Chapter 23 Chemical Thermodynamics
In Chapter 14 we started our study of how heat energy is gained or lost in a reaction.
We now return to our study of energy gain or loss in a reaction. This time we will focus
on another problem, however. What we want to examine now is the problem of why
some reactions are spontaneous, and others are non-spontaneous.
in the 1860's it was proposed that enthalpy, ÄH determined spontaneity. It was
proposed that all spontaneous reactions were exothermic. In the 1870's however Gibbs
proposed that another property called entropy also had to be considered.
23-1 Spontaneity of Reactions
First a review
ÄHRXN = Hproducts - HReactants
So Exothermic means -ÄHRxn To get a negative ÄH reaction the H or the
products must be lower than the H of the reactants.
Remember that you can determine ÄHreation experimentally or through various
tables and that ÄH ~ ÄU (except for gas phase reactions)
Now the new stuff.
What do we mean by spontaneity?
Key definition
A spontaneous process is on that takes place without energy from an external
source.
Another definition would be reaction goes as written (Left to right) without input of
energy
Examples: Water flowing downhill. CH4 + 2O2 6CO2 + 2H2O
Na(s) + 2 H2O 62NaOH(aq) + H2(g)
Note the two chemical reaction are very exothermic and are not reversible
Now how about a couple of chemical reactions that are spontaneous and not
reversible, but are not exothermic:
2 gases mixing, NaCl(s) 6Na+(aq) + Cl-(aq)
Then there are some reactions like
Ba(OH)2(s) + 2NH4NO3(s) 6Ba(NO3)2 (s) + 2H2O(l) + 2NH3(aq)
Can I do as a demo?
That are spontaneous, non-reversible but very endothermic
Back in chapter 14 we talked about the system and the surroundings. Now here
is an interesting observation. In any spontaneous process you cannot restore
both the system and the surroundings to their original state, you can do one or
the other but not both. Why? The why leads us to the next section.
23-2 Second Law of Thermodynamics
Let’s start with the definition of a reversible and irreversible processes
Key definition:
A reversible process is a process in which a change is introduced very slowly
and very gradually, so the process can be reversed at any step along the way,
making the entire process reversible. It is associated with equilibrium processes
An irreversible process is one in which one or more spontaneous unidirectional
steps take place that cannot be reversed, so the process as a whole cannot be
reversed. Thus this would correspond to any non-reversible chemical or physical
reaction
Strictly speaking truly reversible systems are a theoretical ideal that can’t really
be obtained in the real world. All real process are irreversible because you
always lose some energy (usually to friction)
Before we tackle the second law of thermodynamics I need to introduce a new
thermodynamic state function called entropy (S). Entropy, as we will see in the
next section (23-3) is tied the randomness or disorder of a system, it is also tied
to heat transfer in a system.
The second law of thermodynamics expresses this tie between entropy and heat
transfer.
Key Equation
Verbally this equation says that entropy can be found by taking the about of heat
transferred into (or out of) a system and dividing it by the T of the system (in K)
This equation has a $ sign. What is the different between
ÄS = q/T and ÄS >q/T
This definition defines ÄS for both reversible and irreversible processes.
ÄS=q/T only in reversible systems, that is in systems where you have introduced
a change so gradually you could for forward or backward at every step.
ÄS>q/T for any irreversible system, that is a system in which some slightly larger
jump occurred that could not be reversed.
Note that units of S are J@K-1
If we have an isolated or adiabatic system the second law equation simplifies.
Key Definition:
An isolated or adiabatic system is one in which the system cannot exchange
energy with the surroundings.
