Download Spectroscopy and Statistical Thermodynamics: Revisiting the HCl

Chemistry 3303
HCl Stat. Mech.
Winter 2012
Spectroscopy and Statistical Thermodynamics:
Revisiting the HCl/DCl IR Lab
Introduction and Theory
This experiment has been developed by Bret Findley and Steven Mylon of the
Department of Chemistry and Physics at Saint Michael’s College, Colchester, Vermont, USA.
Statistical thermodynamics bridges the micro and macro worlds by using the concepts of
quantum mechanics and spectroscopy to understand and predict thermodynamic properties.
The data from the rotational-vibrational spectrum of HCl that you acquired in a Chemistry 2302
laboratory exercise provides you with the important molecular level information for HCl that will
enable you to predict the thermodynamic properties of the bulk HCl gas.
One of the fundamental concepts in statistical thermodynamics is the molecular partition
function, 𝑞𝑞, which represents the number of states, or microstates, thermally accessible to the
molecule. Calculating 𝑞𝑞 requires information about the states available to the molecule. From
your study of quantum mechanics, you know that ideal gas molecules have several possible
modes of energy: electronic, rotational, vibrational, translational and nuclear spin. The total
number of thermally accessible states, 𝑞𝑞 , is equivalent to the product of the thermally
accessible states in each mode and is given by equation 1. Each mode has a different number
of thermally accessible states associated with it, however, note that most substances at room
temperature have only one accessible electronic and nuclear spin state (the ground state);
consequently, the molecular partition functions for these states equal 1. This leads to a
simplified expression for 𝑞𝑞 .
𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 (at room temperature)
Equation 1
This exercise is divided into three parts. The theory for each part is present below.
Before attempting any of the work in the results section, please read the corresponding theory
section carefully. In section A you will begin by considering the vibrational and rotational energy
levels together. Then you will consider the translational energy levels in section B. In section
C, you will use the results from A and B to calculate some desired bulk thermodynamic
properties, which is the objective the exercise. It is assumed that you have a good working
knowledge of Microsoft Excel, which should be valid since it has been used extensively in
Chemistry 2301 and 2302. Don’t hesitate to consult your lab instructor if you have any
questions about using Excel.
Chemistry 3303
HCl Stat. Mech.
Winter 2012
PART A: Calculating the rotational-vibrational partition function and the
rotational-vibrational contribution to the internal energy
This section requires information about the rotational and vibrational energy levels
available to H35Cl. This information can be determined from the spectroscopic constants (𝜈𝜈 ,
𝜈𝜈 𝑥𝑥 , 𝐵𝐵 , 𝛼𝛼 and 𝐷𝐷 ) that were calculated in “The Infrared Vibrational-Rotational Spectrum of HCl
and DCl” laboratory experiment performed in 2302.
From that experiment, recall that, using the rigid-rotor-linear-harmonic-oscillator model,
the vibrational-rotational energy levels for H35Cl can be calculated using:
𝐸𝐸 𝜈𝜈, 𝐽𝐽 = 𝜈𝜈 𝜈𝜈 +
1
+ 𝐵𝐵 𝐽𝐽 𝐽𝐽 + 1
2
Equation 2
Where 𝜈𝜈 is the fundamental vibrational wavenumber, 𝜈𝜈 is the vibrational quantum number, 𝐵𝐵 is
the equilibrium rotational constant and 𝐽𝐽 is the rotational quantum number. In reality, H35Cl is
neither a harmonic oscillator nor a rigid rotor. Spectroscopists add terms to equation 2 to
correct for anharmonicity, 𝜈𝜈 𝑥𝑥 𝜈𝜈 +
, rotational-vibrational coupling, 𝛼𝛼 𝜈𝜈 +
centrifugal distortion, 𝐷𝐷 𝐽𝐽 𝐽𝐽 + 1 . Thus, equation 2 becomes:
𝐸𝐸 𝜈𝜈, 𝐽𝐽 = 𝜈𝜈 𝜈𝜈 +
1
1
− 𝜈𝜈 𝑥𝑥 𝜈𝜈 +
2
2
+ 𝐵𝐵 𝐽𝐽 𝐽𝐽 + 1 − 𝛼𝛼 𝜈𝜈 +
Equation 3
𝐽𝐽 𝐽𝐽 + 1 , and
1
𝐽𝐽 𝐽𝐽 + 1 − 𝐷𝐷 𝐽𝐽 𝐽𝐽 + 1
2
Use of spectroscopic constants determined previously and this equation will allow for the
calculation of rotational-vibrational energy terms (levels).
