CHM 1046: General Chemistry and
Qualitative Analysis
Unit 13
Thermochemistry
Dr. Jorge L. Alonso
Miami-Dade College –
Kendall Campus
Miami, FL
Textbook Reference:
•Chapter # 15 (sec. 1-11)
Thermochemistry
•Module # 3 (sec.
I-VIII)
Energy
Potential Energy an object possesses by virtue of
its position or chemical composition (bonds).
2 C8H18 (l) + 25 O2 (g)
Chemical
bond
energy
16 CO2(g) + 18 H2O(g)
Energy (-E):
work + heat
Kinetic Energy an object possesses by virtue of its
motion.
1
KE =  mv2
Thermochemistry
2
Chemical
bond
energy
Energy
2 C8H18 (l) + 25 O2 (g)
 gas=2534= +9
9 x 22.4L/ = 202 L
16 CO2(g) + 18 H2O(g)
H = - 5.5 x 106 kJ/
Energy produced by chemical reactions = heat or work.
 Heat (q): spontaneous transfer (flow) of
energy (energy in transit) from one object
to another; causes molecular motion &
vibrations (kinetic energy); measured as
temperature.
 Work (w): Energy (force) used to cause
an object that has mass to move (a
distance) = kinetic energy. W = F x d
Law of Conservation
of Energy:
E = q + w
Thermochemistry
Units of Energy: Joule & calorie
• The joule (J) = SI unit of energy (work)
Energy (Work) = F x d
Joule (J) = N·m
kg m
=  m
s2
kg m2
= 
s2
The Newton (N) is the amount of force that is required to accelerate
a kilogram of mass at a rate of one meter per second squared.
• The calorie (cal) =
0
heat required to raise 1 g of H2O 1 C
1 cal = 4.184 J
Thermochemistry
{Nutritional Calorie = 1000 calories (1kcal)=4,184 J}.
System and Surroundings
• The system
includes the
molecules we want
to study.
Work
+Ew
-Ework
System
Heat
-Eq(heat)
+Eq
Surroundings
• The surroundings
are everything else
(here, the cylinder
and piston).
E = q + w
Thermochemistry
First Law of Thermodynamics
• Energy is neither created nor destroyed.
• Also known as the Law of
Conservation of Energy
Work
+Ew
-Ew
System
Heat
-Eq
+Eq
• In other words, the total
energy of the universe is a
constant; if the system loses
energy, it must be gained by
the surroundings, and vice
versa.
Esys + Esurr = 0
Esys = -Esurr
or
Thermochemistry
Surroundings
E sys = (q + w)surr
Changes in Internal Energy (E)
E, q, w, and Their Signs : energy released/absorbed as either work or
heat, is looked at from the perspective of the
system (the chemicals)
+q
-q
>q
Syst looses heat
Syst gains heat
+w
Work done
on Syst
>w
-w
Work done by Syst
E = q + w
+ E =
- E =
± E =
+q
-q
±
+w
-w
±
Thermochemistry
Why consider Energy Change (ΔE) and
not the total Internal Energy (E)?
Esys + Esurr = 0
Esys + Esurr = ETotal (constant)
• Total Internal Energy (E) = the sum of all kinetic and
potential energies of all components of the system.
How can it be determined?
• Usually we have no way of knowing the total internal energy of
a system; finding that value is simply too complex a problem.
• But when a change occurs we can measure the change
in internal energy:
Thermochemistry
E = Efinal − Einitial
Energy as Work (w) of Gas Expansion
F = PA
Work = - (Force x distance)
w = - (F x h)
w = - (P x A x h)
w = - P V
(@ constant P)
w = - nRT
(@ constant V)
Thermochemistry
{WorkGasExpansion}
Exchange of Heat (H)
by chemical systems
q
q
- H
• When heat is released by
the system to the
surroundings, the process is
exothermic.
+ H
• When heat is absorbed
by the system from the
surroundings, the
process is endothermic.
Thermochemistry
Enthalpies of Reaction (H)
The change in enthalpy,
H, is the enthalpy of the
products minus the
enthalpy of the reactants:
Reactants:
H = Hproducts − Hreactants
This quantity, H, is called
the enthalpy of reaction, or
the heat of reaction.
Products
Notice it is the same value, but of opposite sign for the reverse reaction
Thermochemistry
State Functions
• Are E, q and w all state functions?
A state function depends only on the present state of the system,
not on the path by which the system arrived at that state.
∆E=
q+w
• However, q and w are not state functions.
• Other state functions are P, V and T.
Thermochemistry
Calorimetry
• The experimental methods of measuring of
heats (H) involved in chemical reactions
1 atm
• Methods utilized:
1. Constant Pressure
Calorimetry (open
container in contact
with atmosphere)
2. Constant Volume
Calorimetry: closed
container (bomb)
Thermochemistry
Thermochemical Equations
2 C8H18 (l) + 25 O2 (g)
16 CO2(g) + 18 H2O(g)
H = - 5.5 x 106 kJ/ C8H18
1. H is per mole of reactant, at the indicated states (s, l, g).
2. Direct quantitative relationship between coefficients of
balanced equation and H.
3. Reverse equation has H of opposite sign.
Problem: Calculate H when 25g of C8H18 (l) (MM= 114) are
burned in excess oxygen?
 1     5.5 x 106 kJ 
6
? kJ  25g C8 H18 
  - 1.2 x 10 kJ

