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.
5.1
The Nature
of Energy
Thermochemistry
• The study of energy transformations
involving heat (either absorbed or
emitted) during chemical rxns
Examples:
• Combustion of fossil fuels (like
gasoline)
• Metabolism of food  conversion
to ATP = energy for our body
Vocab: Energy, Work , & Heat
• Energy: The ability to do work or
transfer heat.
Work: energy used to move an
object that has mass.
Heat: energy used which causes the
temperature of an object to rise.
* Energy must be transferred between
matter for work to be done or to
cause a change in temp.
How can matter possess energy?
• Kinetic Energy (KE): energy of motion.
KE depends on mass & velocity
All matter (particles/atoms/molecules)
has mass and is in constant motion,
therefore it possesses KE
Since the motion of particles is directly
related to temperature, thermal energy
(energy a substance possesses because
of its temperature) is a type of KE
How can matter possess energy?
• Potential Energy (PE): in chemistry…
energy based on chemical composition.
The chemical energy of a substance is a type of
PE that depends on the numbers & kinds of
atoms bonded in a substance (its chemical
composition!)
• Chemical energy is based on the strength of
attractive/repulsive forces between bonded atoms
• The weaker a bond, the more stored chemical
potential energy it contains.
• Weak bonds are unstable = high potential energy.
• Strong bonds are stable = low potential energy.
Units of Energy
• The SI unit of energy is the joule (J).
kg•m2
1 J = 1 
s2
• A joule is not a lot of energy, so
kilojoules (kJ) are often used
(1 kJ= 1000 J)
Units of Energy
• An older, non-SI unit still in widespread
use is the calorie (cal)
1 cal = 4.184 J (exactly)
• In food, the energy given off from a
substance during digestion is a nutritional
Calorie (Cal)
1 Cal = 1000 cal = 1 kcal
System and Surroundings
• The system (sys) includes the molecules we
want to study (in rxns, the reactants &
products).
• The surroundings (surr) are everything else
(the glassware, containers, people, air,
everything outside of the system).
• System + Surroundings = Universe
System and Surroundings: Example
2 H2 (g) + O2 (g)  2 H2O (g) + energy
• Sys = H2, O2 & H2O
• Surr = glassware, people, air, etc…
Due to the Law of Conservation of
Mass, no MATTER ever enters or
leaves the system once a chemical rxn
begins
ENERGY, however, can be transferred
between the system and surroundings
as either work or heat
Transferring Energy: Work & Heat
• Work: energy used to move an object over
some distance
w = (force)(distance)
• In chem, the most common form of work is
called Pressure-Volume work (P-V work)
• Comes into play when dealing with
gases because they can cause
pressure inside a system
• This pressure is a FORCE which can
move a piston some DISTANCE =
work 
Example
of P-V
Work
Transferring Energy: Work & Heat
• Heat (2 definitions):
(1) energy transfer between sys. and surr.
as a result of differences in temperature
 Temp = quantitative measure of how
much heat is in a sys at a given time
(2) energy which is transferred from
hotter objects to colder objects.
***Heat ALWAYS flows from
hotter  colder ***
5.2
st
The 1 Law of
Thermodynamics
First Law of Thermodynamics
• Energy is neither created nor destroyed
(aka: energy is conserved)
• The total energy of the universe is
constant:
if the sys loses energy, it must be
gained by the surr, and vice versa.
• Energy can be…
converted from one form to another
transferred between sys. and surr.
Internal Energy: energy of a system
• Internal energy (E) of a system is the
sum of all KE and PE of all its
components
Generally, the actual value of E for a
sys. is not known
What can be determined is the
change in internal energy (E) of
the sys.
Changes in Internal Energy
• E is concerned with
ANY energy transferred
between sys & surr
Includes both heat
(q) or work (w).
• E = q + w
(equation comes from 1st
Law of Thermodynamics)
Changes in Internal Energy
• E = q + w
• Like a bank
account, E of a
system increases
when deposits
are made to the
system!
