Thermodynamics

Lecture 9

The Third Law

Free Energy Functions

NC State University

Thermodynamics

The Third Law

NC State University

The Third Law of Thermodynamics

The third law of thermodynamics states that every substance

has a positive entropy, but at zero Kelvin the entropy is

zero for a perfectly crystalline substance. The third law

introduces a numerical scale for the entropy. Stated

succinctly:

S(0) = 0 for all perfectly ordered crystalline materials.

It is commonly said that all motion ceases at absolute zero.

Certainly all translation and rotation have ceased at absolute

zero. However, the nuclei still vibrate about their equilibrium

positions in so-called zero point motion. However, all

molecules are in their lowest vibrational state and entropy is

zero in any case provided there are no imperfections in the

crystal.

Residual Entropy

The first statistical picture of entropy was that of Boltzmann.

The function W that represents the number of ways we can

distribute N particles into a number of states. Using the

function W the entropy can be expressed as:

S = kB ln W

At zero Kelvin the system is in its lowest energy state. For

a perfect crystal there is only one way to distribute the energy

and W = 1, therefore S = 0.

However, the entropy not equal to zero at T = 0 K if the

substance is not a perfect crystal. Although the residual

entropy in such cases is a small correction to the entropy

calculated for chemical reactions it still leads to an important

concept.

CO: an Imperfect Crystal

The molecule CO has a very small dipole moment and

there is a finite chance that CO will crystallize as

CO:CO:CO instead of CO:OC:CO. For each CO molecule

there are two possible orientations of the molecule, therefore

there are two ways each CO can exist in the lattice.

The number of ways per molecule is w = 2 for each CO.

If we have N CO molecules there are wN ways or 2N ways

that all of the CO can be distributed. Therefore, the entropy

at zero Kelvin is

S = k ln W = k ln(wN) = Nk ln w = nR ln 2.

The entropy at zero Kelvin is known as residual entropy.

There are number of substances that show similar statistical

variations in orientation that lead to a residual entropy.

Because of the multiple hydrogen bond partners possible

In ice, there is a residual entropy of S = R ln(3/2).

The Bernal-Fowler rule allows the rotation of water molecules

within the ice lattice by hydrogen atoms jumping sites. A

single hydrogen atom lies on a line between each oxygen atom.

The angle between oxygen atoms in ice is109°, thus

only a small variation of the H-O-H angle in lattice. There are

six possible configurations of H atoms around O atoms.

Residual entropy of ice

The Temperature

Dependence of Entropy

We have seen calculations for the entropy change for processes.

However, it is also possible to calculate the absolute entropy.

We can begin with the definition dS = dqrev/T. The heat

transferred during a process at constant volume is

dqv,rev = CvdT. Thus, the entropy change at constant volume is:

S = S(T) – S(0) =CV(T)dT

T0

T

We have kept the equation general by showing Cv(T) as a

function of temperature. This calculation of the entropy is

valid only at constant V. At constant P we find an analogous

expression. Starting with the heat transferred, dqp,rev = CpdT.

We have

S = S(T) – S(0) =

CP(T)dT

T0

T

The temperature dependence of

the heat capacity

The third law of thermodynamics shows that

Cp 0 as T 0 K.

It is necessary to treat Cp(T) as a function of temperature for

this reason alone. The Einstein and Debye theories of heat

capacity can be used to determine the functional form of the

heat capacity of at these low temperatures. The Debye law

states that the heat capacity depends on T3 near T = 0 K.

At temperatures close to T = 0, CV,m = aT3

The constant a is an empirical constant. For practical calculation

of the entropy experimental values can be used and the

integrals are evaluated numerically.

Absolute Entropy

The absolute entropy can be calculated from 0 to any

temperature T using the integral of the function CP(T)/T.

If there is a phase transition between 0 and temperature T

we can also calculate the contribution to the entropy from

the transition.

The heat capacity of each phase is different and the heat

capacities are also a function of temperature. Remember

that the heat capacity approaches zero and the temperature

goes to zero so that the Cp is a strong function of T at very

low temperature.

