270
Thermodynamics
271
Test Content
from AP Chemistry Course Description
III. Reactions (35

40%)
E.
Thermodynamics
1.
State functions
2.
First law: change in enthalpy; heat of formation; heat of reaction; Hess's law;
heats of
vaporization and fusion; calorimetry
3.
Second law: entropy; free energy of formation; free energy of reaction; dependence of
change in free energy on enthalpy and entropy changes
4.
Relationship of change in free energy to equilibrium constants
and electrode
potentials
Types of Calculations found on Exam
adapted
from AP Chemistry Course Description
Summary of types of problems either explicitly or implicitly included in the preceding
material.
8. Equilibrium constants and their
applications, including their use for simultaneous equilibria
10. Thermodynamic and thermochemical calculations
THERMODYNAMICS

The study of energy in matter

Thermodynamics allows us to predict whether a chemical reaction occurs or not.

Thermodynamic
s tells us nothing about how fast a reaction occurs.

i. e., thermodynamics can’t explain kinetics and vice versa
STATE FUNCTIONS
A
State Function
is a thermodynamic quantity whose value depends only on the state at the
moment, i. e., the temperature, p
ressure, volume, etc…
The value of a state function is independent of the history of the system.
The fact that internal energy is a state function is extremely useful because it we can measure
the energy change in the system by knowing the initial energy
and the final energy.
272
TYPES OF ENERGY AND ENERGY CHANGES
Two types of energy changes
1.
Heat
–
q

chaotic change in molecular motion

related to temperature

heating increases (or decreases) molecular motion in all directions

not a state function (must know history)

sign convention
+q = heat gained by system

q = heat lost by system
2.
Work
–
w

concerted change in molecular motion

Work = Force
distance
w = F
d

movement against force is work

work increases (or decreases) molecular motion in a specific direction

in gases, w =

p
V

not a state function (must know history)

sign convention
+w = work done on system
(compression)

w = work done by system
(expansion)
Three typ
es of energy
1.
Internal Energy
–
E

Internal Energy
is the sum of kinetic and potential energy in a thermodynamic system.

state function
2.
Enthalpy
–
H

modified form of internal energy

H = E + pV

state function

value is
very close to value of internal energy for most chemical systems.

change in enthalpy for constant pressure process is equivalent to heat

H = H
f
–
H
i
= q
p
2.
Gibbs Free Energy
–
G

modified form of enthalpy

G = H
–
TS
(S
–
entropy)

3 very helpful uses
1.
Predict whether reaction is spontaneous
2.
Give amount of useful work
3.
Relate how completely a reaction will proceed

much more later in the chapter!
273
REVIEW OF FIRST LAW OF THERMODYNAMICS

The change in
the energy of a system is due to heat and/or work.
E = q + w

The first law is a conservation of energy statement.

We will see that although energy is conserved, not all of it will be useful.

i. e., some energy in a process will always be wasted.

Be sure to review first law, especially sign convention of heat and work.
DEFINITIONS OF PROCESSES
Spontaneous Processes

A process that occurs without outside help.
Reversible Processes

A reversible process is a
process that is always in equilibrium.
Examples
1.) Ice melting at O
C
H
2
O (s)
H
2
O (l)

Equilibrium is adjusted by adding or subtracting heat.

Water can go from solid to liquid to solid to liquid etc…
2.) Haber process in a closed
container
N
2
(g) + 3 H
2
(g)
2 NH
3
(g)

Production of ammonia can increase or decrease by adjusting the external pressure. (also
temperature)

Aside: Haber process is extremely important for support of human population. Natural
fertilizer is insuf
ficient. Thank goodness for synthetic fertilizer.
Spontaneous
Pants fall to floor
Raw egg becomes hard boiled in water
Dry ice su
blimates at room temp.
CH
4
+ 2 O
2
CO
2
+ 2 H
2
O
Nonspontaneous
Pants hang themselves in closet
Hard

boiled egg becomes raw egg
CO
2
gas deposits as solid at room temp.
CO
2
+ 2 H
2
O
CH
4
+ 2 O
2
274
Irreversible Processes

A process not in equilibrium
Examples:
1.) Ice melting at 25
C
2.) Precipitation of AgCl
Ag
+
(aq) + Cl

(aq)
AgCl (s)

As a solution of Ag
+
and a solution of Cl

are mixed, the system is out of equilibrium.

Thus, precipitation occurs

The solid AgCl will never redissociate into ions.

