CHEM 304: Problem Set 4 of 5
Return the solutions by Friday, November 16, classtime to Roman Krems
No late submissions will be accepted
For questions Q1-Q5, give the answer in terms of the variables.
Note: Spectroscopic values are often expressed in units of wavenumbers, cm−1 . If ν̃ and B̃ are the
harmonic frequency and rotational constant expressed in wavenumbers, respectively then Θvib =
hν/kB = hcν̃/kB and Θrot = B/kB = B̃hc/kB , in which c = 2.998 × 108 m s−1 is the speed of light.
Q1. Using expressions derived from statistical mechanics, determine the value of ∆H, in kJ, for
the irreversible expansion and heating of 3 mol of oxygen gas from an initial volume of 1 L and
temperature 25 ◦ C to a final volume of V L and temperature T ◦ C. Treat the gas as non-interacting
with ν̃ = 1580 cm−1 and B̃ = 1.43 cm−1 .
Using the supplied values for ν̃ and B̃, along with the formulae given at the beginning of the
document gives Θvib = 2273.3 K and Θrot = 2.0575 K. Since T is much greater than Θrot we can
use the classical expressions for the rotational contributions to enthalpy. However, we need to use
the exact expressions for the harmonic contributions since T is smaller than Θvib . Also,
∆H = ∆U + ∆(P V ) = ∆U + nR∆T = ∆E + nR∆T .
Using the expressions for E from the notes for contributions from various degrees of freedom then
gives
cl
cl
cl
cl
∆H = N elec (T2 ) − N elec (T1 ) + E trans (T2 ) − E trans (T1 ) + E rot (T2 ) − E rot (T1 )
exact
exact
+E vib (T2 ) − E vib (T1 ) + nR(T2 − T1 )
3
Θvib
Θvib
Θvib
Θvib
= nR(T2 −T1 )+nR(T2 −T1 )+nR
+ Θ /T2
+ Θ /T1
−nR
+nR(T2 −T1 )
2
2
e vib − 1
2
e vib − 1
7
1
1
= nR (T2 − T1 ) + Θvib Θ /T2
−
,
2
e vib − 1 eΘvib /T1 − 1
where in the first line the electronic contributions to the internal energy cancel because they are
independent of temperature (since we have assumed in our case that molecules always remain in
their electronic ground states). Substituting the value of Θvib , T1 = 298.15 K, T2 = T + 273.15 K,
and n = 3 mol, and converting to kJ units then gives the desired result.
Q2. Using expressions derived from statistical mechanics, determine the value of ∆S, in J K−1 ,
for the irreversible expansion and heating of 3 mol of chlorine gas from an initial volume of 1 L and
temperature 25 ◦ C to a final volume of V L and temperature T ◦ C. Treat the gas as non-interacting
with ν̃ = 559.7 cm−1 and B̃ = 0.2439 cm−1 .
Using the supplied values for ν̃ and B̃, along with the formulae given at the beginning of the
document gives Θvib = 805.303 K and Θrot = 0.35093 K. Since T is much greater than Θrot we
can use the classical expressions for the rotational contributions to entropy. However, we need to
use the exact expressions for the harmonic contributions since T is smaller than Θvib . Using the
expressions for S from the notes for contributions from various degrees of freedom gives
cl
cl
(V1 , T1 )+
(V2 , T2 ) − Strans
∆S = nR ln(gelec ) − nR ln(gelec ) + Strans
cl
cl
exact
exact
Srot
(T2 ) − Srot
(T1 ) + Svib
(T2 ) − Svib
(T1 )
5
5
T2
T1
3/2
3/2
= nR
+ ln(aV2 T2 ) −nR
+ ln(aV1 T1 )
+nR 1 + ln
−nR 1 + ln
2
2
σΘrot
σΘrot
Θvib /T1
Θvib /T2
+nR Θ /T2
− ln(1 − e−Θvib /T2 ) − nR Θ /T1
− ln(1 − e−Θvib /T1 )
vib
vib
e
−1
e
−1
(
!
