Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 5
Thermochemistry
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.
Thermochemistry
Keys to Success in Ch 5
• Learn the vocab (make flashcards!) –
there’s a lot.
• AND learn how the words are related (a
word web can really help)
• Units and abbreviations are essential!
• Watch your + and - signs
Thermochemistry
Do Now – Jan 18
Use section 5.1 in your book to answer
the questions. Then put answers on
poster paper to share with the class.
1.Define “energy”, “work”, and “heat” and
state the units of each one.
2.There are two types of energy
discussed. What are they and how are
they related?
3.Which of the two types of energy in #2
Thermochemistry
is contained in chemical bonds?
Energy, E
• Chemical reactions and phase changes
either release or absorb energy.
• Energy is the ability to do work or
transfer heat.
Work: Energy used to cause an object that
has mass to move.
Heat: Energy used to cause the
temperature of an object to rise.
What are the units? Where do they come
from?
Thermochemistry
Units of Energy
• The US unit of energy are calories
• The metric units of energy are Joules
kg m2
1 J = 1 
s2
• 1 Joule = 4.184 calories
Thermochemistry
2 kinds of energy
An object has potential energy by virtue of its
position or chemical composition (energy stored in
chemical bonds).
An object has Kinetic energy by virtue of its motion
(molecules or atoms moving in chemistry). Potential
and kinetic energy can be converted into each other
through chemical and physical processes.
How does this energy conversion happen in a chemical
reaction?
Thermochemistry
2 kinds of energy
An object has Kinetic energy by virtue of its position
or chemical composition (energy stored in chemical
bonds).
An object has Potential energy by virtue of its
motion (molecules or atoms moving in chemistry).
Potential and kinetic energy can be converted into
each other through chemical and physical processes
a) Potential energy in a reactant molecules’ bonds is
released and converted to kinetic energy in a
chemical reaction
b) Kinetic energy is converted to potential energy into
make the products’ bonds
Thermochemistry
System and Surroundings
• The system includes the molecules
we want to study, the reactants and
products (here, the hydrogen and
oxygen molecules).
• The surroundings are everything
else (here, the cylinder and piston).
• A closed system can exchange
energy but not matter w/
surroundings
 Can the gas particles escape?
 Can the piston move? Or get hotter?
Thermochemistry
Work, w
• Energy used to
move an object over
some distance.
• w = F  d,
where w is work, F
is the force, and d is
the distance over
which the force is
exerted.
In chemistry work is …
Thermochemistry
Heat, q
• Energy can also be
transferred as heat.
• Heat flows from
warmer objects to
cooler objects.
• In chemistry heat
is…
Thermochemistry
Internal Energy
The internal energy of a system is the sum of all
kinetic and potential energies of all components
of the system; we call it E.
Use Fig. 5.5
Thermochemistry
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• In other words, the total energy of the universe is a
constant; if the system (the chemical reaction in
our case) loses energy, it must be gained by the
surroundings, and vice versa.
• By definition, the change in internal energy of the
system, E, is the final energy of the system minus
the initial energy of the system: E = Efinal − Einitial
• And E for the surroundings will be equal in
amount but opposite in sign!
Thermochemistry
E, q, w, and Their Signs
•E = q+ w is the equation for the 1st law.
•This means that the change in energy E is due to
work w (either done by the system or done on the
system) and/or heat q (either released or absorbed
by the system)
• What is E for a system that does 300J of work on the
surroundings and absorbs 150 J of heat? Is it endergonic or
exergonic? What is E for the surroundings in this case?
Thermochemistry
Enthalpy
• If a process takes place at constant pressure (in an
open container, as most of the reactions we study do)
and the only work done is pressure-volume work
(expanding or contracting gases)
• THEN we can measure heat flow during the process
by measuring the enthalpy H of the system.
H = q
• So, at constant pressure the change in enthalpy is
the heat gained or lost.