However, once precipitate forms, system is in equilibrium. Now the formation of solid
can be consider
ed reversible (e. g. addition or subtraction ammonia causes amount of
solid to change.
AgCl(s)
Ag
+
(aq) + Cl

(aq)
AgCl(s) + 2 NH
3
(aq)
[Ag(NH
3
)
2
]
+
(aq) + Cl

(aq)
ENTROPY

Review notes from solutions chapter about entropy.

Entropy is measure of
how energy can be dispersed or spread out.

Entropy makes systems more disordered.

Increasing the number of ways that a particle in a system can distribute energy, increases the
entropy of the system.
Example: Which has more entropy: 1 mole of gas in
10 L or 1 mole of gas in 20 L?

Larger volume means molecules can be in more places, the gas in the 20 L has more
entropy.

When heat exchange is not involved, spontaneous processes always increase entropy.
Consider diffusion of gas in two connected g
as bulbs.

Gas spontaneous expands, increasing entropy.
Thermodynamic definition of entropy

Entropy is defined via a change in heat (i. e., a change in choatic
motion)

q is often difficult to measure for irreversible processes, but is usually easy to measure for
reversible processes.
open stopcock
1 atm
0 atm
½ atm
½ atm
q
rev
–
heat in a reversible process
275
Example: Calculate the change of entropy when 36.02 g of ice mel
ts at 0.00
C. The molar
enthalpy of melting for ice is 6.008 kJ/mol.
Using the thermodynamic definition of entropy, we need to find q
rev
and T.
T = 0.00
C = 273.15 K
Melting of ice is
a reversible process; thus
Therefore the increase of the entropy of the water is
SECOND LAW OF THERMODYNAMICS
In any spontaneous processes, the total entropy of the universe increases for irreversible
processes and is zero for reversible processes.
For a thermodynamic process, the universe can be broken into two parts.
1.) System
–
contains what we’re interested in
2.) Surroundings
–
everything else
Example: For a
beaker of melting ice
System
–
beaker and ice
Surroundings
–
laboratory and everything else
Example: Measuring the enthalpy of neutralization in a coffee cup calorimeter
System
–
reaction e.g. NaOH + HCl
Surroundings
–
water and calorimeter
or
System
–
reaction, water and calorimeter
Surroundings
–
laboratory
*Defining system and surroundings can be a matter of perspective.*
Since universe can be broken into two parts, 2nd Law can be rewritten as
Note that
S
sys
can be negative (system becomes more ordered), if
S
surr
is more positive.
Note: Always use Kelvin for
thermodynamic calculations.
example of an
isolated system
276
Example: Water freezing at
–
10
C.
System
–
water
Surroundings
–
refrigerator
Water spontaneously decreases entropy, but 2nd law says
S
univ
>
0
Therefore,
S
surr
must be much greater than zero

i. e.,
S
surr
>> 0
Water transfers heat (choatic motion) to air in refrigerator.
Heating of air increases entropy of “surroundings”.
Molecular motion and entropy

As the motion of molecule
increases, its entropy increases.

According to kinetic theory of matter, motion is proportional to temperature.

Therefore, as the temperature of a system increases, its entropy increases.
Types of molecular motion
1. Translational

Particle moves i
n a straight line, i. e., particle is translated.

Accounts for most entropy in gases.

Accounts for substantial amount of entropy in liquids.

Has no contribution to the entropy of solids.
2. Vibrational

Atoms in a bond vibrate as if on a sprin
g.

Accounts for substantial part of entropy of liquid.

vibrations occur within molecules

vibrations occur between molecules (librations)

Accounts for all entropy in solid.
Vibrations of water
3. Rotational

Molecule rotates on axis

Minor contribution to entropy except at low and very high temperatures for
gases and
liquids.
O
H
H
O
H
H
symmetric stretch
O
H
H
H
antisymmetric stretch
O
H
H
bending
O
H
H
O
H
H
277
Molecular motion and degrees of freedom
Degrees of freedom
–
number of independent options for movement
In general for N atoms in a molecule
Translational DOF
3
Rotational DOF
Linear molecule
2
Nonlinear molecule
3
Vibrational DOF
Linear molecule
3N
–
5
Nonlinear molecule
3N
–
6
Total DOF
3N
Miscellaneous notes on entropy and molecular motion
1.)
Entropy increases during phase changes
s
l
l
g
s
g
2.)
Entropy increases when
number of particles increases.
BF
3
(g) + NH
3
(g)
BF
3
NH
3
(s)
[negative
S]
N
2
O
4
(g)
2 NO
2
(g) [positive
S]
3.)
Entropy increases when temperature increases.

