Mr. Nelson
AP Chemistry
 Chemical
reactions involve changes in
energy
 Thermochemistry
 Study of relationships
of chemical reactions
and energy changes
 Thermodynamics
 Study of energy and its transformations
Energy
is the ability to do work or
produce heat and is the sum of all
potential and kinetic energy in a system
ETotal   PE   KE
Potential energy is energy possessed
due to relative position to other objects
 Commonly in chemistry this is energy
stored in bonds
 Kinetic energy is the energy of motion,
usually of particles

 Heat
(q) transfer of energy from object
with higher KE to object with lower KE
 Hence heat flows from hot to cold
 Heat transfers because of temperature (KE)
difference, but temperature is not a measure
of heat directly
 Work (w)
is a force acting over a distance
 Usually used when discussing
gases
 When related to gases, work is a function of
pressure and volume:
E  q(heat )  w( work )
 Energy
is a state function while heat and
work are not
 State function is a property that is
independent of past or future behavior,
only on current conditions (temperature,
pressure, etc)
 Example: Many roads from your home to
school, but you have the same starting point
and ending point
 Signs
of q
 +q means heat is absorbed (endothermic)
 -q means heat is released (exothermic)
 Signs
of w
 +w if work done on the system (i.e.
compression)
 -w if work done by the system (i.e.
expansion)
Energy
cannot be created or
destroyed, only conserved
 Energy of universe is constant
Energy
may only transfers between
system (the experiment) and
surroundings (universe)
SI
unit of energy is Joule
 1 Joule = 1 kg*m2/s2
 Commonly reported in kilojoules
Non-SI
unit of energy is calories
 1 cal = 4.184 J
 Nutrional Calorie (Cal) = 1 kcal =
1000 calories
 Describes
heat gained or lost at constant
pressure
H = qp

Following the derivation, E = H + P∆V or H = E - P∆V
 Enthalpy
is a state function
 Can be calculated from multiple sources:





Calorimetry
Stoichiometry
Heats of formation (tables of standard values)
Hess’ Law
Bond Energies
Enthalpy of
a reaction can be
utilized stoichiometrically based on
coefficients
For endothermic processes ∆H is
positive and should be included as a
reactant in the rxn
For exothermic processes ∆H is
negative and should be included as
a product in the rxn
 Example
Upon adding solid potassium hydroxide pellets to
water the following reaction takes place:
KOH ( s )  KOH ( aq)  43kJ
Answer the following questions regarding the
addition of 14.0 g of KOH to water:
1. Does the beaker get warmer or colder?
2. Is the reaction endothermic or exothermic?
3. What is the enthalpy change for the dissolution
of the 14.0 g of KOH?
 Guidelines
 Enthalpy is an extensive property
 Enthalpy of the reverse reaction is
equal in magnitude but opposite in
sign
 Enthalpy change depends on physical
states of reactants and products
 If the thermochemical equation is
multiplied by a factor of n, then ∆H
must also change by the same factor
 Process
of measuring heat based on observing
the temperature change when a body absorbs
or discharges energy as heat

2
Assumption is that no heat is lost to surroundings
(aka closed system)
types of Calorimetry:
Constant pressure(coffee-cup setup)
 Constant volume (bomb calorimeter)
 Used in industry for determining food calories

 Commonly
used in lab to determine specific
heats of metals
 Can also measure the heat released/absorbed
of many types of reactions:



Neutralizations
Ionization (Dissolution/Dissolving)
Reaction (ppt, combustion, etc.)
 Heat
capacity– energy required to raise the
temperature of an object by 1 degree C (J/°C)
 Specific heat
capacity– same as above but specific
to 1 gram of the substance (J/ g °C)
 Molar
heat capacity – Same as heat capacity, but
specific to one mole of a substance (J/mol K or
J/mol °C)
 Heat
capacities are extensive properties, while
specific heat is an intensive property
 Heat capacity can be determined by
multiplying a substances mass by its specific
heat:
J/°C = J/g°C x g
 Specific heat of water is 4.184 J/g°C and is the
same for all dilute aqueous solutions
 Heat of substance + heat of solution = 0
 Therefore, qsubstance = -qsoln
q=mCp∆T
 q=

