Enthalpy

1
Enthalpy

• In Chapter 2, we looked at enthalpies of reaction.

Remember that enthalpy can be positive or
negative. Examples
MgO(s) H2O(l) ? Mg(OH)2(s) DH -37.1 kJ/mol
C2H4 (g) ? C2H2 (g) H2 (g) DH 174.47 kJ/mol
Reactions with a negative enthalpy are more
common, but chemical reactions can happen on
their own (spontaneously) even if the enthalpy is
positive.
2
Entropy
There must be a second factor, in addition to the
enthalpy, that determines whether or not a
reaction can occur.
This second factor is the entropy. Entropy is a
much more subtle concept than you may have
suspected from General Chemistry. Before we
define entropy, well begin by noticing some
unusual results we get when we apply the
principles from Chapter 2 to a heat engine
containing one mole of gas.
3
Thermodynamics
P1, V1
1
P2, V2
4
2
P3, V3
P4, V4
3
A diagram like the one we used in the last
chapter will help a lot.
We have a four-step process
1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression
4
Thermodynamics
Because of all the work we did in the last
chapter, we already know what the energy, heat,
and work are for each of these steps.
For the energy
1. DU 0
When you add these together, you get zero. Why
does this make sense?
2. DU CV (Tc - Th)
3. DU 0
4. DU CV (Th - Tc)
5
Thermodynamics
For the heat
1. qrev
We can make a simplification here...
2. qrev 0
3. qrev
4. qrev 0
6
Adiabatic Expansion
Recall that
for an adiabatic process.
So for step 2
And for step 4
7
Adiabatic Expansion
Flipping the first equation and setting the two
equal to each other gives
8
Adiabatic Expansion
9
Thermodynamics
For the heat
Using ...
1. qrev
2. qrev 0
3. qrev
4. qrev 0
10
Thermodynamics
For the work
Again using ...
1. wrev
2. wrev CV (Tc - Th)
3. wrev
4. wrev CV (Th - Tc)
11
Thermodynamics
So to sum up our findings so far, for the entire
cycle
DU 0
The heat and work are equal but opposite in sign,
as they must be, since DU 0.
qrev
wrev
12
Thermodynamics
wrev
qrev
Since V2 gt V1, q is a positive number (i.e., the
system absorbs heat) and w is a negative number
(the system performs work).
Nicolas Sadi Carnot came up with all these ideas
in 1824. He was trying to develop a theory of
heat in order to design more efficient steam
engines.
And he DID come up with a nice, easy equation for
engine efficiency, based on the equation for work
above...
13
Thermodynamics
wrev
The efficiency is given by the actual work
performed (given by the equation above) divided
by the maximum possible work, which is the work
that would have been performed if the engine were
always at the high temperature
efficiency
We can also express this using heat instead of
temperature.
Note that the equation with heat features a plus
sign, because qc (unlike Tc) is a negative number.
14
Efficiency
Suppose we build a steam engine in which the two
reservoirs are at 1C and 150C. What is the
maximum efficiency of this engine?
0.35212 35.212
What would have to happen in order for an engine
to have 100 efficiency?
15
Thermodynamics
efficiency
If we play with this equation a little, well
arrive at a very important result.
This is actually a strange and surprising result.
Its also very important, so lets look at it
some more...
16
Thermodynamics
For our engine, DU 0 because energy is a state
function, so U has to return to its initial value
when we go through a cyclic process.
But q is a path function, so we dont expect it
to be zero for a cyclic process (and, as we just
saw, it isnt).
However, we can see that is a state
function. This was very surprising when it was
first discovered, because state functions are
usually fairly obvious (like volume or pressure),
but this one popped up unexpectedly.
17
Thermodynamics
So the sum of the for all the different
temperatures used in the cycle (there could be
more than two) is equal to zero.
This new state function was given the name
entropy (symbol S). If the steps in our cycle
are infinitesimally small (meaning its a
reversible process), we have
18
More Fun With Entropy
We now want to get a more intuitive feel for what
entropy is.
Suppose we have an isolated system (i.e., neither
particles nor energy can be transferred to or
from the surroundings).
The system consists of two chambers with
different temperatures. Particles cant be
exchanged between the two, but energy (as heat)
can.
19
More Fun With Entropy
From the First Law of Thermodynamics, we know
dUA dqA,rev dwA,rev
dUB dqB,rev dwB,rev
dUA dqA,rev
dUB dqB,rev
dUA TAdSA
dUB TBdSB
20
More Fun With Entropy
dUA TAdSA
dUB TBdSB
So, overall, dS
But wait! Since this is an isolated system, we
know that no energy can escape. Thus, dUA -dUB.
21
More Fun With Entropy
What does this prove? Plenty! Suppose the
temperature is higher on the left side. What can
we say about dUA and the term in parentheses?
dUA is negative. So is the term in parentheses.
Thus, dS is positive. The entropy increases!
22
More Fun With Entropy
Now suppose the temperature is higher on the
right side. What can we say about dUA and the
term in parentheses?
dUA is now positive. So is the term in
parentheses.
Thus, once again, dS is positive. The entropy
increases!
23
More Fun With Entropy
The only time dS isnt positive is when the
temperature of both chambers is equal. What
happens in this case?
dUA and the term in parentheses are both zero,
and so is dS.
In this situation, energy is exchanged equally
between both chambers thus, it is a reversible
(or equilibrium) process.
24
More Fun With Entropy
In the other situation (where the temperature is
higher on one side), energy flows only in one
direction, so it is an irreversible (or
spontaneous) process.
As you can tell, the entropy change is higher for
a spontaneous process than for a reversible
process.
Thus, .
25
The Second Law of Thermodynamics
We can also write DS 0
Either of these two equations is better known as
the Second Law of Thermodynamics.
What it means is that, when a spontaneous process
(like a chemical reaction) occurs, the entropy of
a system always increases.
26
Entropy
So we now have a good mathematical definition of
entropy but what property does actually
represent?
Entropy is the degree of disorder or randomness a
system has. The more disordered a system is, the
higher the entropy.
27
Entropy