Number of modes of motion increases
4.)
Entropy increases when a gas is produced i
n a chemical reaction.
HNO
3
(aq) + Rb
2
CO
3
(s)
2 RbNO
3
(aq) + CO
2
(g) + H
2
O (l)
2 HBr (aq) + Cd (s)
CdBr (aq) + H
2
(g)
5.)
Entropy increases when the volume of a gas increased.
Example: Which has more entropy in a 1 liter volume: 1 atm of gas or
2 atm of gas?
The volume with 2 atm of gas has more entropy since it has more molecules and these
molecules have many more states in which they can reside.
MC Question:
Of the following reactions, which involves the largest decrease in entropy?
(A)
CaCO
3
(s)
CaO(s) + CO
2
(g)
(B)
2 CO(g) + O
2
(g)
2 CO
2
(g)
(C)
Pb(NO
3
)
2
(aq) + 2KI(aq)
PbI
2
(s) 2 2KNO
3
(aq)
(D)
C
3
H
8
(g) + 5 O
2
(g)
3 CO
2
(g) + 4 H
2
O(l)
(E)
4 La(s) + 3 O
2
(g)
2 La
2
O
3
(s)
278
Entropy for a phase transition
Since a phase change at the
transition temperature is a reversible process, calculation of the
entropy change of phase transition is a straight forward application of the thermodynamics
definition of entropy.
Example:
R

134a is a refrigerant used automotive air conditioning and ha
s the formula,
CH
2
FCF
3
. A typical air conditioning unit holds 28 oz (800 g) of refrigerant. If the
molar enthalpy of vaporization of R

134a is 22.0 kJ/mol at

26.6
C,
a)
Calculate the entropy of the phase change.
Using the thermodynamic
definition of entropy, we need to find q
rev
and T.
T =

26.6
C = 246.55 K
The evaporation of a liquid at its boiling point is a reversible process; thus
Therefore
the increase of the entropy of the R

134a is
b)
Calculate the molar entropy of the phase change.
To calculate the molar entropy, we need the molar heat, which in this case is the
enthalpy of vaporization.
Note: Always use Kelvin for
thermodynamic calculations.
279
THE THIRD LAW OF THERMODYNAMICS
In other words, if we have a perfect crystal, we must be at absolute zero.
Consequences of the Third Law

Absolute zero is
unattainable.

Entropy of all substances at absolute zero is zero.

At temperature above zero, crystal will not be perfect.

Vibrational motion introduces imperfections

To remove imperfection takes some sort of motion

But introducing motion keep
s crystal imperfect

It’s a no

win situation!
CALCULATION OF ENTROPY CHANGES
The entropy change of chemical reaction can be calculated from a table of standard entropies.

Standard entropies are measured at 1 atm or 1 M and 298 K

Note:
S
is not zero
for elements in standard state. (Different than
H
and
G
)
For the general reaction a A + b B
c C + d D
S
rxn
= c S
(C) + d S
(D)

a S
(A)

b S
(B)
GIBBS FREE ENERGY
Reconsider 2nd Law of Thermo.
Using the
thermodynamic definition of entropy, the entropy change of the surroundings can be
related to the heat of the system.
At constant pressure:
Therefore the second law can be rewritten as
Define Gibbs Free Energy as G = H
–
TS
At constant temperature:
G =
H
–
T
S
The entropy of a perfect crystal is zero at absolute zero (0 K).
280
Return to 2nd Law
Second Law in terms of Gibbs Free Energy:
G
sys
< 0
G is useful to decide if a reaction occurs
G < 0
rxn is spontaneous
G = 0
system is at equilibrium
Recall
S
univ
= 0 for reversible process
G > 0
rxn is nonspontaneous; i.e., rxn does not happen
Negative sign of
G is consistent with idea that nature always chooses lowest energy.

