Notes for Unit 17 of AP Chemistry (Thermodynamics)

AP
CHEMISTRY:
CHAPTERS 6
AND 17
NOTES
THERMODYNAMICS

6.1 The Nature of Energy
• In this chapter, we will
study energy in terms of
chemical potential energy
and how that energy can
change form to accomplish
work. We will focus
specifically on
thermochemistry, which
involves heat and energy
transfer.

ENERGY AND
WORK
• Energy- the capacity to do work or to
produce heat
• Work- force acting over a distance
(Work = Fd)
–It involves a transfer of energy

The 1st LAW OF THERMODYNAMICS:
LAW OF CONSERVATION OF ENERGY
• 1st Law of Thermodynamics- also
known as the Law of Conservation of
Energy.
• States that energy can be converted
from one form to another but it can be
neither created nor destroyed.
–The total amount of energy in the universe
is constant.

Energy can be
classified in two
ways:
• Potential energy- energy due to position or
composition (included chemical potential energy)
– KE = ½ mv2
– m = mass in kg
– v = velocity in m/s
– units are J, since J = (kg.m2)/s2
• Kinetic energy- energy due to the motion of an
object
– Kinetic energy is dependent on the mass and velocity of an
object

Heat
• Heat- (q) involves a transfer of energy
between two objects due to a
temperature difference.
–Heat always moves from warmer matter to
cooler matter.

SYSTEM VS
SURROUNDINGS
• In this chapter, we will describe where heat
moves and will use the terms system and
surroundings.
• In terms of a chemical reaction, the system is
our reaction.
• The surroundings are everything else,
including things like the container the
reaction occurs in, the room it sits in, etc.

TEMPERATURE
• Temperature- a property that reflects
random motions of the particles of a
particular substance
–Exothermic- reaction which releases heat
• Energy flows out of the system
• Potential energy is changed to thermal energy
• Products have lower potential energy than
reactants

•Reaction which absorbs heat
•energy flows into the system
•thermal energy is changed into
potential energy
•products have higher PE than
reactants
ENDOTHERMIC

THE VALUE OF “E”
• Internal energy (E) of a system is the sum of the
kinetic and potential energies of all the particles
in a system.
• Thermodynamic quantities always consist of a
number and a sign (+ or -). The sign represents
the system’s point of view. (Engineers use the
surrounding’s point of view)
– Exothermic -E (systems energy is decreasing)
– Endothermic +E (systems energy is increasing)

CALCULATING THE ENERGY
IN A SYSTEM
•Example: Calculate E if q = -50 kJ
and w = +35kJ.
E = q + w
 E is the change in the system’s internal
energy
 q represents heat
 w represents work usually in J or kJ
E = q + w
= -50 kJ + 35 kJ
= -15 kJ

Work on
Gases
• For a gas that expands or is compressed, work
can be calculated by:
– w = - PV
– units: L.atm
Units come from: 1L.atm = 0.001m3.101,325 Pa
(1 Pa = N/m2), so 101.325 N.m or J

Example: Calculate the work if the
volume of a gas is increased from
15 ml to 2.0 L at a constant
pressure of 1.5 atm.
• w = -PV
• w = -1.5 atm (1.985L)
• w = -3.0 L . atm

PRACTICE QUESTION
One version of the first law of thermodynamics is
expressed as ∆E = q + w
Which gives the sign convention for this relationship
that is usually used in chemistry?
Heat, q
Added to
the
system
Heat, q
added to the
surroundings
Work, w
done on
the
system
Work, w,
done on the
surroundings
A) - + - -
B) + + + +
C) + + + -
D) + - + -
E) + - - -

LET’S TALK LAB…
• Solve for the calorimeter constant and write this on your cup.
Be sure to hang on to your cup!
• Solve for q=mCΔT for the solid you used. The mass used is the solid
and the liquid mass. The specific heat used is for water.
• Discuss your procedures for Thursday with your lab partner. You
determine how you will calculate the amount of heat lost or gained
by an entire cold pack.