• So what factors can influence the value of S ?

Think about what gases, liquids, and solids look
like
gas
liquid
solid
molecules have random locations
molecules have orderly locations
high entropy
low entropy
28
Entropy

• So the entropy depends on the phase of the
  molecules (gas, liquid, or solid).

It also depends on the number of different
compounds present
all molecules the same less random
different kinds of molecule more random
low entropy
high entropy
29
Entropy

• We can use this to predict the entropy change of
  a chemical reaction.

NaHCO3(s) HCl(aq) ? CO2(g) H2O(l) NaCl(aq)
The entropy goes up how can you tell?
DS 67.3 J/mol K
O2(g) 2H2(g) ? 2H2O(g)
What happens to the entropy, and why?
DS -278.3 J/mol K
30
Entropy

• The same logic also applies when considering
  similar molecules with different numbers of
  atoms. For example, consider the molecules

H3C-CH3(g) H2CCH2(g) HCCH(g)
S 229.2 219.3 200.9 J/mol K
Which has the smallest entropy?
In addition, atoms with a greater mass tend to
have higher entropy. Based on this, which of the
following has the smallest entropy?
FeCl3(s) UCl3(s) CrCl3(s)
S 142.3 159.0 123.0 J/mol K
31
Entropy

• The structure of a molecule can also give us a
  clue about the entropy. For example, both of the
  following are liquids with the formula C4H8O

S 246.6 J/mol K 204.3 J/mol K
Which has the smallest entropy?
32
Entropy
So there are several factors that can affect the
entropy

