Chapter 7:
Energy and
Chemical Change
Chemistry: The Molecular Nature
of Matter, 6E
Jespersen/Brady/Hyslop
Thermochemistry
 Study of energies given off by or absorbed by
reactions.
Thermodynamics
 Study of heat transfer or heat flow
Energy (E )
 Ability to do work or to transfer heat.
Kinetic Energy (KE)
 Energy of motion
 KE = ½mv 2
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Potential Energy (PE)
 Stored energy
 Exists in natural attractions
and repulsions
 Gravity
 Positive and negative charges
 Springs
Chemical Energy
 PE possessed by chemicals
 Stored in chemical bonds
 Breaking bonds requires energy
 Forming bonds releases energy
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Your Turn!
Which of the following is not a form of kinetic
energy?
A. A pencil rolls across a desk
B. A pencil is sharpened
C. A pencil is heated
D. A pencil rests on a desk
E. A pencil falls to the floor
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Factors Affecting Potential Energy
Increase Potential Energy
 Pull apart objects that attract
each other
 A book is attracted to the earth by
gravity
 North and south poles of magnets
 Positive and negative charges
 Push together objects that repel
each other
 Spring compressed
 Same poles on two magnets
 Two like charges
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Factors Affecting Potential Energy
Decrease Potential Energy
 Objects that attract each other come together
 Book falls
 North and south poles of two magnets
 Positive and negative charges
 Objects that repel each
other move apart
 Spring released
 North poles on two
magnets
 Two like charges
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6
Your Turn!
Which of the following represents a decrease in
the potential energy of the system?
A. A book is raised six feet above the floor.
B. A ball rolls downhill.
C. Two electrons come close together.
D. A spring is stretched completely.
E. Two atomic nuclei approach each other.
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7
Law of Conservation of Energy
 Energy can neither be created nor destroyed
 Can only be
converted from
one form to
another
 Total energy
of universe
is constant
Total
Energy
=
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Potential
Energy
+
Kinetic
Energy
Chemistry: The Molecular Nature of Matter, 6E
8
Temperature vs. Heat
Temperature
 Proportional to average kinetic energy of object’s
particles
 Higher average kinetic energy means
 Higher temperature
 Faster moving molecules
Heat
 Total amount of energy transferred between
objects
 Heat transfer is caused by a temperature difference
 Always passes spontaneously from warmer objects
to colder objects
 Transfers until both are the same temperature
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Heat Transfer
 Hot and cold objects placed in
contact
hot
cold
 Molecules in hot object moving
faster
 KE transfers from hotter to
colder object
 A decrease in average KE of hotter
object
 An increase in average KE of colder
object
 Over time
 Average KEs of both objects becomes the same
 Temperature of both becomes the same
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10
Units of Energy
Joule (J)
 KE possessed by 2 kg object moving at speed
of 1 m/s.
1
1 m 
1 J  2 kg 

2
1s 
2
1 kg  m
1J
2
s
2
 If calculated value is greater than 1000 J, use
kilojoules (kJ)
 1 kJ = 1000 J
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Chemistry: The Molecular Nature of Matter, 6E
11
Units of Energy
A calorie (cal)
 Energy needed to raise the temperature of
1 g H2O by 1 °C
 1 cal = 4.184 J
(exactly)
 1 kcal = 1000 cal
 1 kcal = 4.184 kJ
A nutritional Calorie (Cal)
 note capital C
 1 Cal = 1000 cal = 1 kcal
 1 kcal = 4.184 kJ
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Your Turn!
Which is a unit of energy?
