Chapter 19: Chemical Thermodynamics
Clues to the Natural Direction of Chemical Processes
Entropy
Gibbs Free Energy
The Thermodynamic Definition of Equilibrium
Failure of the 1st Law
Information: Is the 1st Law enough?

If it takes 5J of energy to push a ball up a hill, the ball will release 5J of energy rolling
back down the hill. This statement simply tells me about conservation of energy

If I hammer a piece of chalk (giving it 100J of energy), I cannot put the chalk together
by hitting the hammer with the chalk sufficient to give the hammer 100J of energy.



Some processes are reversible, some are not
Again, energy is conserved, but this time the system doesn’t take the energy back in
such a way as to restore itself.
The 1st law does not state anything about the natural direction a process takes when
energy is transferred!

Some other variable or set of variables must gives us this information
Information: Spontaneous Processes
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The natural direction of a process is the spontaneous direction it undergoes when left
alone—unaided by human intervention.
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


For example: my keys fall when I release them.
It behooves us to determine what variables give us clues as to the spontaneity of a reaction.
For the key example above, how does the potential energy change as the keys fall? It decreases.
This is always true for gravity. So the keys move to minimize their PE.
Let’s look at enthalpy first (a known form of energy) to see if it tells us the same thing
for chemical situations.
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Clues to Spontaneity
1)

Below are three reactions that occur spontaneously.
NaCl(s)  Na+(aq) + Cl-(aq);
H > 0
CH4(g) + 2O2(g)  CO2(g) + 2HOH(g);
H < 0
N2(g) + 3H2(g)  2NH3(g);
H = -91.8kJ

Here is a non spontaneous reaction
2NH3(g) + CO2(g)  NH2CONH2(aq) + H2O(l);


H < 0
Which statement below best describes the trend here? Explain.
a) An exothermic reaction is always spontaneous
b) An exothermic reaction is never spontaneous
c) An exothermic reaction is a good sign that the reaction is spontaneous
If you drop a plate, it spontaneously breaks into many pieces. This is an endothermic
process, of sorts (energy was added to the plate from the fall).
Weak acids barely break into any ions, but they do dissolve and dissociate
spontaneously.
Would you say that distributing energy into more places is a good indication that a
process occurs spontaneously?


2)
3)
Do the example reactions above show #2 as always or just sometimes true?
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Entropy
Information: A New State Function

To sum up, most processes are spontaneous

if they give off heat (are exothermic)

if they break apart into more “pieces”


because they will end up at a lower energy than when they started
because this spreads the energy out (disperses the energy)
Appears we need to know two properties to decide if a process will occur
spontaneously.



The first is the enthalpy (H), the second is…
Define the second property as the entropy (S) change!
Entropy is a measure of the dispersal of energy in an atom, molecule, rxn or process.


In the plate example, the energy is given to one object (the plate) and then dispersed into many
objects (the pieces, the floor, and the air).
4)
Is the energy dispersal increased or decreased for the spontaneous plate breakage
(does S increase or decrease)?
5)
Would you say that when S increases the system tends to become more ordered or
disordered? Explain.
6)
Guess the sign of S for each of the example reactions on the previous page
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Boltzmann and Entropy
The ‘distribution’ guy, Boltzmann, was an early entropy advocate.
He connected entropy with probability—the more possible states of a system, the
lower the probability of knowing which state the system is in, and the more random
the system is.


S = k·ln(), where  is the number of states the system can take

For a molecule, the different states can be

Translational (velocity; applies to molecules and atoms)
Rotational (requires a bond, so only applies to molecules)
Vibrational (requires a bond, so only applies to molecules)
Different arrangements of the atoms or bonds within a molecule




7)
The more states available to a system, the (higher or lower) the entropy?
8) Which of the two anions below has the greatest entropy?
Which of the two anions below is most stable?
O
O
C
C
O
H3C
O
O
O
S
HO
9) The molecules shown are arranged (left to right)
in order of increasing entropy. Explain.
H
O
HO
H
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O
O
S
O
H3C
O
C
S
O
HO
H
H
H
H
C
H
O
H
H
C
H
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Volume and Temperature Considerations
10)
Imagine a small box with four compartments. You have two balls to place in the box.
There are 6 arrangements for these two balls, as shown below. Suppose I have a
box with 6 compartments. Will the entropy of my box (with two balls in it) be higher
or lower than your box.
1 of the possible
states for my box
Your box has
6 possible states
12) On the given axis, draw a Maxwell-Boltzman distribution
for a pure gas at two different temperatures.
Which curve (low T or high T) has the largest range
of realistically available velocities?
Fraction of
molecules
11) Based upon your answer above, if the volume is increased will the entropy increase,
decrease, or stay the same?
Velocity (m/s)
13) Based upon your answer above, does the entropy increase, decrease, or stay the same
when the temperature is increased?
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Calculations with S: I
Information: The 3rd Law of Thermodynamics

Imagine a pure atomic substance that is a perfect, crystalline solid at 0K. It doesn’t
exist, but imagine it. It would have no kinetic energy (translational, rotational or
vibrational) at this temperature. Thus, there would be no dispersal of energy, since
there is no energy. The entropy of such a solid would be zero.



