Lecture 8Lecture 8 Statistical Thermodynamics...

Chemistry 433Chemistry 433

Lecture 8Lecture 8Statistical Thermodynamics

Thermochemistry ConsiderationsExact Differentials

NC State University

The rotational partition function

h l f b d d f lThe rotational partition function can be derived from rotational

energy levels. The sum over levels has the form

2

qrot = 2J + 1 exp – h2

2IkTJ J+1ΣJ = 0

∞

If the levels are sufficiently closely spaced relative to thermal

energy then the sum can be written as an integral

qrot = 2J + 1 exp – h2

2I J J+1 dJ∞

2I0

The rotational partition function

2Then we make the substitution, let u = J(J+1) = J2 + J.

Then du = (2J+1)dJ so that

qrot = exp – h2

2IkTu du0

∞

= 2IkTh2 = 8π2IkT

h2

is the rotational partition function for a diatomic

0

molecule in the high temperature limit. For most gas

phase molecules at room temperature, the high

temperature limit is valid.

The vibrational partition function

The vibrational partition function is

qvib = exp hω v + 1Σ∞

q = exp – kT v + 2Σv = 0

= exp – hω2kT exp – vhω

kTΣv = 0

∞

2kT kT0

= exp – hω2kT exp – hω

kT

ν

Σv = 0

∞

=exp – hω

2kT

1 – exp – hωkTe p kT

Review of the l ftranslation partition function

We have seen that the translation partition function can also be derived using a Gaussian integral. It is

qtrans = 2πmkT2

3/2

V

We can also write it as

qh2 V

qtrans = VΛ3

Where L is called the thermal wavelength.

Λ = 12 kT 3/22πmkT

h2

The molecular partition functionThe molecular partition function

The molecular partition function is the product of the Individual partition functions rotation, vibration and translation:

q(V T) = e– β(εatrans + εbvib + εcrot +εdelec)Σq(V,T) eΣa,b,c,d

= e– βεatrans e– βεbvibΣb

e– βεcrotΣcΣa e– βεdelecΣd

t ib t l= qtransqvibqrotqelec

qrot = Krot

β , Krot = 8π2I2q β , rot h2

Average energy and heat capacityWe have seen that the internal energy can be written as:

E = ∂ln Q

Many books write the energy as E (physics and statistical

E = – ∂β

mechanics), however, in thermodynamics we often writeit as U. They are the same thing.

∂ln Q

We have seen that the heat capacity is the derivative of

U = – ∂ln Q∂β

We have seen that the heat capacity is the derivative of the energy with respect to temperature.

∂UCv = ∂U∂T v

Heat capacity for a diatomic or polyatomic molecule

The derivative can be rewritten asThe derivative can be rewritten as∂∂β = – kT2 ∂

∂T or ∂∂T = – kβ2 ∂

∂β

Therefore the heat capacity can also be written as

C = – kβ2 ∂U = kβ2 ∂2ln Q

Recalling that Q = qN/N! we can also write the internal

Cv = – kβ ∂β v= kβ

∂β2v

Recalling that Q q /N! we can also write the internalenergy and heat capacity in terms of the molecular partitionfunction.

∂ln q ∂2ln qU = – N ∂ln q∂β v

, Cv = Nkβ2 ∂ ln q∂β2

v

The contribution of rotationThe contribution of rotationThe contributions of the translational, rotational and ib ti l titi f ti t th i t lvibrational partition functions to the internal energy can

be made separately. The fact that they can be writtenas logarithms helps greatly. The rotational contributiong p g yto the energy and heat capacity for a diatomic is:

ln(qrot) = ln(Krot) – ln(β)(q ) ( rot) (β)

U = N ∂lnβ∂β = N

β = NkT , Urot = RT if N = NA∂β v β rot A

Cv = – Nkβ2 ∂2ln β∂β2 = Nk, Cv rot = R if N = NAv β∂β2

v

, v,rot A

Consequences of including rotation

We have seen that the internal energy and heat capacityf t i id lof a monatomic ideal gas are:

Um = 32RT C m = 3

2R

These were derived from the translational partition function(see lecture 4) In this lecture we have formally included

Um 2RT , Cv,m 2R

(see lecture 4). In this lecture we have formally includedThe contribution from rotations for a diatomic molecule.The internal energy and heat capacity of a diatomic

l lmolecule are:Um = 5

2RT , Cv,m = 52R

Rotation of polyatomic moleculesDiatomic molecules have two rotational degrees of freedomand polyatomic molecules have three rotational degrees offreedom. The partition function for a polyatomic can be written as:

rot 8π2IAkT 1/2 8π2IBkT 1/2 8π2ICkT 1/2

qrot = π A

h2π B

h2π C

h2

= KAKBKC1/2

It is immediately obvious that the contribution from rotations

= A B C

β 3

is Urot = 3/2RT and Cv,rot = 3/2 R. The total values are:

Um = 3RT , Cv,m = 3R

The contribution of vibrationsThe contribution of vibrationsThe vibrational partition function has a different form thaneither the rotational or translational One important differenceeither the rotational or translational. One important differenceis that the energy of vibrational modes is not zero at absolute zero. The zero point energy is 1/2hν. To properlyaccount for this zero point energy we will need to introducea reference internal energy, Uo, which is the zero point energy.The vibrational partition function is:p

qvib =exp – βhω

2q =1 – exp – βhω

ln(qvib) = – βhω2 – ln (1 – exp – βhω )2

The contribution of vibrationsThe contribution of vibrationsThe first term is the zero point energy:

Uo =∂ βhω

2∂β = hω

2

The second term is the population of higher vibrational levels:

β

Uvib =∂ ln (1 – exp – βhω )

∂β =exp – βhω hω

1 – exp – βhω

To see how big this term is let’s consider the function in termskT (i.e. 1/β). We will make a substitution x = hν/kT. If vibrational energy levels are larger than thermal energyhν >> kT, then x >> 1.

The contribution of vibrationsThe contribution of vibrationsThe second term is the population of higher vibrational levels:

Uvib =exp – x x

1 – exp – xkT

The function is plotted to theright. Note that the y-axis g yunits of RT are appropriate for a molar energy (kJ/mol) and kT in the aboveand kT in the above expression is appropriate for a single molecule J.

The pressureThe pressureBased on the definition of pressure volume work dU = PdV so thatdU = -PdV so that

Pj N,V = –∂Uj

∂V N

The index j comes from the fact that we are going to averageover an ensemble (collection) of j systems. We are assuming

∂V N

( ) j y gthat the number of molecules and volume of each system isthe same. Using our previous definition for the calculationof a molecular property the average pressure is:of a molecular property the average pressure is:

P = pj N ,V,βΣj

Pj N ,V

Little pj is the probability.

The pressureSubstituting in we find:

P = p N V βΣ ∂Uj = e– βE j(N,V)

Σ ∂Uj

Since the derivative is with respect to volume, the pressure

P = pj N,V,βΣj

– j

∂V = eQ(N,V,β)Σ

j– j

∂V

p , pdepends only the translational partition function. The derivative of the partition function with respect to volume is

Th th b itt

∂Q∂V N,β

= – β∂Uj

∂V Ne– βU j(N,V)Σ

j

The average pressure can then be written as

P = kT ∂QP = Q(N,V,β) ∂V

The statistical formula for pressure

The pressure can be expressed solely terms of theThe pressure can be expressed solely terms of the partition function.

P kT ∂lnQP = kT Q∂V N,β

With this definition in hand we can also write the thermodynamicfunction of enthalpy in terms of the partition function.

