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.
Residual entropy of ice
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.
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:
T
S = S(T) – S(0) =
0
CV(T)dT
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.
T
We have
CP(T)dT
S = S(T) – S(0) =
0
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.
Tfus
S(T) =
0
Cp(T)dT  fusH
+
+
T
Tfus
Tvap
Tfus
Cp(T)dT  vapH
+
+
T
Tvap
T
Tvap
Cp(T)dT
T
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.
Graphical representation
Gas
Liquid
Solid
Tfus
S(T) =
0
Cp(T)dT  fusH
+
+
T
Tfus
Tvap
Tfus
Cp(T)dT  vapH
+
+
T
Tvap
T
Tvap
Cp(T)dT
T
Graphical representation
Gas
Liquid
Solid
Tfus
S(T) =
0
Cp(T)dT  fusH
+
+
T
Tfus
Tvap
Tfus
Cp(T)dT  vapH
+
+
T
Tvap
T
Tvap
Cp(T)dT
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
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.
rS =  Sm(products) –  Sm(reactants)



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).
An example: formation of H2O
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.
The spontaneity of chemical reactions
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:

 r Ssurr
 rH
=
T
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).
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
A
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.
time
Definition of Helmholtz Free Energy
Expressing the change in the Helmholtz free energy we have
A = U – TS
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 – TS  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 = TS 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 – TS  0 (at constant T and P)
A = U – TS  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
>0
<0
<0
>0
S
>0
<0
>0
>0
<0
Description of process
Endothermic, spontaneous for T > H/S
Exothermic, spontaneous for T < H/S
Exothermic, spontaneous for all T
Never spontaneous
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
vapGm = Gm H2O(g) – Gm H2O(l)
where we have expressed G as a molar free energy.
From the definition of free energy we have
vapGm = vapHm – TvapSm
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,
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:
G =
P2
VdP
P1
For one mole of an ideal gas we have:
Gm = RT
P2
P1
dP = RT ln P2
P
P1
Note that we have expressed G as a molar quantity Gm = G/n.
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.
Gm T = G 0 T + RT ln
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:
RT ln
P
= RT ln P – RT ln 1 = RT ln P
1 bar
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 = G 0 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:
Rate of change of G as number of
moles of i changes with all other
 i = G
ni T,P, j  i
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.
2
NO 2
P
K=
PN 2O4
At constant T and P we will write the total Gibbs energy as:
dG =  NO2dn NO2 +  N2O4dn N2O4
dG = 2 NO2dn –  N2O4dn
We use the reaction stoichiometry to obtain the factor 2 for NO2.
Definition of the Gibbs free energy
change for chemical reaction
We now define rxnG:
 rxnG = dG
dn
T,P
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,
Gm T = G 0 T + RT ln
P
1 bar
to a multicomponent system:
 i T =  0i T + RT ln
Pi
1 bar
These two expressions are essentially identical. The
chemical potential, i, is nothing more than a molar free energy.
Application of definitions to the
chemical reaction
We can write the Gibbs energy as:
G = 2NO2 – N2O4
and use the chemical potentials:
 NO 2=  0NO 2 + RT ln PNO 2
 N 2O4=  0N2O4 + RT ln PN2O4
to obtain the following:
G = 2 0NO 2 –  0N2O4 + 2RT ln PNO 2 – RT ln PN2O4
2
P
NO 2
G = G 0 + RT ln
PN2O4
, G 0 = 2 0NO 2 –  0N2O4
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:
2
P
NO 2
0
0
0 = G + RT ln K , G = –RT ln K , K =
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.
G 0 = –RT ln(K)
G 0 = H 0 – TS 0
0
0
H
S
ln(K) = –
+
RT
R
0
H
1 – 1
ln(K 2) – ln(K 1) =
R T1 T2
If we plot ln(K) vs 1/T the slope is -Ho/R. This is a useful
expression for determining the standard enthalpy change.
Van’t Hoff plots
Slope = -Ho/R
Note: Ho > 0
The standard method for obtaining the
reaction enthalpy is a plot of ln K vs. 1/T
van’t Hoff plot for exothermic process
Slope = -Ho/R
Note: Ho < 0
In this example, the slope is positive because the enthalpy
of binding is negative (i.e. binding is exothermic).
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.
Pressure dependence
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
N2O4
Initial
Delta
Final
NO2
Total
0
-x
2x
+x
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:
N2O4
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?
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 = G + Vm P – 1
0
P = 1 – G = 1 – G
M –M
Vm
r d r gr
0
0
Example: the formation of diamond
Plugging the values we find:
0 = G 0 + Vm P – 1
0
0
G
G
P=1–
=1–
M –M
Vm
r d r gr
2900 J/mol
=1–
0.012 kg/mol 1 – 1
3510 2260
9
= 1.5 x 10 Pa = 15000 bar