In an isolated system that is at equilibrium ÄSsystem =0
In an isolated system that is undergoing a spontaneous, irreversible change
ÄSsystem > 0. In fact S will increase up to some maximal value until the system
hits equilibrim, at which point S levels off, and ÄS becomes 0
Figure 23.6 on board
So in an isolated system we are either at equilibrium and ÄS = 0, or, if there is a
spontaneous process going on in the system ÄS > 0
We can use the same logic on the surroundings. If the surroundings are at
equilibrium ÄSsurrounding = 0. If some spontaneous process is evolving then the
surroundings, ÄSsurroundings > 0
Taking the viewpoint of the universe = system + surroundings then
if we are at equilibrium ÄSsys + ÄSsurr =0
If we are spontaneously evolving ÄS sys + ÄS surr >0
The last equation lead us to a non mathematical expression of the second law of
thermodynamics
Key definition
Second law of thermodynamics (alternate definition) The entropy of the universe
is always increasing!
McQuarrie makes an additional point that while you have seen ÄSsys $0, that is
only for a spontaneous, isolated process. ÄSsys can be negative! In fact, as part
of a spontaneous process ÄSsys can be negative as long as ÄSsurr is positive and
larger than ÄSsys
While the mathematical statement of the second law of thermodynamics looks
quite simple, if has many very profound implications. Did you ever wonder why
heat flows from a high temperature region to a low temperature region? Your
book’s author spends about a page proving that that can be derived from the
second law. Further as demonstrated in problem 23-88 the second law can be
used to show that if you try to extract mechanical work from heat energy you only
reach 100% efficiency at 0 K
23-3 Entropy and Disorder
On a molecular level you can consider entropy to be a measure of disorder or
randomness
Thus disorder can come in two forms:
Key Definition
Positional disorder - refers to the distribution of particles in space
and
thermal disorder - refers to the distribution of energy among the particles
Any process that creates a more random arrangement of particles, or that
increases the temperature of the particles (at constant pressure) increases the
entropy of a system
The isothermal expansion of a gas is an entropy driven process.
1 way that can get all 4 molecules on one side of the chamber abcd:0
4 ways to get 3:1 abc:d abd:c acd:b bcd:a
6 ways to get 2:2 ab:cd ac:bd ad:bc bc:ad bd:ac cd:ab
4 ways to get 1:3
1 way to get 0:4
We have 1+4+6+4+1 or 16 states
and the chances of having
all 4 on one side 1/16 =
.06
3 on one 1 on the other 4/16 = .25
Evenly distributed 6/16
.316
1 and 3
.25
0 and 4
.06
This type of logic, calculating probabilities based on position is called positional
probability because it depends on the number of possible configuration in a
given state. Detailed calculations like this are sometimes hard to do and often
very tedious, but you can use the logic to understand physical processes, and it
leads to an area of analysis based on statistics called statistical mechanics.
Can you use positional probability to determine the relative entropy of a solid vs
a liquid vs a gas?
solid<liquid<<gas
Let’s take that thought 1 step further and plot the S of a substance vs T
Figure 23.10
We start with a very important point, When a material is in a perfectly ordered
crystal, and the crystal is at 0 K, we assume it’s entropy is 0. In fact we call this
the third law of Thermodynamics
Key Definition
The third law of Thermodynamics: The entropy of a perfect crystalline substance
at 0 K is 0.
As the T increases, the molecules are still stuck in their lattice, but they begin to
vibrate that leads to thermal disorder, as that thermal energy gets distributed
between different energy levels in the different molecules.
You can see a bigger jump in entropy at the melting point. Here the molecules
become much more random as the begin to move around in the liquid.
As the T increases the molecules move around more an more, and finally at the
vaporization T the molecules can jump into the gas phase and become even
more random
Clicker questions:
Predict the ÄS of the following processes
N2 gas at .01 atm 61.0 Atm (-)
NaCl dissolving
(+)
condensation
(-)
diffusion of a material (+)
In previous chapters we discussed ÄHvap and ÄHfus. We can also define
ÄSvap and ÄSfus. Interestingly the two quantities are closely related.
