Statistical Thermodynamics
Solution Exercise 9
HS 2016
Solution Exercise 9
Problem 1: Entropy from molecular properties
a) We want to calculate the molar entropy and therefore N = NAV . We start from the fact
that entropy can be determined by summing relevant contributions from all energy modes
under consideration, i.e.,
s = str + sel + srot + svib + sns .
(1.1)
The nuclear spin contribution sns is in principle not negligible since its magnitude is of the
order of the other contributions. Assuming only the contribution from the most abundant
isotopes 16 O (I = 0) and 14 N (I = +1), we can calculate sns as
ssn = NAV kB
X
ln(2Ii + 1) = NAV kB ln(2 · 1 + 1) ≈ 1.0986 · NAV kB .
(1.2)
i
Dividing the result by the gas constant R = NAV kB , we obtain dimensionless nuclear spin
contribution sns = 1.0986. In the following, we will ignore this contribution which does not
cause problems, as the contribution is constant under most conditions where experiments
are conducted. Moreover, we neglect the zero-point energy as we had defined the partition
function for an energy scaled by the zero-point energy (see script p. 73). In order to have
dimensionless quantities we divide above equation by the gas constant R
str sel srot svib
s
=
+
+
+
.
R
R
R
R
R
(1.3)
Now, we can evaluate all these contributions. The translational contribution can be directly
calculated using Sackur-Tetrode equation, thus obtaining
"
#
3/2
str
5
2πmkB T
V
= + ln
= 20.1281.
(1.4)
R
2
h2
NAV
Next, the contribution from all internal energy modes, in general, is
∂ ln Zint
sint
=T
+ ln Zint ,
R
∂T
V
(1.5)
where Zint = Zel Zrot Zvib is the internal partition function. The electronic contribution can
be calculated, see script p. 73 and particularly eqn. (7.61), as
X
Zel =
gj e−j /kB T ' gel .
(1.6)
j
We neglect the contribution from all excited states because even the lowest excited state is
separated from the ground state by 3.1 eV which is much higher than kB T (≈ 0.04 eV for T
= 500 K). The degeneracy gel of the electronic ground state is given by the multiplicity and
can be deduced from the term symbol of the molecule. In the case of NO2 , the term symbol
is 2 A1 , therefore Zel = 2 and the electronic contribution is
sel
= ln gel = 0.6931.
R
(1.7)
page 1 of 5
Statistical Thermodynamics
Solution Exercise 9
HS 2016
The rotational partition function can be expressed, eqn. (7.46) in the script, as
s
1
πT 3
Zrot =
,
σ θr,x θr,y θr,z
(1.8)
where σ is the symmetry number and θr,i is the rotational temperature defined as
θr,i =
hc
Bi ,
kB
(1.9)
with Bi being the rotational constant in cm−1 along the principal rotational axis labelled by
the index i, see figure 1-1. We must be careful here because the speed of light c must be
Figure 1-1: Principal rotational axis for NO2 molecule.
expressed in units of in cm·s−1 in order to get correct values. Afterwards we can derive that
s
!
srot
3
1
πT 3
= + ln
.
(1.10)
R
2
σ θr,x θr,y θr,z
Because NO2 belongs to the C2v point group (a point group containing identity, two-fold
symmetry axis and two different mirror plane), the symmetry number is σ = 2 (see Table
7.1 in the script). Using the data from Table 1.1 in the exercise sheet, we get
srot
= 9.9788.
R
(1.11)
The vibrational partition function can be expressed, see eqn. (7.49) in the script, as
Zvib =
m
Y
i=1
1
,
1 − e−θv,i /T
(1.12)
where m = 3N −6 is the number of vibrational modes and θv,i is the vibrational temperature
for i-th mode defined as
hc
θv,i =
ωi ,
(1.13)
kB
page 2 of 5
Statistical Thermodynamics
Solution Exercise 9
HS 2016
Figure 1-2: Modes of vibration for NO2 molecule.
with ωi being the vibrational frequency in cm−1 (see figure 1-2). Note that c must again be
expressed in cm·s−1 .
Now we are ready to calculate the vibrational contribution to the molar entropy, see eqn.
(7.57) in the script, as
3
svib X
=
R
i=1
θv,i /T
−θv,i /T
− ln(1 − e
) .
eθv,i /T − 1
(1.14)
Because NO2 has three vibrational modes with frequencies summarized in Table 1.1, the
vibrational entropy is
svib
= 0.5692.
(1.15)
R
Finally, we get the overall molar entropy of NO2 as
s
sel srot svib tr
s=R·
+
+
+
= 260.8178 J/K · mol.
(1.16)
R
R
R
R
b) Here, it is sufficient just to recalculate all contributions from the previous part at temperature
T = 298.15 K. We get
s298
(1.17)
calc = 240.0818 J/K · mol.
The experimental value can be found, for example, in the CCCBDB database (cccbdb.nist.gov)
=298
sTexp
= 240.166 J/K · mol.
(1.18)
From this, we see that the calculated value s298
calc is about ∆ = 0.0842 J/K·mol smaller than the
experimental value. This surprisingly good agreement between calculated and experimental
value can be explained by the fact that the calculations are performed at low temperatures.