Now the rotational-vibrational partition function, 𝑞𝑞 , can be calculated using:
𝑞𝑞 = 𝑞𝑞 𝑞𝑞 =
Equation 4
𝑔𝑔 𝑒𝑒 This quantity represents the number of rotational-vibrational states that are thermally accessible
at the temperature considered. Here 𝑔𝑔 represents the degeneracy of each energy level, 𝛽𝛽 is
the inverse of the product of the temperature and Boltzman’s constant, 1 𝑇𝑇𝑘𝑘 , and 𝜀𝜀 is the
energy (in Joules) of each level relative to the ground vibrational state:
𝜀𝜀 = ℎ𝑐𝑐 𝐸𝐸 𝜈𝜈, 𝐽𝐽 − 𝐸𝐸 0,0
Equation 5
Thus, 𝜀𝜀 is zero for the ground state. Provided that the ground state is non-degenerate (it is in
this case), 𝑔𝑔 𝑒𝑒 is equal to one for the ground state. This makes sense as the ground state is
Chemistry 3303
HCl Stat. Mech.
Winter 2012
always accessible, even at 0 Kelvin. At higher temperatures, more microstates are accessible
and, therefore, 𝑞𝑞 > 1.
Once you have calculated 𝑞𝑞 at the desired temperature, the probability that each
level is occupied can be calculated using equation 6. Finally the rotational vibrational
− 𝑈𝑈,
(0), can be determined using
contribution to the molar internal energy, 𝑈𝑈,
equation 7 (where 𝑁𝑁 is Avogadro’s Number):
𝑝𝑝, =
𝑔𝑔 𝑒𝑒 𝑞𝑞
Equation 6
𝑈𝑈,
− 𝑈𝑈,
0 =
Equation 7
𝑁𝑁 𝑝𝑝 𝜀𝜀
PART B: Calculating the translational partition function and the
translational contribution to the internal energy
Because of the assumption that HCl behaves as an ideal gas, one can think of each HCl
molecule as moving independently (not affected by other molecules). Another assumption, that
an HCl molecule is a free particle-in-a (one dimensional)-box, allows for its translational energy,
𝐸𝐸 , to be calculated using:
𝐸𝐸 =
𝑛𝑛 ℎ
8𝑚𝑚𝐿𝐿
Equation 8
Where 𝐿𝐿 is the length of the box, and 𝑛𝑛 = 1, 2, … Recall the wavefunction for a particle in a one
dimensional box,
sin
, must have some non-zero values within the box if the particle is not
there; therefore, the ground state energy level is 𝑛𝑛 = 1, not 𝑛𝑛 = 0. The energy relative to the
ground translational state is then: 𝜀𝜀 =
. In principle, the translational partition function can
be calculated in a similar fashion to the method used for 𝑞𝑞 using: 𝑞𝑞 = . At
𝑒𝑒
room temperature the spacing of the translational energy levels for an HCl molecule in one
can be closely approximated by a
dimension is so small that successive terms from 𝑒𝑒
continuous function. Thus, the summation can be replaced by the integral below, which is
evaluated from 𝑛𝑛 = 0 to 𝑛𝑛 = ∞ due to the convenience of its solution:
Chemistry 3303
HCl Stat. Mech.
