CH 6
THERMOCHEMISTRY
AP Chemistry
2014-2015
6.1 THE NATURE OF ENERGY
 Energy is the ability to do work or produce heat. The
sum of all of the potential and kinetic energy in a
system is known as the internal energy of the
system.
 Potential energy is the energy of an object by virtue
of its position (which can mean more than one
thing).
 In chemistry, potential energy is usually the energy stored in
bonds. When bonded atoms are separated, the PE increases
because energy must be added to overcome the coulombic
attraction between each nucleus and the shared electrons.
When atoms bond, this coulombic attraction results in energy
being released and a subsequently lower PE.
Kinetic energy is energy of motion
(translational, rotational, and vibrational
motion of particles in our case),
proportional to Kelvin temperature. KE
depends on the mass and velocity of an
object.
NEED TO KNOW FOR AP EXAM
 Rotational motion
(motion due to the
rotation of bonds) is
typically studied using
microwave spectroscopy
(polar molecules) and
Raman scattering
(nonpolar molecules).
 Vibrational motion
(motion due to bond
vibration, see bottom
figure on right) is
typically studied using
infrared spectroscopy.
The law of conservation of energy states
that any change in the energy of a
system must be balanced by the transfer
of energy either into or out of the
system—this is the First Law of
Thermodynamics—the energy of the
universe is constant.
HEAT AND TEMPERATURE
Heat (q): two systems with different
temperatures that are in thermal contact will
exchange thermal energy (which we call heat).
This transfer is a process and a measurement
of energy.
 +q heat is absorbed
-q heat is released
Temperature (T) is proportional to the average
KE of molecules in whatever you’re looking at.
It is not a measure of energy.
THERMODYNAMIC EQUILIBRIUM
(not in notes)
Heat is transferred from
warmer objects to cooler
ones until thermodynamic
equilibrium is reached—
that is, until both are at
the same temperature.
At thermodynamic
equilibrium, there is zero
net transfer of heat
between the two objects.
WHAT ABOUT HEAT CAPACIT Y
If the masses of these
objects are equal, what
can we say about their
heat capacities?
The ending temperature
will NOT always be the
average of the two
starting temperatures.
ANOTHER EXPLANATION. THANKS NASA.
ENTHALPY
 Enthalpy (H) is the flow of energy at constant
pressure when two systems are in contact. Change
in enthalpy is written ΔH.
 Types





Enthalpy
Enthalpy
Enthalpy
Enthalpy
Enthalpy
of
of
of
of
of
reaction
combustion
formation
fusion
vaporization
ΔH rxn
ΔH comb
ΔH f
ΔH fus
ΔH vap
 Endothermic: +ΔH, net absorption of energy by the system,
energy is a “reactant”
 Exothermic: -ΔH, net loss of energy by the system, energy is a
“product”
In thermochemistry, the system is the area of
the universe you are focusing on, and the
surroundings are everything outside of it.
ENTROPY
Entropy (S) is a measure of the dispersal
of matter and energy (entropy = disorder)
Increased disorder/dispersal: +ΔS (natural
tendency of the universe, we’ll see more on
this)
Increased disorder/dispersal: -ΔS
ENTROPY IN EQUATIONS (NOT IN THE
NOTES)
Entropy increases when…
 Solids are converted to liquids or gases
 Liquids are converted to gases
 The number of moles is greater on the products side
than the reactants side (add the coefficients)
Entropy decreases when the opposite
happens.
 For example: 2 H 2(g) + O 2(g) → 2 H 2O(l)
 Entropy decreases in this equation because a) the number of
moles decreases and b) gases are being “converted” to liquid
form in a roundabout way.
GIBBS FREE ENERGY
Gibbs Free Energy (G) is the criteria for
determining thermodynamic favorability and
calculating the theoretical amount of energy
to do work.
ΔG = ΔH – TΔS
 Spontaneous reaction/process: - ΔG
 Example: a ball rolling down a hill, ice melting on a hot
surface
 Nonspontaneous reaction/process: +ΔG
 Example: a ball rolling up a hill, water freezing into ice
on a hot surface
 Work is a force acting over a distance ( w = -PΔV = -pressure
multiplied by change in volume) in physics where gases are
involved; expressed in J or kJ
 Positive work = decrease in volume = work done on system = energy
added to system
 Negative work = increase in volume = work done by system = energy
lost by system
 1 L atm = 101.3 J = 0.101325 kJ
 STP
0°C = 273 K
torr = etc….see Ch 5
 Standard (lab) conditions 25°C = 298K
1 .0 M for solutions
1 atm = 760
1 atm etc.