LAB DATA
Compound
Name
Heat of Solution
Group 1
Heat of Solution
Group 2 (if
tested)
Avg. Heat of
Solution (if
tested twice)
NaCl
CaCl2
NaC2H3O2
Na2CO3
LiCl
NH4NO3

Enthalpy
• Enthalpy (H) concerns the
heat energy in a system.
• H = q at constant pressure
only
– Reactions that do not involve
gases or where moles of
gases do not change are
considered “at constant
pressure”.
• At constant pressure, the
terms heat of reaction and
change in enthalpy are used
interchangeably.
• H = E + PV
– E is internal
energy
– P is pressure
– V is volume

Enthalpy Change in a
System
• The change in the
enthalpy of a system can
be calculated using:
–H= ΣmH products - ΣnH
reactants
• For an exothermic reaction,
H is negative
• For an endothermic
reaction, H is positive

EXAMPLE: Sulfur dioxide reacts with oxygen. Write the balanced
equation for the reaction of one mole of sulfur dioxide,
calculate the enthalpy value for the reaction, the draw an
energy diagram for this reaction based on your answer. Be sure
to label reactants, products, and the enthalpy of the system.
• Balanced Equation
• Enthalpy Value
• Energy Diagram
H= ΣmH products - ΣnH reactants
H= [-395.7kJ/mol] - [-296.8kJ/mol + ½ 0] = -98.9 kJ/mol

THERMODYNAMIC EQUATIONS
• For many reactions, a value is given alongside
a balanced equation called a thermochemical
equation. This value should be associated
with the moles of each substance given in the
problem.
– 1/8 S8(s) + O2(g)  SO2(g) H = -296.8
kJ
– As you can see, 1/8 mole of sulfur would release
this amount of energy while one mole of sulfur
would release eight times that amount of energy.

Example: For the reaction
2Na + 2H2O  2NaOH + H2 , H = -368 kJ
Calculate the heat change that occurs when 3.5
g of Na reacts with excess water.
• 3.5g Na 1 mol Na -368 kJ =
• 22.99g Na 2 mol Na
• H = -28 kJ (or say 28 kJ are
released)

PLEASE HAVE YOUR NOTES
OUT ON YOUR DESK. 
• Today we will do some quick notes before
you begin your lab. You will have plenty of
time, don’t worry!
• HW 1 and 2 will be checked tomorrow.

Heat Capacity (C)
C = heat absorbed
Increase in temp.
C = J/g°C or
J/mol°C
• Calorimetry- the science of
measuring heat flow in a
chemical reaction.
– It is based on observing the
temperature change when a
body absorbs or discharges
heat.
– The instrument used to
measure this change is the
calorimeter.
CALORIMETRY

Here are some
helpful tables
that are found in
the back of your
text book.

q = mCT
J = (J/g°C )(g)(°C)
• Constant pressure calorimetry- pressure
remains constant during the process
– Constant pressure calorimetry uses a set up called a
coffee cup calorimeter
• The primary reaction to calculate heat changes
in a system is the “Mcat” equation.
– Energy released as heat = (heat capacity) (mass of
solution ) (increase in temp)
CALORIMETRY

Example: A coffee cup calorimeter contains 150 g H2O
at 24.6oc. A 110 g block of molybdenum is heated to
100oc and then placed in the water in the calorimeter.
The contents of the calorimeter come to a temperature
of 28.0oc. What is the heat capacity per gram of
molybdenum?
• qwater = mCΔT and qwater = qMo
• So… mwCw ΔTw = mMoCMo ΔTMo
• (150g)(4.18J/goC)(3.4oC) = (110g)
(CMo)(72oC)
• CMo = 0.27J/goC