Since ÄS $q/T, and if we assume that we melt under reversible conditions then
ÄS =q/T
and P is constant so qp = ÄH
ÄSfus = ÄHfus /Tm
Similarly we can find that for vaporization
ÄSvap = ÄHvap /Tboil
Key Equations:
ÄSfus = ÄHfus /Tm
ÄSvap = ÄHvap /Tboil
Sample calculation
Given the following data for ammonia (Table 15.3 page 524)
Melting T: 195.4 Boiling T 239.8, ÄHfus 5.66 kJ/mol ÄHVap 23.33 kJ/mol
determine ÄSvap and ÄSFus for ammonia
ÄSfus = ÄHfus /Tm = 5660 J/195.4K = 28.97 J/K
ÄSvap = ÄHvap /Tboil = 23330J/239.8K = 97.29 J/K
23-4 Entropy and Molecular structure
The fact that the third law of thermodynamics makes the entropy of a solid 0 at
0K makes entropy a different kind of thermodynamic parameter. We could
never determine the absolute value of U or H because we would have to know
the kinetic and potential energy of every particle in our system. Thus we were
always calculating ÄH or ÄU the change that happens as we go from one state
to another
As we just saw in the last section, knowing that S of a solid is 0 at 0K allows us
to determine its absolute value!
Take a look at Table 23.1
This table gives standard molar values of several thermodynamics parameters.
Note is has absolute values for So , but relative values for ÄHf (Do you remember
what a ÄHfo is?
When you look at this table there are several interesting trends you can see that
are tied to molecular structure:
1. Table 23.2 Noble gases are interesting because they can rotate or
vibrate to increase their entropy. What do you observe? As go down
periodic table S 8. The explanation? As periodic 9, mass 8, as mass 8
quantum theory says that energy levels become more closely spaced.
Thus at any given T, the heavier the gas, the more energy levels taht are
availble, so the more random the distribution of energies and S8. This
also hold for the halogen gases as well (table 23.3)
2. Ssolid < Sliquid < Sgas We have already discussed this
3. The more atoms in a molecule the higher the S. Due to more vibrations
so more energy levels again ethyne, ethene and ethane margin 867
4. The more riding the molecule the lower the S Acetone & oxetane
margin 868
One thing that this table doesn’t show you is the dependence of S on
concentration or pressure. S is very sensitive to concentration or pressure
because both of these quantities affect the positional entropy of the material.
ÄH, on the other hand, is usually not very sensitive to changes in concentration
or pressure
Clicker questions
Which material has the higher S
CH4 or CF4
CH4 or C2H6
pentane or cyclopentane (draw since they don’t know the structure)
23-5 Entropy changes for Reactions
Now that you can find the S for individual compounds, you can determine the ÄS
for different chemical reactions
ÄSrxn = sum So [products] - sum So[reactants]
Key Equation
For the reaction aA + bB 6 cC + dD
ÄSrxn = 3(cS0C + dS0D) - 3(aS0A + bS0B)
Sample calculation
Using the data in table 23.1 determine ÄSrxn for the reactions CH4(g) + 3O2(g)
6CO2(g) + 2H2O
ÄSrxn = [1(213.8) + 2(188.8)]-[1(186.3) + 3(205.2)]
=[213.8 + 377.8]-[186.3+615.6)]
=591.6-801.9
=-219.3 J K-1 mol-1
Which makes sense we have more gas molecules as reactant than as products
to the reaction becomes less random
23-6 ÄGrxn and spontaneity
You have now seen many chemical reactions that are energy favored, that is,
ÄH or ÄU is negative, so the lower energy of the products make the reaction go
forward. You have also seen come processes, like the expansion of a gas, that
are entropy favored , that is, there is no net change in energy but the ÄS of the
precess is positive. Thus we have two different factors to consider to determine
if a process is spontaneous.
These two factors were combined in a single figure that was, in turn, used to
predict spontaneity by Gibbs in 1878. This parameter is now called Gibbs
Energy or Gibbs Free energy, G.