In the high temperature limit there will be strong discrepancies due to the breakdown of most
of the approximations made during the above calculation. For example, we assumed that
the vibrational motion has a harmonic character which is, of course, not correct and at high
temperatures the vibrational anharmonicities will significantly contribute to the vibrational
entropy of the molecule.
Problem 2: Equilibrium constant
a) From classical thermodynamics we know that for each chemical reaction we can write
X
νi µi = 0,
(2.1)
i
page 3 of 5
Statistical Thermodynamics
Solution Exercise 9
HS 2016
where νi is a stoichiometric coefficient, which is according to the definition positive for
products and negative for reactants, and µi is a chemical potential. The connection between
classical and statistical thermodynamics can be made by expressing the chemical potential
µi using the partition function zi . The relationship between the chemical potential and the
free energy is given as
∂fi
,
(2.2)
µi =
∂ni T,V,nj6=i
where we can insert the expression linking the partition function to the free energy associated
with the contribution of the species i
fi = −kB T ln zi .
(2.3)
Using these relations, we obtain
fi = −kB T ln
1 Ni
Z = −kB T (Ni ln Zi − Ni ln Ni + Ni ),
Ni ! i
(2.4)
where in the first step we have used
zi =
1 Ni
Z ,
Ni ! i
(2.5)
which implies that particles are indistinguishable (see in the script section 5.2.1), and in the
second step we applied Stirling’s approximation. This expression can be further modified to
get
Zi
fi = −ni RT − ni RT ln
,
(2.6)
ni NAV
where ni is a molar mass and NAV is the Avogadro number. Therefore the chemical potential
for the species i is
Zi
µi = −RT ln .
(2.7)
Ni
To use the last equation, all partition functions Zi must be referenced to a common origin
of the energy scale. These new partition functions Zi0 are then given as
X
X
0
Zi0 =
gij e−ij /kB T =
gij e−ij /kB T eD0i /kB T = Zi eD0i /kB T ,
(2.8)
j
j
where we defined ij as the particle energy for the jth level as determined relative to the
lowest accessible state of the ith pure component and 0ij as the particle energy for the jth
level of the ith pure component but evaluated relative to a common reference level chosen
at the ground electronic state of any dissociated atom in the gaseous assembly. The explicit
form of 0ij is then
0ij = ij for atoms
(2.9)
0ij = ij − D0i
for molecules,
(2.10)
where D0i is the dissociation energy defined with respect to the vibrational ground state.
The vibrational contribution to the Z in equation (2.8) is shifted by the zero-point energy
as discussed in the script. Using these corrected partition functions Zi0 and the well known
relation from the classical thermodynamics
X
−RT ln K = ∆G◦r =
νi µ◦i ,
(2.11)
i
page 4 of 5
Statistical Thermodynamics
where
◦
Solution Exercise 9
HS 2016
indicates a selected standard state, we get (see eqn. (7.88) in the script)
X
Zi exp(D0i /kB T )
∆G◦r = −RT
νi ln
.
N
i
i
The final form of the equilibrium constant K † is therefore
" #
P
Y Zi νi
†
i νi D0i
,
K =
· exp
Ni std
kB T
i
(2.12)
(2.13)
where std stands for a standard state. This equilibrium constant can be easily related to the
constant Kp , see the script p. 79.
b) From the stoichiometry of the reaction, it follows that the equilibrium constant (2.13) can
be written as
ZNO2 /NNO2
D0,NO2 − D0,O2 − 1/2D0,N2
†
K =
.
(2.14)
· exp
(ZO2 /NO2 )(ZN2 /NN2 )1/2
kB T
To simplify this, we identify the classical enthalpy of formation as
X
◦
∆f Hi,T
νj D0,j .
=0 = −NAV
(2.15)
j
Therefore we get
∆f HN◦ O2 ,T =0
ZNO2 /NNO2
.
K =
· exp −
(ZO2 /NO2 )(ZN2 /NN2 )1/2
RT
†
Partition functions in this expression can be assumed as
1
1
kB T
Zi
= Zi,tr · Zi,int ·
= 3 · Zi,int · ◦
Ni
Ni
Λi
p
(2.16)
(2.17)
where Λi is the de Broglie wavelength. In the second step, we used the ideal gas law.
Finally, we can express the equilibrium constant as
◦ 1/2
Λ3O2
∆f HT◦ =0
p
ZNO2 ,int
3/2
†
K = 3
exp −
ΛN 2
.
(2.18)
ΛNO2 kB T
(ZO2 ,int )(ZN2 ,int )1/2
RT
We can also identify by comparison with the exercise sheet that
Λ3O2
κ1 = 3
ΛNO2
(2.19)
and
◦
∆f Hi,T
=0
.
R
c) κ1 can be easily evaluated. We have just shown that
3/2 3/2
mNO2
46.008 a.u.
=
κ1 =
= 1.724.
mO2
32.000 a.u.
κ2 =
(2.20)
(2.21)
To evaluate κ2 we need to find the standard enthalpy of formation for NO2 at absolute
zero. This value can be obtained from JANAF database (http://kinetics.nist.gov/janaf/) as
∆f HT◦ =0 = 35.927 kJ·mol−1 . Afterwards we get
κ2 =
∆f HT◦ =0
35927 J · mol−1
=
= 4321 K.
R
8.3145 J · mol−1 · K−1
(2.22)
page 5 of 5