1

 114 g  
Thermochemistry
(1) Calorimetry at Constant Pressure
• Most chemical reactions are carried out in
beakers or flasks that are open to the
atmosphere (i.e., at constant pressure, isobaric).
E = q + w
E = qp - PV
1 atm
SOLVE FOR
q = m  c  T
E + PV = qp
qp = Heat of Reaction = H
or Enthalpy of Reaction
Big
Problem:
calorimeter
also
absorbs
heat
E + PV = H
H = E + PV
Thermochemistry
H = Epv  and when the volume is constant
(a) Determining the Heat
Capacity of a Calorimeter
50 mL of H2O @ 650C
are added…….
50 mL
H2O
50 mL
H2O
…..to 50 mL of H2O
@ 250C inside the
calorimeter
The final temperature of the 100 mL of
H2O inside the calorimeter is 420C
Heat lost by
hot water
=
heat gained by
cold water
qhot = qcold
+
+
heat gained by
calorimeter
q
calorimeter
Thermochemistry
Determining the Heat Capacity
of a Calorimeter (Ccal)
qhot = qcold + q
calorimeter
(mcT)H O = (mcT)H O + (Ccal T)
2
Ccal =
2
(mcT)H O - (mcT)H O
2
2
Includes both
the mass and
heat capacity
(T)cal
Ccal

50g4.184J / g 0C 42  650 C  - 50g 4.184J / g 0C 42  250C 

42  25 C 
0
Ccal

-4811.6J  - (3556.4J)

 492 J / C
0
(17 0 C)
Thermochemistry
(b) Calculating the
Heats of Reactions (H)
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
50 mL@25°C
30°C
Heat of
reaction
+
=
50 mL@25°C