• The phase of the compound

• The number of atoms of different types

• The mass of the atoms

• The structure of the molecules

33
Entropy
Now were ready to do some calculations with
entropy! But to do that, we need to get DS in a
more useful form.
Recall that for the reversible, isothermal
expansion of the gas
qrev
So
DS
34
Entropy
Suppose we have 2.50 L of O2 and 1.00 L of H2,
both at 300K and 1 atm, and both acting like
ideal gases. The two gases are in separate
containers.
A valve between the containers is then opened,
allowing the gases to mix. What is the entropy
of this process?
35
Entropy
2.50 L of O2 and 1.00 L of H2, both at 300K and 1
atm
We need to know the number of moles of each gas.
0.10155 mol
0.04062 mol
36
Entropy
2.50 L of O2 and 1.00 L of H2, both at 300K and 1
atm
The total entropy is .
0.28408 J/K 0.42308 J/K
0.70716 J/K
37
Entropy
Lets work with entropy some more, to get a few
more useful equations! From the First Law, we
know dU dqrev dwrev
But we also know that qrev TdS, so dU TdS -
PdV
38
Entropy
dU TdS - PdV

• In Chapter 2, we saw that
• dH dU d(PV)

dU PdV VdP
TdS VdP
Now, since we know that H is related to heat, it
makes sense to express dH as a function of T
(which is proportional to heat) and P (which
appears in the second term above, showing that
pressure is a variable here).
39
Entropy
dH TdS VdP
From our discussion of partial differentials in
Chapter 2, we can write the total derivative of H
So TdS VdP
40
Entropy
TdS VdP
We want to get an expression with entropy by
itself on one side
TdS
dS
Using the definition of the constant-pressure
heat capacity
dS
41
Entropy
dS
Compare this to the total derivative of S
This tells us
and
We now have an equations for the way entropy
varies with temperature or pressure!
42
Entropy
This allows us to write a new equation for DS
This will be very useful shortly!
43
Entropy
So how can we find S (the absolute entropy of a
substance) instead of DS (the entropy change of a
process)?

• This solid has a low entropy, but not zero
  entropy. There is still some disorder in the
  system. Why?

The atoms still have some kinetic energy they
vibrate as though they were held together by
springs. How can we reduce the amount of kinetic
energy the atoms have?
44
The Third Law of Thermodynamics

• The kinetic energy will drop to zero if we
  decrease the temperature to 0 Kelvin absolute
  zero!

When this happens, the atoms stop moving, and we
have a perfect crystal. The crystal is perfectly
ordered, so the entropy is 0.
This is the Third Law of Thermodynamics the
entropy of a perfect crystal is 0 at absolute
zero.
45
Entropy
This allows us to back to the equation we
recently found for DS
Suppose we set the lower temperature limit (T1)
at 0 Kelvin. From the Third Law, we know that S1
must then be zero, so the entropy at temperature
T is
At last, we have an expression for the absolute
entropy of a substance at temperature T!
46
Entropy

• S is called the standard molar entropy. It
  represents the entropy at 298.15 K (25C), which
  is called standard temperature. Remember, the
  entropy is different at other temperatures!

The units for S are . There are a couple of
things to notice about this.
What can we do with this? Wed like to know the
change in entropy for a chemical reaction. As we
did with the enthalpy, we can use
DS Sproducts Sreactants
47
Entropy
DS Sproducts Sreactants
MgO(s) H2O(l) ? Mg(OH)2(s)
DS 63.24 (26.8 69.91)
-33.47
Does this result make sense? Why?
48
Entropy

• Determine the entropy of the reaction

O2(g) 2H2(g) ? 2H2O(g)
DS Sproducts Sreactants
2 188.83 (205.0 2 130.58)
-88.5
Remember to account for the number of moles of
each compound!
49
The Second Law

• What will happen if we mix together solutions of
  NaCl and AgNO3?

