Download Chemical Equilibrium

Homework Solution
Initially, lump A is at 100ºC and lump B is at 50ºC. Calculate the equilibrium
temperature of the combined A+B system for the conditions (a) CVA = 2CVB and
(b) 2CVA = CVB.
CVA(Tf – TiA) = -CVB(Tf – TiB)
If CVA = 2CVB, this reduces to
Tf – TiA = (TiB –Tf)/2
2Tf – 200 = 50 –Tf, or 3Tf = 250 => Tf = 83°C.
If 2CVA = CVB, this reduces to
Tf – TiA = 2(TiB –Tf)
Tf – 100 = 100 –2Tf, or 3Tf = 200 => Tf = 67°C.
(The final temperature ends up closer to the initial temperature of the system
with the greater heat capacity.)
Vibrational Degrees of Freedom: Solids
Simple solids like Ar(s) and metals can be treated as systems with only
vibrational degrees of freedom. Statistical thermodynamics allows the
apparently diverse heat capacity behaviour of solids:-
…to be explained by their characteristic temperature.
High-temperature-limit
 0 at low T
LK Nash “Elements of Statistical Thermodynamics”
Systems with Many Degrees of Freedom
The ideal atomic gas has only one type of degree of freedom - translational. In
order to understand realistic chemical systems, we need to be able to deal with
multiple degrees of freedom.
Recall that the energy of a molecule in a given molecular state, is simply the sum
of the energies of each degree of freedom, translational, rotational, vibrational
and electronic, viz.
emolecule = etranslation + erotation + evibration + eelectronic
Using our definition of the total energy of a system, E 
write it as a sum of components
n e
i
i
we can similarly
i
Emolecule = Etranslation + Erotation + Evibration + Eelectronic
…if the degrees of freedom are independent of each other. That is, if a change
in vibrational state does not cause any change in any of the other allowed
energies. E.g. No Coriolis coupling between rotation and vibrations.
Heat Capacity of Diatomic Gases
Once we can separate independent energies, we can also write the heat
capacity as a sum of independent components
CVmolecule = dE/dT|V= CVtranslation + CVrotation + CVvibration + CVelectronic
Each degree of freedom has its own set of allowed energies, partition
function, characteristic temperature, and heat capacity. For a diatomic gas
there are 3 translations, one rotation, one vibration, and one set of electronic
states. (More complex systems just have more vibrational and rotational
modes.) So far we have
Translation:
Qtrans ~ 0K
CVtrans  23 nR
Vibration:
Qvib ~ hn/kB
CVvib  nR
This is the high-temperature value of
the vibrational heat capacity, which
occurs when T >> Qvib. Like
translation, CV increases from zero
below Qvib to its constant maximum
value.
Non examinable: Where did the heat
capacity come from? (e.g. vibrational)
• Final (non-examinable) slide of lecture 3 had
the following equation:
E
N dq
d ln q
d ln q
N
 Nk BT 2
q d
d
dT
• For vibrations, at high T, we can solve this with:
1
qvib T   1  exp  hn / kBT 
 hn / k BT  ... when k BT  hn
-1
• The definition of heat capacity gives:
CV 
dE
dT
 NkB
constant V
• So if N=1 mol, the molar heat capacity is NAkB=R
Rotations
The rotational energy levels of a diatomic molecule are given by
eJ 
h2
8 I
2
J ( J  1)
Where I is the moment of inertia of the molecule
I
m1m2 2
r
m1  m2
…where m1 and m2 are the masses of the nuclei, and r is the bond length.
The characteristic rotational temperature is given by
Qrot
h2
 2
8 Ik B
Qrot is below 5K for all but the lightest molecules.
Cvrot = nR far above Qrot.
Qrot(H2) = 85K
Qrot(H2O) = 40K
Qrot(SO2) = 3K
Flash Quiz!
• Calculate Qrot for the following diatomic
molecules:
– HD (bond length 74pm)
– HF (bond length 92pm)
– HI (bond length 161pm)
Data:
NA = 6.022×1023 mol-1
kB = 1.38×10-23 JK-1
h = 6.626×10-34 Js
 = 3.1415
MH = 1 g/mol, MD = 2 g/mol, MF = 19 g/mol, MI = 127 g/mol
Answer
• Calculate Qrot for the following diatomic
molecules:
– HD (bond length 74pm)
– HF (bond length 92pm)
– HI (bond length 161pm)
HD
66 K
HF
30 K
HI
9K
Qrot
h2
 2
8 Ik B
I
m1m2 2
r
m1  m2
Heat Capacity versus T
The energy gap between ground and the first excited electron state is typically
very large (i.e. Qelec > 104K) for diatomics and, hence, there is vanishingly small
probability of there being any significant excitation at temperatures of interest.
(This is not so for large molecules and solids.) A large energy gap means a
large Qelec and, assuming that T << Qelec, we have CVelec  0.
Putting these contributions to the heat capacity of a gas of H2 together we have
something which looks as follows:7R/2
Cv
5R/2
3R/2
85K
5000K
Q
Q
temperature
Heat Capacity versus T
The data shown below for diatomic molecules has temperature scaled by the
characteristic vibrational temperature, as we did for solids, but also shows molar
heat capacity scaled by R. This data focuses on vibrations only; The heat
capacity increases from
5R/2 (translation + rotation) towards 7R/2 (translations + rotation + vibration)
LK Nash “Elements of Statistical Thermodynamics”
Chemical Equilibrium
So far we have considered changes in the equilibrium state associated with a
change in energy - i.e. a gas or solid (or a liquid) whose state has been altered
by heat flow or work being done. We are also interested in processes which
change the numbers of various molecular components - i.e chemical reactions.
Often these number changes are associated with energy changes, e.g.
2H2 + O2 = 2H2O
with heat released
What does this sort of process look like statistically?
Consider an isomerization reaction
A⇌B
The species A and B could, for example, stand for the ‘boat’ and ‘chair’
conformers of cyclohexane, or cis and trans conformers of 2-butene. Thanks to
the isomerization process particles have access to both A states and to B
states. To start with we shall assume the molecular states of the two species
resemble the follow simple ‘ladder’ of states.
Chemical Equilibrium
We return briefly to small systems. Consider the system N = 10 an E = 5
quanta.
energy
A
3
3
2
2
1
1
0
0
B
First, let’s start with ten A molecules. We already worked out that the most
probable configuration is n0A = 6, n1A = 3, n2A = 1 with W = 840 microstates.
Now let’s allow the isomerization reaction to occur. The most probable
configuration in this case is
n0A = 3, n0B = 3, n1A = 2, n1B = 1, n2B = 1 with W = 50400 microstates
This is a huge increase in the number of microstates made available by
accessing the energy levels of the B isomer.
Chemical Equilibrium
The number of molecules of each species are obtained by adding up the
occupation numbers of the two sets of molecular states.
N A   niA *  5
i
N B   niB *  5
i
(* denotes most probable configuration.)
We can define the equilibrium constant as
A and B are ideal gases.
PB N B
, assuming that both
K