H = U + PV = – ∂ln Q∂β + V

β∂ln Q∂V

Differential relationships for enthalpy

We have defined a relationship between the enthalpy andWe have defined a relationship between the enthalpy and Internal energy

H = U + PVThe infinitesimal change in the state function H results inThe infinitesimal change in the state function H results in

H + dH = U + dU + (P + dP)(V + dV)Therefore

dH = dU + PdV + VdPNow we substitute dU = δq + δw into this expression

dH = δq + δw + PdV + VdPdH δq δw PdV VdPSince δw = - PdV

dH = δq + VdPAt constant pressure dP = 0 and we haveAt constant pressure, dP = 0, and we have

dH = δqP

A note on using current flow in l i t tcalorimetry measurements

Energy is measured in Joules. Electrical energy is used to delivery heat in calorimetry applications. In electrostaticsthe units of energy are:

Joules = Coulombs * VoltsJoules Coulombs VoltsJ = CV

Coulomb is a unit of charge and volt is a unit of potential.A charge moving through a potential is a little like a waterfallA charge moving through a potential is a little like a waterfall.The problem here is that we need a dynamic descriptionsince “moving charge” is not “static”.

C

V

O


As charge moves through the potential energy (heat) isAs charge moves through the potential energy (heat) is Released.

Power = Amperes * VoltsW IVW = IV

Power has units of energy per unit time. So as the currentflows through a wire at a certain rate, heat energy is addedto the system at that rate.

C

V

O


Over a period of time t the total energy added to the systemOver a period of time, t, the total energy added to the systemis:

Energy = Amperes * Volts * timeE IVtE = IVt

This equation is used in the text to describe a number of calorimetry applications.

C

V

O

The standard stateThe standard stateThe standard state of a substance is its pure form a 1 bar ofpressure The standard state can have any temperaturepressure. The standard state can have any temperature,but the temperature should be specified.

For example, the standard enthalpy of vaporization of waterIs the enthalpy of vaporization at 373 K (the boiling point) and1 bar of pressure.p

Biological implicationsBiological implicationsObviously, the food we eat release heat in our bodies. Thisheat is used both to maintain body temperature and forheat is used both to maintain body temperature and for processes that build up our bodies (anabolic processes).

We often talk about calories in the food we eat. The calorieon a cereal box is equal to 1000 calories in the chemicalnomenclature. 1 calorie = 4.184 Joules. Therefore, the,Intake required for an average man of 12 MJ/day is about 3000 calories per day (in the sense of diet).

Thermodynamic cyclesThe energies and enthalpies of ionic solids are dominated by Coulombic interactions. The lattice enthalpy can be y pycalculated from Coulombic interactions. Using the calculated lattice enthalpy and experimental data one can obtain the enthalpy of formation of an ionic solid by meansobtain the enthalpy of formation of an ionic solid by means of the Born-Haber cycle. This is illustrated on the next slidefor a generic salt of a monovalent ion M+X-.

The reactions needed to obtain M+X- in the solid phase to M+ + X- in the vapor phase are given below. The overall p p gprocess of separating the ions in an ionic solid into the constituent ions is written asM+X- (s) M+ (g) + X- (g)M X (s) M (g) + X (g)Lattice enthalpy

The Born-Haber cycleThe Born Haber cycleThe thermodynamic cycle is illustrated. The scheme showsthat if one knows all of the energies except one (difficult tothat if one knows all of the energies except one (difficult tomeasure) quantity, one can calculate it using Hess’ law.

M+ + X-Lattice Enthalpy

I i ti

M+X-

Enthalpyof Formation

Enthalpyof Vaporization

ElectronAffinity

IonizationPotential

1/2 X2 (g) + M (s) M (s)of Vaporization

Bond DissociationX (g)

Bond Dissociation

The Born-Haber cycleThis reaction is composed of the following stepsM+X- (s) M (s) + 1/2 X2 (g) Dissociation enthalpy M ( ) M ( ) E th l f i tiM (s) M (g) Enthalpy of vaporization1/2 X2 (g) X (g) Bond dissociation M M+ + e- Ionization potential X + e- X- Electron affinityThus, we can express the above in terms of a thermodynamic cycle that allows us to determine thethermodynamic cycle that allows us to determine the dissociation energy using known experimental quantities. This is necessary for ionic solids because it is essentially i ibl t di tl th di i ti th limpossible to directly measure the dissociation enthalpy. The dissociation enthalpy is equal to the enthalpy of formation (but opposite in sign). The above cycle and the reasoning applied will allow us to determine with some the forces between ions due to Coulombic interactions.

