CHE 106
CHE 106: General Chemistry
CHAPTER
FIVE
Copyright © James T. Spencer 1995 - 1999
All Rights Reserved
Chem 106, Prof. J.T. Spencer
1
CHE 106
Chapter Five

ENERGY RELATIONSHIPS
Thermodynamics - The
study of energy and its
transformation.
 Thermochemistry - The
relationships between
chemical reactions and
energy changes.

Chapt. 4.1
Chem 106, Prof. J.T. Spencer
2
CHE 106
Energy
3
Force - “push” or “pull” exerted on
an object (i.e., gravitational force).
 Work - moving a body through a
distance by the application of a force;
Work = Force x distance
W = Fd
 Heat - energy transferred between
bodies of different temperatures.
 Energy - is the capacity to do work or
transfer heat. Also objects can possess
energy because of their motion or
position.

Chapt. 4.1
Chem 106, Prof. J.T. Spencer
CHE 106
Energy

Kinetic Energy - energy of motion
Ek = 1/2 mv2

Ek = kinetic energy
m = mass
v = velocity
Potential Energy “stored”
Potential Energy
energy of
position
Kinetic Energy
Chapt. 4.1
Chem 106, Prof. J.T. Spencer
4
CHE 106
Kinetic and Potential Energy
KINETIC
POTENTIAL
KINETIC
+ DIEHARD
Chem 106, Prof. J.T. Spencer
5
CHE 106
Kinetic and Potential Energy
KINETIC
POTENTIAL
KINETIC
Chem 106, Prof. J.T. Spencer
6
CHE 106
Energy

7
Potential Energy - positional
– electrostatic (electron and proton)
– gravitational (brick on desk)
– chemical (battery or chemical bonding)
– thermal (random motion of molecules)
» a substance is hotter if the average kinetic
energy of its component molecules is
greater.
» energy flows from higher to lower energy
(across a gradient)
Chem 106, Prof. J.T. Spencer
CHE 106
Energy Units


8
SI Unit - joule (J)
– 1 joule = 1 kg m2 s-2
– moving a mass of 2 Kg at 1 m/s has 1 J ok Ek
– relatively small compared with chemical
reactions so the kilojoule (kJ) is used
[1 kJ = 1000 J]
Older (non-SI) units - calorie (cal) - the amount of
energy required to raise the temperature of 1 g of
water by 1° C
– 1 cal = 4.184 J
– 1 kilocalorie (kcal) = 1000 cal
– NOTE the nutritional Calorie (capitalized) is:
1 Cal = 1000 cal = 1 kcal.
Chem 106, Prof. J.T. Spencer
CHE 106
James Prescott Joule
9
•Waterfall sparked his interest;
water should be warmer at bottom
than at top due to loss of potential
energy (attempts to measure
temperature differences failed)
•Invented accurate and reliable
thermometers
•Determined quantitative
relationship between work and heat
•Creditied with general statment
and proof of conservation of energy
established through 40 years of
extensive experimentation includes
not only mechanical energy,added
heat, electrical energy and chemical
energy to work/energy equation
Chem 106, Prof. J.T. Spencer
CHE 106
Energy Transfers



10
The item under consideration or study is called
the system.
All other items (not directly under study) are the
surroundings.
In chemistry - the reagents are ususally
considered the system and everything else (the
flask, etc...) is the surroundings.
A, B and C are the system.
The piston and everything
else are the surroundings.
No mass is lost from the
reactor. Only energy is
transferred between the
system and the surroundings
A(g) + B(g)
C(g)
piston
reactant
gases
Chem 106, Prof. J.T. Spencer
CHE 106
Energy Transfers vs. Money Transfers
11
Energy Transfer
Money Transfer
system
Reactants
Your
Checkbook
surroundings
solution, flask
air, etc...
Pizza Delivery
SU bookstore
Phone Co.,
Parents, etc...
Chem 106, Prof. J.T. Spencer
CHE 106
Energy and the First Law



12
Systems tend to move toward the lowest
possible energy states.
First Law of Thermodynamics - Energy is neither
created nor destroyed but is conserved (despite
conversions).
Internal Energy - Total energy of a system (sum
of both kinetic and potential energies).
– very difficult to measure total energy but
CHANGES in energy can be accurately
measured.
– change in internal energy (E) is the difference
in energy between the final and initial states;
E = Efinal - Einitial
Chem 106, Prof. J.T. Spencer
CHE 106
Internal Energy Changes (E)

Internal Energy Changes;
– Value (numerical portion)
– Unit (J, cal, etc...)
E =
– sign (positive or negartve)
» E is negative when Efinal < Einitial
» E is potitive when Efinal > Einitial
13
Ef - Ei
» In chemical reactions, the initial state refers to
the energy of the reactants and final state
refers to the energy of the products.
» neg E means energy is lost to the surroundings (exothermic) and pos. E means energy
is gained from the surroundings (endothermic).
Chem 106, Prof. J.T. Spencer
CHE 106
Reaction Profiles

14
In an exothermic chemical reaction, the
potential energy stored in the chemical
bonds is converted to thermal energy (random
kinetic energy) as heat.
EXOTHERMIC
ENDOTHERMIC
2 NO
CH4 + O2
PE
PE
H2O + CO2
PE
CH4 + O2
H2O + CO2
N2 + O2
N2 + O2
Chem 106, Prof. J.T. Spencer
2 NO
CHE 106
E, Heat and Work

E can be written as a function of the heat added
to or removed from the system and the work
done on or by the system.
E = q + w

15
q = heat
w = work
Sign conventions! – heat ADDED TO the system is POSITIVE
– heat REMOVED FROM the system is Negative
– work DONE ON the system is POSITIVE
– work DONE BY the system is NEGATIVE
[just remember whose giving and receiving if you’re the system]
Chem 106, Prof. J.T. Spencer
CHE 106
Energy Transfers vs. Money Transfers
Energy Transfers
system
Reactants
16
Money Transfers
Your
Checkbook
Heat to Rx +
Money from Home +
Heat to Solution - Pay Pizza, etc... surroundings
solution, flask
air, etc...
Pizza Delivery
SU bookstore
Phone Co., etc...
Chem 106, Prof. J.T. Spencer
CHE 106
Sign Conventions: Two Cases
HEAT
17
HEAT
WORK
SYSTEM
q = positive (+)
w = positive (+)
E = (+) endothermic
WORK
SYSTEM
q = negative (-)
w = negative (-)
E = (-) exothermic
Chem 106, Prof. J.T. Spencer
CHE 106
Internal Energy Change (E)
18
Example Problem
Calculate the Internal Energy Change (E)
for the following situations.
 (A) A system does 195 kJ of PV work and
absorbs 38 J of heat.
» E = q + w
» E = (38 J) + (-195 kJ)(1000 J)
(1 kJ)
» E = -194,962 J
E = -195 kJ
w = -195 kJ
q = +38 J
Chem 106, Prof. J.T. Spencer
CHE 106
Internal Energy Change (E)
19
Example Problem
Calculate the Internal Energy Change (E)
for the following situations.
 (B) A chemical reaction in a piston chamber
gives off 500 J of heat to its surroundings. The
expanding gas moves the piston upward and
does 240J of work.
w = -240 J
» E = q + w
q = -500 J
» E = (-500 J) + (-240 J)
» E = -740 J
piston
reactant
gases
Chem 106, Prof. J.T. Spencer
CHE 106
Internal Energy Change (E)
20
Example Problem
Calculate the Internal Energy Change (E)
for the following situations.
 (C) The chemical reaction in part (B) was rerun with the piston in a locked position and
the reaction was found to generate 740 J of heat.
» E = q + w
» E = (-740 J) + (0 J)
» E = -740 J
w=0J
q = -740 J
piston
reactant
gases
Chem 106, Prof. J.T. Spencer
CHE 106
Internal Energy Change (E)
Sample exercise: Calculate the
change in the internal energy of the
system for a process in which the
system absorbs 140 J of heat from
the surroundings and does 85 J of
work on the surroundings.
Chem 106, Prof. J.T. Spencer
21
CHE 106
Internal Energy Change (E)
Sample exercise: Calculate the
change in the internal energy of the
system for a process in which the
system absorbs 140 J of heat from
the surroundings and does 85 J of
work on the surroundings.
E = q + w
Chem 106, Prof. J.T. Spencer
22
CHE 106
Internal Energy Change (E)
Sample exercise: Calculate the
change in the internal energy of the
system for a process in which the
system absorbs 140 J of heat from
the surroundings and does 85 J of
work on the surroundings.
E = q + w
140 J
85 J
Chem 106, Prof. J.T. Spencer
23
CHE 106
Internal Energy Change (E)
Sample exercise: Calculate the
change in the internal energy of the
system for a process in which the
system absorbs 140 J of heat from
the surroundings and does 85 J of
work on the surroundings.
E = q + w
140 + (-85)
140 J
85 J
Chem 106, Prof. J.T. Spencer
24
CHE 106
Internal Energy Change (E)
Sample exercise: Calculate the
change in the internal energy of the
system for a process in which the
system absorbs 140 J of heat from
the surroundings and does 85 J of
work on the surroundings.
E = q + w
140 + (-85)
140 J
+55 J
85 J
Chem 106, Prof. J.T. Spencer
25
CHE 106
State Functions