𝑞𝑞 =
𝑒𝑒
Winter 2012
𝑑𝑑𝑑𝑑 =
Equation 9
2𝜋𝜋𝜋𝜋
𝐿𝐿
ℎ 𝛽𝛽
This approximation does not cause any appreciable error, even though there is no 𝑛𝑛 = 0 for the
particle-in-a-box system. In three dimensions, equation 9 becomes:
𝑞𝑞 =
2𝜋𝜋𝜋𝜋
ℎ 𝛽𝛽
Equation 10
𝑉𝑉 =
𝑉𝑉
Λ
Where 𝑉𝑉 is the volume of the box, Λ is the thermal wavelength of HCl, and 𝑚𝑚 is the molecular
mass of 1H35Cl (in kg). The thermal wavelength can be calculated using:
Λ=
ℎ 𝛽𝛽
2𝜋𝜋𝜋𝜋
Equation 11
The separation distance between molecules in one mole of 1H35Cl at STP is approximately 𝑉𝑉 or 2 × 10 m, which is much greater than Λ, ~2×10 m, under these conditions; therefore,
the use of this equation is justified. Note that the thermal wavelength has units of length. This
is equivalent to the thermal de Broglie wavelength 𝜆𝜆 = ℎ 𝑝𝑝 for the moving particle
using the average of the absolute value of its momentum.
Differentiating 𝑞𝑞 with respect to 𝛽𝛽 at constant 𝑉𝑉 allows one to calculate the translational
energy contribution to the molar internal energy of the system by using:
𝑈𝑈, = −
𝑁𝑁
𝑞𝑞
Equation 12
𝜕𝜕𝑞𝑞
𝜕𝜕𝜕𝜕
This form of equation 11 is equivalent to the form presented for 𝑈𝑈,
− 𝑈𝑈,
0 in
equation 7, but proves more conventional for translational energy.
PART C: Calculating Bulk Thermodynamic Properties
The calculation of the molar internal energy 𝑈𝑈,
− 𝑈𝑈,
0
of HCl is the easiest
thermodynamic property to calculate as it is merely the sum of the two contributions calculated
in parts A and B:
Chemistry 3303
HCl Stat. Mech.
Winter 2012
𝑈𝑈
− 𝑈𝑈
0 = 𝑈𝑈,
+ 𝑈𝑈,
− 𝑈𝑈,
0
Equation 13
Calculating the absolute molar entropy of HCl, 𝑆𝑆
, requires the molar internal energy and 𝑄𝑄, the
canonical partition function as in: 𝑆𝑆 = 𝑘𝑘 ln 𝑄𝑄 +
. The system of HCl molecules is a
macroscopic system containing one mole of molecules. In the case of entropy, we are
concerned with the number of ways the system can be arranged at a particular temperature.
Therefore, instead of considering the molecules individually, we consider a collection of
replications of the system, which all have the same volume and number of molecules and are in
thermal equilibrium with each other. Such a collection of macroscopic systems is known as a
canonical ensemble and the number of thermally accessible states for the system is known as
the canonical partition function, 𝑄𝑄. Because each molecule has a probability distribution of
microstates, one might expect the number of thermally accessible states for the system to be
equivalent to the product of molecular partitions for each molecule, 𝑄𝑄 = 𝑞𝑞 ; however,
because HCl molecules are all the same, they cannot be distinguished from one another and
this equation over counts the number of thermally accessible states for the system. To account
for the indistinguishable nature of the molecules in our system, the canonical partition function
takes the form: 𝑄𝑄 =
!
=
!
𝑆𝑆 = 𝑘𝑘 ln
, and the molar entropy equation becomes:
− 𝑈𝑈 0
𝑈𝑈
𝑞𝑞 +
𝑁𝑁 !
𝑇𝑇
Equation 14
Stirling`s approximation for large factorials ln 𝑛𝑛! = 𝑛𝑛 ln 𝑛𝑛 − 𝑛𝑛 is typically used here to reduce
equation 14 to a more convenient form:
𝑆𝑆 = 𝑅𝑅 ln 𝑞𝑞 − ln 𝑁𝑁 + 1 +
Equation 15
− 𝑈𝑈 0
𝑈𝑈
𝑇𝑇
Once the molar internal energy 𝑈𝑈
− 𝑈𝑈
0 and 𝑆𝑆 are calculated, the molar enthalpy,
0 , are easily found using definitions from
𝐻𝐻 − 𝐻𝐻 0 , and Gibbs free energy, 𝐺𝐺 − 𝐺𝐺
introductory thermodynamics:
− 𝐻𝐻
0 = 𝑈𝑈
− 𝑈𝑈
0 + 𝑃𝑃𝑉𝑉
𝐻𝐻
Equation 16
𝐺𝐺
− 𝐺𝐺
0 = 𝐻𝐻
− 𝐻𝐻
0 − 𝑇𝑇𝑆𝑆
Equation 17
However, these quantities are not typically used in thermodynamics. Instead, the much more
and
common values of the standard molar enthalpy and Gibbs free energy of formation, ∆𝐻𝐻 ∆𝐺𝐺 , are used.
