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Chemical Potential 1
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Chemical Potential
Grand Canonical Ensemble
Statistical physics employs three kinds of probability distributions or
ensembles. The micro canonical ensemble applies when the system is
isolated from the environment so that no energy or particles are exchanged.
Usually this amounts to applying S  k ln W to equally probable states
without concern for energy levels. Most familiar is the canonical ensemble
wherein energy is exchanged with the environment. This is characterized by
the Boltzmann distribution p j  exp(  E j) Z . We introduce the grand
canonical ensemble in this chapter. Here both energy and particles can be
exchanged with the environment. The grand canonical ensemble is
characterized by  the energy-per-particle (or per mole) called the chemical
potential.
Chemical Potentials
The fundamental relation,
dU = T dS - P dV
(1)
can apply to more general systems by adding an important work term that pertains to
particle systems. The energy required to introduce dn moles of a particular chemical is
dn, where  is the chemical potential and has units of energy per mole.  is comparable
to P and dn is comparable to dV. An expanded version of Eq. (1) for several chemicals is
(2)
dU = T dS - P dV +  dnj .
We can obtain an expression for the chemical potential µ using the dynamic interpretation
of temperature. Add an increment of energy such that dN particles acquire transition
energy kT in a system of N particles. Let the temperature remain fixed so there is a
redistribution of potential energy per molecule so that the chemical energy for each
molecule is increased by d. The gained energy can then be written in two ways,
gained energy  N d
gained energy  kT dN
Equating these and integrating, we find and a most useful relation:
  kT ln N / N 0   0
(3)
where two constants of integration are 0 and N0. 0 is the chemical potential at standard
conditions listed in reference books (usually 298K and 1 ATM) It does not have any N
dependence, but it may depend on T. N0 is the particle number under these standard
Chemical Potential 2
conditions. Usually, we do not need to know the value of this somewhat arbitrary quantity.
Note that this derivation works only when the particles are relatively independent.
Note also that the ratio N / N 0 can be replaced by the ratio of concentrations c / c0 or the
ratio of pressures p / p0 . Usually the standard quantities N0, c0, or p0 are taken to be the
same for all the chemicals in a reaction and simply cancel out.
1. Solve the differential equation Nd  kTdN . Show the result is equivalent to
Eq.(3).
Here are two important facts about chemical potentials:

Chemical potentials are equivalent to true potential energies. A full expression of
particle energy  consists of the part that depends on concentration and temperature
given by kT ln N / N 0   0 and any external potential energy U.
  kT ln N N 0   0  U

(4)
Just as temperature is an intensive variable that determines thermal equilibrium (so
that net heat does not flow from system 1 to system 2 when T1 = T2), chemical
potential is an intensive variable determining particle or diffusive equilibrium:
1   2
example
Use chemical potentials to derive the barometric pressure equation for gas
molecules of mass m at a temperature T at an altitude y above ground level.
solution
Molecules at sea level must be in equilibrium with those at elevation y. Diffusive
equilibrium requires  g   where the subscript g indicates values at ground level.
kT ln( pg / p0 )  0  kT ln( p / p0 )  0  mgy
and we reduce this to p  pg exp  mgy kT 
2. Free electrons at a temperature T are separated into two compartments by a
potential V. Derive an expression for the ratio of electron concentrations n2/n1
where n2 refers to the compartment at higher potential. Calculate the ratio when
the voltage is 0.1 V and the temperature is 840 K. [ans 4]
3. Sodium ions outside of a cell are found to have a concentration of 0.106 mol/L and
Na+ ions inside the cell have a concentration of 0.149 mol/L. A membrane
potential separates the ions, V  Vinside  Voutside . Calculate the membrane
potential. (The situation is called Donnan equilibrium). [ans. –9.1 mV]
Note: A characteristic resting potential for nerve cell membranes is about 0.1 V. Clearly,
this is not an equilibrium situation. A dynamic process must maintain it.
Chemical Potential 3
Mass Action
The law of mass-action is central to chemical kinetics. It is easily derived using chemical
potentials. Each molecule in an equilibrium reaction is assigned a chemical potential and
the total reactants’ potentials must equal the total products’ potentials. For example, the
reaction H2 + I2  2 HI has a corresponding relation with chemical potentials:
 H 2   I 2  2 HI
Substituting from Eq.(4) we have
kT ln [H2 ][ I2 ]  standard potentials  2kT ln [HI]   2 standard potential
The standard potentials depend only on temperature, so this can be rewritten as
[ HI ]2
K.
[ H 2 ][ I 2 ]
Here K is a function of temperature alone and is called the “equilibrium constant.” By
convention, K is taken to express the ratio of products over reactants.
4. Write a relation between the chemical potentials of molecules in the reaction
2H 2 O  2H 2  O2 .
Show that the concentrations are related by
H 2 2 O 2   K
H 2 O2
where K is a “constant” that depends only on temperature–the equilibrium
constant.
Grand Canonical Distribution
Here we derive the probability that a system with accessible energy levels E1, E2,... and
particle numbers N1, N2,... at temperature T is in any particular state with an energy level
Ei and particle number Nj. Again define the Helmholtz free energy F as the average work
required to assemble the system from its well-separated components at constant
temperature. Denote the average energy exchanged per interaction as kT. We want to
evaluate the probability pj that the system is in level j.
The static equilibrium probability of remaining in a level must also be the probability of
finding the system in that level. At first, we consider the energy available to leave the
state. The total energy of  transitions* is kT and the energy available for transition out
of the state is E j  N j  F . This expression recognizes that the energy committed to
particles, Nj, is unavailable for leaving the state. It follows that the probability the system
in level Ej will leave this level in any one transition is (Ej –Nj -F)/ kT. Each interaction
*
 is used to avoid confusing it with the average particle number N.
Chemical Potential 4
then has a complementary probability 1-(Ej –Nj -F)/ kT of not causing the system to
leave level j. The probability that none of the  interactions causes a transition is pj,

 E j  N j  F 
p j  1 
 .
kT


1
In the limit of large  this becomes the grand canonical probability,
 F  E j  N j 
p j  exp 

kT


(5)
A more conventional derivation is to maximize entropy after adding to the canonical
constraints  p j N j  N . Requiring the result to correspond to the thermodynamic
expression (2),  appears as a Lagrange multiplier. 
5. Substitute the probability (5) into the entropy expression,
S  k  p j ln p j
and show that
F  U  TS  N
(6)
This is the thermodynamic definition of free energy F including a chemical
potential.
Grand Partition Function
The partition function Z is based on the canonical distribution and is the instrument for
many direct applications of statistical mechanics. A similar quantity, the grand partition
function Z is based on the grand canonical distribution and is employed when particle
numbers are involved.
Define the grand partition function through
F = –kT ln Z
6. Use the normalization condition,
p
i
(7)
 1 , and Eq.(7) to determine that the grand
i
partition function is given by
  Ei  N j 
Z   exp 

kT
j


7. Show that the average number of particles N is given by

N
ln Z

1
Recall that lim 1  x    exp( x ).

(8)
(9)