26
State Function - a property of a system that
depends only on its condition or state. The value
of a state function does not depend on the
“history” of a sample, only its present condition.
altitude is pathway
independent (state) while your
work and distance travelled
are dependent on pathway
(NOT state functions)
altitude
Chem 106, Prof. J.T. Spencer
CHE 106
State Functions

27
Example
50 g H20 at
100° C
50 g H20 at
50° C
50 g H20 at
0° C
The internal Energy (E) is the same REGARDLESS
of the pathway therefore it is a state function.
Chem 106, Prof. J.T. Spencer
CHE 106
State Functions

28
Work and heat are not state functions while E is
a state function
The flashlight gives off only heat and
radiant energy (light) while the car gives
off heat and does work. E, however, is
the same regardless of pathway.
A
+ -
A
B
Charged Battery
Heat
DIEHARD
B
Heat
+
radiant
energy
E
Work
Discharged Battery
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy


29
Most chemical reactions occur under constant
pressure (atmospheric) and involve very little
work relative to heat. Define ENTHALPY (H) as
heat absorbed or released under constatnt
pressure.
Relationship between E and H;
E = H - PV

Consider;
(1) E = q + w and,
(2) work only associated with the expansion
of a gas
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy; Const. P work
30
P = F/Area
piston
piston
gases
h
P = 760 Torr (const.)
w = F x (distance)
w = F x h
F = P x Area
w = P x Area x h
V = Area x h then;
w = - P x V
(neg because work
given off to surroundings)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy and Internal Energy
31
is equivalent to heat content and H is the
heat lost or gained at constant pressure.
Enthalpy
Thus;
w = - PV
then at const. pressure,
E = qp + w
E = qp + PV
For many
reactions, the
volume change is
near zero making
PV small. In
these cases, E is
approximately H
H = qp
E = H - PV
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpic Changes
HEAT
SYSTEM
H = (+) endothermic
HEAT
SYSTEM
H = (-) exothermic
Chem 106, Prof. J.T. Spencer
32
CHE 106
Enthalpy:Exothermic Reactions

Since ENTHALPY (H) is;
H = E + PV
or
H = E + PV

33
and E, P and V
are all state
functions, then
H is a state
function
H
Hrxn = Hfinal - Hinitial
EXOTHERMIC
CH4 + O2
H
H2O + CO2
CH4 + O2
H2O + CO2
[RXN = reaction]
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy:Exothermic Reactions

34
Reaction of Zinc with Sulfur:
– Complex Reaction Sequences and Pathways:
Zn(s) + S(s)
ZnS(s) H = -206.0 kJmol-1
additional reactions
Zn(s) + 1/2 O2
ZnO(s) H = -348.3 kJmol-1
S(s) + O2
SO2(g) H = -297.0 kJmol-1
– The reaction requires a thermal initiation
(activation energy to begin), however, once the
reaction begins it is self-sustaining (exothermic)
[Video No. 23; 0:44 m]
Chem 106, Prof. J.T. Spencer
CHE 106
Activation Energy
35
PAGE 503-505 of TEXTBOOK

Even though a reaction is exothermic (H is
negative), sometimes it needs something to start
the reaction which, once started, continues on its
own.
To Start A Reaction
EXOTHERMIC
•Add heat
•Add light
Zn(s) + S(s)
•Add mechanical energy
H
NOT THE WHOLE
ZnS(s)
Zn(s) + S(s) ZnS(s)
PICTURE, SOMETIMES
A REACTIONS NEEDS
A PUSH.
Chem 106, Prof. J.T. Spencer
CHE 106
36
Activation Energy
Activation Energy minimum energy to initiate
a chemical reaction barrier to reaction
Activation E
non-state function
A
Eact
PE
state function
B
Chem 106, Prof. J.T. Spencer
CHE 106
37
Activation Energy

Chemical Reactions - E and H unaffected by
path while Eact can be changed without affecting
E or H. Eact depends on how reactants come
together.
Zn(s) + S(s)
Eact
E
EXOTHERMIC
Zn(s) + S(s)
ZnS(s)
Energy
ZnS(s)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy:Exothermic Reactions


38
Reaction of Potassium Permanganate and
Glycerine [spontaneous combustion]
14 KMnO4(s) + 4 C3H5(OH)3(l)
7 K2CO3(s) + 7 Mn2O3(s) + 5 CO2 + 16 H2O(g)

Reaction is initially slow but the heat it generates
speeds that reaction rate until the reaction
eventually catches fire.
[Video No. 24; 1:16 m]
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy:Exothermic Reactions


39
Explosion of Nitrogen Triiodide (NI3)
Preparation of NI3;
Complex series of steps
overall
NH3(aq) + I2(aq) + H2O
NH3NI3(s) + 3
H2O
(not balanced)

Explosive decomposition (only when dry)
8NH3NI3(s)

5 N2(g) + 6 NH4I(s) + 9 I2(g)
What is the purple cloud after the explosion?
[Video No. 26; 1:14 m]
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy:Exothermic Reactions


Reaction of Sugar (sucrose; C12H22O11)and
Sulfuric Acid (Dehydration)
Reactions:
C12H22O11(s)
12 C(s) + 11 H2O(l)
Hrxn = -918.9 kJ/mole
and

H2SO4(H2O)n + mH2O
H2SO4(H2O)n+m
Hdillution = 40.6 kJ/mole
Large heat of hydration for sulfuric acid drives
this reaction.
[Video No. 27; 1:18 m]
Chem 106, Prof. J.T. Spencer
40
CHE 106
Enthalpy:Exothermic Reactions

Thermite Reaction
Fe2O3(s) + 2 Al(s)



41
2 Fe(s) + Al2O3(s)
H = -849 kJ mol-1
Requires heat input to start reaction but is then
self-suftaining (in a big way).
Reaction has been used for spot welding and for
the preparation of pure metals from their oxides.
The reaction initiated from the heat generated
from the combustion of glycerine and potassium
permanganate.
[Video No. 25; 2:38 m]
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
42

Enthalpy is an extensive property; this means
that enthalpy is dependent on the amount of
reactants used.

The enthalpy change (H) for a reaction is equal
in magnitude but opposite in sign for the reverse
reaction.

The enthalpy change (H) for a reaction depends
on the state of the products and reactants.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