Chemistry 3303
HCl Stat. Mech.
Winter 2012
The calculations involving ∆𝐻𝐻 require 𝐻𝐻
− 𝐻𝐻
0 for 1H2 and 35Cl2 gases. For 1H2,
𝐻𝐻
− 𝐻𝐻
0 can be determined quite simply by using an analogous Excel template created for
HCl and inputting hydrogen`s spectroscopic constants from the literature (ref.1). This is fairly
straightforward, however, due to hydrogen`s symmetry, nuclear spin statistics need to be
considered.
Recall that the Pauli Exclusion Principle states that the total wave function for a system of
fermions (particles with half integral spin) must be antisemmetric. Because the 1H nucleus is a
fermion, 1H2 can occupy rotational states with even 𝐽𝐽 (𝐽𝐽 = 0, 2, 4, … ) only when its nuclear spins
are paired (para-hydrogen). It can occupy rotational states with odd 𝐽𝐽 (𝐽𝐽 = 1, 3, 5, … ) only when
its nuclear spins are parallel (ortho-hydrogen). If this is not immediately clear, ask yourself
whether the rotational wave functions for even 𝐽𝐽 (𝐽𝐽 = 0, 2, 4, … ) are symmetric or antisemmetric
functions. These rotational wave functions turn out to be symmetric and therefore must be
multiplied by the antisymmetric nuclear spin functions for the total wave function to be
antisymmetric as required by the Pauli Exclusion Principle. The rotational wave functions for
odd 𝐽𝐽 (𝐽𝐽 = 1, 3, 5, … ) are antisymmetric requiring multiplication by any two of the three symmetric
spin functions for the total wave function to be antisymmetric. The weighting of ortho- and parahydrogen (3:1) does make a difference in the calculations due to the large spacing between the
rotational-vibrational energy levels of 1H2.
Although similar symmetry exists in 35Cl2 the spacings between rotational-vibrational levels are
so small at room temperature that one may ignore the difference in ortho and para contributions.
Additionally, because there a so many thermally accessible energy levels at room temperature,
a much larger Excel worksheet would be required. While, in principle, this is not a problem,
35
Cl2 is not the focus of this exercise. It is therefore more time-efficient to use 5 2 𝑅𝑅𝑅𝑅 for
− 𝑈𝑈
0 in the calculation of 𝐻𝐻
− 𝐻𝐻
0 and the literature value of 𝑆𝑆
for 35Cl2 at 298.15 K.
𝑈𝑈
You will finally calculate ∆𝐻𝐻 by using the dissociation energy from the ground vibrational
state, 𝐷𝐷 (ref.1) and 𝐻𝐻
− 𝐻𝐻
0 for each substance. In introductory chemistry courses you
were taught that one can calculate ∆𝐻𝐻 using bond or dissociation energies of the molecules
= 𝑁𝑁 1 2 𝐷𝐷, + 𝐷𝐷, − 𝐷𝐷, does give a good result;
involved and indeed ∆𝐻𝐻 − 𝐻𝐻
0 values accounts for the thermal energy in each substance and
however, including 𝐻𝐻
improves the result. Therefore the equation becomes:
= 𝐻𝐻
− 𝐻𝐻
0
∆𝐻𝐻 −1 2
𝐻𝐻
− 𝐻𝐻
0
+ 𝑁𝑁 1 2 𝐷𝐷, + 𝐷𝐷, − 𝐷𝐷,
+ 𝐻𝐻
− 𝐻𝐻
0
Equation 18
Values of 𝐷𝐷 for all three substances could, in principle, be determined using the spectroscopic
constants of each species, but it is more time-efficient to use the literature values in this case.