(See Table 5.1)
E, q, w, and their signs
Table 5.1 - pg. 174 (top)
+
–
For q
+ means system
gains heat
- means system
loses heat
For w
+ means work done
on system
- means work done
by the system
For E
+ means net gain of
energy by system
- means net loss of
energy by system
•E has 3 parts:
a number value, a unit, and sign (+/-)
E Calculation Practice
• If a system absorbs 140 J of heat from
the surroundings and does 85 cal of
work on the surroundings what is the
change in internal energy?
•
•
•
•
E = q + w
E = (+140 J) + (-85 cal)
E = (+140 J) + (-355.6 J)
E = -215.6 J
85 cal 4.184 J
1 cal
= 355.6 J
(a net of 215.6 J is lost by the sys.)
Thermochemistry
Vocab: State Functions
• Internal energy (E) is an example of a
state function:
It is determined only by the system’s
current condition or state, NOT by the
path taken to get to there
• Analogy:
Altitude
state
function
vs.
Distance
not a state
function
Internal Energy is a State Function
• At any given time, a system has a specific
amount of internal energy and it does not
matter HOW (by what means) it is achieved
system
State Functions
• E = q + w
Heat and work are NOT state functions
(which path energy takes, q or w, is
important)
 However, E is a
state function
 Therefore, E only
depends on Efinal
and Einital
Another way to determine E
• Another way to determine E is to compare
a system’s initial and final energy states:
E = Efinal − Einitial
• Generally, in chemical rxn:
initial state = reactants
final state = products
• Energy
(aka: enthalpy)
diagrams
 Compare the energy
of reactants (initial
state) with the
energy of the
products (final state)
in a chemical rxn
 – E : when
reactants release
heat to surroundings
(system loses
energy)
EXOthermic (–E ):
• a process in which a system releases
heat to its surroundings
Heat flows OUT of system and into
the surrounding
–system becomes cooler as heat
EXITS
–surroundings become warmer
* every change in the sys. has an opposite change in the surr.
• Energy
(aka: enthalpy)
diagrams
 + E : when
reactants must
absorb heat from
surroundings
(system gains
energy)
ENDOthermic (+E ):
• a process in which a system absorbs
heat from its surroundings
Heat flows from the surroundings
INTO the system
–system gets warmer as heat
ENTERS
–surroundings become cooler
5.3
Enthalpy
Enthalpy
• Enthalpy (H) is a measure of the total internal
energy of a system at constant pressure
This means that essentially NO work is done
ALL energy transferred is in the form of
HEAT
• Again, H is difficult to determine so, instead, the
change in enthalpy (ΔH) of a system is found
+ ΔH = endothermic ; – ΔH = exothermic
• Enthalpy = state function (indep. of path)
Thermochemistry
5.4
Enthalpies of
Reaction
Thermochemical Equations
• A balanced chemical equation that also
includes the enthalpy change
associated with the rxn
• There are 2 different ways to represent
a thermochem equation:
Energy is either a reactant or product in
the rxn
 ΔH is listed at the end of the rxn
Exothermic Chemical Equations
• Heat is released by the system,
therefore it is a PRODUCT of the rxn
 2 H2 (g) + O2 (g)  2 H2O (g) + energy
 2 H2 (g) + O2 (g)  2 H2O (g) + 483.6 kJ
• Exothermic = negative ΔH
 2 H2 (g) + O2 (g)  2 H2O (g) H = - 483.6 kJ
Endothermic Chemical Equations
• Heat is absorbed by the system,
therefore it is a REACTANT in the rxn
(in other words, heat is needed for the
reaction to proceed)
 2 N2 (g) + O2 (g) + energy  2 N2O (g)