S(T) =Cp(T)dT

T0

Tfus

+ fusH

Tfus

+Cp(T)dT

TTfus

Tvap

+vapH

Tvap

+Cp(T)dT

TTvap

T

Graphical representation

Solid

Liquid

Gas

S(T) =Cp(T)dT

T0

Tfus

+ fusH

Tfus

+Cp(T)dT

TTfus

Tvap

+vapH

Tvap

+Cp(T)dT

TTvap

T

Graphical representation

Solid

Liquid

Gas

S(T) =Cp(T)dT

T0

Tfus

+ fusH

Tfus

+Cp(T)dT

TTfus

Tvap

+vapH

Tvap

+Cp(T)dT

TTvap

T

Absolute Entropies can be used to

calculate Reaction Entropies Entropies are tabulated in order to facilitate the calculation

of the entropy change of chemical reactions. For the

general reaction

aA + bB yY + zZ

the standard entropy change is given by

rSo = ySo[Y] + zSo[Z] - aSo[A] - bSo[B]

where the absolute entropies So are molar quantities.

Note the significant difference compared to enthalpy. There

is no such thing as an absolute enthalpy. Instead, we used

a reference of the elements in their standard states. In

that case the enthalpy of formation was set arbitrarily to zero.

Since entropies are zero and T = 0 K we can use So as an

absolute quantity.

The standard reaction entropy, rS , is the difference

between the standard molar entropies of the reactants

and products, with each term weighted by the

stoichiometric coefficient.

The standard state is for reactants and products at 1 bar

of pressure. The unit of energy used is J/mol-K.

IMPORTANT: Do not confuse entropy and enthalpy.

One common mistake is to set the entropies of elements

equal to zero as one does for enthalpies of formation.

Elements have an entropy that is not zero (unless the

temperature is T = 0 K).

The standard reaction entropy

rS

= Sm

(products) – Sm

(reactants)

We apply the absolute entropies of H2, O2 and H2O to the

calculation of the entropy of reaction for:

2H2(g) + O2(g) 2 H2O(l)

The entropy change is:

S = 2 S (H2O, l) - S(O2, g) - 2S(H2, g)

= 2(70) - 2(131) - 205 J/mol-K

= - 327 J/mol-K

This result is not surprising when you consider that 3 moles

of gas are being consumed. A gas has a greater number

of translational degrees of freedom than a liquid.

On the other hand, it is difficult (at first) to understand how

a spontaneous reaction can have such a large negative

entropy.

An example: formation of H2O

The spontaneity of chemical reactions must be understood

by consideration of both the system and the surroundings.

This is one of the most subtle points of thermodynamics.

The heat dissipated by the negative enthalpy change in the

reaction to make water results in a large positive entropy

change in the surroundings. In fact, the reaction to make

water proceeds with explosive force. The heat dissipated

in the surroundings rH is also an entropy term. The entropy

change in the surroundings is:

The entropy change of the surroundings is:

rS = -(-572 kJ/mol)/298 K = + 1.92 x 103 J/mol-K

which is much larger than the negative entropy change of the

system (-327 J/mol-K).

The spontaneity of chemical reactions

rSsurr

=

rH

T

System and surroundings both

play in role in the entropy

In an isolated system the criterion dS > 0 indicates that a

process is spontaneous. In general, we must consider dSsys

for the system and dSsurr for surroundings. Since we can

think of the entire universe as an isolated system dStotal > 0.

The entropy tends to increase for the universe as a whole.

If we decompose dStotal into the entropy change for the

system and that for the surroundings we have a criterion

for spontaneity for the system that also requires consideration

of the entropy change in the surroundings. The free energy

functions will allow us to eliminate consideration of the

surroundings and to express a criterion for spontaneity solely

in terms of parameters that depend on the system.

Thermodynamics

Free Energy

NC State University

Free Energy at Constant T and V

Starting with the First Law

dU = dw + dq

At constant temperature and volume we have dw = 0 and

dU = dq

Recall that dS dq/T so we have

dU TdS

which leads to

dU - TdS 0

Since T and V are constant we can write this as

d(U - TS) 0

The quantity in parentheses is a measure of the spontaneity

of the system that depends on known state functions.

Definition of Helmholtz Free Energy

We define a new state function:

A = U -TS such that dA 0.

We call A the Helmholtz free energy.

