The field of thermodynamics studies the behavior
of energy (heat) flow. From this study, a number of
physical laws have been established.
The laws of thermodynamics describe some of the
fundamental truths observed in our Universe.
Thermodynamics isn’t just a good
idea, it’s the law!
A way to think about thermodynamics, is to liken it to
gambling. The Universe is the Great Casino.
You have to understand, that no matter what, in the
long run, the house is going to win.
These are the rules:
1. You can’t win. You can’t get
something for nothing.
2. You can’t break even. The house
always wins!
3. You can’t get out of the game.
In the end, you’re going to lose….no matter what.
You can’t make any new money,….the amount of
money is fixed, and finite. Kind of like “table
stakes”.
Every time you place a bet…the house takes it’s
share.
…and so first there was the
second law, then there came
the first law, which was
followed 80 years later by the
zeroth law, and then finally the
third law, which may not
actually be a law at all. So,
what confuses you?
Thermodynamics
First Law: The energy of the universe is
constant. Energy can be neither created nor
destroyed, so while energy can be converted to
another form, the total energy remains
constant.
Energy In
Any Isolated
System
Energy Out
This is merely a statement of
conservation of energy.
For a long time, people tried to design and build
Perpetual Motion Machines
Electric
Generator
Electric
Motor
Cool. I’m going to be rich!
Mechanical
Work Out
Enthalpy – Heat of Reaction
The enthalpy of a substance is a
measure of the energy that is released
or absorbed by the substance when
bonds are broken and formed during a
reaction.
When bonds are formed, energy is
released. In order to break bonds,
energy must be absorbed.
H = Hproducts - Hreactants
Consider a chemical reaction (system).
The Universe
Surroundings
System
When heat leaves the system, the
reaction is said to be exothermic. (H = -)
The Universe
Surroundings
System
Heat
Energy of the
surroundings goes up.
H = (+)
When heat enters the system, the
reaction is said to be endothermic. (H = +)
The Universe
Surroundings
System
Heat
Energy of the
surroundings goes
down. H = (-)
The Second Law
You can’t break even….the house always takes it’s
cut.
This all has to do with something called entropy.
The time has come the
Walrus said, to speak of
many things. Of shoes,
and ships, and sealing wax.
Of Entropy, Enthalpy, and
Free Energy.
Why
doanswer
things is
happen
the way they do,
The
entropy.
and not in reverse?
So, what’s the question?
Glassware breaks spontaneously
when it hits the floor. Yet you can’t
drop broken glass and have it form a
graduated cylinder.
Why is it that….
A parked new car left to itself will become junk
over time…..
The junk car
will never
become like
new over
time
A sugar cube dissolves
spontaneously in hot coffee,
but dissolved sugar never
precipitates out to form a
sugar cube?
x
Entropy
A measure of the disorder of a
system.
Systems tend to change from a
state of low entropy to a state of
higher entropy. That is to say, if left to
themselves, systems tend to increase
their entropy.
Increasing Entropy
Solids
Liquids
Fewer Particles
Solutions
Gases
More Particles
The second law can be stated in
many forms…
Heat can never pass spontaneously from a cold
to a hot body. As a result of this fact, natural
processes that involve energy transfer must have one
direction, and all natural processes are irreversible.
If you do work, if you use energy, if you convert it
from one form to another, you will lose some of it. No
machine can be 100% efficient. This lost energy goes to
increasing the disorder of the Universe.
The Second Law of Thermodynamics
The entropy of the universe
increases in a spontaneous process and
remains unchanged in an equilibrium
process.
S universe = S system + S surroundings
For a spontaneous process:
S universe > 0
For an equilibrium process:
S universe = 0
What does it mean that
for some process we find
that S universe is
negative?
This means that the process is not
spontaneous in the direction
described. Rather, the process is
spontaneous in the opposite direction.
Energy (heat) flows from the warm swimmer to the cold
ocean, never from the cold ocean to the warm swimmer.
Dave’s Hand
2.6 million to one
What are the
odds?
John’s Hand
2.6 million to one
Johns hand is one of a very select
group of hands called a straight flush.
Out of the 2.6 million possible hands,
there are only 40 straight flushes.
Dave has junk. There are over a
million hands that are junk.
In other words, there are very few
combinations of five cards that form a
straight flush, and very many
combinations that result in junk.
Entropy is defined in terms of
microstates and macrostates.
Microstate
Macrostate
John’s Hand
7 8 9 10 J
Straight Flush
Dave’s Hand
4 8  7 3 K
Junk
The more microstates a
macrostate can have,
the higher its entropy.
Hey, wake up loser. The law
of entropy says you’re
probably going to get junk.
I
Consider a tank with 4 air
molecules.
Microstates
I
II
Microstates
I
II
This is the most
likely
III
Microstates
And so it is with air molecules.
Considering the extremely large
number of molecules in a house for
example, the odds are
overwhelmingly high that the
molecules will spread out evenly.
Probably Not!
So, next time you’re
thirsty, go ahead, wander
into the kitchen and get