 2 N2 (g) + O2 (g) + 66.4 kJ 2 N2O (g)
• Endothermic = positive ΔH
 2 N2 (g) + O2 (g)  2 N2O (g) H = + 66.4 kJ
Enthalpies of Reaction
• The change in enthalpy (H ) associated
with a chemical rxn is known as the
enthalpy of reaction or heat of reaction
• The Hrxn can be determined by looking at
the change in enthalpy between the
reactants and products
Hrxn = Hproducts − Hreactants
Exothermic Enthalpy Diagrams
• In an exothermic rxn, H = negative
therefore:
 The products must have less
enthalpy than the reactants
 Hproducts − Hreactants = – H
 In other
words, the rxn
loses energy
as time goes
on
Endothermic Enthalpy Diagrams
• In an endothermic rxn, H = positive
therefore:
 The products must have more enthalpy
than the reactants
 Hproducts − Hreactants = + H
 In other
words, the rxn
gains energy
as time goes
on
Guidelines for using thermochemical
equations and enthalpy diagrams:
1) Enthalpy is an extensive property:
 this means the magnitude of ∆H is
directly proportional to the
AMOUNT of reactant used
 the more reactant used, the more
exo or endothermic the rxn will be
 The EXACT ∆Hrxn can be
calculated using some of
thermochem stoichiometry!
Thermochem. Eq. & Calculations
• Hydrogen Peroxide can decompose to water and
oxygen by the following rxn:
2 H2O2 (l)  2 H2O (l) + O2 (g) + 196 kJ
• Is this rxn endo or exothermic?
• How much heat is released when 5.00 g H2O2
decomposes at constant pressure?
5.00 g H2O2 1 mol H2O2
-196 kJ
= -14.4 kJ
34.02 g H2O2 2 mol H2O2
TWO
Demonstrations
& Example
Calculations!!!
Thermochemistry
5.5
Calorimetry
What Affects Temperature Change?
• When heated, any object will
experience a rise in temperature…
• …however, the magnitude of the
temperature change depends on two
things:
(1) The identity of the substance heated
• Ex) rubber slide vs. metal slide @ park
(2) The mass being heated
• Ex) boiling 1 cup H2O vs. 1 gallon H2O
Heat Capacity
• Definition: the amount of energy
required to raise the temperature of
a substance by 1 K (1C)
This is dependent upon the mass
and identity of the entire object
(and how much heat is require to
change it by 1C)
Ex) puddle on the street
vs. swimming pool
Specific Heat Capacity
(Cp) or (s)
• Definition: the amount of energy
required to raise the temperature of
1 gram of a substance by 1 K (1C)
This removes mass from the
equation so that ONLY the identity
of the substance matters
Cp units = J/gC
Ex) stainless steel = 0.50 J/gC
copper = 0.385 J/gC
Specific Heat of Water
4.184 J/g  C
or
1 cal/g  C
• Must be MEMORIZED!!!
• The Cp value you use depends on if
heat is in cal or J units
• Note: The Cp for other substances
will be included in problems as
needed
Specific Heat Calculations
q
OR
q  m c  T
c  
m T
p
p
Symbol
Meaning
Unit
Cp
Specific heat
J/g•K or cal/g•K
[could be K or ºC]
q
Heat lost (-) or gained (+)
J or cal
ΔT
Temperature change
[ Tfinal – Tinitial ]
K or ºC
m
mass
g
Calorimetry
• Calorimetry is the experimental
measurement of the quantity of heat
transferred during a chemical or
physical rxn
Heat transfer is measured based on
a change in temperature (ΔT) when
two substances are in contact
Calorimetry
• Calorimetry must be done using an
insulated container known as a
calorimeter (the chemical or
physical rxn takes place inside of
this)
The calorimeter’s job is to
conserve all the heat so that none
is transferred OUTSIDE of the
calorimeter
Coffee- Cup
Calorimetry
How does calorimetry work?