heat transferred (J)
Recall: at constant pressure q=∆H
m
= mass (g)
 Cp = specific heat of material at constant
pressure
 ∆T = Tf – Ti (final – initial)
 Example
How much heat is needed to raise 10.0 grams of
aluminum from 22.0 °C to 42.0 °C? (Specific heat
of aluminum is 0.90 J/ g K)
Also, what is the molar heat capacity of aluminum?
 Example
(Specific Heat of Metal)
A lead (Pb) pellet having a mass of 26.47 g at
89.98C was placed in a constant-pressure
calorimeter of negligible heat capacity containing
100.0 mL of water. The water temperature rose
from 22.50 °C to 23.17 °C. What is the specific
heat of the lead pellet?
 Example
(Molar Heat of Neutralization)
A quantity of 1.00 x 102 mL of 0.500 M HCl was
mixed with 1.00 x 102 mL of 0.500 M NaOH in a
constant-pressure calorimeter of negligible heat
capacity. The initial temperature of the HCl and
NaOH solutions was the same, 22.50 °C, and the
final temperature of the mixed solution was 25.86
°C. Calculate the heat change for the
neutralization reaction on a molar basis (that is,
the molar heat of neutralizaton)
 In
order to perform calculations using a constant
volume calorimeter, the bomb calorimeter must
be calibrated so that the heat capacity of the
calorimeter is known
qcal = Ccal∆T
(where Ccal is the heat capacity of the calorimeter)
 The
Ccal is determined by burning a substance with
an accurately known heat of combustion

This is a constant that is generally given!
Finally,
qcal= -qrxn
 Example
(Molar Heat of Combustion)
A quantity of 1.435 g of naphthalene (C10H8) was
burned in a bomb calorimeter. The temperature of
the water rose from 20.28 °C to 25.95 °C. If the
heat capacity of the bomb plus water is 10.17
kJ/°C, calculate the heat of combustion of
naphthalene on a molar basis.
 Example
(Heat Capacity of Bomb Calorimeter)
Camphor (C10H17O) has a heat of combustion of
5903 kJ/mol. When a sample of camphor with
mass of 0.1204 g is burned in a bomb calorimeter,
the temperature increases by 2.28 °C. Calculate
the heat capacity of the calorimeter
 Heat
(or enthalpy) of formation, ∆Hf, is the heat
required to form the elements
 Standard enthalpies of formation (∆H°f) can be
used as reference points for determining the
standard enthalpy of a reaction (∆H°rxn)


Standard state conditions are 1 atm and 25 °C
Standard enthalpies of formation of any element in its
most stable form is zero
 The greater
the heat of formation of a molecule,
the less stable the molecule

Higher ∆H°f implies more energy to form the bonds
 Change of
enthalpy that occurs in a chemical
reaction can be given by:
H rxn   nH  f ( prods)   mH  f (rcts)
 Where n
and m represent coefficients
 Two methods exist for determining the ∆H°rxn for a
reaction


Direct Method (using ∆H°f)
Indirect Method (using Hess’ Law)
 When to

If all of the standard heats of formation are known for
each participant in a chemical equation, then plug
them into the equation on the last slide
 When to



use Direct Method?
use Indirect Method?
The heat of a reaction is independent of the steps it
takes to get there (enthalpy is a state function)
If only heats of reaction for a series of reactions is
known, then these can be manipulated to determine
the heat of the reaction
This is known as Hess’s Law
 Example
(Enthalpy of Reaction)
Calculate the ∆H°rxn for the following:
3 Al (s) + 3 NH4ClO4 (s) → Al2O3 (s) + AlCl3 (s) + 3 NO (g) +
6 H2O (g)

Given the following values:
Substance
∆H°f (kJ/mol)
NH4ClO4 (s)
-295
Al2O3 (s)
-1676
AlCl3 (s)
-704
NO (g)
90.0
H2O (g)
-242
Answer: -2680 kJ/mol (exo)
 Example
(Enthalpy of Formation)
Sometimes all values are not found in the table of
thermodynamic data. For most substances it is
impossible to go into a lab and directly synthesize a
compound from its free elements. The heat of
formation for the substance must be found by working
backwards from its heat of combustion. Find the ∆Hf of
C6H12O6 (s) from the following information:
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + 2800 kJ