A. Pascal
B. Newton
C. Joule
D. Watt
E. Ampere
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Heat
 Pour hot coffee into cold cup




Heat flows from hot coffee to cold cup
Faster coffee molecules bump into wall of cup
Transfer kinetic energy
Eventually, the cup and the coffee reach the
same temperature
Thermal Equilibrium
 When both cup and coffee reach same
average kinetic energy and same temperature
 Energy transferred through heat comes from
object’s internal energy
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Internal Energy (E )
 Sum of energies of all particles in system
E = total energy of system
E = potential + kinetic = PE + KE
Change in Internal Energy
E = Efinal – Einitial
  means change
 final – initial
 What we can actually measure
 Want to know change in E associated with given
process
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E, Change in Internal Energy
 For reaction: reactants  products
 E = Eproducts – Ereactants
 Can use to do something useful
 Work
 Heat
 If system absorbs energy during reaction
 Energy coming into system has a positive sign (+)
 Final energy > initial energy
Ex. Photosynthesis or charging battery
 As system absorbs energy
 Increase potential energy
 Available for later use
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Kinetic Molecular Theory
 Kinetic Molecular Theory tells us
Temperature
 Related to average kinetic energy of particles in
object
Internal energy
 Related to average total molecular kinetic
energy
 Includes molecular potential energy
Average kinetic energy
 Implies distribution of kinetic energies among
molecules in object
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Temperature and Average Kinetic
Energy
In a large collection of gas molecules
 Wide distribution of kinetic energy (KE)
 Small number with KE = 0
 Collisions can momentarily stop a molecule’s
motion
 Very small number with very high KE
 Most molecules intermediate KEs
 Collisions tend to average kinetic energies
 Result is a distribution of energies
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Distribution of Kinetic Energy
1: lower temperature
2: higher temperature
At higher temperature, distribution shifts to higher kinetic
energy
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Kinetic Energy Distribution
Temperature
 Average KE of all atoms and molecules in object
 Average speed of particles
 Kelvin temperature of sample
 T (K) Avg KE = ½ mvavg2
 At higher temperature
 Most molecules moving at higher average speed
 Cold object = Small average KE
 Hot object = Large average KE
Note: At 0 K KE = 0
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so
v=0
Chemistry: The Molecular Nature of Matter, 6E
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Kinetic Theory: Liquids and Solids
 Atoms and molecules in liquids and solids
also constantly moving
 Particles of solids jiggle and vibrate in place
 Distributions of KEs of particles in gas, liquid
and solid are the same at same
temperatures
 At same temperature, gas, liquid, and
solid have
 Same average kinetic energy
 But very different potential energy
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Chemistry: The Molecular Nature of Matter, 6E
21
Your Turn!
Which statement about kinetic energy (KE) is true?
A.Atoms and molecules in gases, liquids and solids
possess KE since they are in constant motion.
B.At the same temperature, gases, liquids and solids
all have different KE distributions.
C.Molecules in gases are in constant motion, while
molecules in liquids and solids are not.
D.Molecules in gases and liquids are in constant
motion, while molecules in solids are not.
E. As the temperature increases, molecules move more
slowly.
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E, Change in Internal Energy
 E = Eproducts – Ereactants
 Energy change can appear entirely as heat
 Can measure heat
 Can’t measure Eproduct or Ereactant
 Importantly, we can measure E
 Energy of system depends only on its current
condition
 DOES NOT depend on:
 How system got it
 What energy the system might have sometime in
future
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State of Object or System
 Complete list of properties that specify object’s
current condition
For Chemistry
 Defined by physical properties
 Chemical composition
 Substances
 Number of moles
 Pressure
 Temperature
 Volume
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State Functions
 Any property that only depends on object’s
current state or condition
 Independence from method, path or mechanism
by which change occurs is important feature of
all state functions
 Some State functions, E, P, t, and V :




Internal energy
Pressure
Temperature
Volume
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E
P
t
V
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25
State of an object
 If tc = 25 °C, tells us all we need to know
 Don’t need to know how system got to that
temperature, just that this is where it currently is
 If temperature increases to 35 °C, then
change in temperature is simply:
 t = tfinal – tinitial
 Don’t need to know how this occurred, just need to
know initial and final values
 What does t tell us?
 Change in average KE of particles in object
 Change in object’s total KE
 Heat energy
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Defining the System
System
 What we are interested in studying
 Reaction in beaker
Surroundings
 Everything else
 Room in which reaction is run
Boundary
 Separation between system and surroundings
 Visible
Ex. Walls of beaker
 Invisible
Ex. Line separating warm and cold fronts
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Three Types of Systems
Open System
 Open to atmosphere
 Gain or lose mass and energy
across boundary
Open system
 Most reactions done in open
systems
Closed System
 Not open to atmosphere
 Energy can cross boundary,
but mass cannot
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Closed system
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Three Types of Systems
Isolated System
 No energy or matter can cross
boundary
 Energy and mass are constant
Ex. Thermos bottle
Adiabatic Process
 Process that occurs in isolated
system
 Process where neither energy nor
matter crosses the
system/surrounding boundary
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Isolated system
Chemistry: The Molecular Nature of Matter, 6E
29
Your Turn!