43
Enthalpy is an extensive property - the
magnitude of H depends on the amount of
material involved in the reaction.
Problem: Consider the reaction:
2 N2(g) + O2(g)
2N2O(g) H = +163.2 kJ
(1) Is the reaction exothermic or endothermic?
(2) Calculate the amount of heat transferred
when 12.8 g of N2O(g) forms (const. P).
(3) How many grams of N2(g) must react to
produce a H of 1.00 kJ.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
44
Problem: Consider the reaction:
2 N2(g) + O2(g)
2N2O(g) H = +163.2 kJ
(1) Is the reaction exothermic or endothermic?
ENDOTHERMIC
(2) Calculate the amount of heat transferred
when 12.8 g of N2O(g) forms (const. P).
(12.8 g N2O)(1 mol N2O) (+163.2 kJ) = +23.7 kJ
(44 g N2O) (2 mol N2O)
Energy transferrred TO system FROM surroundings.
Value of H depended on amount of material present
EXTENSIVE PROPERTY
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
Problem: Consider the reaction:
2 N2(g) + O2(g)
2N2O(g) H = +163.2 kJ
(3) How many grams of N2(g) must react to
produce a H of 1.00 kJ.
(1.00 kJ) (2 mol N2) 28 g N2
(163.2 kJ) 1 mol N2
= 0.343 g N2
Chem 106, Prof. J.T. Spencer
45
CHE 106
Enthalpy
46
Sample exercise: Hydrogen peroxide
can decompose water and oxygen by
the reaction:
2H2O2  2H2O + O2 H = -196 kJ
Calculate the value of q when 5.00
g of H2O2 decomposes at constant
pressure.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
47
Sample exercise: Hydrogen peroxide
can decompose water and oxygen by
the reaction:
2H2O2  2H2O + O2 H = -196 kJ
Calculate the value of q when 5.00
g of H2O2 decomposes at constant
pressure.
5.00 g H2O2
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
48
Sample exercise: Hydrogen peroxide
can decompose water and oxygen by
the reaction:
2H2O2  2H2O + O2 H = -196 kJ
Calculate the value of q when 5.00
g of H2O2 decomposes at constant
pressure.
5.00 g H2O2
1 mol H2O2
34 g H2O2
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
49
Sample exercise: Hydrogen peroxide
can decompose water and oxygen by
the reaction:
2H2O2  2H2O + O2 H = -196 kJ
Calculate the value of q when 5.00
g of H2O2 decomposes at constant
pressure.
5.00 g H2O2
1 mol H2O2
34 g H2O2
-196 kJ
2 mol H2O2
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
50
Sample exercise: Hydrogen peroxide
can decompose water and oxygen by
the reaction:
2H2O2  2H2O + O2 H = -196 kJ
Calculate the value of q when 5.00
g of H2O2 decomposes at constant
pressure.
5.00 g H2O2
1 mol H2O2
34 g H2O2
-196 kJ
= -14.4 kJ
2 mol H2O2
Chem 106, Prof. J.T. Spencer
CHE 106
Thermochemistry
51
Hindenburg exploded in
1937 near Lakehurst, NJ
upon arrival
7,062,000 cubic feet of H2;
2H2 + O2
2H2O
H = -484 kJ
(7,062,000 ft3 H2)(28.3 L) (1 mol )(-484 kJ) = 2.16 billion kJ
1 ft3
22.4 L 2 mol H2
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

52
The enthalpy change (H) for a reaction is
equal in magnitude but opposite in sign for
the reverse reaction - H is a state function
so it must be equal in magnitude and opp. in
sign for the forward and reverse reactions
Problem - Oxygen may be generated on small scales
in the laboratory by the thermal decomposition of
potassium chlorate:
2KClO3(s)
2 KCl(s) + 3O2(g) H = -89.4 kJ
For this reaction, calculate the H of formation for;
(1) 6.45 g of O2(g)
(2) 9.22 g of KClO3
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
53
Problem - Oxygen may be generated on small scales
in the laboratory by the thermal decomposition of
potassium chlorate:
2KClO3(s)
2 KCl(s) + 3O2(g) H = -89.4 kJ
For this reaction, calculate the H of formation for;
(1) 6.45 g of O2(g)
(6.45 g O2)(1 mol O2) (-89.4 kJ) = -6.01 kJ
(32 g O2) (3 mol O2)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
54
Problem - Oxygen may be generated on small scales
in the laboratory by the thermal decomposition of
potassium chlorate:
2KClO3(s)
2 KCl(s) + 3O2(g) H = -89.4 kJ
For this reaction, calculate the H of formation for;
(2) 9.22 g of KClO3
(9.22 g KClO3) (1 mol KClO3) (+89.4 kJ) = +3.36 kJ
(122.5 g KClO3)(2 mol KClO3)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

55
The enthalpy change (H) for a reaction
depends on the state of the products and
reactants - the states of the products and
reactants must be given.
Problem: Which of the following has the highest
enthalpy at a given temperature and pressure.
(1) H2O(s)
(2) H2O(l)
(3) H2O(g)
H2O(s) < H2O(l) < H2O(g)
remember that it takes
heat to convert H2O(s) into
H2O(g) so H2O(s) has the
lowest enthalpy (heat
content).
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

56
When solutions containing silver ions and
chloride ions are mixed, silver chloride
precipitates:
Ag+(aq) + Cl-(aq)
AgCl(s) H = -65.5
kJ
(1)Calculate the H for the formation of
0.200
mol AgCl from this reaction.
2.50 g of
(2) Calculate the H for formation of
AgCl.
AgCl
(3) Calculate H when 0.350 mol of
dissloves in water.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

57
When solutions containing silver ions and
chloride ions are mixed, silver chloride
precipitates:
Ag+(aq) + Cl-(aq)
AgCl(s) H = -65.5
kJ
(1) Calculate the H for the formation
of
0.200 mol AgCl from this
reaction.
(0.200 mol AgCl) (-65.5 kJ) = 13.1 kJ
(1 mol AgCl)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

58
When solutions containing silver ions and
chloride ions are mixed, silver chloride
precipitates:
Ag+(aq) + Cl-(aq)
AgCl(s) H = -65.5
kJ
(2) Calculate the H for formation of
2.50 g of
AgCl.
(2.50 g AgCl) (1 mol AgCl) (-65.5 kJ) = 1.14 kJ
(143.5 g AgCl)(1 mol AgCl)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy

When solutions containing silver ions and
chloride ions are mixed, silver chloride
precipitates:
Ag+(aq) + Cl-(aq)
AgCl(s) H = -65.5
kJ
(3) Calculate H when 0.350 mol of
AgCl
dissloves in water.
kJ

59
(0.350 mol AgCl) (+65.5 kJ) = +22.9
(1 mol AgCl)
NOTE: THE SIGN IS SWITCHED ON REVERSING
THE REACTION.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy
60

Enthalpy is an extensive property.

The enthalpy change (H) for a reaction is equal
in magnitude but opposite in sign for the reverse
reaction.

The enthalpy change (H) for a reaction depends
on the state of the products and reactants.

State Function.
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
61
Calorimetry - Measurement of heat flow.
Depending on conditions, H or E
may be calculated directly.
 Lord Kelvin - “I often say that when you
can measure what you are speaking about
and express it in numbers you know
something about it; but when you cannot
express it in numbers your knowledge is a
meager and unsatisfactory kind: it may be
the beginning of knowledge but you have
scarcely advanced to the stage of
science.”

Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
62
– Heat Capacity (C) - amount of heat required to
raise the temperature of a body by 1 ° K. Depends
on nature and amount of material in object.
C=
heat absorbed
increase in temperature
– Molar Heat Capacity - the heat capacity of 1 mol of
a substance (or the heat required to raise 1 mol of
substance by 1° K).
Molar Heat Capacity =
quantity of heat
(mol of subst.)(temp. change)
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
63
– Specific Heat Capacity - the energy (heat)
required to raise 1 gram of a substance by 1° K.
Specific heat =
heat transferred
(g of substance)(temp. change)
= q
mass T
– OR q = (Specific Heat)mT
– Constant Pressure calorimetry measures H
[since H = qp]
Constant Volume calorimetry measures E
[since pressure changes and Vol is const.]
– Units are typically J/gram K or J/gram °C
Chem 106, Prof. J.T. Spencer
CHE 106
Specific Heats
Substance
Specific Heat (J °C-1 g-1)
H2O(l)
H2O(s)
4.18
2.03
Al(s)
C(s)
Fe(s)
Hg(l)
CCl4(l)
CaCO3(s)
0.89
0.71
0.45
0.14
0.86
0.85
Chem 106, Prof. J.T. Spencer
64
CHE 106
Calorimetry

65
A swimming pool, 10.0 m by 4.0 m, is filled to a
depth of 3.0 m with water at 20.2 °C. How much
energy is required to raise the water temperature
to 30.0°C?
q = (Specific Heat)mT
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry

66
A swimming pool, 10.0 m by 4.0 m, is filled to a
depth of 3.0 m with water at 20.2 °C. How much
energy is required to raise the water temperature
to 30.0°C?
Vol of water = (10.0 m)(4.0m)(3.0m) = 120 m3
mass of water = (120 m3)(100 cm)3(1 mL) (1.0g)
(1 m)3 (1 cm3)(1 mL)
= 1.2 x 108 g
T = 30.0° C - 20.2° C = 9.8° C
q = (specific heat)(m)T
q = (4.18 J)(1.2 x 108 g)(9.8° C) = 4.9 x 109 J
°C g
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry Problem

67
How much energy is required to raise the
temperature of an 8.50 x 102 g block of aluminum
from 22.8°C to 94.6°C? What is the molar heat
capacity of aluminum?
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry Problem

