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STATISTICAL THERMODYNAMICS OF IODINE SUBLIMATION NAME DATE PRELIMINARY REPORT To be completed and turned in at least 24 hours before you carry out the lab. Assignment: Your goal in this lab will be to find the energy of sublimation of iodine by measuring the vapor pressure of I2 at a series of different temperatures. The energy of sublimation is defined as how much energy, , it takes to go from the vibrational ground state of an iodine molecule in a crystal to the vibrational ground state of an iodine molecule in the gas phase. You will also find the enthalpy and entropy of sublimation of iodine using partition functions and the Clausius-Clapeyron equation. References: This experiment and analysis is based on Experiment 48: Statistical Thermodynamics of Iodine Sublimation in Shoemaker which you may find to be a useful reference. A scanned copy of the Shoemaker pages is posted on Moodle under Iodine_ShoemakerReference. List of materials needed: Brief outline of the procedure: List data to be collected: Any potential hazards (chemical, electrical, visual, etc.) associated with the lab:

Questions 1. For this experiment, you will determine the vapor pressure of I2 by using a uv-vis spectrometer to measure the baseline corrected absorbance at 520 nm of I2 vapor. How will you translate the absorbance measurements to pressure values? Pay attention to units. 2. What are the equations that define the translational partition function, rotational partition function, and vibrational partition function? 3. What modes of motion (translational, vibrational, rotational) are important when describing crystalline I2? Why? 4. How will you use the Clausius-Clapeyron equation to obtain the enthalpy of sublimation?

E 00

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In this lab you will have the opportunity to apply statistical mechanical machinery to studying the phase equilibrium between I2(s) crystals and I2(g). You will find that the experimental part of the lab is straightforward, but the calculations are pretty complex. You will have to be organized and careful if you hope to successfully complete this lab. Your goal in this lab will be to find the energy of sublimation of iodine. In other words, how much energy, , does it take to go from the vibrational ground state of an iodine molecule in a crystal to the vibrational ground state of an iodine molecule in the gas phase? You will also find the enthalpy and entropy of sublimation of iodine using partition functions and the Clausius-Clapeyron equation. Statistical Mechanical Background In the temperature and pressure ranges you will use in this experiment, iodine exists in an equilibrium between its gas and solid phases:

I2(s) I2(g) If the two phases are in equilibrium then the chemical potentials for each of the two phases must be equal:

mg = ms

We can use the partition functions for I2 in each phase and the machinery of statistical mechanics to determine these chemical potentials. This will then let us continue on to work out the energetics of the sublimation. The chemical potential can be easily written as a derivative of the Helmholtz energy taken with respect to the amount of material at constant temperature and volume.

We can write the Helmholtz energy as a function of the partition function as

A = -kB T ln Q

Combining these two equations yields:

(1)

where n is the number of moles and N is the number of molecules.

E 00

= An

T ,V

= An

T ,V

= kBT lnQ

n

T ,V

= RT lnQN

T ,V

3

Now we consider each of the two phases. First, lets look at the solid phase. Solid Phase Crystalline I2 can be reasonably approximated as a collection of independent harmonic oscillators. Each of the oscillators will have its own molecular partition function, qi. Since we have assumed the individual oscillators are independent the partition function for the solid is just the product of the partition functions for the individual oscillators.

and

(2)

What should M, the number of oscillators be? Each atom can move in three dimensions and there are two iodine atoms in each iodine molecule. Therefore the number of oscillators will be:

M = 3 x 2 x N = 6N.

We should subtract off six modes of motion for the three translational and three rotational degrees of freedom of the crystal, but since the number of atoms is large, 6N 6 6N. Because the oscillators in the crystal will vary in their vibrational frequencies it is best to define a quantity qS which is the geometric mean of the qi for the crystal.

(3)

Since the only motion we have to consider in the crystal is vibrational motion, the partition function qi will take the form of a vibrational partition function:

and

(4)

QS = qii=1

M

lnQS = lnqii=1

M

qS = qii=1

M

1M

or lnqS =1M

lnqii=1

M

qi =1

1 exp iT

lnqi = 1 exp iT

4

i is the vibrational temperature of each of the modes of vibration of the iodine crystal.