• Basically this means we can measure the energy
change of a chemical reaction by measuring the heat
Thermochemistry
change, which is easy, and ignore work.
Exchange of Heat between
System and Surroundings
• When heat is absorbed by the system from
the surroundings, the process is endothermic
and H is positive.
• When heat is released by the system to the
surroundings, the process is exothermic and
H is negative.
Thermochemistry
Enthalpies of Reaction
The change in enthalpy, H,
is the enthalpy of the
products minus the enthalpy
of the reactants:
H = H(products) −
H(reactants)
An enthalpy diagram shows this
graphically
Write the equation for the
combustion of methane. Is this
reaction endo or exo? Which
has more internal energy, the
reactants or the products
Thermochemistry
Enthalpies of Reaction
This quantity, H, is called the enthalpy of reaction, or the
heat of reaction.
For the reaction pictured below:
2H2 (g) + O2 (g)  2 H2O (g), H =-483.6 kJ
A thermochemical equation shows the heat as a reactant
(if endo) or product (if exo)
2H2 (g) + O2 (g)  2 H2O (g) + 483.6 kJ
,
Thermochemistry
The Truth about Enthalpy
1. Enthalpy is an extensive property (it
depends on amount!!!).
2. H for a reaction in the forward direction is
equal in size, but opposite in sign, to H for
the reverse reaction.
3. H for a reaction depends on the state
and temperature of the products and the
state of the reactants.



The solid phase of a substance has less enthalpy
then the liquid which has less than the gas
Higher temp = higher enthalpy
Thermochemistry
More particles = more enthalpy
Thermochem Example
The thermochemical equation for the
formation of water is:
2H2 (g) + O2 (g)  2 H2O (g) + 483.6 kJ
How much energy is absorbed or
released in the formation of 4.5g of
water?
Thermochemistry
Thermochem Example
The thermochemical equation for the formation of
water is:
2H2 (g) + O2 (g)  2 H2O (g) + 483.6 kJ
How much energy is absorbed or released in the
formation of 4.5g of water?
Use stoich with H as a part of the mole ratio!
4.5 g H2O /(18.02 g/mol) = 0.25 mol H2O
0.25 mol (483.6 kJ/2 mol H2O) = 60kJ released
The 60kJ is released (I know the energy is
released because it is a product in the equation)
Thermochemistry
Enthalpy
• When the system changes at constant
pressure, the change in enthalpy, H, is
H = (E + PV)
• This can be written
H = E + PV
Thermochemistry
State Functions
Usually we have no way of knowing the
internal energy of a system; finding that value
is simply too complex a problem.
Thermochemistry
State Functions
• However, we do know that the internal energy
of a system is independent of how the system
got to where it is now.
 In the system below, the water could have reached
room temperature from either direction.
Thermochemistry
State Functions
• Therefore, internal energy E is a state function.
• It depends only on the present state of the
system, not on the path by which the system
arrived at that state.
• And so, change in energy E depends only on
Einitial and Efinal.
Thermochemistry
State Functions
• However, q and w are
not state functions.
• Whether the battery is
shorted out or is
discharged by running
the fan, its E is the
same.
 But q and w are different
in the two cases.
Thermochemistry
Work
When a process
occurs in an open
container, commonly
the only work done is a
change in volume of a
gas pushing on the
surroundings (or being
pushed on by the
surroundings).
Thermochemistry
Work
We can measure the work done by the gas if
the reaction is done in a vessel that has been
fitted with a piston.
w = −PV
Thermochemistry
Enthalpy
• Since E = q + w and w = −PV, we
can substitute these into the enthalpy
expression:
H = E + PV
H = (q+w) − w
Thermochemistry
Endothermicity and
Exothermicity
• A process is
endothermic,
then, when H
is positive.
Thermochemistry
Endothermicity and
Exothermicity
• A process is
endothermic when
H is positive.
• A process is
exothermic
when H is
negative.