100.0 ml of 0.100M silver nitrate is mixed with 100.0 ml
of 0.200M sodium chloride. Both solutions start at
room temperature (25.0°C) and, once combined into a
coffee cup calorimeter, the final temperature reading
after mixing is 26.8°c. Find the heat of reaction in
kj/mol of silver nitrate.
• AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
• q = mCΔT = (200.0g)(4.18 J/g°C)(1.8°C)
• q = 1500 J or 1.5 kJ
• (0.1000L)(0.100M AgNO3) = 0.0100 mol AgNO3 in the
reaction
• 1.5kJ/0.0100mol = 150 kJ/mol

E
I
E
I
I
I
EXTENSIVE VS INTENSIVE
PROPERTIES
• In terms of
calorimetry, we can
describe certain
properties of the
reaction as an:
– Extensive property-
this depends on the
amount of substance
(ex. Heat of reaction)
– Intensive property-
doesn’t depend on
the amount of
• Let’s Practice!
• The number of calories of energy
you derive from eating a banana
• The number of calories of energy
made available to your body
when you consume 10.0 g of
sugar
• The density of your blood
• The mass of iron present in
blood
• The electrical resistance of a
piece of 22-gauge copper wire.
• The melting point of copper
wire. I

CALORIMETRY CAN ALSO BE DONE
IN A CLOSED, RIGID CONTAINER.
• This is called constant volume
calorimetry
– Ex. Flashbulb in a camera or a bomb
calorimeter
– No work can be done since the
volume doesn’t change
– Heat evolved = T x heat capacity of
calorimeter (energy required to
change the temp 1oC)

HESS’S LAW
• Hess’s Law- States that the change in
enthalpy from products to reactants, or H, is
the same whether the reaction occurs in one
step or in several steps.
– H is not dependent on the reaction pathway.
– The sum of the H for each step equals the H for
the total reaction.
– If a reaction is reversed, the sign of H is reversed.
– If the coefficients in a reaction are multiplied by
an integer, the value of H is multiplied by the
same integer.

N2 + 2O2  2NO2
N2 + O2  2NO
2NO + O2  2NO2
HESS’S LAW APPLIED

Example: given the following reactions and
their respective enthalpy changes, calculate
H for the reaction: 2C + H2  C2H2.
• 2C + 2O2  2 CO2 H = 2(-393.5) kJ
• H2 + ½ O2  H2O H = -285.9 kJ
• 2CO2 + H2O  C2H2 + 5/2 O2 H = +1299.6 kJ
• 2C + H2 C2H2  H = 226.7kJ
C2H2 + 5/2 O2  2CO2 + H2O H = -1299.6 kJ/mol C2H2
C + O2  CO2 H = -393.5 kJ/mol C
H2 + ½ O2  H2O H = -285.9 kJ/mol H2

Example: the heat of combustion of C to CO2 is -
393.5 kJ/mol of CO2, whereas that for
combustion of CO to CO2 is -283.0 kJ/mol of CO2.
Calculate the heat of combustion of C to CO.
• First, write the equations given, as well as your goal
equation.
• Goal: C + ½O2  CO
• Given: C + O2  CO2 H = -393.5 kJ
CO + ½ O2  CO2 H = -283.0 kJ
• Now, rearrange to find the goal equation!
• C + O2  CO2 H= -393.5 kJ
• CO2  CO + 1/2 O2 H= +283.0 kJ
• C + 1/2O2  CO H= -110.5 kJ

6.4 STANDARD
ENTHALPIES OF
FORMATION

ΔHF°
• Standard enthalpy of formation (𝛥𝐻𝑓
0
)
• The change in enthalpy that accompanies the
formation of one mole of a compound from
its elements with all substances in their
standard states at 25oC.
• The degree sign on a thermodynamics
function indicates that the process it
represents has been carried out at standard
state conditions.