Key equation:
For a reaction run at constant temperature the Gibbs energy change is given by
the equation:
ÄGrxn = ÄHrxn -TÄSrxn
A reaction is spontaneous if ÄG is <0
A reaction is not spontaneous and will not occur without an outside source of
energy if ÄG is > 0
A reaction is at equilibrium and no further change will occur when ÄG =0
This nicely explains the melting point and boiling points of materials
In section 23-3 we had ÄSfus = ÄHfus /Tm
ÄSvap = ÄHvap /Tboil
if ÄG = ÄH-TÄS
you can see that at the melting point we are at equilibrim so the entropy term and
the enthalpy term just match each other so ÄG is 0. However if T is < the
melting temp the entropy term is less favorable than the enthalpy term so the
material will not melt. Conversely, if we are above the MP the TÄS term gets
larger thant he enthalpy term so the reaction (melting) occurs spontaneously.
If you think about this a bit would can come up with some general rules for
spontaneity, as shown in table 23.4
23-7 ÄGrxn and the reaction quotient
In chapter 19 we used Q to determine if a reaction will go forward (be
spontaneous) or backward (nonspontaneous).
Q>K too many products -non spontaneous
Q<K not enough products - spontaneous
Q=K at equilibrium
So what is the tie between Q and ÄGrxn?
Key Equation
ÄGrxn = RT ln(Qx /Kx)
where Qx = Qp or Qc and Kx = Kp or Kc
You can see how this works. If Q>K then ln of a # >1 is + and ÄGrxn is + and
unfavorable. If Q=K, ln(1) =0 and ÄGrxn=0 (at equilibrium). If Q<K Q/K <1,
ln(Q/K) is negative and ÄGrxn is negative so reaction is spontaneous.
And remember just because ÄGrxn is negative, it is no guarantee that the reaction
will occur. Kinetics can get it the way. For instance ÄGrxn favors Cgraphite over
Cdiamond. Yet those sparkly diamonds will not turn into graphite for millions of
years because the energy barrier of that reaction is so high.
Sample calculation
The reaction : AgCl(s) WAg+(aq) + Cl-(aq) has a K of 1.8x10-10 M2 at 298K. If I
add 5 mLs of 1 uM AgNO3 to 4 mLs of 1 mM NaCl, what is the ÄGrxn? Will a reaction
occur? What will ÄGrxn be after equilibrium is achieved?
First what are the concentrations of Ag+ and NO3- when I mix 5 mLs with 4 mLs
M1V1 = M2V2
1uM(5)=X(4+5)
X = 1(5)/9 = .556uM Ag+ and NO3How about NaCl
1 mM (4) = X(9); X =1(4)/9 = .444 mM
Now do we worry about Na+ or NO3-? (No spectator ions
so Q = .555 uM x .444 mM (remember what a u is?)
= 2.46x10-10 M2
Q/K = 2.46x10-10 / 1.8x10-10 = 1.37
ÄGrxn = RT ln(Q/K)
=8.314(298)ln(1.37)
+778 J/KAmol
ÄG is + reaction is non-spontaneous as written so reverse reaction, ions 6solid is
spontaneous so solid will form.
Other uses for ÄGrxn
So we can predict which direction a reaction will go from ÄGrxn. Big wow. We
could do that with Q. Is there anything we can do with it?
1. In the next section we will show how you can calculate K from ÄGrxn.
2. ÄGrxn represents the energy that can be released from a reaction to do work.
The absolute value of a - ÄG is the maximum energy that can be used to
perform work (like running an electric motor from a battery). The value of
a + ÄG is the minimum energy that must be supplied force a nonspontaneous reaction to occur
23-8 Relation between ÄGrxn and ÄGrxno
The thermodynamic equilibrium constant K
Since chapter 19 we have been using Kc and Kp, K’s defined in terms of
concentration or pressure, and the units of K varied with the units
(molarity or pressure) and the number of moles of products and reactants.