Thermochemistry
Calorimetry
Since we cannot
know the exact
enthalpy of the
reactants and
products, we
measure H through
calorimetry, the
measurement of
heat flow.
Q = mcT
Thermochemistry
Heat Capacity and Specific Heat
• The amount of energy required to raise the
temperature of a sample of substance by 1 K
(1C) is its heat capacity.
• Heat capacity depends on both the mass of the
sample and it’s identity.
• Specific heat capacity (specific heat) is the
amount of energy required to raise the
temperature of 1 g of a substance by 1 K.
• Specific heat capacity ( s or c) is characteristic of
a substance. The higher it is the more energy a
substance requires to heat it. Compare waterThermochemistry
and metal – which would have a higher c?
Heat Capacity and Specific Heat
Specific heat, s (or c), is
Specific heat =
c =
q
m  T
heat transferred
mass  temperature change
This equation can be used
to solve for any of the four
variables!
EXAMPLE A piece of iron metal requires 220
J to raise its temperature from 22C to 40C.
Thermochemistry
What is the mass of the iron?
Specific heat example:
A piece of iron metal requires 220 J to
raise its temperature from 22C to 40C.
What is the mass of the iron?
q
Rearrange
c =
m  T
To solve for m, m =q/(c  T)
= 220 J/[(0.45J/g*C)*(40C-22C)]
= 27g Fe
Thermochemistry
Constant Pressure Calorimetry
We can measure H for a reaction or
dissolving occurring in aqueous
solution using a calorimeter and
calculating with the specific heat
equation:
H =qH2O = m  c  T
You must measure the mass and
temp. change in the lab and use the
specific heat of water (4.184J/g*C)
The heat lost (or gained) by the water is
the heat gained (or lost) by the
chemical system being studied.
Thermochemistry
qH2O = -qrxn
10mL of 1M NaOH solution and 10mL of 1MHCl
solution at 22°C are added to 100mL of water in a
coffee-cup calorimeter at 22°C.
The final temperature of the system is
37°C.
What is qrxn? What is qH2O?
Is this reaction endo or exo? EXPLAIN! How many
moles of HCl were used?
What is the enthalpy change (H ) per mole of HCl?
Thermochemistry
10mL of 1M NaOH solution and 10mL of 1MHCl
solution at 22°C are added to 100mL of water in a
coffee-cup calorimeter at 22°C.
The final temperature of the system is
37°C.
What is qrxn? What is qH2O?
qH2O = m  c  T = 120g X 4.184J/(g*C) * (37-22)C
= 7524J
qrxn = -qH2O = -7524J
Is this reaction endo or exo? EXPLAIN! qrxn Is
negative
How many moles of HCl were used?
1M*0.01L = 0.01moles
What is the enthalpy change (H ) per mole of HCl? -7524J/0.01moles=-752000J/mol = -752kJ
Thermochemistry
Bomb Calorimetry
•
Reactions can be carried out
in a sealed “bomb,” such as this
one, and measure the heat
absorbed by the water.
• Because the volume in the
bomb calorimeter is constant,
what is measured is really the
change in internal energy, E,
not H.
• For most reactions, the
difference between E and H
is very small so we ignore it.
• This is how the calories in food
are determined
Thermochemistry
Hess’s Law
 H is well known for many reactions, and it is
inconvenient to directly measure H for
every reaction in which we are interested.
• However, we can estimate H for any
reaction using H values that are already
known and the additive properties of
enthalpy.
Thermochemistry
Hess’s Law
“If a reaction is carried out in
a series of steps, H for the
overall reaction will be equal
to the sum of the enthalpy
changes for the individual
steps.”
H1 = H2 + H3
What happens when you add
equations 2 and 3 together?
TRY IT!
So H depends only on the
intial reactants and the final
Thermochemistry
products. Not how you get
Using Hess’s Law
IF you are given an overall equation and a set of
equations to arrange so they add up to the
overall equation, you need to reverse them or
multiply them to get the overall equation.