STANDARD STATE CONDITIONS
• for gases, pressure is 1 atm
• for a substance in solution, the concentration is 1 M
• for a pure substance in a condensed state (liquid or
solid), the standard state is the pure liquid or solid.
• for an element, the standard state is the form in which
the element exists under conditions of 1 atm and 25oC.
• Values of 𝛥𝐻𝑓
0
are found in Appendix 4 and at the end of
these notes
• 𝛥𝐻𝑓
0
reaction = 𝛥𝐻𝑓
0
products - 𝛥𝐻𝑓
0
reactants

Example: The standard enthalpy
change for the reaction
CaCO3(s)  CaO + CO2(g) is 178.1 kJ.
Calculate the 𝛥𝐻𝑓
0
for CaCO3(s).
• Solve for the heat of a single compound by
using 𝛥𝐻𝑓
0
reaction = 𝛥𝐻𝑓
0
products - 𝛥𝐻𝑓
0
reactants
• 178.1 kJ = [-635 kJ + -393.5] - [x]
• X = -1207 kJ

6.5 Present Sources of
Energy
• In this section, we will discuss some sources
of energy, including fossil fuels, and their
effects on the environment.

PETROLEUM
• Petroleum- A thick, dark liquid composed of
hydrocarbons chains of 5-25 carbons
– Refining petroleum involves the process of
pyrolytic cracking, or distilling the fractions of
petroleum from the main sample based on their
molecular mass and boiling point.

GASOLINE
• Gasoline C5-C12
– Gasoline, when first used in car engines, caused a
dramatic knocking sound and was thus treated
with tetraethyl lead, (C2H5)4Pb, an antiknock
agent. This introduced lead into the atmosphere
as the fuel was spent and increased the amount of
ingested lead in the human and animal
populations until 1960 when “leaded” gas was
finally phased out.
– kerosene & jet fuel C10-C18
– heating and lubricating oil and diesel fuel C15-
C

• Natural gas- This
substance is usually
found alongside
petroleum and is
composed mostly of
methane. It also
contains ethane,
propane, and butane.
NATURAL GAS

Coal
• Coal- formed from the remains
of plants buried under pressure
for many years. Cellulose,
CH2Ox, gradually loses its H and
O.
• Coal develops through 4 stages:
– lignite (least valuable)
– subbituminous
– bituminous (high sulfur)
– anthracite (most valuable)
• Coal provides 20% of our energy
in the U.S.

CO2 and the
Environment
• Effects of CO2 on the Climate
• Greenhouse effect - H2O and CO2 molecules in the
atmosphere reflect IR radiation and send it back to
earth thus raising the earth’s temperature.
• The CO2 concentration has increased by about 16% in
the past 100 years because of increase in the use of
fossil fuels.

NEW
ENERGY
SOURCES
• Coal gasification- treating coal with oxygen and steam at high
temperatures to break down many of the C-C bonds and form
C-O and C-H bonds.
• The products are syngas (CO + H2) and methane gas. Syngas
may be converted to methanol.
– CO(g) + 2H2(g)  CH3OH(l)

NEW ENERGY SOURCES
• Hydrogen as a Fuel
• H2 (g) + ½O2(g) 
H2O(l)
• Ho = -286 kJ
– ~2.5 times the energy of
natural gas
– 3 problems: production
(too expensive), transport
(too volatile), and storage
(large volume,
decomposes to H atoms
on metal surfaces, makes
metal brittle-forms metal
hydrides)

OTHER ALTERNATIVES:
• oil shale
• ethanol
• gasohol
• seed oil
(sunflower)

CHAPTER 17:
Spontaneit y, Entropy
and Free Energy

17.1 & 17.2
SPONTANEOUS
PROCESSES AND
ENTROPY AND THE
SECOND L AW OF
THERMODYNAMICS

1st Law of
Thermodynamics
• The first Law of Thermodynamics states that energy is
neither created nor destroyed; it is constant in the
universe. We can measure energy changes in
chemical reactions to help determine exactly what is
happening and how it is occurring.
• Keep in mind that thermodynamics deals with the
reactants and products while kinetics deals with how
reactants become products.