 Add a ° to the symbol—for example ΔH f ° is enthalpy of formation at
298 K and 1 atm
 New convention is that ΔH° is expressed in kJ/mol rxn , where “mol rxn ” is
“moles of reaction”
 Energy is often defined as the ability to do work
w = heat + work
ΔE = q +
EXERCISE 6.1 INTERNAL ENERGY
 Calculate the ΔE for a system undergoing an
endothermic process in which 15.6 kJ of heat flows
into the system and where 1.4 kJ of work is done on
the system. (answer = 17.0 kJ)
ΔE = q + w
Endothermic, so q = +15.6 kJ
Work is done on the system, w = + 1.4 kJ
ΔE = 15.6 kJ + 1.4 kJ = 17.0 kJ
EXERCISE 6.2 PV WORK
 Calculate the work associated with the expansion of a gas
from 46 L to 64 L at a constant external pressure of 15 atm.
(answer = -270 Latm = -27 kJ)
w = -PΔV
P = 15 atm
ΔV = 64 L - 46 L = 18 L
w = - 15 atm * 18 L = -270 Latm (work done BY system)
-270 Latm * (101.3 J/1 Latm) = -27360 J = -27 kJ
EXERCISE 6.3 INTERNAL ENERGY, HEAT,
AND WORK
 A balloon is being inflated to its full volume by heating the air
inside it. In the final stages of this process, the volume of the
balloon changes from 4.00 x 10 6 L to 4.50 x 10 6 L by the
addition of 1 .3 x 10 8 J of energy as heat. Assuming that the
balloon expands against a constant pressure of 1 .0 atm,
calculate the ΔE for this process . (answer = 8.0 x 10 7 J)
ΔE = q + w
Addition of heat, so q = +1.3 x 108 J
Work is done by the system
w = -PΔV = - 1.0 atm * (4.50 x 106 – 4.00 x 106) L
= - 5.00 x 105 Latm = - 5.07 x 107 J
ΔE = 1.3 x 108 J - 5.07 x 107 J = 8.0 x 107 J
6.2 ENTHALPY AND CALORIMETRY
 We can only measure the change in enthalpy, ΔH, the
difference between the potential energies of the
products and the reactants
 ΔH is a state function (does not depend on path, only
on beginning and end)
 ΔH = q at constant pressure (ex. atmospheric
pressure)
 Enthalpy can be calculated from stoichiometry,
calorimetry, tables of standard values, Hess’s Law,
and bond energy
EXERCISE 6.4 ENTHALPY
 Upon adding potassium hydroxide pellets to water,
the following reaction takes place:
 KOH(s)  KOH(aq) + 43 kJ/mol
 Suppose 14.0 g of KOH are added to water.
 Does the beaker get warmer or colder?
 Is the reaction endothermic or exothermic?
 What is the enthalpy change for the dissolution of the 14.0 g
of KOH? (answer = -10.7 kJ/mol rxn , what is the sign?)
14.0 g KOH
1 mol KOH
56.11 g
-43 kJ ____ = -10.7 kJ/molrxn
1 mole KOH
Note: for problems like this, the correct units are kJ/molrxn
CALORIMETRY
 Calorimetry is the process of measuring heat based on
observations of temperature change when a body absorbs or
discharges heat. All of the bullet points below discuss
constant-pressure calorimetry.
 In coffee-cup calorimetry, we use a, well, disposable coffee cup and
reactants that begin at the same temperature, then we look for a
change in temperature. This is constant-pressure calorimetry (since
the container is open), so q = ΔH. After all the data is collected
(mass or volume, initial and final temperatures) we can calculate the
energy released or absorbed. In order to do this we need to know
things like the specific heat capacities of each substance (unless one
is an unknown, of course).
 Heat capacity is the energy required to raise the temperature of
something by 1°C.
HEAT CAPACIT Y
 Specific heat capacity (Cp) is the energy required to raise
one gram of a substance by 1°C, if the
process/experiment is carried out at constant pressure.