Once the enthalpy of formation is calculated for 1H35Cl, the molar Gibbs free energy of formation
can be calculated using:
Chemistry 3303
HCl Stat. Mech.
Winter 2012
∆𝐺𝐺 = ∆𝐻𝐻 − 𝑇𝑇∆𝑆𝑆
Equation 19
where ∆𝑆𝑆
is the change in entropy for the formation reaction.
Experimental
There is no conventional laboratory work for this experiment. You will use your results
from a previous HCl/DCl IR experiment (performed in Chemistry 2302) to calculate some bulk
thermodynamic properties for HCl (or DCl). In theory, this exercise could be completed for all of
the isotopic species studied in the original experiment (H35Cl, D35Cl, H37Cl and D37Cl), however,
for the sake of time you will just consider one.
During this 3-hour lab period use Excel to perform the calculations outlined in the results
section. You will need your laboratory report entitled “The Infrared Vibrational-Rotational
Spectrum of HCl and DCl” from Chemistry 2302. Please bring this report with you on the day
that you are scheduled to complete this lab.
Even though you are not performing a conventional laboratory experiment, please don’t
hesitate to ask your instructor for help with any of the calculations.
Results
The procedure below should only be performed for the 1H35Cl isotopic species. Even if
your calculations in the previous lab were for DCl (deuterium chloride) you should have obtained
spectroscopic constants for 1H35Cl from your partner and included them as part of your report.
Use these constants, where appropriate, in the following exercise. If you do not have these
constants they may be obtained from Herzberg or Sime (refs.1 or 3). DO NOT use
spectroscopic constants for DCl in this experiment!!
To complete the exercise, please note that in addition to the spectroscopic constants
calculated in your 2302 lab report, you will need to use data for H 2 and 35Cl2 that can be
obtained from the references listed (ref.1).
PART A: Calculating the rotational-vibrational partition function and the
rotational-vibrational contribution to the internal energy
1. Begin setting up your Excel worksheet for hydrogen chloride by creating cells for values
that you will use throughout this exercise as shown in Table 1. You will need the various
constants (the speed of light, Boltzmann’s constant, etc.) listed and the spectroscopic
constants (𝜈𝜈 , 𝜈𝜈 𝑥𝑥 , 𝐵𝐵 , 𝛼𝛼 and 𝐷𝐷 ) from the lab you completed previously. If you did not
calculate one or more of these constants in your previous lab you may use the literature
values (refs. 1 or 3).
Chemistry 3303
HCl Stat. Mech.
HYDROGEN CHLORIDE
Constants
𝐶𝐶 (m/s)
𝑒𝑒 (C)
𝑘𝑘 (J/K)
𝑅𝑅 (J/K
mol)
ℎ (J/s)
𝑁𝑁 (1/mol)
299792460
1.60218E-19
1.38065E-23
8.31447
6.62608E-34
6.02214E23
Spec. Constants (cm1
)
𝜈𝜈
𝜈𝜈 𝑥𝑥
𝐵𝐵
𝛼𝛼
𝐷𝐷
Winter 2012
Other Relevant Data
𝑇𝑇 (K)
𝛽𝛽 (1/J)
𝑚𝑚 (g/mol)
𝑚𝑚 (g/mol)
Table 1 – Initial Excel spreadsheet setup for hydrogen chloride
2. Set up an additional table for calculations relevant to the rotational-vibrational internal
energy. See the suggested column headings in Table 2. Then use equation 3 and your
spectroscopic constants to calculate the vibrational-rotational energy terms of H35Cl,
𝐸𝐸 𝜈𝜈, 𝐽𝐽 . Do this for the first 16 rotational levels in the ground vibrational state and the first
10 rotational levels in the first excited vibrational state. While the choice to stop at these
rotational levels is somewhat arbitrary, it probably makes little difference to the results if
these numbers were, for instance, 18 and 12 or even higher. Why is this true at the
temperature considered?
