Download Statistical Thermodynamics of lodine Sublimation The sublimation of

Statistical Thermodynamics of lodine Sublimation
The sublimation of iodine involves the phase coexistence between solid, I2 (s),
and its vapor, I2 (g). The solid forms black crystals, which when gently heated above
room temperature produce a purple vapor. This vapor, I2 (g), can thus be detected
spectroscopically, and so the partial pressure of iodine as a function of temperature can
be determined by measuring the absorbance at the proper wavelength.
Question: In what range of wavelengths do you expect to measure absorbance to best
determine the concentration/partial pressure of iodine vapor?
The system I2(s)-I2(g) provides an opportunity for the application of statistical
mechanics to derive thermodynamic information from spectroscopic data. For the gas
phase, the vibrational frequency of the I2 molecule, needed in formulating the vibrational
partition function, can be obtained from the absorption spectrum in the visible region; the
rotational partition function in the gas phase will be calculated from the known
internuclear distance in the iodine molecule. For the crystalline phase, the microstates
correspond to the normal modes of vibration called phonons, which are completely
analogous to the travelling waves on a string in one dimension. Each crystalline solid has
its own characteristic set of vibrational frequencies, just as every bell rings with its own
characteristic set of harmonics or pitches. With the above information and statistical
mechanical theory, the molar energy difference ∆ E˜ 00 between the vibrational ground
states of crystalline and gaseous iodine can be determined from a measurement of vapor
pressure at one temperature. From the fully defined partition functions for both
crystalline and gaseous iodine, the entropy and enthalpy changes attending sublimation of
iodine can be calculated at any temperature T. The value of ∆ H˜ sub obtained in this way
will then be compared with the value obtained using macroscopic thermodynamics via
the Clausius-Clapeyron equation, given by
ln p = constant -
∆H˜ sub
RT
Question: What assumptions are made by the Clausius-Clapeyron equation? You may
need to dig out your thermodynamics texts, if necessary. How is it related to the
Clapeyron equation?
Question: Using statistical mechanics, we will determine ∆ H˜ sub based on a measurement
at a single temperature. At how many temperatures do measurements have to be made to
determine ∆ H˜ sub if using the Clausius-Clapeyron equation?
In two weeks, we will perform this wet laboratory to determine the solid-vapor
phase equilibrium of iodine, and to measure its sublimation energy and enthalpy. In the
dry lab this week, however, we will develop the statistical mechanics necessary to make a
prediction of these thermodynamic, macroscopic quantities.
THEORY
The discussion here will focus on the application of statistical mechanics to the phase
equilibrium
I2(s) = I2(g)
(1)
It is required for equilibrium that the chemical potential of I2 be the same in the two
phases:
µs = µg
(2)
The basic question is: How can these chemical potentials be determined?
Question: What are the two other conditions required for phase equilibrium in this onecomponent system?
Statistical Mechanical Background
We will present a brief review of the basic statistical mechanical concepts needed in this
experiment.
Given the canonical partition function Q for a system,
Q = ∑ e − Ei / kT
i
(3)
where Ei is the energy of the ith quantum state of the entire system, the Helmholtz free
energy is given by the equation
A = -kT ln Q
Question: Does the partition function, Q, increase or decrease with the addition of an
extra energy level, Ei, to the sum? How about the free energy, A? Does it matter how
high or low, or positive or negative Ei is?
For a one-component system the chemical potential per mole is given by
(4)
=
 A
 lnQ 
 ln Q 
= −kT
= −RT
 n  T ,V
 n  T ,V
 N  T,V
(5)
where n is the number of moles in the system and N is the number of molecules in the
system (=N0n, where N0 is Avogadro's number).
Question: What are the equations for P and E that are analogous to Eq (5)? Do these
agree with the equations for the ensemble averaged P and E? What is special about the
set {µ, P and E} in the canonical ensemble?
In order to evaluate the canonical partition function Q for a gas, we shall consider
the system to be composed of an aggregate of essentially independent particles
(molecules). As we shall see later, a crystal may be considered to a good approximation
as an aggregate of independent harmonic oscillators. Each of these has its own partition
function:
qi = ∑ e
−
( i)
j
/ kT
(6)
j
