CHE-20028: PHYSICAL & INORGANIC CHEMISTRY STATISTICAL THERMODYNAMICS: LECTURE 2 Dr Rob Jackson...

CHE-20028: PHYSICAL & INORGANIC CHEMISTRY

STATISTICAL THERMODYNAMICS: LECTURE 2

Dr Rob Jackson

Office: LJ 1.16

[email protected]

http://www.facebook.com/robjteaching

che-20028: Statistical Thermodynamics Lecture 2

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Statistical Thermodynamics: topics for lecture 2

• Summary from lecture 1• The molecular partition function• Calculation of thermodynamic properties from

Statistical Thermodynamics– Internal energy– Heat capacity– Residual entropy– Entropy


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The molecular partition function- 1

• What contributes to the energy of a molecule?– Vibrational (V), translational (T) and rotational (R)

modes of motion– Electron distribution (E)– Electronic and nuclear spin (S)

Si

Ei

Ri

Ti

Vii EEEEEE


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• Remembering from lecture 1 that the partition function q is given by:

• We can substitute for Ei in this expression:

i

i )kT/Eexp(q

)kT/EkT/EkT/EkT/EkT/Eexp(q Si

Ei

i

Ri

Ti

Vi


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• We then use some mathematical trickery to simplify the expression (ea+b = eaeb):

• The total energy is the sum of contributions from the different terms, and the total partition function is the product of these contributions:

i i

Si

Ei

i

Ri

i i

Ti

Vi )kT/Eexp()kT/Eexp()kT/Eexp()kT/Eexp()kT/Eexp(q

SERTV qqqqqq


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Note on electron and spin partition functions

• While it is not possible to obtain expressions for these, as we have done for the other terms, we note the following:– For closed shell molecules, excited states are so

high in energy that only the ground state is occupied, and qE= 1

– Electron spin makes an important contribution when there are unpaired electrons, since the electron can occupy either spin state, and then qS=2


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Calculation of internal energy from the partition function - 1

• It can be shown that the internal energy is obtained from the derivative of the partition function with respect to temperature:

• Note that E is the energy with respect to the lowest energy state of the molecule

dT

dq

q

NkTE

2


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Calculation of internal energy from the partition function - 2

• So, in general terms we should write the internal energy U as:

U = E + U(0), where U(0) is the zero point energy

e.g. for a harmonic oscillator U(0) = ½ h• We can use these expressions to calculate

the internal energy for some example systems:


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The internal energy of a monatomic gas - 1

• Only need to consider qT (and assume qE=1)• Remember that qT = aT3/2

(where a= (2mk)3/2 V/h3) (lecture 1 slide 23)

• So – (from slide 6)

• And so

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2

3 /T

aTdT

dq

NkTaTxaT

NkTE /

/T

2

3

2

3 2123

2


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The internal energy of a monatomic gas - 2

• So if N=NA, ET = (3/2)RT

U = U(0) + (3/2)RT• Example: calculate the internal energy of

Argon gas at 300 K U = U(0) + (3/2) x 8.314 x 300 J mol-1

= 3741.3 J mol-1

• What would be the result for another monatomic gas?


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The internal energy of a diatomic gas - 1

• Neglecting vibrational motion, we just need to calculate the energy contribution of rotational motion, ER:

• From lecture 1 slide 20, qR = bT • (where b=k/(hB)) dqR/dT = b

• If N=NA, ER = RT

NkTbbT

NkTE R

2


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The internal energy of a diatomic gas - 2

• So to summarise:U = U(0) + ET + ER = U(0) + (3/2)RT + RT

U = U(0) + 5/2RT

• Note: both expressions, for the monatomic gas and the diatomic gas, assume the gas to be perfect.

• The vibrational contribution has been neglected for the diatomic gas.


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How can we make comparisons with experimental values?

• Heat capacities can be calculated, and compared with experimental values.

• Remember that CV= dU/dT at constant volume.– For a monatomic gas:

U = U(0) + (3/2)RT, dU/dT= (3/2)R= 12.47 J mol-1K-1

– For a diatomic gas:

U = U(0) + (5/2)RT, dU/dT= (5/2)R= 20.79 J mol-1K-1


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Heat capacities

• Good agreement for inert gases and for nitrogen; agreement gets worse for halogens with increasing departure from perfect gas behaviour.