SPONTANEOUS… WHAT??
• Spontaneous process-
occurs without outside
intervention (ex.
Rusting)
– This may be fast or slow
(CO2 sublimes at room
temperature vs. iron
rusting when in the
presence of oxygen)

ENTROPY
• Entropy (S)- a measure of randomness or disorder
– This is associated with probability. (there are more ways for
something to be disorganized than organized)
– Entropy increases going from a solid to a liquid to a gas and
when solutions are formed.
– Entropy increases in a reaction when more atoms or molecules
are formed.
– Entropy increases with increasing temperature.
High
Entropy!
Low
Entropy!

ENTROPY INCREASES…
• SLG
• More molecules
produced
• Temperature
increases

2ND LAW OF
THERMODYNAMICS
• 2nd Law of
Thermodynamics- In any
spontaneous process
there is always an
increase in the entropy
of the universe. The
energy of the universe is
constant but the entropy
of the universe is

17.3 & 17.4
ENTROPY AND
FREE ENERGY

FREE ENERGY
• Free energy (G)- the amount of energy
available to do work.
– Free energy change is a measure of the
spontaneity of a reaction. It is the maximum
work available from the system.
• A spontaneous reaction carried out as
constant temperature and pressure has a
negative G. For example, when ice
melts H is positive (endothermic), S is
positive and G = 0 at 0˚C.
G = H -TS

G = H -TS
S H G Spontaneous?
Yes, always
Yes, at high T
Yes, at low T
No, never

17.5 ENTROPY
CHANGES IN
CHEMICAL
REACTIONS

3RD LAW OF
THERMODYNAMICS
• Third Law of Thermodynamics- The entropy
of a perfect crystal at 0 K is zero.
• Any substance that is not at 0 Kelvin must
have a value for entropy! This sets entropy
apart for enthalpy. Enthalpy values are just
changes, while entropy values are absolute
and cannot drop below zero.

Ex. Given the following standard
molar entropies, calculate So for
the reaction:
2Al(s) + 3MgO(s)  3Mg(s) + Al2O3(s)
• Mg(s) = 33.0 J/K Al2O3(s) = 51.0 J/K
• Al(s) = 28.0 J/K MgO(s) = 27.0 J/K
• So
reaction = So
products - So
reactants
• So = [3(33.0J/K) + 51.0J/K] – [2(28.0J/K) +
3(27.0J/K)]
• So = 13.0 J/Kmol.rxn

17.6 FREE
ENERGY AND
CHEMICAL
REACTIONS

The Standard Free Energy
Change
• Standard free energy change (G°) -The
change in free energy that occurs if the
reactants in their standard states are
converted to products in their standard
states.
– G° =Gf
o
products - Gf
o
reactants at standard
conditions
– Gf
o for a free element in its standard state is
zero.

EX. GIVEN THE EQUATION N2O4(G)
 2NO2(G) AND THE FOLLOWING
DATA, CALCULATE GO.
• Gf
o for N2O4(g) = 97.82 kJ/mol,
• Gf
o for NO2(g) = 51.30 kJ/mol
• Go = [2(51.30kJ/mol)] – [ 97.82 kJ/mol]
• = 4.78 kJ/mol rxn

THE GIBBS-HELMHOLTZ
EQUATION
• The Gibbs-Helmholtz equation works
like the Gibbs Free Energy equation, just
at standard state conditions.
–Go =Ho -TSo
–(When working this, change the units for S
to kJ).

Ex. For the given reaction and the
following information, calculate Go
at 25°C.
2PbO(s) + 2SO2(g)  2PbS(s) + 3O2(g)
Ho = [2(-100.0) + 3(0)]- [2(-218.0) + 2(-297.0)]
= 830.0 kJ/mol rxn
So = [2(91.0) + 3(205.0)] – [2(70.0)+2(248.0)]
= 161.0 J/K.mol rxn = 0.1610 kJ/mol
Go = Ho - T So
Go= 830.0 kJ/mol -298K(0.1610 kJ/mol)
= 782 kJ/mol rxn
Ho (kJ/mol) So (J/mol.K)
PbO (s) -218.0 70.0
SO2 (s) -297.0 248.0
PbS (s) -100.0 91.0
O2 (g) ? 205.0