Units = J/gK = J/g°C
 Molar heat capacity is the energy required to raise one
mole of a substance by 1°C, if the process/experiment is
carried out at constant pressure. Units = J/ moleK =
J/mole°C
THE M-CAT EQUATION
q released or gained at constant
pressure = mCpΔT
The specific heat of water (liquid state) =
4.184 J/g°C = 1.00 cal/g°C; we need this
value because the heat lost by the
substance is equal to the heat gained by
the water (or vice-versa)
UNITS OF ENERGY
calorie = amount of heat needed to raise
the temperature of 1 of water by 1 1°C
Kilocalorie = 1000 cal = 1 Calorie (for
calories, capital/lower-case c mean
different things)
Joule = SI unit of energy, 1 cal = 4.184 J
EXERCISE 6.5 COFFEE-CUP CALORIMETRY
 In a cof fee cup calorimeter, 100.0 mL of 1 .0 M NaOH and
100.0 mL of 1 .0 M HCl are mixed. Both solutions are
originally at 24.6°C. After the reaction, the final temperature
is 31 .3°C. Assuming that all solutions have a density of 1 .0
g/cm 3 and a specific heat capacity of 4.184 J/ g°C, calculate
the enthalpy change for the neutralization of HCl by NaOH.
Assume that no heat is lost to the surroundings or the
calorimeter. (answer = -5.6 kJ/mol rxn )
The energy gained by the water is equal to the energy lost by
the reaction, but opposite in sign. So
ΔHrxn = - mwater * Cwater * ΔTwater
= - (200.0 g)(4.184 J/g°C)(31.3-24.6°C)
= -5600 J/molrxn = -5.6 kJ/molrxn
EXERCISE 6.6 ENTHALPY
 When one mole of methane is burned at constant pressure,
890 kJ/mol of energy are released as heat. Calculate ΔH for
a process in which a 5.8 g sample of methane is burned at
constant pressure. (answer = heat flow = -320 kJ/mole)
5.8 g CH4
1 mol CH4 -890 kJ ____ = -320 kJ/molrxn
16.04 g
1 mole CH4
Note: for problems like this, the correct units are kJ/molrxn
EXERCISE 6.7 CONSTANT-PRESSURE
CALORIMETRY
 When 1 .00 L of 1 .00 M barium nitrate solution at 25.0 °C is mixed
with 1 .00 L of 1 .00 M sodium sulfate solution at 25.0 °C in a
calorimeter, the white solid BaSO 4 forms and the temperature of the
mixture increases to 28.1 °C. Assuming that the calorimeter only
absorbs a negligible quantity of heat, that the specific heat capacity
of the solution is 4.1 84 J/ g°C, and that the density of the final
solution is 1 .0 g/mL, calculate the enthalpy change per mole of
BaSO 4 formed. (answer = -26 kJ/mol rxn )
The energy gained by the water is equal to the energy lost by
the reaction, but opposite in sign. So
ΔHrxn = - mwater * Cwater * ΔTwater
= - (2000.0 g)(4.184 J/g°C)(28.1-25.0°C)
= -26000 J/molrxn = -26 kJ/molrxn
6.3 HESS’S LAW
 Take moles into account
 A consequence of Hess’s Law is the summation of
individual reactions to yield a total reaction. The
basic idea and an example are shown below.
EQUATIONS, ENTHALPY OF REACTION,
AND HESS’S LAW
 First decide how the given equations need to be
arranged to “add up” and “cancel out”. Make sure
your reactants and products end up on the correct
sides.
 If an equation has to be reversed, also reverse the
sign of its ΔH° (more correctly, its ΔH rxn °)
 If an equation has to be multiplied by a factor to
obtain the correct coefficients, also multiply the
ΔH rxn ° by this factor (it’s per mole)—same goes for
division
EXERCISE 6.8 HESS’S LAW
 Calculate the ΔH for the overall reaction 2 H 3 BO 3 (aq) 
B 2 O 3 (s) + 3 H 2 O(l) given the following equations. (answer =
14.4 kJ/mol rxn )
The first equation needs to be doubled—look at the boric acid, H3BO3.
The third equation needs to be halved—look at the diboron trioxide, B2O3.