NaCl(aq) AgNO3(aq) ?
NaNO3(aq) AgCl(s)
DS Sproducts Sreactants
207 96.11 (115.5 221.93)
-34.32
But wait! We said that entropy tends to increase
during a chemical reaction, but here it
decreases. Does this reaction break the Second
Law of Thermodynamics?
50
The Second Law
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
DS -34.32

• Remember what the Second Law says

A spontaneous process always results in an
increase of the entropy of a closed system.
The key here is those last four words. Our
reaction is not a closed system! Why not?
51
The Second Law
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
DS -34.32
reaction
environment closed system
The system here includes both the reaction and
its environment (the beaker, the air in the lab,
etc.).
The entropy of the reaction decreases here this
is allowed only if the entropy of the environment
increases even more.
52
The Second Law
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
DS -34.32
What could make the entropy of the environment
increase?
reaction
environment
heat
Simple! The entropy of the environment increases
if it gets hotter. Remember, dS .
In other words, this reaction must give off heat
so that the environment gets warmer.
53
The Second Law
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
DS -34.32
How do we determine whether this reaction really
does give off heat?
DH Hproducts Hreactants
-446.2 -127.0 (-407.1 -100.2)
-65.9
So the reaction does give off heat.
54
Phase Transitions

• Theres one more subtlety we need to mention
  about entropy. At 0 Kelvin, our substance is
  guaranteed to be a solid. However, if T is large
  enough, the substance will eventually become a
  liquid or gas.

gas
liquid
solid
high entropy
low entropy
At the freezing and boiling points, the entropy
suddenly increases dramatically!
55
Entropy
Therefore, it makes sense to split this integral
into several pieces. Suppose our substance is a
gas at the final temperature, T
The entropy change during freezing
The entropy change during boiling
The entropy change from 0 Kelvin to the freezing
point, Tfus
The entropy change between the freezing and
boiling points
The entropy change from the boiling point, Tvap,
to T
56
Real Systems

• So far, weve looked at the entropies of isolated
  systems. Unfortunately, things are rarely that
  simple in a real experiment.

Usually, our experiments are not isolated.
However, with a little care, we can keep a couple
of properties constant. As long as we do that,
the equations arent too complex.
In most experiments, we either keep volume and
temperature or pressure and temperature constant.
57
Real Systems

• One thing were interested in is whether or not a
  chemical reaction will take place spontaneously
  under given conditions.