• Two substances (at different temperatures)
are placed together in a calorimeter…
 When in contact with one another, both
substances will seek thermal equilibrium
• Heat flows from hotter  colder until equal
both are at the SAME final temperature (Tf)
Due to the insulation, NO heat leaves the
calorimeter… it is only transferred
• Therefore, ALL the heat lost by one
substance is gained by the other
– q (lost) = + q (gained)
Calorimetry Calculations
 qlost  + qgained
 Cp  m T    Cp  m T 
 Cp  m   Tfinal  Tinitial     Cp  m   Tf  Ti  




Signs of T and q:
• + q = heat is absorbed/gained when …
Substance getting hotter
 (10º30º) then T = 30º –10º = + 20º
• – q = heat is lost/released when…
Substance getting cooler
(75º50º) then T = 50º –75º = - 25º
Calorimetry Calculation Examples
A 46.2 g sample of copper is heated to 95.4ºC
and then placed in a calorimeter containing
75.0 g of water at 19.6ºC. The final
temperature of both the water and the copper
is 21.8ºC. Given that the specific heat of
water is 4.184 J/gºC, what is the specific
heat of copper?
Cu  temp decreases … heat is released
H2O  temp increases … heat is absorbed
Calorimetry Calculation Examples
- qCu = qwater
- (mCpΔT)Cu = (mCpΔT)water
-(46.2 g)(CpCu)(21.8ºC-95.4ºC) =
(75.0 g)(4.184 J/gºC )(21.8ºC-19.6ºC)
(CpCu)(3400.3 gºC) = 690.4 J
Cp (copper) = 0.203 J/gºC
Calorimetry Calculation Examples
Suppose we mix 90.0 grams of hot water (90.0ºC)
with 10.0 grams of cold water (10.0ºC) inside a
calorimeter. What is the final temperature, TF , of
the system?
Whot  temp decreases … heat is released
Wcold  temp increases … heat is absorbed
Calorimetry Calculation Examples
- qhot = qcold
- (mCpΔT)hot = (mCpΔT)cold
- (90.0 g)(4.184 J/gºC )(TF - 90.0ºC) =
(10.0 g)(4.184 J/gºC )(TF -10ºC)
- (376.56)(TF - 90.0ºC) = (41.84)(TF - 10.0ºC)
- (376.56)(TF ) + (33,890.4) = (41.84)(TF) - (418.4)
34,308.8 = (418.4)(TF)
82ºC = TF
Thermochemistry
5.6
Hess’s Law
Some features of
thermochemical equations:
1) If any reactants or products change their state
of matter, ∆H changes as well.
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ
• CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ
Some features of
thermochemical equations:
2) If you reverse a thermochemical reaction,
the sign of ∆H must also be reversed.
• Forward Rxn:
H2 (g) + Cl2 (g)  2 HCl (g) ∆H = -183 kJ
• Reverse Rxn:
2 HCl (g)  H2 (g) + Cl2 (g) ∆H = +183 kJ
Some features of
thermochemical equations:
3) Because enthalpy is an extensive property,
multiplying a thermochemical equation by a
constant also multiplies the ∆H by that
constant
• H2 (g) + Cl2 (g)  2 HCl (g) ∆H = -183 kJ
• 2 H2 (g) + 2 Cl2 (g)  4 HCl (g) ∆H = -366 kJ
• ½ H2 (g) + ½ Cl2 (g)  HCl (g) ∆H = -91.5 kJ
Some features of
thermochemical equations:
Basically…
WHATEVER you
do to chemical
equation, you must
also do to ΔH!!!
Hess’s Law
• Some rxns cannot be performed in a
lab (very unstable, expensive, etc.)
but ΔH can still be determined…
• Because H is a state function
(independent of path), the total H of
a rxn can be determined by
combining the H for multiple
chemical rxns (which are easier/more
convenient to perform)
Hess’s Law
• Hess’s Law states
that “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 all the
individual steps.”
Hess’s Law: Basic Idea
• Say we need the ΔH for the rxn:
A+RK+T
ΔHrxn = ?