3/2
V2 T2
Θvib /T2
T2
−Θvib /T2
= nR ln
)
+ Θ /T2
− ln(1 − e
+ ln
3/2
T1
e vib − 1
V1 T1
Θvib /T1
−Θvib /T1
)
− Θ /T1
− ln(1 − e
e vib − 1
(
! )
5/2
V2 T2
Θvib /T2
Θ
/T
vib
1
= nR ln
+ Θ /T2
− ln(1 − e−Θvib /T2 ) − Θ /T1
− ln(1 − e−Θvib /T1 )
,
5/2
vib
vib
e
e
−
1
−
1
V1 T1
where in the first line a = (2πmk)3/2 /(N h3 ) has been used to simplify the notation for the translational contribution to the entropy, after having substituted P = N kB T /V . Substituting the value
of Θvib and using V1 = 1 L, V2 = V L, T1 = 298.15 K, T2 = T + 273.15 K, and n = 3 mol in the
formula above then gives the value of ∆S for the process.
Q3. In a gaseous sample containing 5 mol of HCl at T ◦ C and 1 bar, how many moles of molecules
are in the J = 3 rotational state? Treat the gas as non-interacting with B̃ = 10.593 cm−1 .
The number of moles of molecules in the J = 3 rotational state is nPJ=3 , in which PJ=3 is the
probability a particle is in the J = 3 state. This probability can be calculated from the partition
function for rotational motion as follows
PJ =
gJ e−βEJ
,
qrot
in which gJ = 2J + 1 is the degeneracy of the rotational state J, and EJ = BJ(J + 1) is the
rotational energy. Here βEJ = J(J + 1)Θrot /T and Θrot = 15.2413 K (calculated from the given
value of B̃). Since T is more than 20 times larger than Θrot , the classical approximation for qrot
should give reasonably accurate values. In this case, HCl is a heteronuclear, diatomic molecule so
σ = 1. Substituting all the values gives
PJ=3 =
7Θrot e−12Θrot /T1
.
T1
Substituting for Θrot , T1 = T + 273.15 K and then multiplying by n = 5 mol gives the desired result.
Q4. In a gaseous sample containing 4 mol of ClF at T K and 1 bar, how many moles of molecules
are in the first excited vibrational state? Treat the gas as non-interacting with Θvib = 1132 K.
The first excited vibrational state has quantum number n = 1, so the number of moles of molecules
in this state is 4 mol × Pn=1 , in which Pn=1 is the probability a particle is in the n = 1 state. This
probability can be calculated from the partition function for vibrational motion as follows
Pn =
gn e−βEn
,
qvib
in which gn = 1 is the degeneracy of the vibrational state n, and En = hν(n+1/2) is the vibrational
energy. Here βEn = (n+1/2)Θvib /T . We need to use the exact expression for the partition function.
Substituting all the values gives
Pn=1 =
(1 − e−Θvib /T ) −3Θvib /2T
e
= e−Θvib /T (1 − e−Θvib /T ) .
e−Θvib /2T
Substituting for Θvib and T and then multiplying by 4 mol gives the desired result.
Q5. Treating the gas as non-interacting with ν̃ = 325.3 cm−1 and B̃ = 0.0821 cm−1 , estimate the
value of CV,m /R for Br2 at T ◦ C.
Using the supplied values for ν̃ and B̃, along with the formulae given at the beginning of the
document gives Θvib = 468.045 K and Θrot = 0.11813 K. At the given temperature, we can use
the classical contributions for all degrees of freedom except vibrations. Adding together these
contributions gives
2
CV,m
CV,trans,m CV,rot,m CV,vib,m
3
Θvib
eΘvib /T1
=
+
+
= +1+
.
R
R
R
R
2
T1
(eΘvib /T1 − 1)2
Substituting the values for Θvib and T1 = T + 273.15 K gives the desired result.
Q6. Calculate the value of kB T when T = 25 ◦ C. Using online or printed resources, find approximate estimates for the energy barriers for i) single bond rotation, for example methyl rotations
in ethane, ii) hydrogen bond formation, and iii) carbon-carbon single bond cleavage. What is the
physical significance of the value of kB T ? Use your calculated value of kB T to describe the ease
with which each of the three listed chemical processes will occur at room temperature.