At constant T and V the Helmholtz free energy will decrease

until all possible spontaneous processes have occurred. At

that point the system will be in equilibrium.

The condition for equilibrium is dA = 0.

A

time

Definition of Helmholtz Free Energy Expressing the change in the Helmholtz free energy we have

A = U – TS

for an isothermal change from one state to another.

The condition for spontaneous change is that A is less

than zero and the condition for equilibrium is that A = 0.

We write

A = U – TS 0 (at constant T and V)

If A is greater than zero a process is not spontaneous.

It can occur if work is done on the system, however. The

Helmholtz free energy has an important physical interpretation.

Noting the qrev = TS we have

A = U – qrev

According to the first law U – qrev = wrev so

A = wrev (reversible, isothermal)

A represents the maximum amount of reversible work that

can be extracted from the system.

Definition of Gibbs Free Energy

Most reactions occur at constant pressure rather than

constant volume.

Using the facts that dqrev TdS and dwrev = -PdV we have:

dU TdS – PdV

which can be written dU - TdS + PdV 0.

The = sign applies to an equilibrium condition and

the < sign means that the process is spontaneous. Therefore:

d(U - TS + PV) 0 (at constant T and P)

We define a state function G = U + PV – TS = H – TS.

Thus, dG 0 (at constant T and P)

The quantity G is called the Gibb's free energy.

In a system at constant T and P, the Gibb's energy will

decrease as the result of spontaneous processes until

the system reaches equilibrium, where dG = 0.

Comparing Gibbs and Helmholtz The quantity G is called the Gibb's free energy. In a system

at constant T and P, the Gibb's energy will decrease as the

result of spontaneous processes until the system reaches

equilibrium, where dG = 0.

Comparing the Helmholtz and Gibb's free energies we see

that A(V,T) and G(P,T) are completely analogous except

that A is valid at constant V and G is valid at constant P.

We can see that

G = A + PV

which is exactly analogous to

H = U + PV

the relationship between enthalpy and internal energy.

For chemical processes we see that

G = H – TS 0 (at constant T and P)

A = U – TS 0 (at constant T and V)

Conditions for Spontaneity We will not use the Helmholtz free energy to describe

chemical processes. It is an important concept in the

derivation of the Gibbs energy. However, from this point

we will consider the implications of the Gibbs energy for

physical and chemical processes.

There are four possible combinations of the sign of

H and S in the Gibbs free energy change:

H S Description of process

>0 >0 Endothermic, spontaneous for T > H/S

<0 <0 Exothermic, spontaneous for T < H/S

<0 >0 Exothermic, spontaneous for all T

>0 >0 Never spontaneous<0

Gibbs energy for a phase change For a phase transition the two phases are in equilibrium.

Therefore, G = 0 for a phase transition.

For example, for water liquid and vapor are in equilibrium

at 373.15 K (at 1 atm of pressure). We can write

where we have expressed G as a molar free energy.

From the definition of free energy we have

The magnitude of the molar enthalpy of vaporization is

40.7 kJ/mol and that of the entropy is 108.9 J/mol-K.

Thus,

vapGm = vapHm – TvapSm

vapGm = Gm H2O(g) – Gm H2O(l)

vapG = 40.65 kJ mol– 1– 373.15 K 108.9 J K – 1 mol

– 1= 0

Gibbs energy for a phase change However, if we were to calculate the free energy of

vaporization at 372.15 K we would find that it is +1.1 kJ/mol

so vaporization is not spontaneous at that temperature.

If we consider the free energy of vaporization at 374.15 K

it is -1.08 kJ/mol and so the process is spontaneous (G < 0).

State Function Summary At this point we summarize the state functions that we have

developed:

U (internal energy)

H = U + PV (enthalpy)

S (entropy)

A = U - TS (Helmholtz free energy)

G = U + PV - TS = H - TS (Gibbs free energy)

Please note that we can express each of these in a differential

form. This simply refers to the possible changes in each

function expressed in terms of its dependent variables.

dH = dU + PdV + VdP

dA = dU - TdS - SdT

dG = dH - TdS - SdT

The internal energy expressed

in terms of its natural variables We can use the combination of the first and second laws to

derive an expression for the internal energy in terms of its

natural variables. If we consider a reversible process:

dU = dq + dw

dw = -PdV (definition of work)

dq = TdS (second law rearranged)

Therefore, dU = TdS - PdV

This expression expresses the fact that the internal energy

U has a T when the entropy changes and a slope -P when

the volume changes. We will use this expression to derive

the P and T dependence of the free energy functions.