PA N A
Notice that the equilibrium configuration involves an equal number of A and B
molecules. This is not surprising given that we have made them with identical
states.
Now we consider what happens when the energy levels of the two species are
not the same.
Chemical Equilibrium
energy
3
4
2
3
1
2
0
1
0
A
B
In this system the ground state of B lies 1 quantum of energy below the ground
state of A. This means that the overall ground state of the combined set of
energy levels is e0B. The most probable configuration of this new system is
n0A = 2, n0B = 6, n1A = 1, n1B = 1 with W = 2520 microstates
Note that, when adding up the number of quanta, you must count e0A as being at 1
quantum since it is 1 quantum above the overall ground state e0B. That is, e0A has the
same energy as e1B. We proceed likewise for the excited states of A.
The equilibrium numbers of the two species now reflect the difference in the
molecules. We now find more molecules of type B than of type A, in the most
probable distribution.
N A   niA *  3
i
N B   niB *  7
i
Chemical Equilibrium
So our statistical approach to chemical equilibrium is
1.
2.
determine the equilibrium configuration over the combined reactant and
product states,
then calculate the relative number of product and reactant molecules.
CHEMICAL EQUILIBRIUM IS DETERMINED SIMPLY BY THE SYSTEM
ADOPTING THE CONFIGURATION WITH THE GREATEST NUMBER OF
MICROSTATES.
Our next (and final) task is to put all this into a more general form. The most
probable configuration is given by
ni*  n0* exp[(e i  e 0 ) / k BT ] 
N
exp[(e i  e 0 ) / k BT ]
q
In this expression, n0* is the occupation number of the lowest energy state of A
or B (which is e0B in the previous example). Similarly, q is the partition function
considering both A and B states, and N = NA + NB is the total number of
molecules in the system.
Chemical Equilibrium
We can now express the equilibrium ratio of products to reactants (equal to the
equilibrium constant for ideal gases) as
KP 
NB

NA
B
n
i*
i
n
A
i
i
*

B
exp[

(
e

i  e 0 ) / k BT ]
i
 exp[(e
A
i
 e 0 ) / k BT ]
i
Notice that the same ground state (which is e0B in the previous example) is used
in both the numerator and denominator of this expression. Continuing with this
convention, the numerator is now simply the molecular partition function for B,
qB. The denominator just requires a little rearrangement:-
 exp[(e
i
A
i
 e 0B ) / k BT ]  exp[(e 0A  e 0B ) / k BT ] exp[(e iA  e 0A ) / k BT ]
i
 exp[(e 0A  e 0B ) / k BT ]  qA
Chemical Equilibrium
We substitute this into out equilibrium constant expression to get
KP 
N B qB

exp[ (e 0B  e 0A ) / k BT ]
N A qA
Chemical equilibrium, and particularly the equilibrium constant, is determined by
the ratio of the number of thermally accessible states of reactants and products,
i.e. qB/qA times the ratio of occupation numbers of the ground states of the
molecules.
The final necessary step in getting from molecular states to the equilibrium
constant is the calculation of the molecular partition function for molecules with
multiple degrees of freedom. To do this we will assume, as we did for heat
capacity, that the different degrees of freedom do not affect one another.
emolecule = etranslation + erotation + evibration + eelectronic
Summary
You should now be able to
• Explain the temperature dependence of the heat capacity in
terms of characteristic temperatures and degrees of freedom
• Calculate characteristic temperatures for translation, rotation
and vibration from appropriate input data
• Define chemical equilibrium in terms of occupation numbers
and microstates
• Write down and explain the general definition of the equilibrium
constant for an isomerization reaction
Next Lecture
• Explicit calculation of KP for isomerization reactions
• Using the partition function to calculate other thermodynamic
quantities