For our first non-isolated system, lets consider
the case where we hold the volume and temperature
constant.
As we often do, we start with the First Law dU
dq dw For this system, what do we know about
the variables in this equation?
58
Constant Volume Temperature
dU dq dw
dU dq
Because DV 0.
Were trying to determine if the reaction is
spontaneous, so it makes sense to consider the
entropy of this system. We know that, by
definition, entropy is
dS
59
Constant Volume Temperature
dU dq
dS
Solving the second equation for dq and
substituting into the equation for dU gives
us dU TdS Keep in mind that this equation
applies only to systems with constant V and T!
When does dU TdS? When is dU lt TdS?
60
Constant Volume Temperature
dU TdS
dU TdS 0
Since T is a constant, we can write this as d(U
TS) 0
What is this equation telling us?
If V and T are constant, a process (like a
chemical reaction) is spontaneous if d(U TS)
0.
61
Constant Volume Temperature
d(U TS) 0
This result is so important, the term in
parentheses gets its own variable, A, called the
Helmholtz free energy.
A U TS so dA 0
So the value of A decreases during a spontaneous
process. Once the process stops (for example,
when a chemical reaction reaches equilibrium), A
stops changing (dA 0).
62
Constant Volume Temperature
A U TS
Since we know that T is constant, we can
write DA DU TDS
And since we know that DA 0 DU TDS 0
Note that if DA 0, that doesnt mean that the
process is impossible just that it isnt
spontaneous! What will have to happen in order
to make the process occur?
63
Constant Volume Temperature
DU TDS 0
What happens as we raise the temperature?
The reaction is more spontaneous. How is this
reflected in the behavior of the system?
64
Constant Pressure Temperature
Its much more common to have reactions at a
constant pressure rather than a constant volume.
dU dq dw
dU dq PdV
Because dw -PdV.
dS
65
Constant Pressure Temperature
dS
dU dq PdV
Solving the second equation for dq and
substituting into the equation for dU gives
us dU TdS PdV Keep in mind that this
equation applies only to systems with constant P
and T!
This expression isnt as simple as the one for
constant V and T, but its even more useful.
66
Constant Pressure Temperature
dU TdS PdV
dU TdS PdV 0
d(U TS PV) 0
This result is so important, the term in
parentheses gets its own variable, G, called the
Gibbs free energy.
G U TS PV so dG 0
So the value of G decreases during a spontaneous
process. Once the process stops (for example,
when a chemical reaction reaches equilibrium), G
stops changing (dG 0).
67
Constant Pressure Temperature
G U TS PV
You may recall that one equation for energy
is DU DH PDV
Thus, U H PV. Substituting this for U in the
first equation gives
G H TS DG DH TDS
This result should be familiar from General Chem!
68
Constant Pressure Temperature
DG DH TDS
As we said earlier, DG 0, so that also means
that DH TDS 0
As usual, the two sides of the equation are only
equal for a reversible process.
This equation has some interesting consequences!
69
Constant Pressure Temperature
DH TDS 0
Consider the reaction NaCl(aq) AgNO3(aq) ?
NaNO3(aq) AgCl(s)
At standard temperature, DH -65.9 and DS
-34.32 .
DH TDS -65900 298.15(-34.32)
-65900 10232.5
-55667.5
So DG lt 0, which means this reaction is
spontaneous.
70
Constant Pressure Temperature
DH TDS 0
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
At 298 K, DH -65.9 , DS -34.32
, DG -55.67
What happens as we raise the temperature?
DS is negative. Since a negative entropy change
makes a reaction less spontaneous, the reaction
will become increasingly unlikely as we raise the
temperature.
71
Constant Pressure Temperature
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
At 298 K, DH -65.9 , DS -34.32
, DG -55.67
Lets calculate DH TDS at 485 K. What else do
we need to know?
CP NaCl 50.5, AgNO3 93.1, NaNO3 92.9, AgCl
50.8
What does this tell you?
If all the compounds in our reaction have the
same phase, we can ignore the effects of the heat
capacity!
72
Constant Pressure Temperature
NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)
At 298 K, DH -65.9 , DS -34.32
, DG -55.67
Lets calculate DH TDS at 485 K.
-65900 485 (-34.32) -49254.8
So the reaction is less spontaneous at this
temperature, as expected.
73
Constant Pressure Temperature
2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)
At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG
-48.9 kJ
Now lets calculate DG at 700 K. We do this the
same way we did it in the last example, right?
Wrong! This time, the compounds have different
phases. We must determine the new DH and DS at
this temperature. How do we do that?
74
Constant Pressure Temperature
2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)
At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG
-48.9 kJ
First, lets calculate DH. Whats the equation
we want?
-89800 7555
-82245 J
75
Constant Pressure Temperature
2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)
At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG
-48.9 kJ
Now lets calculate DS. Whats the equation we
want?
-137.1 16.1
-121.0 J/K
76
Constant Pressure Temperature
2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)
At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG
-48.9 kJ
Finally, lets calculate DH TDS at 700 K.
-82245 700 (-121) 2455 J
What does this tell us?
77
Measuring Entropy