A closed system can __________
A.include the surroundings.
B.absorb energy and mass.
C.not change its temperature.
D.not absorb or lose energy and mass.
E.absorb or lose energy, but not mass.
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Heat (q)
 Cannot measure heat directly
 Heat (q) gained or lost by an object
 Directly proportional to temperature change (t) it
undergoes
 Adding heat, increases temperature
 Removing heat, decreases temperature
 Measure changes in temperature to quantify
amount of heat transferred
q = C × t
 C = heat capacity
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Heat Capacity (C )
 Amount of heat (q) required to raise
temperature of object by 1 °C
Heat Exchanged = Heat Capacity × t
q = C × t
 Units for C = J/°C or J°C
–1
 Extensive property
 Depends on two factors
1. Sample size or amount (mass)
 Doubling amount doubles heat capacity
2. Identity of substance
 Water vs. iron
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Learning Check: Heat Capacity
A cup of water is used in an experiment. Its heat
capacity is known to be 720 J/ ˚C. How much
heat will it absorb if the experimental
temperature changed from 19.2 ˚C to 23.5 ˚ C?
q = C ´ Dt
(
´ ( 4.3 °C)
q = 720 ´ 23.5 - 19.2 °C
J
°C
q = 720
J
°C
)
q = 3.1 × 103 J
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Learning Check: Heat Capacity
If it requires 4.184 J to raise the temperature of
1.00 g of water by 1.00 °C, calculate the heat
capacity of 1.00 g of water.
q
C 
t
C 1.00 g water =
Jespersen/Brady/Hyslop
4.184 J
1.00 C
= 4.18 J/°C
Chemistry: The Molecular Nature of Matter, 6E
34
Your Turn!
What is the heat capacity of 300. g of an object
if it requires 2510. J to raise the temperature of
the object by 2.00˚ C?
A. 4.18 J/˚ C
B. 418 J/˚ C
C object =
C. 837 J/˚ C
2510 J
2.00 C
= 1255 J/°C
D. 1.26 × 103 J/˚ C
E. 2.51 × 103 J/°C
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Specific Heat (s)
 Amount of Heat Energy needed to raise
temperature of 1 g substance by 1 °C
C=s×m
or
 Intensive property
C
s 
m
 Ratio of two extensive properties
 Units
 J/(g °C) or J g1 °C1
 Unique to each substance
 Large specific heat means substance releases
large amount of heat as it cools
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Learning Check
 Calculate the specific heat of water if it the heat
capacity of 100. g of water is 418 J/°C.
C
s=
m

418 J/ C
s 
 4.18 J/g °C
100. g
 What is the specific heat of water if heat
capacity of 1.00 g of water is 4.18 J/°C?

4.18 J/ C
s 
 4.18 J/g °C
1.00 g
 Thus, heat capacity is independent of amount of
substance
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
The specific heat of silver 0.235 J g–1 °C–1. What
is the heat capacity of a 100. g sample of silver?
A. 0.235 J/°C
B. 2.35 J/°C
C. 23.5 J/°C
C = s ´m
C = 0.235
D. 235 J/°C
E. 2.35 × 103 J/°C
Jespersen/Brady/Hyslop
J
g C
´ 100. g
Chemistry: The Molecular Nature of Matter, 6E
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Using Specific Heat
Heat Exchanged = (Specific Heat  mass)  t
q = s  m  t
Units = J/(g °C)  g  °C = J
 Substances with high specific heats resist changes
in temperature when heat is applied
 Water has unusually high specific heat
 Important to body (~60% water)
 Used to cushion temperature changes
 Why coastal temperatures are different from inland
temperatures
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Chemistry: The Molecular Nature of Matter, 6E
40
Learning Check: Specific Heat
Calculate the specific heat of a metal if it takes
235 J to raise the temperature of a 32.91 g
sample by 2.53 °C.
q = m ´ s ´ Dt
q
235 J
s=
=
m ´ Dt 32.91 g ´ 2.53 °C
J
s = 2.82
g °C
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
The specific heat of copper metal is 0.385 J/(g ˚C).