So the units of every K were different, and Kp was usually different than
Kc.
This is confusing and messy. The actual K we should use is called the
thermodynamic equilibrium constant K (no subscript!)
Key equation
K = Kc/Qoc = Kp/Qop
First. What are Qok or Qop?
They are the standard state reaction quotients, or, a reaction quotient (Q) in
which you use the standard state of the material (1M for concentrations, 1 bar for
a pressure, etc)
What does this calculation get you?
1. Units drop out- the units of Qox cancel the units of Kx, so K is unitless
2. Since standard states are 1M and 1 bar. It is your original Kx/1, so there
is no real calculation!
This is why I wasn’t too preoccupied with units of K. Under normal
circumstances you use the thermodynamic K which is unitless!
Clicker questions
The value of Kc for the equation CH3COO-(aq)+ H2O(l)W CH3COOH(aq) + OH-(aq)
is 5.6x10-10M @ 298K. What is K
K = Kc/Qoc = 5.6x10-10 M/1.0M = 5.6x10-10
Given Kp = 1.90 bar for the equation C(s) + CO2(g) W2CO(g)
What is K
1.90 bar/1.0 bar = 1.90
The Standard Gibbs Free Energy Change
if ÄGrxn = RT ln(Qc /Kc)
then ÄGo rxn = RT ln(Qoc /Kc)
=RT ln (1/K)
=-RTlnK
Key equation
ÄGo rxn = -RT ln K, where ÄGo rxn is called the standard Gibbs free energy change
and refers to the energy gained or lost when the reaction where all reactants and
products are initially at standard state conditions evolve to equilibrium.
Sample question
The reaction CH3COOH + H2O WCH3COO- + H3O+ has a K of 1.75x10-5 at 25oC,
what is the ÄGo rxn for this reaction.
ÄGo rxn = -RT ln K,
= -8.3145 x 298 x ln(1.75x10-5)
= -2478(-10.95)
=+27140K or 27.1 kJ
Make sure you can co the other direction (ÄG6K) as well!
It is important to remember the difference between ÄGrxn and ÄGorxn.
ÄGrxn is the free energy gained or lost when the system in its current state
evolves to equilibrium. ÄGorxn is the free energy gained or lost when the system
at standard state evolves to equilibrium
If you know the current state of the system, and the ÄGorxn you can calculate the
ÄGrxn using the equation:
Key Equation:
ÄGrxn=ÄGo rxn +RTlnQ
where Q is the thermodynamic reaction quotient, Qc/Qoc
Sample calculation
Continuing with our acetic acid example previously we found that for the reaction
CH3COOH + H2O WCH3COO- + H3O+ has a K of 1.75x10-5 at 25oC
ÄGo rxn =+27140K or 27.1 kJ if [CH3COOH]=[CH3COO-]=[H+] = 1M
But what is ÄGrxn if [CH3COOH]=[CH3COO-]=[H+] = .0001M
ÄGrxn=ÄGo rxn +RTlnQ
=27140 + 8.314(298)ln(.0001[.0001)/.0001]
=27140-22819
=4,329J
it is still unfvorable, but much closer to equilibrium. Lets try one more, if
[CH3COOH]=[CH3COO-]=[H+] = .00001M
ÄGrxn=ÄGo rxn +RTlnQ
=27140 + 8.314(298)ln(.00001[.00001)/.00001]
=27140-28524
=-1384 J The reaction is now spontaneous!
The above example show that ÄG for any reaction depends on the current
conditions. So while you can find a ÄGo and you think it tells you if a reaction is
spontaneous or non-spontaneous, it really doesn’t you always have to plug in
your actual conditions to see how they work.
23-9 Gibbs Energies of Formation
Back in chapter 14 we worked with ÄHof. There is an equivalent
that
tabulates the free energy gained or lost when a compound in it standard state is
formed from its elements.