So far, weve determined the entropy by either
looking at a very small system or using the heat
capacities over a wide range of temperatures.
Often, neither of these techniques is practical.
How can we determine the entropy of a process
when not all the heat capacities are known?
The Helmholtz free energy can help, even if were
not at constant V and T!
78
Measuring Entropy
Remember our equation for the Helmholtz free
energy
A U TS
dA dU TdS SdT
Neither of these equations looks very promising
if we want to know S, because both equations
contain terms that are hard to measure (U, A, and
their derivatives).
However, remember this equation we learned for dU
of a reversible process dU TdS PdV
79
Measuring Entropy
Combining the two equations gives
dA PdV SdT
Now, this looks more helpful! dV and dT are easy
to measure. The only problem is dA. How can we
get rid of it?
The work we did with partial derivatives will
help! Whats the total derivative of A? dA
80
Measuring Entropy
dA PdV SdT
dA
-P
-S
We now have an equation for S! Unfortunately, we
still have A in the equation. We can get rid of
it if we differentiate the first equation by T
and the second one by V.
Recall that the terms on the left are called
cross derivatives. What do we know about them?
81
Measuring Entropy
An equation like this one, which we get by
setting cross derivatives equal to each other, is
called a Maxwell Relation. This particular
equation is very useful!
dS
DS
This is very handy! Its the first easy-to-use
equation for entropy weve had that works even
for real gases!
82
Measuring Entropy
DS
Suppose we used this equation on an ideal gas.
What would the term in parentheses look like?
P
DS
This is exactly what we found earlier for ideal
gases!
83
Measuring Entropy
But wait! Why stop with the Helmholtz free
energy? We can get another helpful expression
for entropy if we start with the Gibbs free
energy
G U TS PV
dG dU TdS SdT PdV VdP
Once again, we substitute our equation for dU of
a reversible process dU TdS PdV
dG SdT VdP
84
Measuring Entropy
dG SdT VdP
Once again, this looks helpful! dT and dP are
easy to measure, but how can we get rid of dG?
Use the total derivative of G. dG
-S
V
85
Measuring Entropy
-S
V
Well now get rid of the G by differentiating the
first equation by P and the second one by T.
So...
This is another Maxwell Relation!
86
Measuring Entropy
dS
DS
87
Measuring Entropy
DS
Suppose we used this equation on an ideal gas.
What would the term in parentheses look like?
V
DS
88
Measuring Free Energy
There are a couple more useful equations to
notice. As we saw earlier, if we know the
enthalpy and entropy of a process, the Gibbs free
energy is easy to calculate. But what if we
dont?
We use an equation we got in passing earlier
V
DG
Notice that this only applies to an isothermal
process!
89
Measuring Free Energy
DG
For an ideal gas, this becomes
DG
Compare this to the equation we just got for
entropy
DS
Why does the similarity make sense?
90
Measuring Free Energy
We just derived a useful equation by starting
with the expression for the variation with Gibbs
energy with pressure. We can get another
important equation if we start with the
relationship between G and temperature.
-S
This relationship would apply to a static
(unchanging) system. For a system that does
undergo change (like a chemical reaction), wed
have
-DS
91
Measuring Free Energy
-DS
Lets think about this for a second. Another
expression we have that involves DG, DS, and T is
DG DH TDS
DS
If we combine this equation with the one at the
top of the slide, we get
92
Measuring Free Energy
We can get a very important result by taking the
derivative of the second term
Multiplying through by T gives
93
Measuring Free Energy
But at the top of the previous slide, we also had
So we can set the right side of the top equation
equal to the left side of the lower one
94
The Gibbs-Helmholtz Equation
This is an important result! This is the
Gibbs-Helmholtz Equation, and is very helpful
when we look at how temperature can affect the
spontaneity of a chemical reaction.
95
Activity
A very useful quantity we can measure for any
solution is its activity. The activity is simply
the ratio of the vapor pressure of a component in
a solution versus the vapor pressure of the pure
compound
For an ideal solution, the activity is equal to
the mole fraction of A. For a nonideal solution,
its not.
96
Activity
Why is the activity useful? If we take the ratio
of the activity and the mole fraction, we get a
number called the activity coefficient
If we have an ideal solution, gA will be equal to
1. Thus, the activity coefficient can be used to
give us an idea of how nonideal a solution is
the farther gA is from 1, the less ideal it is.
97
Activity
The benzene in a benzene/water solution has a
vapor pressure of 608 mmHg when the mole fraction
of benzene is a 0.191. At the same temperature,
pure benzene would have a vapor pressure of 225
mmHg. Calculate the activity and the activity
coefficient.
2.70
Unsurprisingly, the activity is not equal to the
mole fraction, as it would be for an ideal
solution.
98
Activity Coefficients
14.15
So this solution is very far from ideality! Can
you think of a reason why?
Water and benzene are not miscible! Weve been
working with an example in which the two
components are barely soluble in one another.
Miscible liquids yield activity coefficients much
closer to 1.