Substance
∆H°f (kJ/mol)
CO2 (g)
-393.5
H2O (l)
-285.8
Answer: -1276 kJ.mol for glucose
 Guidelines




First decide how to rearrange equations so reactants
and products are on appropriate sides of the arrows
Manipulate the equations using the thermochemical
equation guidelines (mentioned previously)
Check to ensure that everything cancels out to give
you the exact equation you want
Note: It is often helpful to begin your work backwards
from the answer that you want!
 Given the
following equations:
H3BO3 (aq) → HBO2 (aq) + H2O (l)
∆Hrxn = -0.02 kJ/mol
H2B4O7 (aq) + H2O (l) → 4 HBO2 (aq)
∆Hrxn = -11.3 kJ/mol
H2B4O7 (aq) → 2 B2O3 (s) + H2O (l)
∆Hrxn = 17.5 kJ/mol
Find the ∆H for this overall reaction:
2 H3BO3 (aq) → B2O3 (s) + 3 H2O (l)
Answer: 14.4 kJ/mol (endothermic)
 Study
of energy changes in chemistry
 Involves three major players: ∆H, ∆S, ∆G
 One of the major objectives of thermodynamics is
to predict whether or not a reaction will occur
when reactants are brought together
(spontaneous vs. nonspontaneous)
 First Law of Thermodynamics (stated previously) is
that energy cannot be created or destroyed, only
conserved
 Enthalpy can
be used to predict spontaneity of a
reaction
 In
general, lower energy states are preferred,
meaning exothermic reactions are favorable (-∆H)
 Enthalpy (∆H)
is not the only predictor of
spontaneity
 Entropy (∆S) can be used, and is the measurement
of disorder in a system
 Second Law of Thermodynamics states that the
entropy (or disorder) of the universe is constantly
increasing

Examples: Ice cube melting, your room at the end of a
week. Nature tends towards chaos!
S system  S surroundings  Suniverse
+


∆S = more disorder in the system
Increase in disorder is favorable, indicates higher
chance to be spont.
∆S = less disorder in the system
Decrease in disorder is not favorable, indicates lower
chance to be spont.
 Thermodynamics
can predict spontaneity, but
does NOT predict the rate (speed) of the
reaction!

Talking state functions: we do not look at pathways,
only beginning and end states!
 Entropy,
like enthalpy, is a state function
 Entropy can be calculated for reactions exactly like
enthalpy:
S rxn   nS  ( prods)   mS  (rcts)
 Units are
J/K*mol (where ∆H is kJ/mol)
 Entropy
increases as one goes from solid to a
liquid
Ssolid

Sliquid 
Sgas
 Entropy
increases as a substance divides into
parts

Dissolving solids into liquids (exception: carbonates!)
 Entropy
will increase in reactions in
which number of product molecules are
greater
 Entropy increases with temperature
 States
that the entropy of a perfect
crystal at 0 K is zero
 This implies no movement, and no
randomness exists
 Not many perfect crystals out there, so
entropy values rarely ever zero
 For
determination of standard values of
entropy (as in ∆S°), this allows an
absolute standard to exist which all
other values can be based on
 To
ultimately decide the spontaneity of a reaction,
we use Gibb’s free energy (G)
 For a constant temperature process, change in
free energy can be given by:
G  H  TS
 The most
important thermodynamic equation,
and one of the most beneficial equations in
chemistry!
 Free energy
is the energy available to do
work
 - ∆G means forward reaction is
spontaneous
 + ∆G means forward reaction is nonspontaneous
 Reverse reaction would be spontaneous!

∆G = 0 means reaction is at equilibrium
(more on this later)
 Standard
free energies of reaction can be
calculated using the following equation:
Grxn   nG f ( prods)   mG f (rcts)
 ∆G
has units of energy (J or kJ)
 ∆G is a state function (like ∆H and ∆S)
 Both
enthalpy and entropy must be known to
predict spontaneous reactions
G  H  TS
∆H
∆S
Result
Negative (-)
Positive (+)
Always spontaneous
Positive (+)
Positive (+)
Spontaneous at high
temperatures only
Negative (-)
Negative (-)
Spontaneous at low
temperatures
Positive (+)
Negative (-)
Never spontaneous