So, S = 0 J/K·mol at 0K
Unlike H, we have a definite zero point and do not need to define the standard states
as having these values.
This is why tables don’t show a Sfº, but instead Sfº

Since S is a state function, we can determine with a Hess’s Law type calculation
14)
Consider the phase change: Fe (CO )5 (l )  Fe (CO )5 (g )
J
K  mole
J
0
Fe (CO )5 (g ); S  445.2
K  mole
The following data can be obtained from tables: Fe (CO )5 (l );
What is the Srxn for this phase change?
15)
S 0  338.0
Does this value for the change in entropy make sense? Explain.
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Calculations with S: II
16)
Consider a substance that can exist at all three phases. Rank the entropies of each
phase from lowest to highest. Explain.
17)
Which of these processes has the largest positive standard entropy of reaction?
Explain your reasoning.
A(s)  A(g)
A(g)  A(s)
A(l)  A(g)
A(s)  A(l)
18) The following path produces compound P from reacting A and B. Determine the
standard entropy of reaction for A + 2B  P
Q  ½A
Srxn = -22.2 J/K·mol
Q + 2B  R
Srxn = 83.4 J/K·mol
A + 2R  2P
Srxn = -32.6 J/K·mol
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Heat, Entropy and Temperature
Information: The Clausius Inequality

A decrease in enthalpy (H < 0) is a good sign that a process is spontaneous

An increase in entropy is a good sign that a process is spontaneous

On a heating curve, we notice that the temperature is constant while a phase change
occurs. This is evidence that all the heat is going into changing the entropy of the
substance. Clausius put it mathematically as: T S  q
T S   H

Hold pressure constant
Where the equality holds when a system is at equilibrium
melting
T(K)

Heat added
In the Clausius Inequality, we might be able to determine the conditions that assure us
of spontaneity.

Note: the melting of ice (a phase change) is spontaneous at T = 25ºC & P = 1 atm.
24) Rearrange the Clausius Inequality so that all the
terms are on one side of the inequality. Keep H put.
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Spontaneity Defined (finally)
Information: A reaction that is spontaneous at all T

Consider the fermentation of sugar, which has been
shown experimentally to occur at all temperatures: C6H12O6(s)  2C2H5OH(l) + 2CO2(g)

For this reaction, H = -68kJ/mol and S = 513 J/Kmol

These are experimentally determined values for T = 25º
25) In (24), you found how S and H are related to zero (0). The question we need to
answer is “does a value less than zero mean we have a spontaneous or nonspontaneous
process”? Use the data above to determine what sign of H – TS represents a
spontaneous process.
26) For a process at constant T & P to be spontaneous, H – TS must be _____________.
27) Is the reverse of the fermentation reaction ever spontaneous?
28) So, for a process at constant T and P to be nonspontaneous as
written, the value of H – TS must be _______________.
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Gibbs Free Energy (G)
Information: Definition of the Gibbs Free Energy

For conditions of constant T and P, a reaction can be deemed spontaneous as written if
H – TS gives a negative number.

Define G = H-TS, where G is the Gibbs Free Energy, then G = H – TS

We now only need to know the value of G to determine spontaneity
29) Imagine the reaction A  B. At 50ºC, Hrxn = -55kJ/mol and Srxn = -110J/K·mol.
a) What is Grxn for this process at 50ºC?
a) Is this reaction spontaneous as written?
b) Which direction will this reaction proceed: left or right?
29) Imagine we had written the above reaction as B  A.
a) What is the value of Hrxn?
b) What is the value of Srxn?
c) What is the value of Grxn as it is written now?
d) Which direction will this reaction proceed: left or right?
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Connecting G to Equilibrium I
31) What is the sign of G for a reaction that proceeds spontaneously in the
direction written?
Information: Q and G reveal the same information.

When G < 0, the process is spontaneous as written.

When G > 0, the process is spontaneous in the opposite direction as it is
written.

When G = 0, the process is spontaneous in both directions. This is
equivalent to saying that the system is in equilibrium.
32)Which of the following tell us a system is shifting toward the product side to
reach equilibrium? (Hint: see (29) and (30)).



Q<K
Q>K
Q=K
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Connecting G to Equilibrium II
Information: The standard thermodynamic quantities

The standard enthalpy and entropy of reaction represent these changes for an
equilibrium process at P = 1 atm.

Gº = Hº – TSº can immediately be written.