where (ij ) is the energy of the ith oscillator in the jth quantum state of that oscillator.
Since the oscillators are only very weakly interacting, the canonical partition function of
the solid is a simple product of the partition functions of the individual oscillators:
Qs = Π qi
i
ln Qs = ∑ ln qi
(7)
i
Since many of these oscillators differ from each other in the values of their frequencies,
energy levels, and partition functions, it is convenient to define a new quantity qs which is
the geometric mean of all of the qi for the crystal:
M 
qs ≡  ∏ qi 
 i =1 
1/ M
ln qs ≡
1 M
∑ ln qi
M i =1
(8)
where M is the number of oscillators. Then, for the crystal,
ln Qs = M ln qs = 3tN ln qs
(9)
where t is the number of atoms in a molecule. Since ln qs can be shown to be
independent of N, we find from Eq. (5) that
s
= −3tRT ln qs
Question: Does qs vary with temperature? If so, is it decreasing or increasing? How
does the chemical potential vary with temperature?
(10)
For a one-component ideal gas, the partition function for an individual molecule
is qg. Therefore, under all ordinary conditions, we may write for a gas
qg N
Qg =
N!
(11)
where the division by N! takes into account the fact that the individual molecules are
indistinguishable. With the aid of the Sterling approximation for ln (N!) = N ln N – N,
we obtain
ln Qg = N ln qg − N ln N + N
(12)
Using Eq. (5) again, we obtain
g
= −RT ln
qg
N
(13)
We will now develop the expressions for the partition functions qs and qg to substitute
into Eqs. (10) and (13).
Gaseous I2
The ideal diatomic gas partition function is developed in the textbook (which you can
refer to for further details), but we simply recap the main points here. The partition
function qg is very well approximated as a product of terms arising from translational,
rotational, vibrational, and electronic degrees of freedom:
qg = qtrans qrot qvib qel
(14)
The translational partition function is given by
qtrans
3 / 2
 2 mkT 
=
V
h2 
(15)
where m is the molecular mass, k is the Boltzmann constant, T is the absolute
temperature, h is Planck’s constant, and V is the volume within which the molecule is
constrained to move.
For a molecule as massive as I2, the rotational energy levels are very closely
spaced and the partition function has the simple form
qrot =
kT
T
=
hcB0
Θrot
(16)
Here σ is the symmetry number of the molecule (an integer which accounts for
indistinguishability), c is the velocity of light, and B0 is the rotational constant defined by
B0 ≡
h
8
2
Ic
(17)
Question: What units does B0 have?
where I is the moment of inertia of the molecule,
I = r02
(18)
(The reduced mass µ is not to be confused with the chemical potential.) In I2 the
interatomic distance r0 is 0.2667 nm, and the rotational constant B0 is 0.037315 cm-1. The
quantity Θrot is the rotational characteristic temperature, given by
Θrot =
hcB0
k
(20)
Question: What units does Θrot have?
The factor hc/k has the value 1.43877 cm K. Since the I2 molecule is end-for-end
symmetric, σ = 2.
For the vibrational partition function the molecule is regarded as a quantu2m
mechanical harmonic oscillator, for which
q = (1− e −h
0
/ kT
) −1 = (1− e −Θ vib / T )−1
(21)
where ν0 is the molecular vibration frequency and Θvib is the vibrational characteristic
temperature,
h
hc ˜ 0
(22)
Θvib = 0 =
k
k
For the I2 molecule, ˜0 has the value 213.3 cm-1.
Equation (21) as written applies to an oscillator for which the reference energy is
the energy of the vibrational ground state (v=0); i.e., the v=0 state in a gas molecule has
been assigned zero energy. For the present situation, in which I2 molecules in the vapor
phase are in equilibrium with crystalline iodine, it is more convenient to take the
reference energy to be that of an I2 molecule in the crystal when the crystal is in its
ground vibrational state. (It should be noted that the location of the energy zero is
arbitrary—as long as it is consistent between the two phases, solid and gaseous.)
Accordingly, the energy of the vibrational ground state of an I2 molecule in the ideal gas
phase is taken to be ∆ 0 , which is the energy required to remove a molecule form the
crystal at absolute zero temperature. Thus, we should write for the I2 molecule in the gas
phase
qvib = (1− e −Θ vib / T )−1 e −∆
0
/ kT
(23)
It remains to deal with qel. The excited electronic states of I2 are separated from
the ground electronic state by an energy difference that is very large compared to kT.
Therefore,
qel = 1
(24)
0
˜
Let us now introduce ∆ E0 = N0 ∆ 0 , the energy needed to sublime 1 mol of
crystalline I2 into the ideal gas phase at absolute zero, and replace V by its ideal-gas
equivalent NkT/p. We can then combine Eqs. (13) to (16), (23), and (24) to obtain
g
  2 mkT  3 / 2kT T