How many J of heat are necessary to raise the
temperature of a 1.42 kg block of copper from
25.0 ˚C to 88.5 ˚C?
A. 547 J
q = m ´ s ´ Dt
B. 1.37 × 104 J
C. 3.47 × 104 J
D. 34.7 J
E. 4.74 × 104 J
Jespersen/Brady/Hyslop
Dt = (88.5 - 25.0) °C
J
q = 1420 g ´ 0.385
´ 63.5 °C
g °C
Chemistry: The Molecular Nature of Matter, 6E
42
Direction of Heat Flow
 Heat is the energy transferred between two
objects
 Heat lost by one object has the same magnitude as
heat gained by other object
 Sign of q indicates direction of heat flow
 Heat is gained, q is positive (+)
 Heat is lost, q is negative (–)
q1 = –q2
Ex. A piece of warm iron is placed into beaker of cool
water. Iron loses 10.0 J of heat, water gains 10.0 J of
heat
qiron = –10.0 J
qwater = +10.0 J
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43
Your Turn!
A cast iron skillet is moved from a hot oven to a
sink full of water. Which of the following is
false?
A. The water heats
B. The skillet cools
C. The heat transfer for the skillet has a
negative (–) sign
D. The heat transfer for the skillet is the same
as the heat transfer for the water
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Chemistry: The Molecular Nature of Matter, 6E
44
Ex. 1 Using Heat Capacity
A ball bearing at 260˚ C is dropped into a cup
containing 250. g of water. The water warms from
25.0 to 37.3 ˚C. What is the heat capacity of the ball
bearing in J/˚ C?
Heat capacity of the cup of water = 1046 J / ˚C
qlost by ball bearing = –qgained by water
1. Determine temperature change of water
t
water
= (37.3 ˚ C – 25.0 ˚C)
= 12.3 ˚C
2. Determine how much heat gained by water
qwater = Cwater  twater = 1046 J/˚C  12.3 ˚C
= 12.87 ×103 J
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Chemistry: The Molecular Nature of Matter, 6E
45
Ex. 1 Using Heat Capacity (cont)
A ball bearing at 260 ˚C is dropped into a cup
containing 250. g of water. The water warms from
25.0 to 37.3 ˚C. What is the heat capacity of the ball
bearing in J/˚C? C of the cup of water = 1046 J /˚ C
3. Determine how much heat ball bearing lost
qball bearing = – qwater
= –12.87 × 103 J
4. Determine T change of ball bearing
t ball bearing = (37.3 ˚C – 260 ˚C)
5. Calculate C of ball bearing
= –222.7 ˚ C
q –12.87 ´ 103 J
= 57.8 J/˚ C
C=
=
Dt
-222.7 °C
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46
Ex. 2 Specific Heat Calculation
How much heat energy must you lose from a
250. mL cup of coffee for the temperature to fall
from 65.0 ˚C to a 37.0 ˚C? (Assume density of
coffee = 1.00 g/mL, scoffee = swater = 4.18 J g1
˚C1)
q = s  m  t
t = 37.0 – 65.0 ˚ C = – 28.0 ˚C
q = 4.18 J g1 ˚ C1250. mL1.00 g/mL(– 28.0 ˚ C)
q = (–29.3  103 J)
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= –29.3 kJ
Chemistry: The Molecular Nature of Matter, 6E
47
Ex. 3 Using Specific Heat
If a 38.6 g piece of gold absorbs 297 J of heat,
what will the final temperature of the gold be if the
initial temperature is 24.5 ˚C? The specific heat of
gold is 0.129 J g–1 ˚C–1.
Need to find tfinal
t = tf – ti
First use q = s  m  t to calculate t
Dt =
q
s ´m
=
297 J
0.129 J g-1 C-1 ´ 38.6 g
= 59.6 ˚ C
Next calculate tfinal
59.6 °C = tf – 24.5 °C
tf = 59.6 °C + 24.5 °C
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= 84.1 °C
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48
Your Turn!