How should you think about the vibrations in the crystal? We will present a short summary here, but you can consult books on solid state physics for a more complete discussion. One way is to think of the iodine crystal as a giant molecule with 6N 6 normal modes. Lets start by thinking about the vibrations in a single unit cell. Each unit cell in an iodine crystal has two molecules in it for a total of four atoms. Four atoms in three dimensions give us twelve modes of motion. For reasons which are a bit too technical to go into here, we can use the vibrational frequencies for the two molecules in a unit cell as the frequencies we need to calculate the partition function for the crystal. If we had two individual I2 molecules there would be six translational modes, four rotational modes, and two I I vibrations. In the unit cell, however, the I2 molecules are constrained from translating and rotating the way they would in the gas phase. This results in six lattice modes, four librational modes, and two internal bond-stretching vibrations. The lattice modes correspond to motions in which the center of mass of each of the molecules moves. If both molecules move in the same direction (either the x-, y-, or z-direction) the motion is called an acoustic vibration. If the two molecules move out of phase (i.e. in opposite directions) then the mode is called an optic vibration. There are three acoustic vibrations and three optic vibrations. The four librations are oscillating or rocking motions. These correspond to the rotations in a free molecule, but because the molecules are constrained from freely rotating because of the adjoining unit cells, the molecules merely rock back and forth. Finally, the two internal vibrations are coupled to each other rather than being identical vibrations in each of the two molecules. One vibration is a symmetric stretch in which both molecules vibrate in phase with each other. The other vibration is an antisymmetric stretch in which one bond lengthens while the other shortens. The table below gives the phonon vibrational frequencies for an I2 crystal.

VIB =h 0kB

=hc 0kB

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Discrete phonon frequencies in I2 crystals

Mode number Type of Mode Frequency (cm-1) 1 acoustic 21.0 2 acoustic 26.5 3 acoustic 33.0 4 optic 41.0 5 optic 49.0 6 libration 51.5 7 libration 58.0 8 optic 59.0 9 libration 75.4 10 libration 87.4 11 symmetric stretch 180.7 12 anti-symmetric stretch 189.5

We can now calculate the partition function and chemical potential for the solid. Start with equation (3), remembering that M = 6 N.

We have unit cells so we multiply by and see that 12 is the number of degrees of freedom

per unit cell.

and substitute in the expression from equation 4.

If we do the sum in equation 3 explicitly for the case of iodine with M = 6N we see that

lnqS =1

6Nlnqi

i=1

6N

N2

N2

lnqS =1

6NN2

lnq j =1

12lnq j

j=1

12

j=1

12

lnqS = 1

12ln 1 exp i

T

j=1

12

lnQS = 6N lnqS

6

We can then find the chemical potential for the solid as:

Gas Phase When an iodine molecule is in the gas phase its energy can be approximated as a sum over four different terms.

Since the energy terms are separable we can write the partition function as a product of the partition functions associated with each of the motions of the molecule:

As discussed in your textbook, the translational partition function is given by:

where m is the mass of a single molecule, kB is Boltzmanns constant, T is the temperature in Kelvin, h is Plancks constant, and V is the volume of the container. The rotational partition function is given by:

The constants are defined as before. In addition, B0 is the rotational constant. The rotational constant for I2 is 0.037315 cm-1. Be aware that for consistent units you should put the speed of light in cm s-1 units. s is the symmetry number for I2. The vibrational partition function is based on the assumption that the vibrations in I2 can be modeled as a quantum mechanical harmonic oscillator. The vibrational partition function is derived by summing over the infinite series of vibrational levels in the harmonic oscillator. The details are given in your text.

S = RT lnQS

N= 6RT lnqS

=RT2

ln 1 exp j

T

j=1

12

=RT2

ln 1 exp j

T

j=1

12

Etot = Etrans + Erot + Evib + Eelec

qg = qtransqrotqvibqelec

qtrans =2mkBT

h2

32V

qrot =kBT

hcB0

7

is the vibrational frequency for I2 in units of s-1. is the vibrational wavenumber in units of

cm-1. The vibrational wavenumber, ,for I2 is 213.3 cm-1.

Notice that in evaluating the vibrational partition function we assumed that the zero of energy we assumed that the vibrationless ground state of the I2(g) molecule is at zero energy. It turns out that it is better to define the reference energy as the energy of an I2 molecule in a crystal when the crystal is in its ground vibrational state. In order to accommodate this definition of the reference state we need to add an amount of energy that we will call to the energy of the gas phase molecule. is the amount of energy it takes to remove a molecule of I2 from the crystal at a temperature of absolute zero. We will also need to define the amount of energy to remove a mole of molecules from the crystal. This quantity will just be multiplied by Avogadros constant. . Incorporating this change in reference state, the vibrational partition function becomes:

Lastly, the electronic ground state is non-degenerate and there are no low-lying excited electronic states, so the electronic partition function equals one.

Now for a one-component gas of indistinguishable molecules the partition function of the gas is related to the molecular partition function by:

We can apply Stirlings approximation to eliminate the factorial term.