System going from high energy t
o low energy must release energy.
Magnitude of
G is also important.
G = w
max

As
G gets more negative, more work can be done by the system.
Note:
When

10 kJ/mol <
G < 10 kJ/mol, reactions do not go to completion but may
favor left side (

G)
or right side (+
G) of the chemical equation.
Consider combustion of fuels
CH
4
(g) + 2 O
2
(g)
CO
2
(g) + 2 H
2
O (g)
G
=

801.14 kJ/mol =

49.92 kJ/g
C
8
H
18
(l) + 25/2 O
2
(g)
8 CO
2
(g) + 9 H
2
O (g)
G
=

5214.1 kJ/mol =

45.65 kJ/g
1 mol of
octane is able to do more work than 1 mol of methane.
1 g of methane is able to do more work than 1 g of octane.
281
CALCULATING STANDARD GIBBS FREE ENERGY CHANGES
For the general reaction a A + b B
c C + d D
G
rxn
= c
G
f
(C) + d
G
f
(D)

a
G
f
(A)

b
G
f
(B)
G
f
= 0; for elements in their standard state by definition.
Temperature dependence of Gibbs free energy
Recall for spontaneous processes,
usually
H is negative and
S is positive.
However, the only reliable measure of spontaneity is
G.
Putting negative
H and positive
S into equation for
G yields negative
G.
Gibbs

Helmholtz Equation
G =
H
–
T
S
Consider possibility when
S is negative such as
4 Fe (s) + 3 O
2
(g)
2 Fe
2
O
3
(s)
S
=

0.5437 kJ/mol
K
H
=

822.16 kJ/mol
G
=

822.16 kJ/mol
–
(298.0 K)(

0.5437 kJ/mol
K) =

660.1 kJ/mol
In this reaction, heat evolved (increasing disorder of universe) compensates for system
becoming more ordered.
In increasing the
temperature, the entropy has a greater effect on the system.
We can increase temperature until the entropy effects become more important than enthalpy
effects.
From the preceding, we see that the spontaneity of a reaction is affected by temperature.
Ex
ample: At what temperature is the rusting of iron nonspontaneous?
We need to find T where entropy effects are balanced with enthalpy effects
G =
H
–
T
S = 0
As long as we keep iron above 1512 K, it won’t rust.
282
Example:
For the reaction CaCO
3
(s)
CaO (s) + CO
2
(g), The enthalpy of reaction is 178.3
kJ/mol and the entropy of reaction is 160.5 J/mol
K.
a) Calculate the Gibbs free energy at 450 K using the Gibbs
–
Helmholtz equation.
G =
H
–
T
S =
178.3 kJ/mol
–
450 K(0.1605 kJ/mol
K) = 106.1 kJ/mol
b) Calculate the Gibbs free energy at 2110 K using the Gibbs
–
Helmholtz equation.
G =
H
–
T
S = 178.3 kJ/mol
–
2110 K(0.1605 kJ/mol
K) =
–
160.4 kJ/mol
c) Calculate the temperature where the
reaction becomes spontaneous.
G = 0
Temperature dependence of free energy can be summarized in table.
H
S
G

+

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潦reaⰠH
2
NCONH
2
. The student places 91.95 g of water at 25°C into a coffee cup calorimeter
and immerses a thermometer in the water. After
50 s, the student adds 5.13 g of solid urea, also
at 25°C, to the water and measures the temperature of the solution as the urea dissolves. A plot
of the temperature data is shown in the graph below.
283
a)
Determine the change in temperature of the solution that results from the dissolution of the
urea.
Δ
T
= 21.8 − 25.0 =
−3.2
C
b)
According to the data, is the dissolution of urea in water an endothermic process or an
exothermic process? Justify your answer.
The process is endothermic. The
decrease in temperature indicates that the process for
the dissolution of urea in water requi
res energy
.
c)
Assume that the specific heat capacity of the calorimeter is negligible and that the specific
heat capacity of the solution of urea and water is 4.2 J g
−1
°C
−1
throughout the experiment.
(i) Calculate the heat of dissolution of the urea in joules.
Assuming that no heat energy is lost from the calorimeter and given that the calorimeter
has a negligible heat capacity, the sum of the heat of dissolution,
q
soln
and the change in
heat energy of the urea

water mixture must equal zero.
q
soln
+
mc
Δ
T
= 0
⇒
q
soln
= −
mc
Δ
T m
soln
= 5.13 g + 91.95 g = 97.08 g
q
soln
= −(97.08 g)(4.2 J g
−1
°C
−1
)(−3.2°C) =
1.3 × 10
3
J
(ii) Calculate the molar enthalpy of solution,
, of urea in kJ mol
−1
.
molar mass of urea = 4(1.0) + 2(14.0) + 12.0 + 16.0 = 60.0 g mol
−1
d)
Using the information in the table below, calculate the value of the molar
entropy of
solution,
, of urea at 298 K. Include units with your answer.
Accepted Value
of urea
14.0 kJ mol