yourself a drink of water.
There’ll be plenty of air
to breath when you get
there. Probably!
In this house we obey the
laws of thermodynamics.
“If someone points out to you that your pet
theory of the universe is in disagreement with
Maxwell’s equations – then so much the worse for
Maxwell’s equations.
If it is found to be contradicted by
observations – well, these experimentalists do
bungle things sometimes.
But if your theory is found to be against the
second law of thermodynamics I can give you no
hope. There is nothing for you do do but collapse
in deepest humiliation”
Sir Arthur Eddington - 1928
Entropy is given the symbol S
S =  npSproducts -  nrSreactants
Two factors determine the
spontaneity of a reaction…
1. Heat of Reaction - Enthalpy
2. Disorder - Entropy
Two factors determine the
spontaneity of reactions…
1. Heat content - - - Enthalpy
Reactions with a (-) H tend
to be spontaneous.
2. Disorder - - - Entropy
Reactions with a (+) S tend
to be spontaneous
Typically, the entropy is expected to
increase for processes in which…
Liquids or solutions are formed from solids
Gases are formed from either solids or
liquids.
The number of molecules increases during a
chemical reaction.
The temperature of a substance is increased.
To Calculate the Change of
Entropy (So)
So =  np S
products
-  nrS reactants
Cool, we now have two separate
means to predict spontaneity.
Enthalpy, for which -H means
the process tends to be
spontaneous,
-andEntropy, for which +S means
the process tends to be
spontaneous.
But what if they
contradict each
other in predicting
spontaneity?
Gibb’s Free Energy
Combines the concepts of enthalpy and entropy
in predicting spontaneity of a reaction.
Defined as the maximum amount of energy that can
be taken out of a reaction to do useful work.
G = H - TS
G = H - TS
T is in Kelvin
If G is (+) the reaction is not spontaneous
If G is (-) the reaction is spontaneous
If G is (0) the reaction is at equilibrium
Gibb’s Free Energy
Can be calculated two ways…
G = H - TS
OR
G = np  Gproducts - nr  Greactants
Consider Melting Ice…
H2O (solid) + Heat  H2O (liquid)
H = (+) Endothermic
S = (+) entropy increases
G = H – T S
= (+) – T(+)
Must be < 0 to occur
 The reaction will occur spontaneously at some
temperature T and above.
Example
Consider Freezing Water…
H2O (liquid)  H2O (solid) + Heat
H = (-) Reaction is exothermic
(This favors spontaneity)
S = (-) Entropy decreases going
from a liquid to a solid
(This favors non-spontaneity)
G = H - T S = (-H) – T(-S)
= (-H) + T(S)
At some temperature T and below, the
S becomes larger, making G (-)
The reaction will occur
spontaneously at some temperature T
and below. This temperature is of
course 0o C.
Another Example…
Will the following reaction occur
spontaneously at room temp (25oC)?
CH4(g) + 2O2(g)  CO2(g) + 2H2O (g)
1. Calculate S
S = npSproducts - nrSreactants
CH4 + 2O2  CO2 + 2H2O (gas)
S = [S(CO2) + 2S(H2O)] – [S(CH4) + 2S(O2)]
S = [ (213.6) + 2 (188.7)] – [ (186.2) + 2 (205)]
S = -5.2 joules/K mole
Next, calculate H
H = npH products - nrH
reactants
H = [H(CO2) + 2H(H2O)] – [H(CH4) + 2H(O2)]
H = [(-395.3) + 2(-241.8)] – [(-74.86) + 2(0)]
H = - 804.0 kJ/mole = -804,000 J/mole
Note: Be careful about the units.
Now, Calculate  G
G = H - T S
= -804,000 – (298k)(-5.2)
= -802,450 J/mole = -802 kJ/mole
Or, Calculate
way
G another
G =  npGproducts -  nrGreactants
G = [G(CO2) + 2G(H2O)] – [G(CH4) + 2G(O2)]
G = [(-394.4) +2(-228.6)] – [(-50.8) + 2(0)]
G = -800.8 kJ
The other laws…
The Third Law: It isn’t possible to reach absolute
zero temperature. It is important to understand
that all motion does not cease at this temperature,
rather this is the lowest temperature that energy
can be extracted from the substance.
The Zeroth Law: If A is in thermal equilibrium
with C, and B is also in thermal equilibrium with C,
then A is in thermal equilibrium with B. This
establishes temperature as the fundamental
variable in thermodynamics.
Gibbs Free Energy
ΔGo = ΔHo -TΔSo
The “o” superscript indicates all substances are in
their standard states – that is, at standard
conditions (25oC, 1 atm pressure)
ΔG = ΔH -TΔS
OK, what if it’s not at standard
conditions?
Lets consider pressure….
ΔG = ΔGo + RT ln(P)
Where R is the gas constant…
R = 8.3145 J/K·mol
From this…
ΔG = ΔGo + RT ln(P)
ΔG = ΔGo + RT ln(Q)
Where Q is the reaction quotient.
See sec.
16.7
Now, lets consider the reaction
when it’s at equilibrium….
At equilibrium……
Q = k where k is the
equilibrium constant
ΔG = 0
ΔG = ΔGo + RT ln(Q)
Now, lets consider the reaction
when it’s at equilibrium….
At equilibrium……
Q = k where k is the
equilibrium constant
ΔG = 0
ΔG = ΔGo + RT ln(Q)
0
Now, lets consider the reaction
when it’s at equilibrium….
At equilibrium……
Q = k where k is the
equilibrium constant
ΔG = 0
ΔG = ΔGo + RT ln(Q)
0
k
ΔGo = -RT ln(k)
ΔGo = -RT ln(k)
ΔGo = ΔHo -TΔSo
G   RT ln(k )  H  TS
o
o
o
H
TS
ln(k )  

RT
RT
o
H o 1 S o
ln(k )  

R T
R
o
This is a linear equation with the form y = mx + b
H 1 S
ln(k )  

R T
R
o
y =
o
m x + b
This is a linear equation with the form y = mx + b
H 1 S
ln(k )  

R T
R
o
y =
o
m x + b
ln(k)
1/T
This is a linear equation with the form y = mx + b
H 1 S
ln(k )  

R T
R
o
y =
o
m x + b
Slope
H o

R
ln(k)
1/T
This is a linear equation with the form y = mx + b
H 1 S
ln(k )  

R T
R
o
Y intercept
o
S
R
y =
o
m x + b
Slope
H o

R
ln(k)
1/T