.

Discharge Process:

( )

( )b

a

O.2HPbSO2eSOH2HPbO

;2ePbSOSOPb

24422

44

+→+++

+→+

−+

−−−

Negative electrode:

Positive electrode:

Because of () the negative electrode acts as a sink for SO4-- ions,

keeping the internal chemical potential (SO4-- ) of the sulfate ions at the

surface of the negative electrode lower than inside the electrolyte.

(-) electrode (+) electrode

- 2eV- (SO4--)

(SO4--)

x

(H+) + eV+

(H+)

x

The electrochemical potentials for SO4-- and H+ before the development of

internal potential barriers that stop the diffusion and the chemical reaction.

Because of (b) the positive electrode acts as a sink for H+ ions, keeping

the internal chemical potential (H+) of the hydrogen ions lower at the

surface of the positive electrode than inside the electrolyte. The chemical

potential gradients drive the ions towards the electrodes, and they drive

the electrical currents during the discharge process.(-) electrode (+) electrode

x

V+ V+

- V-

(x)

The electrostatic potential (x) after the formation of the barrier.

If battery terminals are not connected, a electrical potential (x) develops

until they equalize the chemical potential steps. Diffusion stops when:

at - electrode -2q V- = (SO42-)

at + electrode +q V+ = (H+).

The two potentials V- and V+ are called half-cell potentials.

The total electrostatic potential difference developed across one full cell of

the battery, as required to stop the diffusion reaction, is

V = V+ - V- = 2.0 volt.

This is the open-circuit voltage or EMF (electromotive force) of the battery.

There is a small electron current through the electrolyte, however it is

negligible! Otherwise a battery would discharge quickly. Electrons

dominantly flow through external connection.

An alternative way to calculate the chemical potential.

Now the number of quantum states has to depend on the number of

particles

dNN

dVV

dUU

dVUNUNV ,,,

∂∂+

∂∂+

∂∂= σσσσ

( ) ( ) ( )

( )( )

( )( ) .

,

UN

UN

NN

U

UN

NN

UU

∂∂+

∂∂=

∂∂+

∂∂=

σδδσ

δδσ

δσδσδσ

τ

τ

τ

τ

τττ

Assume isotropic process , select in such a way

that call these values respectively then

when

( ) ( ) ( )τττ δδδσ NU ,,

0=dV dNdUd ,,σ

,0=τd

0=τd

After division by ( ) ,τδN

VUVV

VUVNV

NN

U

N

NN

U

UN

,,,

,,

1

,

∂∂+

∂∂=

∂∂⇒

∂∂+

∂∂

∂∂=

∂∂

στστ

σσσ

ττ

τ

τ

τ

( )NVN

F

N

U

N VUVV

,,,,,

τµτστττ

≡

∂∂≡

∂∂+

∂∂−⇒

.,VUN

∂∂−= στµ

and on comparison with above equation

By the original definition of chemical potential

Thermodynamic identity

We can generalize the statement of the thermodynamic identity to

include systems in which the number of particles is allowed to change.

or

dNpdVdUd

dUN

dVV

dUU

dVU

p

NUNV

µστ

σσσσ

τµ

ττ

−+=⇒

∂∂+

∂∂+

∂∂=

− ,,

1

,

workChemical workMechanicalheat Thermal

dNpdVddU µστ +−=

Grand Canonical Ensemble and Grand Potential

Volume is fixed Volume is fixed

Can exchange energy withreservoir

Can additionally exchangeparticles with reservoir

Concept of thermal equilibrium Concept of chemical/diffusiveequilibrium

Definition of temperature(energy fluctuates)

Definition of chemical potential(particle number fluctuates)

Boltzmann factor Gibbs factor

Canonical partition function Gibbs sum(Grand Canonical partition function)

Free energy Grand potential

Canonical Ensemble Grand Canonical Ensemble

Gibbs Factor

U,V,NState S

Exchange energy and particles

Reservoir

NN

VV

UU

−−−

tot

tot

tot

System

U and N fluctuate, V is fixed

What is the probability of

finding the system in a

microstate S with some

given values of U and

N?

• Derivation follows exactly along the lines of the canonical case. The

probability is proportional to the multiplicity!

( )

( ) NNUU

NNUUg

reservoirg

systemgreservoirg

systemreservoirgSP NU

−−=

−−=

=

=

+

tottotr

tottotr

microstate fixedin is systemeffectsbounbary no

,

,exp

,

1).(

)().(

)( )(

σ

( )

( )

( ) .exp . constant

1,exp

,exp

tottotr

tot

r

1

tot

rtottotr

Taylor

τµ

τµ

τσ

σσσ

τµ

τ

NU

NUNU

NU

UU

NU

−−=

+−=

∂∂−

∂∂−=

−

Hence: The probability of finding the system in a particular state S, in

which it has energy U and particle number N is proportional to the

Gibbs factor

P(SU,N) ( ) .exp τ

µNU −−

In order to get an equality, and not just a proportionality, we need to

find the normalization factor. It is called the “Gibbs sum” or “grand

canonical partition function”:

( ) ( )( )

( )( )

( )µττ

µ

τµµτ

,,

exp

S-exp:,,

,

NN

0

VZ

NU

SP

NHVZ

NU

N SN

−−

=

−= ∑∑∞

=

( ) ( )µττµτ ,,ln:,, VZV −=Ω

So that we have

The logarithm of Z is again a thermodynamic potential:

It is called Grand Potential.

Relation of the grand potential with other

thermodynamic potential:

( ) ( )( )

( ) ( )

( )( ) ( )

( )∑

∑

∑∑

∑∑

−−−=

−−=

−−=

−−==Ω−

UN

N,U

N UN

N S

N

NU

NUNU

NUUg

NSHZN

,

exp

exp ,exp

exp

expexp

τµτσ

τµσ

τµ

τµ

τ

For large systems only the largest term in the sum will contribute

significantly

−−−≈ τµτσ NU

UN ,system large

minexp

.min

min,

NF

NU

N

UN

µ

µτσ

−=

−−≈Ω

Taking the logarithm, we find

Example: What is the average number of particles in our system

( )

( )( )

( )( )

( )( )

( )( )∑∑

∑∑

∑∑

∑∑

∑∑

∞

=

∞

=

∞

=

∞

=

∞

=

−−

−−

∂∂

=

−−

−−

=

=⟩⟨

0

0

0

0

0

exp

exp

exp

exp

N S

NN

N S

NN

N S

NN

N S

NN

N SN

N

N

N

N

N

NSH

NSH

NSH

NSHN

SPNN

τµ

τµ

µτ

τµ

τµ

Consider

( )( ) ( )

( )( ) ( )

.log

since , exp

exp

1

,, ,,ln

,,ln ,,

,,

1

ZN

Z

Z

ZN

VVZ

VZVZ

VZ

N S

sN

λλ

λµλ

µτελ

τµλ

µτ

µµτµττ

µ

µτµ

τµτ

µτµ

τ

λτ

∂∂=⟩⟨⇒

∂∂

∂∂=

∂∂

−=⇒

≡

∂∂=⟩⟨⇒

∂Ω∂−=−

∂∂−=

∂∂=∂

∂

=

∑∑

Also