heat gained by
NaCl solution
100 mL of NaCl(aq) @
+
heat gained by
calorimeter
H(qrxn) = qNaCl soln
+ qcal
qrxn = (mc*T)NaCl + (CcalT)
qrxn = - Hrxn
soln
Thermochemistry
= heat of neutralization
rxn
* How do you determine the cNaCl soln?
How do you determine the c
NaCl soln
?
50 mL of H2O @ 650C
are added…….
50 mL
H2O
50 mL
NaCl soln
The final temperature of the 100 mL of
solution inside the calorimeter is 410C
…..to 50 mL of
NaCl solution @
250C inside the
calorimeter
Heat lost by = heat gained by + heat gained by
hot water
salt water soln
calorimeter
qh = q
+ qcal
(mcT)H2O = (mc*T)NaCl + (CcalT)
NaClsoln
soln
Thermochemistry
(2) Calorimetry at Constant
Volume (Bomb Cal.)
E = q + w
w = P V
V =0
w=0
E = qv
q = m  c  T
Thermochemistry
Bomb calorimeter
Bomb Calorimetry
• Because the volume
in the bomb
calorimeter is
constant, what is
measured is really
the change in
internal energy, E,
not H.
E = qv
• For most reactions,
the difference is
very small.
H = E + nRT
Thermochemistry
Review of Thermochemical Equations
E = q + w
E = qp - PV
E = qp - nRT
since ∆H= qp
1 atm
E = H - PV
H = E + PV
E = q + w
E = qv
Thermochemistry
Comparing H and E
H = E + PV
2 C8H18 (l) + 25 O2 (g)
16 CO2(g) + 18 H2O(g)
H = - 5.5 x 106 kJ/
E = H - PV
(useful at const P)
E = H - nRT
(useful at const V)
Problem: calculate E @ 25°C for above equation @ const. V
 gas=34-25 = +9 = 4.5 C8H18
R= 8.314J/K.
E  5.5 x 106 kJ /   (4.5)(8.314 J / K)(1kJ / 103 J )( 298 K )
E  5.5 x 10 kJ /   11kJ /   5.5 x 10 kJ / 
6
6
Thermochemistry
Constant-Volume Calorimetry
Problem: When one mole of CH4 (g) was combusted at 250C in a bomb
calorimeter, 886 kJ of energy were released. What is the enthalpy change for
this reaction?
CH4 (g) + 2 O2 (g)
H = E + PV
= E + nRT
CO2(g) + 2 H2O(l)
E = qv = - 886 kJ
R = 8.31 J/ .K
n = 1 – (1+2) = -2 
 1 kJ 
H   886 kJ  (-2η) (8.31 J/η  K) 
 ( 25  273)K
 1000 J 
ΔH  - 886 kJ - 4.95 kJ   891 kJ
Thermochemistry
Energy in Foods
Most of the fuel in the food we eat comes from carbohydrates & fats.
Thermochemistry
Methods of determining H
1. Calorimetry (experimental)
2. Hess’s Law: using Standard
 )
Enthalpy of Reaction (Hrxn
of a series of reaction steps
(indirect method).
3. Standard Enthalpy of
Formation (Hf ) used with
Hess’s Law (direct method)
4. Bond Energies used with
Hess’s Law
Experimental
data combined
with
theoretical
concepts
Thermochemistry
(2) Determination of H
using Hess’s Law
 Hrxn is well known for many reactions, but it is
inconvenient to measure Hrxn for every reaction.
 However, we can estimate Hrxn for a reaction of
interest by using Hrxn values that are published for
other more common reactions.
 ) of a
 The Standard Enthalpy of Reaction (Hrxn
series of reaction steps are added to lead to reaction
of interest (indirect method).
 Standard conditions (25°C and 1.00 atm pressure). Thermochemistry
(STP for gases T= 0°C)
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.”
- 1840, Germain Henri Hess
(1802–50), Swiss Thermochemistry
Calculation of H by Hess’s Law
3 C(graphite) + 4 H2 (g)  C3H8 (g)
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
H= -104
H= +104
H=-1181
H=-1143
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
•
Appropriate set of Equations with their H values are obtained (or given), which
containing chemicals in common with equation whose H is desired.
•
These Equations are all added to give you the desired equation.
•
These Equations may be reversed to give you the desired results (changing the sign
of H).
•
You may have to multiply the equations by a factor that makes them balanced in
relation to each other.
Thermochemistry
•
Elimination of common terms that appear on both sides of the equation .
Calculation of H by Hess’s Law
C3H8 (g)  3 C(graphite) + 4 H2 (g) H= +104
3 C(graphite) + 3 O2 (g)  3 CO2 (g) H=-1181
4 H2 (g) + 2 O2 (g)  4 H2O (l)
H=-1143
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Hrxn =
+ 104 kJ
-1181 kJ
- 1143 kJ
- 2220 kJ
Thermochemistry
Calculation of H by Hess’s Law
Calculate heat of reaction
W + C (graphite)  WC (s)
Given data:
2 W(s) + 3 O2 (g)  2 WO3 (s)
½(2 W(s) + 3 O2 (g)  2 WO3 (s) )
W(s) + 3/2 O2 (g)  WO3 (s) )
ΔH = ?
ΔH = -1680.6 kJ
½(ΔH = -1680.6 kJ)
ΔH = -840.3 kJ
C (graphite) + O2 (g)  CO2 (g)
C (graphite) + O2 (g)  CO2 (g)
ΔH = -393.5 kJ
ΔH = -393.5 kJ
2 WC (s) + 5 O2 (g)  2 WO3 (s) + CO2 (g)
ΔH = -2391.6 kJ
½(2 WO3 (s) + CO2 (g)  2 WC (s) + 5 O2 (g))
WO3 (s) + CO2 (g)  WC (s) + 5/2 O2 (g)
W + C (graphite)  WC (s)
½ (ΔH = +2391.6 kJ)
ΔH = + 1195.8 kJ)
ΔH = - 38.0
Thermochemistry
Hess’s Law
Problem: Chloroform, CHCl3, is formed by the following reaction:
Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g)
Determine the enthalpy change for this reaction (ΔH°rxn), using
the following:
2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g)ΔH°f = – 103.1 kJ/mol
CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g)
ΔH°rxn = – 890.4 kJ/mol
2 HCl (g) → H2 (g) + Cl2 (g)
ΔH°rxn = + 184.6 kJ/mol
C (graphite) + O2 (g) → CO2(g)
ΔH°rxn = – 393.5 kJ/mol
H2 (g) + ½ O2 (g) → H2O (l)
ΔH°rxn = – 285.8 kJ/mol
answers: a) –103.1 kJ
d) + 305.2 kJ
b) + 145.4 kJ
e) – 305.2 kJ
c) – 145.4
kJ
Thermochemistry
f) +103.1 kJ
Methods of determining H
1. Calorimetry (experimental)
2. Hess’s Law: using Standard
 ) of
Enthalpy of Reaction (Hrxn
a series of reaction steps
(indirect method).
3. Standard Enthalpy of
Formation (Hf) used with
Hess’s Law (direct method)
4. Bond Energies used with
Hess’s Law
Experimental
data combined
with
theoretical
concepts
Thermochemistry
(3) Determination of H using
Standard Enthalpies of Formation (Hf)
Enthalpy of formation, Hf, is defined as the enthalpy change for the
reaction in which a compound is made from its constituent elements in their
elemental forms.
∆Hf= -393.5 kJ/ 
Standard Enthalpy of formation Hf are measured under standard
C + O2  CO2
conditions (25°C and 1.00 atm pressure).
Thermochemistry
Calculation of H
CH4(g) + O2(g)  CO2(g) + H2O(g)
C + 2H2(g)  CH4(g)
C(g) + O2(g)  CO2(g)
2H2(g) + O2(g)  2H2O(g)
ΔHf = -74.8 kJ/ŋ
ΔHf = -393.5 kJ/ŋ
ΔHf = -241.8 kJ/ŋ
We can use Hess’s law in this way:


H = nHf(products)
- mHf(reactants)
where n and m are the stoichiometric coefficients.
n CO2(g) + n H2O(g)
-
n CH4(g) + n O2(g)
H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)]
= - 560.5 kJ
Thermochemistry
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = nHf(products) - mHf(reactants)
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
Table of Standard Enthalpy of formation, Hf
Thermochemistry
(4) Determination of H
using Bond Energies
• Most simply, the strength of a bond is
measured by determining how much energy
is required to break the bond.
• This is the bond enthalpy.
• The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is measured to be 242 kJ/mol.
Thermochemistry
Average Bond Enthalpies (H)
• Average bond enthalpies are positive, because
bond breaking is an endothermic process.
NOTE: These are
average bond
enthalpies, not
absolute bond
enthalpies; the
C—H bonds in
methane, CH4,
will be a bit
different than
the
C—H bond in
chloroform,
CHCl3.
Thermochemistry
Enthalpies of Reaction (H )
• Yet another way to
estimate H for a
reaction is to compare
the bond enthalpies of
bonds broken to the
bond enthalpies of the
new bonds formed.
• In other words,
Hrxn = (bond enthalpies of bonds broken) 
(bond enthalpies of bonds formed)
Thermochemistry
Hess’s Law:
Hrxn = (bonds broken)  (bonds formed)
CH4(g) +
Cl2(g) 
CH3Cl(g) +
HCl(g)
Hrxn = [D(C—H) + D(Cl—Cl)  [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)]  [(328 kJ) + (431 kJ)]
= (655 kJ)  (759 kJ)
= 104 kJ
Thermochemistry
Bond Enthalpy and Bond Length
• We can also measure an average bond
length for different bond types.
• As the number of bonds between two atoms
Thermochemistry
increases, the bond length decreases.
2003 B Q3
Thermochemistry
Thermochemistry
2005 B
Thermochemistry
Thermochemistry
2002
Thermochemistry
Thermochemistry
Thermochemistry
2002 #8
Thermochemistry
Thermochemistry
2003 A
Thermochemistry