Calculation of Erot-vib (J/mol)
𝜈𝜈
𝐽𝐽
𝐸𝐸 𝜈𝜈, 𝐽𝐽 𝜀𝜀 2𝐽𝐽 + 1
0
0
0
1
etc. etc.
𝑔𝑔 𝑒𝑒 Prob. 𝐸𝐸 (J)
𝐸𝐸 (J/mol)
Table 2 – Suggested headings for Excel calculations associated with the
rotational-vibrational contribution to the internal energy
3. Use equation 5 to calculate the energy 𝜀𝜀 , in Joules for each rotational-vibrational energy
level relative to the ground rotational vibrational state.
4. Include the degeneracies of each energy level as a separate column in Excel. Recall
that vibrational states are non-degenerate, but rotational levels for a diatomic molecule
are 2𝐽𝐽 + 1 -fold degenerate.
5. Once you have calculated both the degeneracy and the energy of each term relative to
the ground state, you are now prepared to calculate the rotational-vibrational partition
function, 𝑞𝑞 , using equation 4. The partition function is calculated by summing over
the 𝑔𝑔 𝑒𝑒 terms in a separate column (see Table 2) and then sum these values in a cell
somewhere else on the same worksheet.
6. Use equation 6 and your answers from step 5 to calculate the probability, 𝑝𝑝, , that
each vibrational-rotational energy level 𝑖𝑖 is occupied at 298.15 K. Use a separate
column labeled “Prob.” as shown in Table 2. Consider your calculated results for the
probability that each state is occupied. Why is it not necessary for you to calculate the
probabilities of more states?
Chemistry 3303
HCl Stat. Mech.
Winter 2012
7. Multiply the individual energies, 𝜀𝜀 , by the probability that this energy level is occupied.
This yields the contribution of each energy level for an average single molecule to the
total molecular energy. As suggested in Table 2, the column labeled “𝐸𝐸 (J)” is created
for the case of a single HCl molecule. In the next column, multiply your values by
Avogadro’s number to obtain the contribution of each energy level to the molar internal
energy in J/mol. Finally, sum the entire ensemble of energy levels to determine the
entire vibrational and rotational contribution to the molar internal energy. Include this
last result in a separate table, perhaps with the rotational-vibrational partition function as
in Table 3 below.
Summary for 𝑼𝑼𝒐𝒐𝒎𝒎,𝒓𝒓𝒓𝒓𝒓𝒓𝒗𝒗𝒗𝒗𝒗𝒗 − 𝑼𝑼𝒐𝒐𝒎𝒎,𝒓𝒓𝒓𝒓𝒓𝒓𝒗𝒗𝒗𝒗𝒗𝒗 (𝟎𝟎)
𝑞𝑞
𝑈𝑈,
− 𝑈𝑈,
(0) (J/mol)
𝑅𝑅𝑅𝑅 (J/mol)
% difference
Table 3 – Suggested table for the results of the rotational – vibrational
contribution to the internal energy
Part B: Calculating the translational partition function and the translational
contribution to the internal energy
1. Before performing the calculations for this section, set up a table similar to Table 4.
Calculation of 𝑼𝑼𝒐𝒐𝒎𝒎,𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
Λ (m)
𝑉𝑉 (m3)
𝑞𝑞
𝑞𝑞
𝜕𝜕𝑞𝑞
𝜕𝜕𝜕𝜕 (J)
𝑈𝑈,
(J/mol)
3 𝑅𝑅𝑅𝑅 (J/mol)
2
% difference
Table 4 – Suggested Excel table for the results involving the
translational contribution to the molar internal energy
2. First calculate the thermal wavelength, Λ, at room temperature (298.15 K) as shown in
equation 11 and the molar volume at 1.00 bar (1.00 bar = 10 5 Pa) assuming ideal gas
behavior. Then use equation 10 to calculate 𝑞𝑞 .
3. You can now calculate the total molecular partition function using
𝑞𝑞 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 𝑞𝑞 . Recall that 𝑞𝑞 𝑞𝑞 = 1 at the temperatures we are
considering.