68
How much energy is required to raise the
temperature of an 8.50 x 102 g block of aluminum
from 22.8°C to 94.6°C? What is the molar heat
capacity of aluminum?
q = (sp. heat)gT
specific heat Al = 0.900 J°C-1g-1
q = ( 0.900 J°C-1g-1)(8.50 x 102 g Al)(71.8°C)
q = 54.9 kJ or 54,900 J
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry Problem

69
How much energy is required to raise the
temperature of an 8.50 x 102 g block of aluminum
from 22.8°C to 94.6°C? What is the molar heat
capacity of aluminum?
q = (sp. heat)gT
specific heat Al = 0.900 J°C-1g-1
q = ( 0.900 J°C-1g-1)(8.50 x 102 g Al)(71.8°C)
q = 54.9 kJ
Molar heat capacity = (0.9 J) (27.0 g) = 24.3 J°C-1mol-1
g°C (mol Al)
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat

70
Ti metal is used as a structural material in many
high tech applications including jet engines.
– What is the specific heat of Ti if it takes 89.7 kJ
to raise a 33.0 Kg block by 5.20 °C?
– What is the molar heat capacity of Ti?
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat

71
Ti metal is used as a structural material in many
high tech applications including jet engines.
– What is the specific heat of Ti if it takes 89.7 kJ
to raise a 33.0 Kg block by 5.20 °C?
– What is the molar heat capacity of Ti?
q = (sp. heat)gT
sp. heat = qrxn /gT
sp. heat Ti = [(89.7 kJ)(1000 J)/(1kJ)]
[(33.0 Kg)(1000g)/(1 Kg)](5.20 °C)
= 0.523 J °C-1 g-1
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat

72
Ti metal is used as a structural material in many
high tech applications including jet engines.
– What is the molar heat capacity of Ti?
sp. heat Ti = 0.523 J °C-1 g-1
mol heat cap. = (0.523 J) (47.6 g) = 24.9 J °C-1 mol-1
g °C mol Ti
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
73
Sample exercise: Large beds of rocks
are used in some solar-heated
homes to store heat. Calculate the
quantity of heat absorbed by 50.0 kg
of rocks if their temperature
increases by 12.0°C. Specific heat of
rocks is 0.82 J/g-K.
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
74
Sample exercise: Calculate the
quantity of heat absorbed by 50.0 kg
of rocks if their temperature
increases by 12.0°C. Specific heat of
rocks is 0.82 J/g-K.
E = mcT
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
75
Sample exercise: Calculate the
quantity of heat absorbed by 50.0 kg
of rocks if their temperature
increases by 12.0°C. Specific heat of
rocks is 0.82 J/g-K.
E = mcT
50,000 g (0.82 J/g-K) (12 K)
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
76
Sample exercise: Calculate the
quantity of heat absorbed by 50.0 kg
of rocks if their temperature
increases by 12.0°C. Specific heat of
rocks is 0.82 J/g-K.
E = mcT
50,000 g (0.82 J/g-K) (12 K)
4.9 x 105 J
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
77
Sample exercise: What temperature
change would these rocks undergo if
they absorbed 450 kJ of heat?
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
78
Sample exercise: What temperature
change would these rocks undergo if
they absorbed 450 kJ of heat?
E = mcT
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
79
Sample exercise: What temperature
change would these rocks undergo if
they absorbed 450 kJ of heat?
E = mcT
450,000 J = 50,000 g (0.82 J/g-K) (x)
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat
80
Sample exercise: What temperature
change would these rocks undergo if
they absorbed 450 kJ of heat?
E = mcT
450,000 J = 50,000 g (0.82 J/g-K) (x)
T = 11 K which is the same as
11°C
Chem 106, Prof. J.T. Spencer
CHE 106
Constant Pressure Calorimetry



Measure H from Const.
Pressure Calorimeter.
The calorimeter contains
a known volume of
solution and prevents
heat loss/gain to the
environment.
Heat from the reaction is
transferred to the
calorimeter solution.
Thus heat lost from the
reaction = heat gained by
the solution
81
H = qrxn = -qsoln
H = - (sp. heat)gT
Constant P Calorimeter
Thermometer
Stirring Rod
Styrofoam Cup
(with reactant
solutions)
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
82
Consider mixing 50 mL of a 1.0 M HCl solution
and 50 mL of a 1.0 M NaOH solution, both at
25°C.
Net Ionic Equation:
Start Finish
H+(aq) + OH-(aq)
H2O(l)
31.9°C
known or given;
30°C
heat Capacity H2O(l) = 4.18 Jg-1°C-1
T = 6.9°C
H = qrxn = -qsoln
H = - (sp. heat)gT
vol of total soln = 100.0 mL
25°C
m of total soln = 100.0 g (from density)

Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry Problem

How much energy is released in the problem?
H = qrxn = -qsoln
H = - (sp. heat)gT
H = - (4.18 Jg-1°C-1)(100 g)(6.9°C)
H = - 2,880 J = - 2.88 kJ
known or given;
heat Capacity H2O(l) = 4.18 Jg-1°C-1
T = 6.90 °C
vol of total soln = 100.0 mL
m of total soln = 100.0 g (from density)
Chem 106, Prof. J.T. Spencer
83
CHE 106
Calorimetry Problem

84
How much energy is released in the sample
problem on a molar basis?
H = qrxn = -qsoln
H = - (sp. heat)gT
H = - (4.18 Jg-1°C-1)(100 g)(6.9°C) = - 2.88 kJ
moles H+ or Cl- = (0.050L)(1.0 mol/L) = 0.050 mol
H = - (2.88 kJ)/(0.050mol) = -58 kJ mol-1
known or given;
heat Capacity H2O(l) = 4.18 Jg-1°C-1
T = 6.90 °C
vol of total soln = 100.0 mL
m of total soln = 100.0 g (from density)
Chem 106, Prof. J.T. Spencer
CHE 106
85
Heat of Dilution

Heat of dilution of sulfuric acid (H2SO4)
Reaction:
H2SO4(conc) + n H2O
H2SO4(dil)
(where n = moles H2O/mole H2SO4)
10 mL
H2SO4
(conc)
20 mL
H2SO4
(conc)
100 mL
H2O
30 mL
H2SO4
(conc)
100 mL
H2O
100 mL
H2O
Measure Temperature Changes
Demonstration 1.6
Chem 106, Prof. J.T. Spencer
CHE 106
86
Heat of Dilution

Heat of dilution of sulfuric acid (H2SO4)
10 mL
H2SO4
(conc)
20 mL
H2SO4
(conc)
100 mL
H2O
30 mL
H2SO4
(conc)
100 mL
H2O
100 mL
H2O
Initial
Temp
25° C
25° C
25° C
Final
Temp
50° C
73° C
95° C
Demonstration 1.6
Chem 106, Prof. J.T. Spencer
CHE 106
Heat of Dilution

87
Heat of dilution of sulfuric acid (H2SO4)
Reaction:
H2SO4(conc) + n H2O
H2SO4(dil)
(where n = moles H2O/mole H2SO4)
mL H2SO4
10 mL
20 mL
30 mL
mL H2O
100 mL
100 mL
100 mL
n
T (°C)
31.0
15.6
10.5
25.0
48.0
70.0
Hrxn
-11.0 kJ
-26.0 kJ
-30.0 kJ
Hrxn values are dependent upon the amount of
material present
Demonstration 1.6
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat

88
Problem: A coffee cup
calorimeter contains 125 g
H = qrxn = -qsoln
of water at 24.2°C. After
H = - (sp. heat)gT
KBr (10.5 g) at 24.2 °C is
added the temperature
Constant P Calorimeter
becomes 21.1°C.
Thermometer
– What is the heat of
solution of KBr (given
Stirring Rod
that no heat is
transferred to the
surroundings from the
Styrofoam Cup
calorimeter and the
(with reactant
specific heat of the
solutions)
solution = 4.18 Jg-1C-1)?
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry: Specific Heat