Plugging this equation into equation (1) lets us find the chemical potential for the iodine in the gas phase:

qvib = 1 exph 0kBT

1

= 1 exp hc 0

kBT

1

0

0

0

0

0

0

E 00 = N00

qvib = 1 exphc 0

kBT

1

exp 0kBT

qelec =1

Qg =qg

N

N!

lnQg = N lnqg N lnN + N

g = RT lnqgN

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Plugging in the expression for the molecular partition function gives us the following expression:

P in this equation is pressure. We have eliminated the volume from this expression by using the

ideal gas law and letting .

Equilibrium Between Solid and Gas When the solid and gaseous iodine are in equilibrium with each other their chemical potentials are equal. We can take the expressions for the chemical potential of the solid and the gas and set them equal to each other. Following some rearrangement we find the following impressive equation:

(5)

Determining the pressure at one temperature allows you to solve this equation for . You can do this calculation at each temperature to see how well the values agree with each other. If you determine the pressure at several temperatures you can make a plot to determine . You can check the quality of your plot by comparing the intercept from your plot to the constant term on the right side of the equation. You can also use your data and the Clausius-Clapeyron equation to obtain the enthalpy of sublimation,

H sub and entropy of sublimation,

S sub . The Clausius-Clapeyron equation tells us that:

lnP = constant H sub

RT

The equations for the molar entropy of the solid is given by:

S S = ST

P

qg = qtransqrotqvibqelec( )

g = E 00 RT ln 2mkBT

h2

32 kBT

PT

rot1 exp hc

0kBT

1

V = NkBTP

lnP lnT

72 1 exp

jT

12

j=1

12

1 exp hc 0

kBT

= ln 2mkBh2

32 kB

rot

E 00

RT

E 00

E 00

9

S S =R2

jT

exp jT

1

ln 1 exp jT

j=1

12

and for the gas phase:

S g = gT

P

S g = E 0

0 gT

+72

R + R

vibT

exp vib T

1

where

vib is defined the same way in the gas phase that it is in the solid phase, but with the vibrational wavenumber of 213.3 cm-1. The expression for mg is given above. The enthalpy of sublimation at some temperature, T, is given by:

H sub = T S sub = T S g S S( )

Experimental Procedure You will measure the vapor pressure of iodine in 5 increments from 20 C to 70 C. You will accomplish this by measuring the absorbance of I2 with a UV/vis spectrometer and then apply Beers Law. Iodine has a strong electronic absorption from the ground electronic state to the second excited state:

X1g+ B3

0+ u . The absorption maximum for this transition is at 520 nm. There is also an excitation to the lowest excited electronic state with a maximum at 700 nm, but it is much weaker and we can neglect it. In addition to the iodine, the cell you are using also has an atmosphere of air. The air collides with the iodine molecules and broadens and smoothes the rotational lines in the spectrum. This makes it easier to get a value of the absorbance without worrying about whether or not you are at the peak of a sharp rotational feature. Make sure that you allow time for the temperature to stabilize. You should take three readings at each temperature so you can get an idea of the reproducibility of your data.

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Analysis of Experimental Results There are two key things to remember when carrying out your calculations. Make sure your spreadsheet is organized. Make sure you pay close attention to your units. For each of your scans determine the difference between the absorbance at 520 nm and 700 nm. There is weak absorbance at 700 nm so you are using that as a background to find a net absorbance. Beers Law tells us how to relate absorbance to pressure.

A = dc

We substitute in for the concentration term using the ideal gas law:

NV

=P

RT

A = dPRT

P = ARTd

d is the pathlength of your UV cell. e is the molar absorption coefficient at 520 nm. The table

below1 gives molar absorption coefficients in units of

lmol cm

.

T (C) (l/molcm) 20 691 30 682 40 672 50 663 60 654 70 646

Once you have determined the pressure at each temperature, you can use a plot of the Clausius-Clapeyron equation to find

H sub . You can also calculate

E 00 at each temperature using equation 5. How well do your values

agree with each other? You can also determine

E 00 graphically using equation 5. How does

this value agree with the others? How close is the value of your y-intercept to the value you would expect based on equation 5? Calculate the molar entropies for both the gas and the solid at 320 K and find the molar heat of sublimation. How do your values compare to the literature values2 for these quantities?

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Uncertainty Analysis You can estimate the uncertainty in your

E 00measurements at the 6 different temperatures by

finding the standard deviation of these results. What assumption are you making when you make this approximation? When determining

E 00 graphically, make sure to include the uncertainty in

your slope (and also your intercept). Do your results agree with accepted values within your uncertainty? References 1. Shoemaker 2. D. A. Shirley and W. F. Giauque, J. Am. Chem. Soc. 31, 4778 (1959)

Questions