EXAMPLE: Given this information:
2 CO(g)  2 C(s) + O2(g) H1 = 221 kJ
CO(g) + 2 H2(g)  CH3OH(g) H2 = 91 kJ
Find H for 2 C(s) + O2(g) + 4 H2(g)  2 CH3OH(g)
Thermochemistry
To get the final equation, reverse the first
equation and change sign of H1 :
2 C(s) + O2(g)  2 CO(g) -H1 = 221 kJ
Double the second equation and double H2
2[CO(g) + 2 H2(g)  CH3OH(g)] 2 H2 = 2(91 kJ)
2 C(s) + 4 H2(g) + O2(g)  2 CH3OH(g)
 Hrxn = -H1 + 2 H2
Thermochemistry
 Hrxn = 221 kJ + 2(91 kJ) = 403 kJ
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this reaction as occurring in 3 steps which
add up to the overall equation
C3H8 (g)  3 C(graphite) + 4 H2 (g) H1 = + 103.85 kJ
3 C(graphite) + 3 O2 (g)  3 CO2 (g) H2 = -1181kJ
4 H2 (g) + 2 O2 (g)  4 H2O (l) H3 = -1143 kJ
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Then H = H1 + H2 + H3 = 103.85 – 1181 -1143
= -2220
Thermochemistry
Thermochemistry
Enthalpies of Formation
• The enthalpy (or heat) of formation of a substance,
Hf, is the heat released or absorbed when making a
1 mole of a compound or element “from scratch”
• “From scratch” means from its elements in their
naturally occurring elemental forms and phases at
25C.
• By definition the heat of formation of a pure element in
its normal form and phase at 25C is zero (e.g. O2 (g)
and H2 (g) but NOT O (g) or H2 (l))
EXAMPLES OF FORMATION REACTIONS:
For CO -- C(s) + 1/2O2(g)  CO(g), H = 110.5 kJ
For water -- H2 (g) + 1/2O2 (g)  H2O (l) H = -285 kJ
Thermochemistry
What is the formation reaction for C2H6(g)?
Standard Enthalpies of Formation
Standard enthalpies of formation, Hf, are measured
under standard conditions (25°C and 1.00 atm

pressure).
An element in its standard form at these conditions has an
enthalpy of formation of zero.
Watch states carefully when looking up these values! H2O
(g) has a different from H2O(l).
Thermochemistry
Calculation of H using Hf
We can use heat of formation to calculate the
enthalpy change H for any reaction using:
H = nHf(products) - mHf(reactants)
where n and m are the stoichiometric
coefficients. 

Basically you add up all of the heats of
formation of each product times its coefficient
in the equation. Do the same for the
reactants. Subtract the reactant sum from the
product sum and that is H!
Thermochemistry
Calculation of H
What is H for this reaction?
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
What is H for this reaction?
CH4 (g) + 2O2(g)  2 H2O (g) + CO2 (g)
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H= [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0
kJ)]
= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
= (-2323.7 kJ) - (-103.85 kJ)
= -2219.9 kJ
What is H for this reaction?
CH4 (g) + 2O2(g)  2 H2O (g) + CO2 (g)
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H =
=
=
=
[3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]
[(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]
(-2323.7 kJ) - (-103.85 kJ)
-2219.9 kJ
What is H for this reaction?
CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (g)
H= [(-393.5 kJ) + 2(-285.8 kJ)] - [1(-79.8 kJ) + 2(0 kJ)]
= (-965.1 kJ) - (-79.8kJ)
= -885.3kJ
Thermochemistry
Energy in Foods
Most of the fuel in the
food we eat comes
from carbohydrates
and fats.
Thermochemistry
Fuels
The vast majority
of the energy
consumed in this
country comes
from fossil fuels.
Thermochemistry