And, in a similar manner you can calculate the ÄGorxn as
Key equation
Example calculation:
determine the ÄGorxn and the K for the reaction: C2H4(g) + H2(g) 6C2H6(g) at
1000oC
ÄGorxn = Products - Reactants
=[1(68.4) + 1(0)]-[1(-32.0)]
=68.4 + 32.0 =100.4 kJ (+ ÄG, not-spontaneous)
ÄGorxn =-RTlnK
100400 = -8.314(1273)lnK
-9.49=lnK
K=e-9.49 = 7.59x10-5
23-10 The van’t Hoff Equation
in section 23-6 we examined the equation
ÄG = ÄH -TÄS
Here we saw that if both ÄH and ÄS were + the reaction is unfavorable at low T
but becomes favorable at T then the TÄS term gets large enough to overpower
the unfavorable + ÄH. This means that K <1 at low T and K >1 at high T, so K
must depend on T.
Let’s see if you can figure out the exact relationship.
ÄGorxn = ÄHorxn - TÄSorxn
but ÄGorxn=-RTlnK so
-RTlnK =ÄHorxn - TÄSorxn
lnK = ÄHorxn/-RT - TÄSorxn/-RT
= - ÄHorxn/RT + ÄSorxn/R
So if you plot lnK vs 1/T
Y=mX + B
m = slope = - ÄHorxn/R
b = intercept = ÄSorxn/R
Thus by measuring K as a function of T, you can measure both ÄH and ÄS of a
reaction. In lab you did this for several temperatures to determine the ÄH for a
chemical reaction. On a test you don’t have time to plot ln(K) vs 1/T for several
points, so instead we use a 2 point equation called the van’t Hoff equation.
Key equation:
van’t Hoff equation:
This actually fits quite well with something you learned back in chapter 19
At that time we said that if you had an exothermic reaction
A6B + heat, then as T8, the reaction should go in reverse and K9
Let’s assume T2 > T1
if we have an exothermic reaction then ÄH is negative, and the right hand side of
the equation becomes negative at T2 increases. The ln of a negative number is
<1 so this woudl say that at T 8, k9 so this agree with what we were doing earlier.
With a little work you should be able to figure out what happens with an
endothermic reaction and see if this also check with what we did with Le
Chatliers P.
Example problems
The value of ÄHorxn for the reaction
2H2O W H3O+ + OH- is 55.8 kJ/mol
if the K for this reaction is 1x10-14 at 250C, what is the K for this reaction at 15oC?
We’ll call T2 = 25oC = 298 K
T1 = 15oC = 288K (it doesn’t matter which is 1 and which is 2!)
Ln(1x10-14/X) = 55800/8.314(298-288)/(298*288)
Ln(1x10-14/X) = 6712(10)/(85824)
=.782
1x10-14/X = e.782
-14
1x10-14
/X =2.19
1x10 = X(2.19)
X = 1x10-14/2.19 = 4.57x10-15
The Clapeyron-Clausius Equation
Back in chapter 15 when we talked about the vapor pressure of liquids we talked
about how the VP of a gas rises exponentially with temperature (figure 15.23).
At that time I put the explanation off until this chapter.
The explanation is now trivial. The vapor pressure of a gas over a liquid simply
depends on the reaction lW g which, in tur n depends on the ÄG, ÄH and ÄS of
the phase transition.
Since K = VP/1 and this will depend on the ÄH of the reaction as we just saw
above you can derive the Clapeyron-Clausius Equation in exactly the same way
Key equation:
Clapeyron-Clausius:
Sample calculation
You know the vapor pressure of water is 1 atm @ 100oC. If the vapor pressure
of water is 1.43 atm @ 110oC, what is the ÄHo vap of water?
Ln(1.43/1.00) = X/8.314(383-373)/(383x373)
.358=X/8.314(10/142859
.358(8.314) = X (7x10-5)
2.97/(7x10-5) =X; X=42400 = 42.4 kJ