Kinetics point of view: use Q to determine motion toward equilibrium, but use
K to describe equilibrium.

Thermodynamics point of view: use G to determine motion toward
equilibrium, but use Gº to describe equilibrium.

Kinetics point of view:




If K < 1, reactants are favored at EQ
If K > 1, products are favored at EQ
If K = 1, neither side is favored at EQ
Thermodynamics point of view:



If Gº < 0, products are favored at EQ
If Gº > 0, reactants are favored at EQ
If Gº = 0, neither side is favored at EQ
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Connecting G to Equilibrium III

It has been determined that the full story connecting the kinetic and
thermodynamic view of equilibrium can be summed up as:
G  G 0  RT ln(Q )
33) At equilibrium, G = 0 and Q = K, substitute these results into the equation above.
34) Solve for the standard Gibbs Free energy of reaction by rearranging your answer
from (33).
35) Move all terms other than the natural log to one side, and undo the natural log on
both sides of your answer to (34).
36) For some process at 25ºC, the
value of Gº is 81.34kJ/mol.
Confirm that Kc = 5.61×10-15.
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A Physical Picture of Gibbs Free Energy
We can view the previous ideas in graphical form by
plotting G vs progress of reaction
37) What is the sign of Gº for this process?
Free Energy 

Gº
38) Will K be equal to, greater than, or less than 1
for this process?
Reactants
Products
Equilibrium
Imagine a different reaction that happens to
have the opposite Gº as the above example.
Free Energy 

39) Complete the graph on the far right. Mark
the equilibrium point and designate the
standard change in Gibbs Free energy.
Reactants
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Products
Chemical Thermodynamics
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Calculating Gº Using Hess’s Law Type Calculations
40) Determine G0 for the conversion of liquid water to water vapor using the following
known reactions:
0
2H2(g) + O2(g)  2H2O(g);
G = -457.2kJ/mol
2H2(g) + O2(g)  2H2O(l);
G0= -474.2kJ/mol
41) Determine G0 for the conversion of liquid water to water vapor using the following
standard Gibbs Free Energies of formation:


H2O(g) has Gf0= -228.6kJ/mol
H2O(l) has Gf0= -237.1kJ/mol
42) Do these numbers make sense? Explain!
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The Possibilities of Grxn
43) Imagine an exothermic reaction that increases the entropy of the process. What can we
say about the value of G for the process?
44) Imagine an exothermic reaction that decreases the entropy of the process. What can we
say about the value of G for the process?
45) Imagine an endothermic reaction that increases the entropy of the process. What can
we say about G for the process?
46) For the problems above that are spontaneous, would you say they are enthalpy driven,
entropy driven, or both?
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A Big Calculation
47) Use the data to the right (usually found in tables) to
determine the standard enthalpy, entropy
and Gibbs Free Energy of reaction.


 J K  mol 
H f kJ
0
S0
2SO2(g) + O2(g)  2SO3(g)
mol
SO 2
O2
SO 3
296.8
0
395.6
248.1
205.0
256.6
48) Does this process favor the reactants or products at equilibrium?
49) Kc for this process is 6.44×1024, which was calculated from the Gº. Does knowing this
value let you say anything about the spontaneity of the reaction? (We don’t know G).
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One Final Calculation
50) A reaction is known to have an enthalpy of reaction of 227 kJ
and an entropy of reaction of 227 J/K.
a) At room T, will this be spontaneous?
a) If not, at which T will it be spontaneous?
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Case Study 1


The Second Law of Thermodynamics tells us that any spontaneous process (under any
set of conditions) must be accompanied by a positive increase in the entropy of the
universe: Suniv > 0 when process is spontaneous.
If two pieces of metal are placed in contact, they will soon come to an equilibrium
temperature. Let’s let this happen for two blocks of identical mass. Let’s place these
metal blocks in a rigid, adiabatic container (does not let heat energy in or out).


Imagine the first 500J of energy to be transferred.
Denote on the diagram which direction the energy will move?
Is this a spontaneous process? How do you know?
T0 = 250K
T0 = 500K

What is the sum of qleft + qright?

What is the sum of Sleft + Sright?

From your answer above, does this computation confirm the second law of thermodynamics?
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Case Study 2

Student X places a large rubber band around his head. His head becomes warmer.


Is stretching a rubber band a spontaneous process? How do you know?
Which of these correctly expresses the enthalpy associated with stretching a rubber band? How
do you know?
Hstretch < 0

Hstretch > 0
Which of these correctly expresses the entropy associated with stretching a rubber band? How
do you know?
Sstretch < 0

Hstretch = 0
Sstretch = 0
Sstretch > 0
Student Y now places his rubber band over a hook on the wall and attaches 200g of
mass to the bottom. This stretches the rubber band 4cm from it’s equilibrium length. If
Student Y heats the rubber band with a hair dryer, what happens?
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