0
˜
= ∆E0 − RT ln  
(1− e −Θ vib / T ) −1 
2

h
p Θrot


(25)
Crystalline I2
The partition function for the crystalline state of I2 consists solely of a vibrational part:
the crystal does not undergo any significant translation or rotation, and the electronic
partition function is unity for the crystal as it is for the gas.
The geometric mean partition function for the crystal can be expressed as
1/ M
M
−Θ i / T −1 
(26)
qs =  ∏ (1− e
) 
 i =1

where ΘI is defined in terms of the ˜i in the same way as Θvib is defined in terms of ˜0 in
Eq. (22). Since the number of iodine atoms is 2N for a crystal containing N molecules of
I2 and since each atom contributes three degrees of freedom, the number of modes of
vibration for the crystal is
M = 3 × t × N − 6 = 6N − 6 ≅ 6N
(27)
The subtracted number 6 represents the 3 translational and 3 rotational degrees of
freedom of the crystal as a whole and will henceforth be ignored.
We now present a brief discussion of the vibrations occurring in a crystal. The
crystal can be thought of as a gigantic molecule with a huge number of normal modes. In
the case of the I2 crystal, each primitive (smallest) unit cell contains two molecules.
These two molecules in the unit cell are easily distinguished because their spatial
orientations are different. As a consequence of this crystal structure, there are 3 x 4 = 12
mechanical degrees of freedom associated with each unit cell. In the gas phase there
would be three translations, two rotations, and one vibration for each of the two I2
molecules. In the crystal, however, only vibrations occur: six lattice modes, four
librational modes, and two internal vibrations (bond-stretching modes), which we will
describe now.
Let us consider first the center-of-mass motions for each of the two I2 molecules
in a unit cell. These types of motion account for six degrees of freedom and give rise to
two kinds of lattice vibration. When both I2 molecules in a given cell move in phase with
each other (say, for example, both are displaced in the +x direction at the same time),
there are three so-called acoustic vibrations. When the two I2 molecules in a given cell
move out of phase (say, one is displaced in the +x direction while the other is displaced in
the –x direction), there are three optic vibrations. (The name optic mode comes from the
behavior of ionic crystals such as Na+ Cl-. When Na+ and Cl- in a given cell move out of
phase with each other, there is an oscillating electric dipole. Optical absorption will
occur for light having frequency equal to that of the optic lattice mode.)
The four librations (torsional oscillations or rocking motions) arise because the
crystal-field potential prevents the I2 molecule form rotating as it would in the gas phase.
The librational motions for each I2 occurs about two axes (α, β) perpendicular to the I-I
bond direction. The librations of the two I2 molecules in the same unit cell are
coupled—giving rise to symmetric (angle displacements in phase) and antisymmetric
(angle displacements out of phase) librations.
Finally, there are two I-I bond-stretching vibrations that are essentially the same
as the gas-phase stretching mode. As expected, these vibrations are coupled to produce a
symmetric stretch in-phase vibration, and an antisymmetric stretch out-of-phase
vibration. In the latter case, one I2 bond is stretching while the other is being compressed.
As a result of interactions in the crystalline phase, these vibrations have lower
frequencies than the gas-phase vibration at 213.3 cm-1.
Up to this point, we have considered the 12 normal modes of motion of the two I2
molecules in the primitive unit cell. If each unit cell were completely independent from
all the other unit cells, we would have an Einstein crystal, as described in McQuarrie in
Chapter 17. This Einstein crystal consists of N/2 primitive unit cells, each yielding the
same set of 12 oscillator frequencies νj. In fact, however, motions of the I2 molecules in
any given unit cell are coupled to those of molecules in other unit cells, which is what
allows vibrations (such as sound waves) to travel through a crystal as a wave. Depending
on the wave speed and direction, the energies and oscillation frequencies can vary. These
variations, however, are not that great, except for the long-wavelength acoustic
vibrations, and the Einstein model actually provides quite a good description of the
vibrational frequencies of a real crystal, except at very low temperatures where the
acoustic vibrations are dominant. For I2, these 12 Einstein frequencies (average
vibrational modes) are 21.0, 26.5, 33.0, (the three acoustic vibrations), 41.0, 49.0, (two
optical modes), 51.5, 58.0, (two librations), 59.0, (an optical mode), 75.4, 87.4, (two
more librations), 180.7, 189.5 (the symmetric and antisymmetric stretches, respectively)
cm-1.
We can now formulate the desired expressions for qs and µs. Using Eqs. (8) to
(10), together with the fact that t=2 for I2 and each unit cell contains two I2 molecules, we
find
1 6N
1 N 12
1 12
(30)
ln qs =
ln
q
=
ln
q
=
∑ i 6N 2 ∑
∑ ln q j
j
6N i =1
12 j =1
j =1
where 12 is the number of degrees of freedom per unit cell. Finally, for I2(s), we obtain
ln qs = −
1 12
−Θ / T
ln(1 − e j )
∑
12 j =1
(31)
and
s
RT 12
−Θ / T
= −6RT ln qs =
ln(1− e j )
∑
2 j =1

RT  12
=
ln ∏ (1− e −Θ j / T )
 j =1

2
(32)