89
Problem: A coffee cup calorimeter contains 125 g of
water at 24.2°C. After KBr (10.5 g) at 24.2 °C is
added the temperature becomes 21.1°C and the
specific heat of the solution = 4.18 Jg-1·C-1)?
Hsoln(KBr) = ??
KNOWN
Reaction: KBr(s) + H2O(l)
K+(aq) + Br-(aq)
total mass = 125 g H2O + 10.5 g KBr = 135.5 g
T = Tfinal - Tinitial = 21.1 - 24.2 = - 3.1 °C
ENDOTHERMIC REACTION
q = - (specific heat)gT
heat gained by system = heat lost by calorimeter
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
90
Reaction: KBr(s) + H2O(l)
K+(aq) + Br-(aq)
total mass = 125 g H2O + 10.5 g KBr = 135.5 g
T = Tfinal - Tinitial = 21.1 - 24.2 = -3.1 °C
ENDOTHERMIC REACTION
q = - (specific heat)gT
Hsoln(KBr) = ??
q = - CgT
heat gained by system = heat lost by calorimeter
heat lost by solution = q = -(4.18J)(135.5 g)(-3.1°C)
g °C
= 1800 J
heat gained by KBr = 1800 J = 170 J g-1
10.5 g
(170 J) (199 g KBr) (1 kJ) = 20.2 kJ mol-1
g KBr mol KBr 1000J
Chem 106, Prof. J.T. Spencer
CHE 106
Calorimetry
Sample exercise: When 50.0 mL of
0.100 M AgNO3 and 50.0 mL of 0.100
M HCl are mixed in a constant
pressure calorimeter, the temperature
mixture increases from 22.30°C to
23.11°C. The temperature increase is
caused by this reaction:
AgNO3 + HCl  AgCl + HNO3
Calculate H for this reaction.
Chem 106, Prof. J.T. Spencer
91
CHE 106
Calorimetry
Sample exercise:
Given: 100 mL = 100 g and 0.005 mol
T = 0.81°C
specific heat is 4.18 J/g-K
AgNO3 + HCl  AgCl + HNO3
E = -mcT
Chem 106, Prof. J.T. Spencer
92
CHE 106
Calorimetry
Sample exercise:
Given: 100 mL = 100 g and 0.005 mol
T = 0.81°C
specific heat is 4.18 J/g-K
AgNO3 + HCl  AgCl + HNO3
E = -mcT
-100 g (4.18 J/g-K) (0.81 K)
Chem 106, Prof. J.T. Spencer
93
CHE 106
Calorimetry
Sample exercise:
Given: 100 mL = 100 g and 0.005 mol
T = 0.81°C
specific heat is 4.18 J/g-K
AgNO3 + HCl  AgCl + HNO3
E = -mcT
-100 g (4.18 J/g-K) (0.81 K)
-339 J
Chem 106, Prof. J.T. Spencer
94
CHE 106
Calorimetry
Sample exercise:
Given: 100 mL = 100 g and 0.005 mol
T = 0.81°C
specific heat is 4.18 J/g-K
AgNO3 + HCl  AgCl + HNO3
-339 J
0.005 mol
Chem 106, Prof. J.T. Spencer
95
CHE 106
Calorimetry
Sample exercise:
Given: 100 mL = 100 g and 0.005 mol
T = 0.81°C
specific heat is 4.18 J/g-K
AgNO3 + HCl  AgCl + HNO3
-339 J = -68,000 J/mol
0.005 mol
Chem 106, Prof. J.T. Spencer
96
CHE 106
Constant Volume Calorimetry
Bomb Calorimeter

Electrical leads for
igniting sample
Stirrer
Thermometer
O2 Inlet
Wire in contact
with sample
Sample
holder
Water
Insulated Container
Bomb
97
Combustion
Reactions studied in bomb
built to withstand great
pressures in a
sealed system.
– filled with O2
– known amt of
H2O
– sample
ignited by
elec. current
Chem 106, Prof. J.T. Spencer
CHE 106
Constant Volume Calorimetry

98
Bomb Calorimetry - when a sample is burned in
an O2 atomsphere the heat is transferred to the
water surrounding the bomb.
qevolved = -Ccalorimeter x T

Need to know the heat capacity of the calorimeter
(Ccalorimeter) which is determined using a standard
with a known Hcombustion (i.e., benzoic acid)
C7H6O2(s) + O2(g)

CO2(g) + H2O(l) H = -26.4
kJ/g
Measure E rather than H because the bomb is
constant volume (no work) not constant pressure
although the correction for E to H is v. small and
not important here (thus we use H of combustion).
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

99
The combustion of 0.1624 g of benzoic acid
(C7H7O2) raises the temperature of a calorimeter
by 2.71°C [comb. benzoic acid = - 26.72 kJ/g].
– What is the heat capacity of the calorimeter?
– If 0.2138 g of vanillin (C8H8O3) is burned in the
calorimeter and the temperature increases by
3.28°C, what is the heat of combustion of
vanillin (per gram and per mole)?
qevolved = -Ccalorimeter x T
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

100
The combustion of 0.1624 g of benzoic acid
(C7H7O2) raises the temperature of a calorimeter
by 2.71°C [comb. benzoic acid = - 26.72 kJ/g].
– What is the heat capacity of the calorimeter?
Heat loss by combustion = heat gained by calorimeter
Heat loss = (26.72 kJ/g)(0.1624 g) = 4.339 kJ
Heat gain = 4.339 kJ = Ccal x T
Ccal = 4.339 kJ = 1.60 kJ °C-1
2.71°C
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

101
Ccal = 1.60 kJ °C-1
– If 0.2138 g of vanillin (C8H8O3) is burned in the
calorimeter and the temperature increases by
3.28°C, what is the heat of combustion of
vanillin (per gram and per mole)?
qevolved = -Ccalorimeter x T
Heat Loss = Heat Gain
Heat gain by calorimeter = (1.60 kJ/°C)(3.28°C) = 5.25 kJ
Heat Loss = - 5.25 kJ of heat from vanillin combustion
Heat of Combustion = - 5.25 kJ = - 24.6 kJ/g
0.2138 g
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

102
Convert the calculated heat of combustion from
kJ per gram to kJ per mole.
Heat of Combustion = - 5.25 kJ = - 24.6 kJ/g
0.2138 g
Heat of Comb. = -24.6 kJ (152 g) = -3740 kJ/mol
g
mol
Remember to keep track of signs. Exothermic
reactions are negative (the temperature of the
calorimeter goes up) while endothermic reactions are
positive (temperature of the calorimeter goes down)
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

103
A 0.1964 g sample of quinone (C6H4O2) is burned
in a bomb calorimeter that has a heat capacity of
1.56 kJ/°C. The temperature of the calorimeter
increases by 3.2 °C. Calculate the heat of
combustion of quinone per gram and per mole.
KNOWN
Ccal = 1.56 kJ/°C
T = 3.2°C
mass of quinone = 0.1964 g
heat gained by calorimeter = heat lost by quinone
heat gained by calorimeter = q = Ccal T
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

104
Calculate the heat of combustion of quinone
per gram and per mole.
Ccal = 1.56 kJ/°C
T = 3.2°C
mass of quinone = 0.1964 g
heat gained by calorimeter = heat lost by quinone
heat gained by calorimeter = Ccal T
heat gained by calorimeter = (1.56 kJ/C)(3.2°C)
= 4.99 kJ
heat lost by quinone = - 4.99 kJ = - 25.4 kJ/g
0.1964 g
-(25.4 kJ) (108 g) = -2740 kJ mol-1
g
1 mol
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry

105
Problem - Under const. V conditions, the heat of
combustion of glucose is -15.57 kJ/g. A 2.500 g
sample is burned and the temperature increased
from 20.55° C to 23.25°C.
– What is the total heat capacity of the
calorimeter?
– If the calorimeter contains 2.70 Kg of water,
what is the heat capacity of the dry
calorimeter?
Heat of combustion for glucose = -15.57 kJ/g
mass of glucose = 2.50 g
T = Tfinal - Tinitial = 23.25°C - 20.55°C = 2.70°C
heat gained by calorimeter = heat lost by system
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
106
– What is the total heat capacity of the
calorimeter?
Heat of combustion for glucose = -15.57 kJ/g
mass of glucose = 2.50 g
T = Tfinal - Tinitial = 23.25°C - 20.55°C = 2.70°C
heat gained by calorimeter = heat lost by system
Heat gained by calorimeter = (+15.57 kJ/g)(2.500 g)
= + 38.93 kJ
q = Ccal T
Ccal = q = +38.93 kJ = 14.4 kJ °C-1
T 2.70°C
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
107
– It the calorimeter contains 2.70 Kg of water,
what is the heat capacity of the dry
calorimeter?
Heat of combustion for glucose = -15.57 kJ/g
mass of glucose = 2.50 g
T = Tfinal - Tinitial = 23.25°C - 20.55°C = 2.70°C
heat gained by calorimeter = heat lost by system
Total Heat Capacity = Ccal (dry) + Cwater
Ccal (dry) = Ctotal - Cwater
Cwater = (4.18 kJ/Kg°C) (2.70 Kg water) = 11.29 kJ/°C
Ccal (dry) = (14.4 kJ/°C) - (11.29 kJ/°C) = 3.11 kJ/°C
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
108
Sample exercise: A 0.5865 g sample
of lactic acid is burned in a
calorimeter whose heat capacity is
4.812 kJ/°C. The temperature
increases from 23.10°C to 24.95°C.
Calculate the heat of combustion
of lactic acid per gram.
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
109
Sample exercise: A 0.5865 g sample
of lactic acid is burned in a
calorimeter whose heat capacity is
4.812 kJ/°C. The temperature
increases from 23.10°C to 24.95°C.
Calculate the heat of combustion
of lactic acid per gram.
E = - CcalT
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
110
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
E = - CcalT
- 4.812 kJ/°C (1.85°C)
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
111
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
E = - CcalT
- 4.812 kJ/°C (1.85°C)
- 8.90 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
112
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
- 8.90 kJ
0.5865 g
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
113
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
- 8.90 kJ = -15.2 kJ/g
0.5865 g
Chem 106, Prof. J.T. Spencer
CHE 106
Bomb Calorimetry
114
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
-15.2 kJ
g
90 g HC3H5O3
1 mol HC3H5O3
Chem 106, Prof. J.T. Spencer
CHE 106
115
Bomb Calorimetry
Sample exercise: A 0.5865 g sample of
lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/°C.
The temperature increases from
23.10°C to 24.95°C. Calculate the heat
of combustion of lactic acid per gram.
-15.2 kJ
g
90 g HC3H5O3 =
1 mol HC3H5O3
-1370 kJ
mol
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
116

H values for very many reactions have been
measured and tabulated.

H is a state function (not pathway dependent)
THEREFORE the magnitude of H depends only
on the amount of material undergoing the change
and the initial and final states.

Hess’s Law - if a reaction is carried out in a series
of steps, H for the reaction will be equal to the
sum of the H’s for the individual steps.
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
117
H does not depend upon the number of steps
necessary to carry out a transformation (i.e., 11
small steps or 1 large steps to reach the top of a
ramp). [11 steps of 1 ft. each = 1 step of 11 ft]
 If you had only a 2 ft. ruler and wanted to measure
the altitude (11 ft. total) you couldn’t measure it
DIRECTLY but by measuring each step and adding
the measurements you could easily determine alt.

altitude
11 steps
1 step
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
118
 In a chemical reaction, H for an unknown reaction
is found by summing the known steps which lead
to the net unknown reaction.
Hess’s Law - H for the reaction will be equal to the
sum of the H for the individual steps.
D
H1
C
? H(A+B=C)
H2
E
C
H3
H(A+B=C)
A+B
H(A+B=C) = H1 + H2 + H3
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
119
Sum the three
known reactions
which add up to
the unknown
reaction
A+B
D
E
D
E
C
H1
H2
H3
A+B
C
H(A+B=C)
D
H1
C
? H(A+B=C)
H2
E
C
H3
H(A+B=C)
A+B
H(A+B=C) = H1 + H2 + H3
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

120
Example : Haber process for the formation of
ammonia (ca. 17 million tons annually).
3 H2(g) + N2(g)
2NH3(g)
H°= -92.2 kJ
The reaction DOES NOT occur in one step,
instead;
(1) 2 H2(g) + N2(g)
(2) N2H4(g) + H2(g)
N2H4(g) H°= ? kJ
2NH3(g) H°= -187.6 kJ
(3) Net: 3 H2(g) + N2(g)
2NH3(g) H°= -92.2 kJ
Hess’s Law: H°(rxn 3) = H°(rxn 1) + H°(rxn 2)
H°(rxn 1) cannot be measured directly but
H°(rxn 1) = H°(rxn 3) - H°(rxn 2)
= (-92.2 kJ)- (187.6
= +95.4kJ
Chem kJ)
106, Prof.
J.T. Spencer
CHE 106
Hess’s Law
121
H2(g) + N2H4(g)
2
Reactants
1
3 H2(g) + N2(g)
3
Products
(1) 2 H2(g) + N2(g)
(2) N2H4(g) + H2(g)
(3) Net: 3 H2(g) + N2(g)
2NH3(g)
N2H4(g)
2NH3(g)
H°= -? kJ
H°= -187.6 kJ
2NH3(g) H°= -92.2 kJ
Hess’s Law: H°(rxn 3) = H°(rxn 1) + H°(rxn 2)
H°(rxn 1) = H°(rxn 3) - H°(rxn 2)
= (-92.2 kJ)- (187.6 kJ) = +95.4kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

122
Using Hess’s Law, the H for a reaction which
cannot be directly measured may be determined
if enough is known about related reactions.
– Example - What is the enthalpy of combustion
of carbon(s) going to carbon monoxide (CO)?
The enthalpy of combustion for carbon to carbon
monoxide cannot be directly measured (can’t do the
reaction cleanly enough). However the following
reactions are known;
C(s) + O2(g)
CO2(g)
H = -393.5 kJ
CO(g) + 0.5 O2(g)
CO2(g)
H = -283.0 kJ
Adding the reactions correctly yields the H C to CO
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
123
Example - What is the enthalpy of combustion of
carbon(s) going to carbon monoxide (CO)?
GIVEN
C(s) + O2(g)
CO2(g)
H = -393.5 kJ
CO(g) + 0.5 O2(g)
CO2(g)
H = -283.0 kJ
(1) C(s) + O2(g)
CO2(g) H = -393.5 kJ
(2) CO2(g)
CO(g) + 0.5 O2(g)
H = +283.0 kJ
(3) C(s) + 0.5 O2(g)
CO(g)
H = -110.5 kJ
H(rxn 3) = H(rxn 1) + H(rxn 2)
Note: Rxn 2 is reversed from given reaction
(H change sign)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law


124
Water gas is a very important mixture of CO and
H2 prepared by passing steam over hot charcoal at
1000°C. Calculate H° for the water-gas rxn.
Given:
(1) C(s) + H2O(g)
CO(g) + H2(g) (Water Gas Rxn)
(2) C(s) + O2(g)
CO2(g) H° = -393.5 KJ
(3) 2 H2(g) + O2(g)
2H2O(g) H° = -483.6 kJ
(4) 2 CO(g) + O2(g)
2 CO2(g) H = -566.0 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

Given:
(1) C(s) + H2O(g)
(2) C(s) + O2(g)
(3) 2 H2(g) + O2(g)
(4) 2 CO(g) + O2(g)
C(s) + H2O(g)
125
CO(g) + H2(g) (Water Gas Rxn)
CO2(g) H° = -393.5 KJ
2H2O(g) H° = -483.6 kJ
2 CO2(g) H = -566.0 kJ
CO(g) + H2(g)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

Given:
(1) C(s) + H2O(g)
(2) C(s) + O2(g)
(3) 2 H2(g) + O2(g)
(4) 2 CO(g) + O2(g)
C(s) + O2(g)
C(s) + H2O(g)
126
CO(g) + H2(g) (Water Gas Rxn)
CO2(g) H° = -393.5 KJ
2H2O(g) H° = -483.6 kJ
2 CO2(g) H = -566.0 kJ
CO2(g)
H° = -393.5 KJ
CO(g) + H2(g)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

Given:
(1) C(s) + H2O(g)
(2) C(s) + O2(g)
(3) 2 H2(g) + O2(g)
(4) 2 CO(g) + O2(g)
C(s) + O2(g)
1/2[2H2O
C(s) + H2O(g)
127
CO(g) + H2(g) (Water Gas Rxn)
CO2(g)
H° = -393.5 KJ
2H2O(g) H° = -483.6 kJ
2 CO2(g) H = -566.0 kJ
CO2(g)
2H2 + O2
H° = -393.5 KJ
H° = 483.6 kJ]
CO(g) + H2(g)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law