• Why does this happen?

Gas Cv/J K-1 mol-1

He

12.5

Ar

12.5

Xe

12.5

N2

20.8

F2

23.2

I2

28.6

Gas Cv/J K-1 mol-1

He

12.5

Ar

12.5

Xe

12.5

N2

20.8

F2

23.2

I2

28.6

Experimental heat capacities


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Entropy and Statistical Thermodynamics

• If we can calculate entropy, this is an important step to being able to calculate further thermodynamic properties

• We are already familiar with the connection between entropy and the disorder of a system

• Boltzmann suggested that the entropy, S, of a system should be given by the expressionS = k ln W(W is the number of different ways the molecules in a system can be arranged to give the same energy)


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Boltzmann's grave in the Zentralfriedhof, Vienna, with bust and entropy

formula.


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Calculating entropy - 1

• The formula makes sense because S = 0 if there is only one way of achieving a given energy (if W = 1, S = ln (1) = 0).

• Similarly the formula predicts a high entropy if there are many arrangements with the same energy (i.e. if W is large).

• In most cases W = 1 when T= 0 because there is only one way to achieve zero energy: put all the molecules into the lowest energy level.


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Calculating entropy - 2

• In this case S = 0 when T = 0, which agrees with the third law of thermodynamics (the entropy of all perfectly crystalline materials is zero at 0 K).

• There are, however, cases where this is not observed.

• This occurs when there is more than one energetically equivalent arrangement of molecules when T = 0.


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Residual entropy - 1

• There are several examples of molecules that can have more than one energetically equivalent arrangement at 0 K– Examples include linear molecules where they

may be no energetic difference between the arrangementAB AB AB AB and the arrangement AB BA AB AB

– e.g. solid carbon monoxide, where there is no energetic difference between the arrangementsCO CO CO CO and CO CO OC CO


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• In this case we say that the substance has a residual entropy.

• For CO, there are 2 orientations of equal energy, CO or OC, and if there are N molecules, the number of ways of getting the same overall energy is 2N

So S = k ln W = k ln 2N = N k ln 2 = R ln 2

• Where we assume that we have one mole of molecules.


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• In this case, S = 8.314 x ln 2= 5.76 J K-1 mol-1

• The experimental value is 5.0 JK-1 mol-1

• Other examples of molecules having a residual entropy include N2O and H2O.

• The summarised expression is:

S = R ln (no. of orientations of equal energy)• Example questions are included in the

problem sheet.


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The entropy of a monatomic gas

• To calculate the entropy of a monatomic gas we use an equation called the Sackur-Tetrode* equation:

• All the symbols have their usual meanings, and p = 105 Pa, (noting that kB = k).

* Derived in 1912, anniversary in 2012.

2/3

2BB

2/5

m h

Tmk2

p

TkelnRS

Sackur-Tetrode equation example

• Calculate the molar entropy of argon gas (M = 39.948 g mol-1) at 298 K and 105 Pa

Divide the calculation into sections and rewrite the equation as: Sm = R ln (A B3/2)

A = e5/2 kT/p = 12.182 x 1.381x10-23 x 293 /105 = 5.014 x 10-25

B = 2mkT/h2 = 2 x (39.948 x 1.661 x 10-27) x 1.381 x 10-23 x 298)/(6.626 x 10-34)2

= 3.908 x 1021, so B3/2 = 2.443 x 1032

S = R ln (5.014 x 10-25 x 2.443 x 1032) = 154.84 J mol-1 K-1

(Experimental value is 154.6 J mol-1 K-1)


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Advice on calculations

• Don’t try to do the whole calculation in one go!

• This is because most calculators have a maximum index value of 10+/- 99.

• Don’t try to calculate h3 (or h4) in one go!• If your calculation comes out as zero, it’s

probably because the calculator can’t handle the numbers. Break down the calculation.


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Summary

• The molecular partition function has been defined.

• Internal energy and heat capacity have been obtained from the partition function.

• Residual entropy has been introduced and calculated.

• The Sackur-Tetrode equation, for calculating entropy, has been introduced and used.

CHE-20028: PHYSICAL & INORGANIC CHEMISTRY STATISTICAL THERMODYNAMICS: LECTURE 2 Dr Rob Jackson...

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