The Gibbs energy expressed

in terms of its natural variables To find the natural variables for the Gibbs energy we begin

with the internal energy:

dU = TdS - PdV

and substitute into:

dH = dU + PdV + VdP

to find:

dH = TdS + VdP (S and P are natural variables of enthalpy)

and using:

dG = dH - TdS - SdT

we find:

dG = -SdT + VdP (T and P are natural variables of G)

Once again we see why G is so useful. Its natural variables

are ones that we commonly experience: T and P.

The variation of the Gibbs

energy with pressure

We have shown that dG = VdP - SdT. This differential can be

used to determine both the pressure and temperature

dependence of the free energy. At constant temperature:

SdT = 0 and dG = VdP.

The integrated form of this equation is:

For one mole of an ideal gas we have:

Note that we have expressed G as a molar quantity Gm = G/n.

G = VdPP1

P2

Gm = RT dPPP1

P2

= RT lnP2

P1

The variation of the Gibbs

energy with pressure We can use the above expression to indicate the

free energy at some pressure P relative to the pressure

of the standard state P = 1 bar.

G0(T) is the standard molar Gibb’s free energy for a gas.

As discussed above the standard molar Gibb’s free energy

is the free energy of one mole of the gas at 1 bar of pressure.

The Gibb’s free energy increases logarithmically with pressure.

This is entirely an entropic effect.

Note that the 1 bar can be omitted since we can write:

Gm T = G0

T + RT ln P1 bar

RT ln P1 bar

= RT ln P – RT ln 1 = RT ln P

The pressure dependence of G for

liquids and solids If we are dealing with a liquid or a solid the molar volume

is more or less a constant as a function of pressure.

Actually, it depends on the isothermal compressibility,

k = -1/V(V/P)T, but k is very small. It is a number of the

order 10-4 atm-1 for liquids and 10-6 atm-1 for solids. We

have discussed the fact that the density of liquids is

not strongly affected by pressure. The small value of k

is another way of saying the same thing.

For our purposes we can treat the volume as a constant

and we obtain

Gm T = G0

T + Vm P – 1

Systems with more than one

component Up to this point we have derived state functions for pure

systems. (The one exception is the entropy of mixing).

However, in order for a chemical change to occur we must

have more than one component present. We need generalize

the methods to account for the presence of more than one

type of molecule. In the introduction we stated that we would

do this using a quantity called the chemical potential. The

chemical potential is nothing more than the molar Gibbs

free energy of a particular component. Formally we write it

this way:

i = Gni T,P, j i

Rate of change of G as number of

moles of i changes with all other

variables held constant.

Example: a gas phase reaction

Let’s consider a gas phase reaction as an example. We will

use a textbook example:

N2O4 (g) = 2 NO2 (g)

We know how to write the equilibrium constant for this reaction.

At constant T and P we will write the total Gibbs energy as:

dG = NO2

dnNO2+ N2O4

dnN2O4

dG = 2NO2dn – N2O4

dn

We use the reaction stoichiometry to obtain the factor 2 for NO2.

K =PNO2

2

PN2O4

Definition of the Gibbs free energy

change for chemical reaction We now define rxnG:

This is rxnG but it is not rxnGo! Note that we will use rxnG

and G interchangably.

If we now apply the pressure dependence for one component,

to a multicomponent system:

These two expressions are essentially identical. The

chemical potential, i, is nothing more than a molar free energy.

rxnG = dGdn T,P

Gm T = G0

T + RT ln P1 bar

i T = i0 T + RT ln

Pi

1 bar

Application of definitions to the

chemical reaction We can write the Gibbs energy as:

and use the chemical potentials:

to obtain the following:

G = 2NO2– N2O4

NO2= NO2

0 + RT ln PNO2

N2O4= N2O4

0 + RT ln PN2O4

G = 2NO2

0 – N2O4

0 + 2RT ln PNO2– RT ln PN2O4

G = G0

+ RT lnPNO2

2

PN2O4

, G0

= 2NO2

0 – N2O4

0

Note the significance of G and Go

The change G is the change in the Gibbs energy function.