Barriers for single bond rotation are about 3 − 10 kJ mol−1 , for hydrogen bond formation about
20 − 40 kJ mol−1 , and for carbon-carbon single bond cleavage about 350 kJ mol−1 . At T = 25 ◦ C,
kB T = 1.38065 × 10−23 J K−1 × 298 K = 411.4 × 10−23 J. Multiplying this by Avogadro’s number to
give consistent units then gives a value of 2.48 kJ mol−1 . The probability a system is in a particular
microstate is determined by the factor exp[−βEi ] = exp[−Ei /kB T ]. This probability is large if Ei
is close to kB T but becomes exponentially smaller as Ei increases. If a particular chemical process
requires energy, the system must at some point find itself in a microstate of energy comparable
to or greater than the energy needed for the process if it is to occur. This probability is again
determined by the ratio Ei /kB T .
Q7. Consider the contributions of translational motion to the partition function for a system of
non-interacting, identical particles. The classical translational energy is
trans =
m 2
m
(vx + vy2 + vz2 ) = v 2 .
2
2
Using a classical mechanical approach and treating the velocity components vx , vy , and vz or the
speed v as continuous variables, the contribution of the translational states to the partition function
is
Z ∞
Z ∞
Z ∞
m
2
2
2
(v + vy + vz )
qtrans =
dvx
dvy
dvz exp −
2kB T x
−∞
−∞
−∞
Z ∞
m 2
2
v dv exp −
= 4π
v
2kB T
0
r Z
4π 2 ∞ √
trans
=
trans dtrans exp −
m m 0
kB T
3/2
2πkB T
,
=
m
in which the integration is expressed in terms of Cartesian velocity components in the first case, in
spherical polar variables of velocity (yielding only the speed v dependence) in the second case, and
in terms of the translational energy in the third case. Using these expressions one can form three
probability distributions as
3/2
m
− m (v 2 +v 2 +v 2 )
p(vx , vy , vz ) =
e 2kB T x y z ,
2πkB T
3/2
m
− m v2
v 2 e 2kB T ,
p(v) = 4π
2πkB T
r
2
trans − trans
p(trans ) = √
e kB T ,
kB T
πkB T
where p(vx , vy , vz ) is the probability for finding a particle with its velocity components equal to vx ,
vy , and vz , p(v) is the probability for finding a particle with speed v, and p(trans ) is the probability
for finding a particle with translational energy trans . These are three different forms of the famous
Maxwell-Boltzmann distribution for a system at equilibrium.
a) Submit a single graph with p(trans ) plotted as a function of trans for i) kB T = 1 and ii)
kB T = 5 with the x-axis varying from 0 to 15, and the y-axis from 0 to 0.5. With an “X”
mark the value of the average particle energy (trans = 3kB T /2) for each curve.
0.5
kT = 1
kT = 5
p(εtrans)
0.4
0.3
0.2
0.1
0
0
3
6
εtrans
9
12
15
b) Why does trans = 3kB T /2 not correspond to the peak of the distribution? Do most particles
in a system have an energy equal to trans = 3kB T /2?
As seen in the figure, the distributions are not symmetric but are broader towards higher
energy. This causes the average energy values to move higher and thus appear to the right
of the peak in the distributions, as seen by the location of the X’s. Most of the particles in
the system do not have an energy equal to the average. In fact, a significant fraction of the
particles have energies less than the average. For the Maxwell-Boltzmann distribution, the
median energy is at about 1.2kB T which is less than the average.
c) How does the shape of the probability distribution change as temperature is increased? How
does the width of the distribution compare with the average energy? What does this imply
about the fluctuations of particle energies? How is this different from the fluctuations of the
total energy of the system?
As temperature increases, the distribution broadens and shifts to higher energies. Because
the distribution is normalized (has area always equal to 1) the values of the probabilities fall
as the distribution broadens. The distribution is noticeably flatter at the higher temperature,
thus corresponding to a higher entropy state, as expected.