Both the first and third equation “run” the same way as the goal equation—
boric acid is a reactant in both, diboron trioxide is a product in both.
The second equation needs to be reversed and halved—reversed so that the
metaboric acid (HBO2) from the first equation is used up and so that the
tetraboric acid (H2B4O7) needed for the third equation is created.
Here’s how that looks:
Note that the changes made to each equation are reflected in their changes in
enthalpy. Equation 1 was doubled, so its enthalpy change was doubled.
Equation 2 was reversed and halved, so its enthalpy because positive, and was
cut in half. Equation 3 was halved, so its enthalpy was halved.
If we add the three new equations together and cancel terms that appear on
both sides, we get
And if we add the enthalpy changes for each reaction, we get
6.4 STANDARD ENTHALPIES OF
FORMATION
 ΔH f ° = standard enthalpy of formation, which is
the enthalpy change associated with the
production of one mole of a compound from its
elements in their standard states. Zero for
elements in their standard states.
 Remember…standard states = 25°C, 1 atm, 1 M
 We can use standard enthalpies of formation and
Hess’s Law to calculate the enthalpy change of a
reaction.
EXERCISE 6.9 ΔH F ° AND ΔH RXN ° I
Given the information below, calculate
the ΔH rxn ° for the following chemical
reaction.
3 Al(s) + 3NH 4 ClO 4 (s) 
Al 2 O 3 (s) + AlCl 3 (s) + 3 NO(g) + 6 H 2 O(g)
ΣΔHf°products= (1 mol)(-1676 kJ/mol)+(1 mol)(-704 kJ/mol) + (3 mol)(90.0
kJ/mol) + (6 mol)(-242 kJ/mol) = -3562 kJ
ΣΔHf°reactants = (3 mol)(0 kJ/mol) + (3 mol)(-295 kJ/mol) = -885 kJ
ΔHrxn= -3562 kJ– (-885 kJ) = -2677 kJ/mol
EXERCISE 6.10 ΔH F ° AND ΔH RXN ° II
Occasionally, a value cannot be found in a 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 of the substance must be
calculated by working backwards from its heat of combustion.
Calculate the ΔH f ° of solid glucose given the following
combustion reaction and data table. (answer = -1276 kJ/mole)
ΣΔHf°products= (1 mol)(-1676 kJ/mol)+(1 mol)(-704 kJ/mol) + (3 mol)(90.0
kJ/mol) + (6 mol)(-242 kJ/mol) = -3562 kJ
ΣΔHf°reactants = (3 mol)(0 kJ/mol) + (3 mol)(-295 kJ/mol) = -885 kJ
ΔHrxn= (6ΔHf°CO2 + 6ΔHf°H2O) - (ΔHf°C6H12O6 + 6ΔHf°O2)
-2800 = ((6*-393.5) + (6*-285.8)) – (ΔHf°C6H12O6 + (6*0))
ΔHf°C6H12O6 = +2800 + ((6*-393.5) + (6*-285.8)) = -1275.8  -1276 kJ/mol
 Bond energies
 Energy must be added/absorbed to break bonds in order to overcome
the coulombic attractions between each nuclei and shared electrons.
Energy is released when bonds are formed because the resulting
coulombic attractions between the newly bonded atoms lower the
potential energy, causing a release. TL, DR: it takes energy to break
bonds, and energy is released when bonds form.
 ΔH = sum of the energies required to break old bonds plus the sum of
the energies released in the formation of new bonds (negative, since
energy is lost from the system).
 Notice that the first term is related to the reactants, and the second
term is related to the products.
EXERCISE 6.11 BOND ENERGIES
 Calculate the change in energy
that accompanies the following
reaction given the data below.
(answer = -544 kJ/mol)
Broken = reactants, formed = products
ΔH = ((1 mol*432 kJ/mol) + (1 mol * 154 kJ/mol)) – (2 mol* 565 kJ/mol)
= - 544 kJ/mol
MORE THINGS ABOUT ENERGY CHANGES
Genuinely matters for AP Exam as well as
reality
Fusion (melting), vaporization, and sublimation
require an input of energy to overcome attractive
forces between their particles (IMFs)—these
processes do not break bonds
Freezing, condensation, and deposition release
energy as IMFs form since the particles achieve a
lower state mainly due to a decrease in
temperature—these processed do not form bonds