Dependence of internal energy h i l d t ton changes in volume and temperature

We have seen from the first law that the internal energychange in the system is equal to the work done and thechange in the system is equal to the work done and theheat transferred. Work depends on volume changes andheat transfer leads to changes in temperature. Thus, it islogical that the initial description of internal energy dependson volume and temperature. Thus, for example the depend-ence of U on volume at constant temperature is:p

U′ = U + ∂U∂V T

dV

And similarly the dependence of U on temperature at constant volume is:

T

∂UU′ = U + ∂U∂T V

dT


Putting these together we have:g g

U′ = U + ∂U∂V T

dV + ∂U∂T V

dT

which can be written as the total derivative:

dU = ∂U dV + ∂U dT

This called the total derivative. The idea is that it includes the

dU = ∂U∂V T

dV + ∂U∂T V

dT

variables in a space. Here the space is V,T space. We shallsee that these are not the only possible variables that candescribe the dependence of Udescribe the dependence of U.


The partial derivatives are slopes in the space.p p p

∂U∂V T

= πV , ∂U∂T V

= CV

We have already seen CV and we know it as the heat capacityAt constant volume. However, πV is new. πV is the change in

fthe internal energy when the volume of a substance is changed at constant temperature.

If the intermolecular interactions are zero (ideal gas), then πVwill be zero. However, real interactions between moleculescan give rise to non-zero π Joules set out to measure πcan give rise to non-zero πV. Joules set out to measure πVin an expansion experiment, but was not successful.

Expansion coefficient and i th l ibilitisothermal compressibility

In order to proceed with the next level of analysis we intro-p yduce the expansion coefficient:

α = 1V

∂V∂T

The expansion coefficient is a measure of the change in molar volume (i.e. also inverse density) of a material with

V ∂T P

temperature. Many substances expand as the temperatureis increased, hence the name, expansion coefficient.The isothermal compressibility is p y

a measure of the degree to which a substance will have a

κ = – 1V

∂V∂P T

a measure of the degree to which a substance will have ahigher density (smaller molar volume) at high pressure.

Heat capacity relationshipsp y p

The heat capacity at constant pressure and constant volumep y pare related by:

CP – CV = nR

The above expression applies only to an ideal gas. We haveseen the utility of that expression in that it applies to monatomic, diatomic and polyatomic gasses.

For liquids and solids we have the formula:q

CP – CV = α2TVκ

The Joules-Thompson coefficientpThe differential of the enthalpy can be expressed as:

Where μ is the Joules-Thomson coefficient. The enthalpyh b itt

dH = – μCPdP + CPdT

change be written:

dH = ∂H∂P dP + ∂H

∂T dTTherefore:

∂H∂P = – ∂T

∂P∂H∂T = – μCP

∂P T ∂T P

The derivation of the above expression relies on the permutation relation:

∂P T ∂P H ∂T P

∂T∂P H

∂P∂H T

∂H∂T P

= – 1

Modern measurement of the J l Th ffi i tJoules-Thompson coefficient

It is relatively difficult to perform measurements underconstant H (isenthalpic or adiabatic) conditions Thereforeconstant H (isenthalpic or adiabatic) conditions. Therefore,the modern measurements are at constant temperature.In other words,

∂H∂P T

= μT

is measured. Based on the expression on the previous slide:

μ = ∂T ∂H = μCμT = – ∂T∂P H

∂H∂T P

= – μCP

Lecture 8Lecture 8 Statistical Thermodynamics...

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