It has three possible ranges of value:

G < 0 (process is spontaneous)

G = 0 (system is at equilibrium)

G > 0 (reverse process is spontaneous)

On the other hand Go is the standard molar Gibbs energy

change for the reaction. It is a constant for a given chemical

reaction. We will develop these ideas for a general reaction

later in the course. For now, let’s consider the system at

equilibrium. Equilibrium means G = 0 so:

0 = G0

+ RT ln K , G0

= –RT ln K , K =PNO2

2

PN2O4

Temperature dependence of Go

The van’t Hoff equation We use two facts that we have derived to determine the

temperature dependence of the free energy.

If we plot ln(K) vs 1/T the slope is -Ho/R. This is a useful

expression for determining the standard enthalpy change.

G0

= –RT ln(K)

G0

= H0

– TS0

ln(K) = – H0

RT+ S

0

R

ln(K2) – ln(K1) = H0

R1T1

– 1T2

Van’t Hoff plots

The standard method for obtaining the

reaction enthalpy is a plot of ln K vs. 1/T

Slope = -Ho/R

Note: Ho > 0

van’t Hoff plot for exothermic process

In this example, the slope is positive because the enthalpy

of binding is negative (i.e. binding is exothermic).

Slope = -Ho/R

Note: Ho < 0

Pressure dependence

We can see from the gas phase form of the equilibrium

constant that the equilibrium concentrations of species

depend on pressure. This dependence is “inside” the

equilibrium constant. The equilibrium constant is not

changed by the species are affected for a given value

of K.

The pressure dependence of free energy contained

in the the term dG = VdP is a different dependence,

which can result in a shift in the equilibrium constant

itself.

Pressure dependence of species

We can see from the gas phase form of the equilibrium

constant that pressure of species depend on pressure.

For the general gas phase reaction,

we can write the equilibrium constant as

And the free energy is

From Dalton’s law

Pressure dependence of species

If we substitute these mole fractions and total pressure into

the equilibrium constant we have

Which depends on the total pressure unless z – c – d = 0.

This expression shows that, in general, the free energy

depends on the total pressure. This means that for the

fixed pressure may affect the proportion of products

to reactants.

N2O4(g) 2 NO2(g)

In the last lecture we treated this problem using

If there is a constraint on the total pressure then

We must use Dalton’s law to calculate the mole

Fractions.

Pressure dependence

N2O4(g) 2 NO2(g)

The equilibrium constant is

For a given initial pressure of N2O4 we have

Initial 0

Delta -x 2x +x

Final

Pressure dependence

N2O4 NO2 Total

To solve a gas phase problem you need to know

whether the pressure can change or not. If it is

fixed, then Ptot has a value that does not change.

The mole fractions are,

For example, if we assume that the initial pressure

is maintained, then the above values should be

substituted into the equilibrium constant

Using the same values as yesterday (K = 0.11 and

an initial pressure of 1 atm), the calculated value

for x = 0.163. Therefore, the pressures are:

The percent dissociation is 28%.

If the total pressure is reduced to 0.1 atm, the value

of x = 0.046 and the pressure of NO2 is 0.063. This

Corresponds to a dissociation of 63%.

Can this trend be explained in simple terms?

N2O4

Shift in equilibrium due to pressure:

the formation of diamond Graphite and diamond are two forms of carbon. Given that

the free energy of formation of diamond is:

C(s, graphite) C(s, diamond) rGo = + 2.90 kJ/mol

and the densities are:

r(graphite) = 2.26 and r(diamond) = 3.51

calculate the pressure required to transform carbon into

diamond.

Solution: Graphite will be in equilibrium with diamond when:

0 = G0

+ Vm P – 1

P = 1 – G0

Vm

= 1 – G0

Mrd

– Mrgr

Example: the formation of diamond

Plugging the values we find:

0 = G0

+ Vm P – 1

P = 1 – G0

Vm

= 1 – G0

Mrd

– Mrgr

= 1 – 2900 J/mol

0.012 kg/mol 13510

– 12260

= 1.5 x 109

Pa = 15000 bar