Chemistry 3303
HCl Stat. Mech.
Winter 2012
4. Differentiate 𝑞𝑞 with respect to 𝛽𝛽 at constant 𝑉𝑉. Then use your result to calculate the
translational energy contribution to the molar internal energy by using equation 12.
Compare your result with 3 2 𝑅𝑅𝑅𝑅, the result you would expect from what you learned in
the thermodynamics unit in Chemistry 2301.
Part C: Calculating Bulk Thermodynamic Properties
1. Set up a table in Excel similar to the one suggested below. Note that the bottom three
quantities, Δ𝑆𝑆 , Δ𝐻𝐻 and Δ𝐺𝐺 , all require information about H2 and Cl2 before they can
be calculated.
Bulk Thermodynamic Properties
Calculated
𝑈𝑈 − 𝑈𝑈 (0) (J/mol)
𝑆𝑆
(J/mol)
𝐻𝐻
− 𝐻𝐻
(0) (J/mol)
𝐺𝐺 − 𝐺𝐺 (0) (J/mol)
Δ 𝑆𝑆 (J/mol K)
Δ𝐻𝐻 (kJ/mol)
Δ𝐺𝐺 (kJ/mol)
Literature
% difference
N/A
N/A
N/A
N/A
Table 5 – Suggested Excel table for summary of results related to calculation of bulk thermodynamic
properties
2. Calculate 𝑈𝑈
− 𝑈𝑈
(0) using equation 13. Compare your result with 5 2 𝑅𝑅𝑅𝑅, the values
expected from classical thermodynamics and calculate the percent difference.
3. Calculate the molar entropy, 𝑆𝑆
, using equation 15 and compare it to the value reported
in Table A.8 located in appendix A of Mortimer (your textbook). Calculate the percent
difference between the calculated and textbook values.
− 𝐻𝐻
(0) and 𝐺𝐺
− 𝐺𝐺
(0) as indicated in equations 16 and 17.
4. Calculate 𝐻𝐻
5. Calculate 𝐻𝐻
− 𝐻𝐻
(0) and 𝑆𝑆
for H2 using the following steps:
a. Copy your entire worksheet for H35Cl and paste it into a new worksheet in the
same workbook (i.e. keep it part of the same Excel file)
b. In your first table, change the heading to “Hydrogen Gas” and replace the
spectroscopic constants of HCl with those of H2 as found in Herzberg (. Also
change the mass labels and some of the quantities under “Other Relevant Data”
as needed.
c. In a table similar to Table 2 of your Excel worksheet, create an additional column
in order to properly account for ortho and para hydrogen as shown below.
Chemistry 3303
HCl Stat. Mech.
Calculation of Erot-vib (J/mol)
2𝐽𝐽
𝜈𝜈 𝐽𝐽
𝐸𝐸 𝜈𝜈, 𝐽𝐽
𝜀𝜀 +1
0
0
0
1
etc. etc.
Winter 2012
Para H2 Ortho H2 Prob. 𝐸𝐸 (J)
0.25×𝑔𝑔 𝑒𝑒 0.75×𝑔𝑔 𝑒𝑒 𝐸𝐸 (J/mol)
Table 6 – Table for calculations involving the rotational-vibrational contribution to the internal energy for H2
gas
d. As mentioned previously, para-hydrogen can only occupy states with even 𝐽𝐽’s
and ortho-hydrogen can only occupy states with odd 𝐽𝐽’s. The weighting of para
to ortho states is 1:3; therefore, an additional weighting factor should be included
(see Table 6). One can incorporate this weighting into the calculation of 𝑞𝑞 :
𝑞𝑞 = 𝑞𝑞 𝑞𝑞 =
0.25× 𝑔𝑔 𝑒𝑒 +
Equation 20
0.75×𝑔𝑔 𝑒𝑒 e. Also include the weighting factor, 𝑤𝑤 = 0.25 𝑜𝑜𝑜𝑜 0.75, to calculate the probablility,
𝑝𝑝, , that each vibrational-rotational energy level 𝑖𝑖 is occupied at 298.15 K
as:
𝑝𝑝, =
𝑤𝑤 𝑔𝑔 𝑒𝑒 𝑞𝑞
Equation 21
f.