1
of urea

6.9 kJ mol

1
284
e)
The student repeats the experiment and the time obtains a result for
of urea that is
11 percent below the accepted value. Calculate the value of
that the student
obtain
ed from the second trial.
Error = (0.11)(14.0 kJ mol

1
) = 1.54 kJ mol

1
14.0 kJ mol

1
–
1.54 kJ mol

1
=
12.5 kJ mol

1
f)
The student performs a third trial of the experiment but this time adds urea that has been
taken directly from a refrigerator at 5
C. What effect, if any, would using the cold urea
instead of urea at 25
C have on the experimentally obtained value of
? Justify your
answer.
There would be
an increase in the obtained value for
bec
ause the colder urea
would have caused a larger negative temperature change
.
Concentration dependence of Gibbs Free Energy
Standard concentrations are
p
= 1 atm
c
= 1 M
The Gibbs free energy for the partial pressure of a gaseous substance is
The Gibbs free energy for the molar concentration of an aqueous substance is
Note: The higher the concentration, the
higher the free energy
285
For a general reaction: a A + b B
c C + d D

Where
G and
G
0
are stoichiometric sums and Q is reaction quotient

As reaction goes to right, Q increases; thus
G
increases

As reaction goes to left, Q decreases; thus
G decreases
Example:
For the reaction N
2
O
4
(g)
2 NO
2
(g) at 298 K the standard free energy of reaction
is 5.40 kJ/mol.
a) calculate the free energy of the reaction when p(N
2
O
4
) = 1 atm and
p(NO
2
) = 1
atm
b) calculate the free energy of the reaction when p(N
2
O
4
) = 0.905 atm and p(NO
2
) = 0.115
atm
286
c)
calculate the free energy of the reaction when p(N
2
O
4
) = 0.444 atm and p(NO
2
)
= 0.224 atm
System is at equilibrium!
**At equilibrium,
G=‰⨪
0 =
G
+ RT ln K
G
㴠

RTnK
Thus the free energy is related to the equilibrium constant
Ah ha! Free energy becomes even more useful.
N
2
(
g
) + 3 F
2
(
g
)
2 NF
3
(
g
)
;
FR
Question:
The following questions relate to
the synthesis reaction represented by the
chemical equation in the box above.
a)
Calculate the value of the standard free energy change,
, for the reaction.
b)
Determine the temperature at which the
equilibrium constant, K
eq
, for the reaction is equal
to 1.00. (Assume that
, and
are independent of temperature.)
When K
eq
= 1, then
c)
Calculate
the standard enthalpy change,
, that occurs when a 0.256 mol sample of
NF
3
(
g
) is formed from N
2
(
g
) and F
2
(
g
) at 1.00 atm and 298 K.
One point is earned for multiplying
H
0
by the number of moles
of NF
3
.
One point is earned for recognizing that 2 mol
NF
3
are produced
for the reaction as it is written.
One point is earned for the answer (including kJ or J).
287
The enthalpy change in a chemical reaction is the differen
ce between energy absorbed in
breaking bonds in the reactants and energy released by bond formation in the products.
d)
How many bonds are formed when two molecules of NF
3
are produced according to the
equation in the box above?
There are six N
–
F bonds
formed.
e)
Use both the information in the box above and the table of average bond enthalpies below
to calculate the average enthalpy of the F
–
F bond.
Bond
Average Bond Enthalpy
(kJ mol

1
)
N
N
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丠
–
=
c
=
㈷2
=
䘠
–
=
c
=
?
=
=
=
=
=
=
bxam灬eW= =
=
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=
CaCl
3
(s)
CaO (s) + CO
2
(g) at 25
C, given a table of Gibbs free energies
G
f
(CaCO
3
(s)) =

1128.76 kJ/mol
G
f
(CaO(s)) =

604.17 kJ/mol
G
f
(CO
2
(g)) =

394.4 kJ/mol
G
= (
–
604.17 kJ/mol) + (
–
394.4 kJ/mol)
–
(
–
1128.76 kJ/mol) = 130.2 kJ/mol
K
p
= p
CO
2
= 1.5
10

23
atm
Example: Calculate
G
for the equilibrium at 25
C
AgCl (s)
Ag
+
(aq) + Cl

(aq)
K
sp
= 1.8
10

10
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