• We know the following:
Rxn #1) A + B  C + D
Rxn #2) G + C  K + B
Rxn #3) D + R  G + T
ΔH1 = x kJ
ΔH1 = y kJ
ΔH1 = z kJ
A + R  K + T ΔHrxn = x + y + z kJ
Thermochemistry
5.7
Enthalpies of
Formation
Types of Enthalpy Processes
• Using Hess’s Law, ΔH for many types of
processes can be calculated:
Enthalpy of vaporization (ΔHvap)
• converting liquid to gas
Enthalpy of fusion (ΔHfus)
• melting a solid into a liquid
Enthalpy of combustion (ΔHc)
• Burning a substance in air
Enthalpy of formation (ΔHf)
• When a compound has been formed from its
component elements
Standard Enthalpy of Formation (ΔHfº)
• the enthalpy change when one mole of a
compound is formed from its elements in
their standard states (at 298K/25ºC and
atmospheric pressure)
• ΔHfº for the MOST STABLE form of any
element is zero:
 Metals  solids (except for Mercury)
 Diatomics (BrINClHOF)  gases [except
for Bromine (l) and Iodine (s)]
Carbon  graphite solid is most stable form
Standard Enthalpies of Formation
The ΔHfº (in kJ/mol) for many elements/
compounds are listed in your book in
Appendix C – pg. 1123
See In-Class Examples WS
Standard Enthalpy of Formation, ΔHfº
• Example 1a)
Write the standard enthalpy of formation
equation for ethanol, C2H5OH (l)
* The rxn must be for the formation of
only ONE mole of the compound*
____________________________________________
2 C(graphite) + 3 H2(g) + ½ O2(g)  C2H5OH (l)
ΔHfº = -277.7 kJ
Standard Enthalpy of Formation, ΔHfº
• Write the standard enthalpy of formation
equation for each of the following compounds:
• Example 1b)
1/
NH3 (g)
3/ H (g)  NH (g)
N
(g)
+
2 2
2 2
3
• Example 1c)
ΔHfº = -46.19 kJ/mol
NaHCO3 (s)
Na(s) + 1/2 H2(g) + Cgraphite(s) + 3/2 O2(g)  NaHCO3(s)
ΔHfº = -947.7 kJ/mol
Using Enthalpies of Formation
(Hfº) to Calculate Enthalpies of
Reaction (Hrxnº)
• The Hrxn for any chemical
reaction can be determined by
combining the enthalpy of
formation reactions(Hfº) using
Hess’s Law.
Calculation of Hrxn using ΔHfº
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
• Using Hess’s Law, write the standard
formations of each reactant and product in the
above rxn and sum the ΔHfº values to
determine the Hrxnº
3 C(graphite) + 4 H2 (g)  C3H8 (g)
ΔHfº = -103.85 kJ/mol
O2 (g) is in its elemental form
ΔHfº = 0 kJ/mol
C(graphite) + O2 (g)  CO2 (g)
ΔHfº = -393.5 kJ/mol
H2 (g) + 1/2 O2 (g)  H2O (g)
ΔHfº = -241.82 kJ/mol
Using Hess’s Law…
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
C3H8 (g)  3 C(graphite) + 4 H2 (g)
ΔHfº = +103.85 kJ
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
ΔHfº = -1180.5 kJ
4 H2 (g) + 2 O2 (g)  4 H2O (g)
ΔHfº = -967.28 kJ
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) ΔHºrxn = - 2043.93
kJ
Calculation of Hºrxn
• Because enthalpy is a state function Hºrxn only
depends on the initial and final states of the
chemical rxn.
 Reactants = initial state
 Products = final state
• Therefore, the standard enthalpy of rxn can be
determined using Appendix C & the equation:
Hºrxn = [nHfº(products)] – [nHfº(reactants)]
* n = # of moles reactant & product (coefficients)