The rest of the calculations for 𝐻𝐻
− 𝐻𝐻
(0) and 𝑆𝑆
of H2 follow the same pattern
as for HCl and should require very little work.
− 𝐻𝐻
(0) and 𝑆𝑆
for 35Cl2 in order to calculate
6. Additionally, you will need to calculate 𝐻𝐻
and ∆𝐺𝐺,
. However, as described above, it is preferable to use some
∆𝐻𝐻,
approximations in order to save time:
− 𝑈𝑈
0
a. Use 5 2 𝑅𝑅𝑅𝑅 for 𝑈𝑈
equation 16.
and then calculate 𝐻𝐻
− 𝐻𝐻
0 for 35Cl2 using
b. Obtain the literature value for 𝑆𝑆
of 35Cl2 at 298.15 K.
7. Obtain the literature values for the dissociation energy of HCl, H2 and Cl2 from the
ground vibrational state, 𝐷𝐷 , by looking them up in the literature. Then use equation 18
.
to calculate ∆𝐻𝐻,
Chemistry 3303
HCl Stat. Mech.
Winter 2012
8. Calculate the change in entropy for the formation reaction and then use equation 19 to
.
calculate ∆𝐺𝐺,
Discussion
Using the data obtained in a previous lab exercise, you have determined vibrational and
rotational partition functions for HCl. You have used the particle-in-a-box model to arrive at the
translational partition function for HCl. Armed with these partition functions, you were then able
to calculate thermodynamic quantities such as the internal energy and absolute entropy. You
did the same for H2 using method described above except with literature values for the
spectroscopic constants. Finally, you were able to calculate the bulk thermodynamic properties
∆𝐻𝐻 and ∆𝐺𝐺, for H35Cl by employing these methods.
In your discussion, comment on your results and answer the following questions:
1) Compare the values of 𝑞𝑞 and 𝑞𝑞 . Which mode of motion has the greater
number of thermally accessible states? Which mode of motion (translational, rotational
or vibrational) has the fewest thermally accessible states? Briefly explain your answers.
2) Why can the electronic partition function for H35Cl be ignored in our treatment.
3) Why can the summation for the translational partition function be replaced by
integration?
4) How will changes in temperature affect your results? Comment on which partition
functions are most affected by incremental changes (±100 K) changes in temperature.
5) For the sake of time, we only considered H35Cl instead of combinations of other
isotopes. How would changes in atomic isotopes affect your results? Which partition
functions would be affected most?
6) By making the jump from the microscopic to the macroscopic world, we compared our
calculated results for the H35Cl molecules to bulk thermodynamic data found in typical
physical chemistry texts. To which of these isotopes do these literature values
correspond? How might changes in isotope affect the dissociation energies involved in
the calculation of ∆𝐻𝐻 ? You might want to consider the Morse potentials and their zero
point energy levels for the species involved.
7) As you know from the thermodynamics unit of physical chemistry, the molar heat
𝜕𝜕𝑈𝑈
capacities at constant volume, 𝐶𝐶, , and constant pressure, 𝐶𝐶, , are
𝜕𝜕𝜕𝜕 and
𝜕𝜕𝐻𝐻
𝜕𝜕𝜕𝜕 respectively. Use your excel worksheets to estimate these values of 𝐶𝐶,
and 𝐶𝐶, at 298.15 K for H35Cl. Compare your result of 𝐶𝐶, with that found in your
textbook.
Chemistry 3303
HCl Stat. Mech.
Winter 2012
References
1. Herzberg, Gerhard. Molecular Spectra and Molecular Structure, Volume 2. D. Van
Nostrand Company, New York, 1953.
(on reserve for Chem 2302 at QEII Library)
2. Mortimer, Robert G. Physical Chemistry, 3rd ed., Elsevier Academic Press, San Diego,
2008. (One copy is available in C-3041)
3. Sime, Rodney J. Physical Chemistry: Methods, Techniques, Experiments, Saunders
College Publishing, Philadelphia, 1990.