Given:
(1) C(s) + H2O(g)
(2) C(s) + O2(g)
(3) 2 H2(g) + O2(g)
(4) 2 CO(g) + O2(g)
C(s) + O2(g)
1/2[2H2O
1/2[ 2CO2
C(s) + H2O(g)
128
CO(g) + H2(g) (Water Gas Rxn)
CO2(g)
H° = -393.5 KJ
2H2O(g) H° = -483.6 kJ
2 CO2(g) H = -566.0 kJ
CO2(g)
2H2 + O2
2CO + O2
H° = -393.5 KJ
H° = 483.6 kJ]
H = 566.0 kJ]
CO(g) + H2(g)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
129
- H° for the reaction:
(1) C(s) + H2O(g)
CO(g) + H2(g) (Water Gas Rxn)
Question
(2) C(s) + O2(g)
CO2(g)
H° = -393.5 KJ
(3) H2O(g)
H2(g) + 1/2O2(g) H° = +241.8 kJ
(4) CO2(g)
CO(g) + 1/2 O2(g) H = +283.0 kJ
(1) C(s) + H2O(g)
CO(g) + H2(g) H = +131.3 kJ
H(rxn
1) = H(rxn 2) - 1/2 H(rxn 3) - 1/2 H(rxn 4)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law


Calculate H for 2F2(g) + 2H2O(l)
Given:
(1) H2(g) + F2(g)
2HF(g)
(2) 2H2(g) + O2(g)
2H2O(l)
130
4HF(g) + O2(g)
H = -537 kJ
H = -572 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law


Calculate H for 2F2(g) + 2H2O(l)
Given:
(1) H2(g) + F2(g)
2HF(g)
(2) 2H2(g) + O2(g)
2H2O(l)
131
4HF(g) + O2(g)
H = -537 kJ
H = -572 kJ
H forUnk Rx = 2H(1) - H(2)
2H2(g) + 2F2(g)
4HF(g)
H = 2(-537 kJ)
2H2O(l)
2H2(g) + O2(g) H = +572 kJ
2H2(g) + 2F2(g) + 2H2O(l)
4HF(g) + 2H2(g) + O2(g)
H(Unk Rx)= 2H(1) - H(2) = 2(-537kJ) + 572 kJ = -502kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
132
Sample exercise: Carbon occurs in two
forms, graphite and diamond. The
enthalpy of combustion of graphite is
-393.5 kJ/mol, and that of diamond is
-395.4 kJ/mol. Calculate H for the
conversion of graphite to diamond:
C(graphite)  C(diamond)
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
133
Sample exercise: Carbon occurs in two
forms, graphite and diamond. The
enthalpy of combustion of graphite is
-393.5 kJ/mol, and that of diamond is
-395.4 kJ/mol. Calculate H for the
conversion of graphite to diamond:
C(graphite)  C(diamond)
C(graphite) + O2  CO2 H = -393.5 kJ
C(diamond) + O2  CO2 H = -395.4 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
134
Sample exercise: Carbon occurs in two
forms, graphite and diamond. The
enthalpy of combustion of graphite is
-393.5 kJ/mol, and that of diamond is
-395.4 kJ/mol. Calculate H for the
conversion of graphite to diamond:
C(graphite)  C(diamond)
C(graphite) + O2  CO2 H = -393.5 kJ
C(diamond) + O2  CO2 H = -395.4 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
135
Sample exercise: Carbon occurs in two
forms, graphite and diamond. The
enthalpy of combustion of graphite is
-393.5 kJ/mol, and that of diamond is
-395.4 kJ/mol. Calculate H for the
conversion of graphite to diamond:
C(graphite)  C(diamond)
C(graphite) + O2  CO2 H = -393.5 kJ
CO2  C (diamond) + O2 H = 395.4 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
136
Sample exercise: Carbon occurs in two
forms, graphite and diamond. The
enthalpy of combustion of graphite is
-393.5 kJ/mol, and that of diamond is
-395.4 kJ/mol. Calculate H for the
conversion of graphite to diamond:
C(graphite)  C(diamond)
C(graphite) + O2  CO2 H = -393.5 kJ
CO2  C (diamond) + O2 H = 395.4 kJ
C(graphite)  C(diamond) H = +1.9
kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
137
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
O3  3/2 O2
H = -142.3
O2  2O
H = 495.0
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
138
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
* 3/2 O2  O3
H = 142.3
O2  2O
H = 495.0
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
139
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
* 2O  O2
H = -495
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
140
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
*1/2[ 2O  O2
H = -495]
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
141
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
*O  1/2O2
H = 247.5
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
142
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
O  1/2O2
H = -247.5
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
143
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
O  1/2O2
H = -247.5
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
144
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
O  1/2O2
H = 247.5
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
145
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
O  1/2O2
H = 247.5
NO + O  NO2
Chem 106, Prof. J.T. Spencer
CHE 106
Hess’s Law
146
Sample exercise: Calculate H for the
reaction:
NO + O  NO2
given: NO + O3  NO2 + O2 H = 198.9
3/2O2  O3
H = 142.3
O  1/2O2
H = -247.5
NO + O  NO2
H = 304.1
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Formation

Many types of thermodynamic data have been
measured and tabulated and given names to
indicate reaction type;
– Heat of Vaporization (H for liquid to gas)
– Heat of Fusion (H for melting solids)
– Heat of Combustion (H reaction with O2)
– Heat of Formation (H compound from
elements) labeled Hf
Chem 106, Prof. J.T. Spencer
147
CHE 106
Enthalpy of Formation

148
Many types of thermodynamic data have been
measured and tabulated and given names to
indicate reaction type;
– Heat of Vaporization (H for liquid to gas)
– Heat of Fusion (H for melting solids)
– Heat of Combustion (H reaction with O2)
– Heat of Formation (H compound from
elements) labeled Hf
Heat of formation (Hf) is usually given for
reactants and products in standard states (since
H depends on the state of these items). When in
standard state, the denotation is H°f
2C(s) + 3H2(g) + 1/2O2
C2H5OH(l) H°f = -277.7 kJ

Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Formation


149
Since H is a state function, H°f values may be
determined by adding appropriate reactions
using Hess’s Law.
By definition, the standard enthalpy of formation
of the most stable form of any element is zero.
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Formation



150
Since H is a state function, H°f values may be
determined by adding appropriate reactions
using Hess’s Law.
By definition, the standard enthalpy of formation
of the most stable form of any element is zero.
Determine the standard enthalpy change for a
given unknown reaction by using H°f values and
Hess’s Law.
H°rxn = •nH°f (products) - •mH°f
(reactants)
[m and n are the stoichiometric coefficients of the
balanced equation]
Chem 106, Prof. J.T. Spencer
CHE 106
Standard Ethalpies of Formation
Compound
Acetylene
Ammonia
Carbon (graphite)
Carbon (diamond)
Methane
Ethane
Propane
Glucose
Water
Water vapor
Sucrose
Formula
C2H2(g)
NH3(g)
C(s)
C(s)
CH4(g)
C2H6(g)
C3H8(g)
C6H12O6(s)
H2O(l)
H2O(g)
C12H22O11(s)
151
H°f (kJ/mol)
+226.7
-46.19
0.0
+1.88
-74.85
-84.68
-103.85
-1260.0
-285.8
-241.8
-2221.0
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction
C3H8(g) + 5 O2(g)
3 CO2(g) + 4 H2O(l)
3 C(graphite) + 4 H2(g) + 5O2(g)
C3H8(g) + 5O2(g)
Reactants
H1
Reactants
Elements
Elements
H2
152
H3
Products
H1 = Hrxn = H2 + H3
H2 = -H°f(C3H8) + 5H°f(O2)
3 CO2(g) + 4 H2O(l)
Products
H3 = 4H°f(H2O) + 3 H°f(CO2)
H°f(O2) = 0 (strd. state)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction
C3H8(g) + 5 O2(g)
153
3 CO2(g) + 4 H2O(l)
H°rxn = •nH°f (products) - •mH°f (reactants)
•H°f (reactants) = [H°f (C3H8) + 5H°f (O2)]
= [1(-103.85 kJ/mol) + 5(0 kJ/mol)]
= -103.85 kJ
•H°f (products) = [4H°f (H2O) + 3 H°f (CO2)]
= [4(-285.8 kJ/mol) + 3(-393.5 kJ/mol)]
= -2324 kJ
H°rxn = (- 2324 kJ) - (- 103.85 kJ) = - 2220 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction
154
Fermentation of Glucose:
C6H12O6(s)
2 C2H5OH(l) + 2 CO2(g)
H°rxn = •nH°f (products) - •mH°f (reactants)
•H°f (reactants) = [H°f (C6H1O6)]
= [1(-1260 kJ/mol)]
= -1260 kJ
•H°f (products) = [2H°f (C2H5OH) + 2 H°f (CO2)]
= [2(-277.7 kJ/mol) + 2(-393.5 kJ/mol)]
= -1342.4 kJ
H°rxn = (- 1342 kJ) - (- 1260 kJ) = - 82 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction

155
Oxyacetylene welding torches burn acetylene
gas, C2H2. Calculate H° (kJ) for the combustion
of acetylene.
2C2H2(g) + 5O2(g)
2H2O(g) + 4CO2(g)
H°rxn = •nH°f (products) - •mH°f (reactants)
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction

156
Oxyacetylene welding torches burn acetylene
gas, C2H2. Calculate H° (kJ) for the combustion
of acetylene.
2C2H2(g) + 5O2(g)
2H2O(g) + 4CO2(g)
•H°f (reactants) = [2H°f (C2H2) + 5 H°f (O2)]
= [2(+226.7 kJ/mol) + 5(0)] = +453.4 kJ
•H°f (products) = [2H°f (H2O) + 4 H°f (CO2)]
= [2(-241.8 kJ/mol) + 4(-393.5 kJ/mol)]
= -2058 kJ
H°rxn = (- 2058 kJ) - (+ 453.4 kJ) = - 2511 kJ
[-2511 kJ for every 2 moles of C2H2!!]
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion
of 1 mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) +
3H2O(l)
Chem 106, Prof. J.T. Spencer
157
CHE 106
158
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion
of 1 mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) +
3H2O(l)
-277.7
0
-393.5
-285.83
Chem 106, Prof. J.T. Spencer
CHE 106
159
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion of 1
mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-277.7
3(0)
2(-393.5)
3(-285.83)
Chem 106, Prof. J.T. Spencer
CHE 106
160
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion of 1
mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-277.7
3(0)
2(-393.5)
3(-285.83)
H = Products - Reactants
Chem 106, Prof. J.T. Spencer
CHE 106
161
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion of 1
mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-277.7
3(0)
2(-393.5)
3(-285.83)
H = Products - Reactants
[2(-393.5) + 3(-285.83)] - [-277.7 + 3(0)]
Chem 106, Prof. J.T. Spencer
CHE 106
162
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion of 1
mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-277.7
3(0)
2(-393.5)
3(-285.83)
H = Products - Reactants
[2(-393.5) + 3(-285.83)] - [-277.7 + 3(0)]
[-1644.49] - [-277.7]
Chem 106, Prof. J.T. Spencer
CHE 106
163
Enthalpy of Reaction
Sample exercise: Using the standard
enthalpies of formation, calculate the
enthalpy change for the combustion of 1
mol of ethanol:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
-277.7
3(0)
2(-393.5)
3(-285.83)
H = Products - Reactants
[2(-393.5) + 3(-285.83)] - [-277.7 + 3(0)]
[-1644.49] - [-277.7]
-1367 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Enthalpy Calculations




164
When a reaction is reversed, the magnitude of
H remains the same but the sign changes.
When the balanced equation for a reaction is
multiplied by an integer, the value for H must
also be multiplied by the same integer.
The change in enthalpy for a given reaction can
be calculated from the enthalpies for the
formation of the reactants and products:
H°rxn = •nH°f (products) - •mH°f
(reactants)
Elements in their standard states are not
included in the H°rxn calculations since H°f for
an element in its standard state is zero.
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels


165
The energy released when 1 gram of material is
burned (combustion) is called its Fuel Value (kJ/g).
Metabolic energy comes primarily from the
controlled combustion of proteins, carbohydrates,
and fats (such as glucose).
protein; [H°(glucose) = -2,816 kJ/mol]
which is roughly 17 kJ per gram
carbohydrate; [H°(glucose) = -2,816 kJ/mol]
which is roughly 17 kJ per gram
fat; [H°(tristearin) = -75,520 kJ/mol
which is roughly 38 kJ per gram
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels

166
Combustion fuels (non-metabolic energy);
wood fuel value (kJ/g) = 18
coal fuel value (kJ/g) = 31
gasoline fuel value (kJ/g) = 48
natural gas fuel value (kJ/g) = 49
hydrogen fuel value (kJ/g) 142
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
167
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
168
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
169
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.62 g carbo 17 kJ
1 g carbo
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
170
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.62 g carbo 17 kJ
= 10.54 kJ
1 g carbo
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
171
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.22 g protein 17 kJ =
1 g protein
10.54 kJ carbo
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
172
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.22 g protein 17 kJ = 3.74 kJ
1 g protein
10.54 kJ carbo
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
173
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.015 g fat
38 kJ
=
1 g fat
10.54 kJ + 3.74 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
174
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.015 g fat
38 kJ
= 0.57 kJ
1 g fat
10.54 kJ + 3.74 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
175
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.015 g fat
38 kJ
= 0.57 kJ
1 g fat
10.54 kJ + 3.74 kJ + 0.57 kJ =
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels
176
Sample exercise: Dry red beans
contain 62% carbohydrate, 22%
protein, and 1.5% fat. Estimate the
fuel value of these beans.
*Assume 1 gram
0.015 g fat
38 kJ
= 0.57 kJ
1 g fat
10.54 kJ + 3.74 kJ + 0.57 kJ = 14.85 kJ
Chem 106, Prof. J.T. Spencer
CHE 106
Fuels: Hydrogen

Hydrogen is a very attractive alternative energy
source
– Very clean (burns to water only).
– Very high fuel value (142 kJ/g).
– can be made from superhgeated steam and
coal or from natural gas;
C(s) + H2O(g)
CH4(g) + H2O

177
CO(g) + H2(g)
CO(g) + 3 H2(g)
Drawback is its explosive flammability leading to
problems (engineering and chemical solutions)
Chem 106, Prof. J.T. Spencer
CHE 106
Hydrogen
178
Chem 106, Prof. J.T. Spencer
CHE 106
10 reasons for
179
Hydrogen





GLOBAL ADVANCEMENT- many countries use U.S.invented technology for hydrogen production and have been
expanding it for years. The United States has stubbornly
clung to oil.
NATIONAL SECURITY- The U.S. could be energy selfsufficient with hydrogen.
JOBS -Converting to a hydrogen-based economy would
create thousands of permanent scientific and industrial jobs.
SUPPLY -Someday, fossil fuels will run dry. Hydrogen is
renewable and, therefore, unlimited.
CLEAN AIR - Pollution from cars and airplanes has created
smog clouds across the country. Magnificent vistas like the
Grand Canyon are disappearing in toxic haze. Hydrogen
emits no toxins.
Chem 106, Prof. J.T. Spencer
CHE 106
10 reasons for
180
Hydrogen





DEFICIT -The government spends billions of dollars every year to
subsidize oil exploration and to militarily defend access to oil.
CLEAN WATER - Huge oil spills like the Exxon Valdez are
becoming common, killing countless waterfowl. The effects on our
food chain are unknown. If hydrogen were spilled, it would
evaporate immediately. The only by-product of hydrogen fuel is
water.
WILDERNESS -Mass consumption of oil requires continued
drilling into pristine wilderness areas, wreaking havoc on some of
the world's greatest ecosystems.
HEALTH -Increasing pollution from cars and airplanes makes
people sick. Hydrogen is clean and efficient.
ECONOMY - U.S. trade balance sheets show that oil imports
drain $1 billion-from the U.S. economy weekly.
Chem 106, Prof. J.T. Spencer
CHE 106
Hydrogen Fuel
181
“In a world first, Daimler-Benz has developed a fuel cell
vehicle with an onboard facility for generating hydrogen.
The revolutionary new vehicle, which is based on the
Mercedes-Benz A-class, represents a decisive
breakthrough in the quest to develop an automobile drive
system with extremely low emissions.”
Chem 106, Prof. J.T. Spencer
CHE 106
Chapter Five: Summary









Energy
Kinetic and Potential Energy
Energy Units and Conversions
First Law of Thermodynamics
Heat (q), Work (w), and Internal Energy (E)
State Functions
Enthalpy; Calculations and Exothermic and
Endothermic
Activation Energy
Calorimetry
Continued next slide
Chem 106, Prof. J.T. Spencer
182
CHE 106
Chapter Five: Summary (Con’t)

Hess’s Law
Enthalpy of Formation (standard states)

Enthalpy of reaction from H°f

Fuel Thermochemistry

Chem 106, Prof. J.T. Spencer
183