17.7 THE
DEPENDENCE OF
FREE ENERGY
ON PRESSURE

ENTROPY AND PRESSURE/
VOLUME
• In an ideal gas, enthalpy does
not depend on pressure.
Entropy, however, does depend
on pressure because it depends
upon volume.
– Gases in large volumes have
greater entropy than in a small
volume.
– Gases at a low pressure have
greater entropy than at a high
pressure.
• Because entropy depends on

17.7 The Dependence of Free
Energy on Pressure
G = G° + RT ln (Q)
• Q = reaction quotient (partial pressure of
products/reactants raised to the power of their
coefficients)-only pressures of gases are
included.
• T = temperature in Kelvin
• R = gas constant 8.3145 J/K.mol
• G° = free energy change at 1 atm (be sure to
change to Joules!)
 If we incorporate PV = nRT with the equation
for free energy, we end up with an equation
to calculate free energy in relation to
temperature and pressure variables.

Ex. Calculate G at 298K for the
following reaction if the reaction mixture
consists of 1.0 atm N2, 3.0 atm H2, and 1.0
atm NH3.
• N2(g) + 3H2(g)  2NH3(g) Go = -33.32 kJ/mol
• G = Go + RT lnQ
• G = -33,320J/mol + 8.314J/Kmol(298K) ln
(1.0𝑎𝑡𝑚)2
(1.0 𝑎𝑡𝑚)1(3.0𝑎𝑡𝑚)3
• G = -41,500J/mol rxn or
• -41.5 kJ/mol rxn

17.8 FREE
ENERGY AND
EQUILIBRIUM

FREE ENERGY AND
EQUILIBRIUM
• The equilibrium point in
terms of kinetics occurs
when the forward and
reverse reactions were
occurring at an equal rate.
• In terms of free energy,
the equilibrium point
occurs at the lowest value
of free energy available to
the reaction system.

FREE ENERGY AND
EQUILIBRIUM
• These two definitions are the same!
– G = Gproducts - Greactants = 0
• If a process has just shifted from nonspontaneous to
spontaneous, then at the point where it changes, the
value for G is zero. If G is zero, then Ho = TSo.

Ex. Given for the reaction Hg(l)  Hg(g)
that Ho = 61.3 kJ/mol and So = 100.0
J/mol.K, calculate the temperature of the
normal boiling point of Hg.
• Go =Ho -T So
• G = 0 at phase change
• 0 = 61.3 –T(0.1000)
• T = 613K or 340oC

THE “RAT LINK” EQUATION
• We can utilize the previous two equations (G =
Gproducts - Greactants = 0) and (G = Go + RT ln (Q)) to
form an equation that describes the relationship
between free energy and the value of the equilibrium
constant.
• Go = -RT ln(K) (the “rat link” equation)
– When Go = 0, free energy of reactants and products are
equal when all components are in their standard states.
During a phase change, G = 0.
– When Go < 0, Go products < Go reactants The reaction is
not at equilibrium, K > 1 since pressure of products is > 1 and
the pressure of reactants is < 1.
– When Go > 0, Go reactants < Go products The reaction is not
at equilibrium, K < 1 since pressure of products is < 1 and the

Ex. Calculate the approximate standard
free energy for the ionization of
hydrofluoric acid, HF (Ka = 1.0 x 10-3), at
25oC.
• Go = -RT ln K
• Go = -8.314(298)ln (1.0 x 10-3)
• = 1.7 x 104J or 17 kJ

FREE ENERGY AND
EQUILIBRIUM
• We can use G =Go +
RT ln(Q) to calculate
the direction that a
reaction will shift to
reach equilibrium.
– Free energy is energy
available to do useful
work. Wmax = G

Notes for Unit 17 of AP Chemistry (Thermodynamics)

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