What is the heat capacity of a container if 100. g
of water (s = 4.18 J/g °C) at 100.˚ C are added
to 100. g of water at 25 ˚C in the container and
the final temperature is 61˚ C?
A. 35 J/˚C
B. 4.12 × 103 J/˚ C
C. 21 J/˚C
D. 4.53 × 103 J/˚C
E. 50. J/˚C
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Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
What is the heat capacity of a container if 100. g of
water (s = 4.18 J/g ˚C) at 100. ˚C are added to 100. g
of water at 25 ˚C in the container and the final
temperature is 61˚C?
qlost by hot water = m × t × s
=(100. g)(61 °C – 100.˚C)(4.18 J/g ˚C) = –16,302 J
qgained by cold water = (100. g)(61 °C – 25 ˚C)(4.18J/g ˚C)
= 15,048 J
qlost by system = 15,048 J + (–16,302 J) = –1254 J
qcontainer = –q lost by system = +1254 J
q
1254 J
C =
=
Dt (61 - 25) C
Jespersen/Brady/Hyslop
= 35
J/°C
Chemistry: The Molecular Nature of Matter, 6E
Chemical Bonds and Energy
Chemical bond
 Attractive forces that bind
 Atoms to each other in molecules, or
 Ions to each other in ionic compounds
 Give rise to compound’s potential energy
Chemical energy
 Potential energy stored in chemical bonds
Chemical reactions
 Generally involve both breaking and making
chemical bonds
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Chemical Reactions
Forming Bonds
 Atoms that are attracted to each other are moved
closer together
 Decrease the potential energy of reacting system
 Releases energy
Breaking Bonds
 Atoms that are attracted to each other are forced
apart
 Increase the potential energy of reacting system
 Requires energy
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Exothermic Reaction
 Reaction where products have less chemical
energy than reactants
 Some chemical heat energy converted to kinetic
energy
 Reaction releases heat energy to surroundings
 Heat leaves the system; q is negative ( – )
 Heat energy is a product
 Reaction gets warmer, temperature increases
Ex.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) + heat
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Endothermic Reaction
 Reaction where products have more chemical
energy than reactants





Some kinetic energy converted to chemical energy
Reaction absorbs heat from surroundings
Heat added to system; q is positive (+)
Heat energy is a reactant
Reaction becomes colder,
temperature decreases
Ex. Photosynthesis
6CO2(g) + 6H2O(g) + solar energy 
C6H12O6(s) + 6O2(g)
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Bond Strength
 Measure of how much energy is needed to
break bond or how much energy is released
when bond is formed.
 Larger amount of energy equals a stronger bond
 Weak bonds require less energy to break than
strong bonds
 Key to understanding reaction energies
Ex. If reaction has
 Weak bonds in reactants and
 Stronger bonds in products
 Heat released
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Why Fuels Release Heat
 Methane and oxygen have weaker bonds
 Water and carbon dioxide have stronger bonds
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Chemistry: The Molecular Nature of Matter, 6E
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Your Turn!
Chemical energy is
A. the kinetic energy resulting from violent
decomposition of energetic chemicals.
B. the heat energy associated with combustion
reactions.
C. the electrical energy produced by fuel cells.
D. the potential energy which resides in chemical
bonds.
E. the energy living plants receive from solar
radiation.
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Heat of Reaction
 Amount of heat absorbed or released in chemical
reaction
 Determined by measuring temperature change
they cause in surroundings
Calorimeter
 Instrument used to measure temperature changes
 Container of known heat capacity
 Use results to calculate heat of reaction
Calorimetry
 Science of using calorimeter to determine heats of
reaction
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Heats of Reaction
 Calorimeter design not standard
 Depends on
 Type of reaction
 Precision desired
 Usually measure heat of reaction under one of
two sets of conditions
 Constant volume, qV
 Closed, rigid container
 Constant pressure, qP
 Open to atmosphere
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
59
What is Pressure?
 Amount of force acting on unit area
force
Pressure 
area
Atmospheric Pressure
 Pressure exerted by Earth’s atmosphere by virtue of
its weight.
 ~14.7 lb/in2
 Container open to atmosphere
 Under constant P conditions
 P ~ 14.7 lb/in2 ~ 1 atm ~ 1 bar
Jespersen/Brady/Hyslop
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Comparing qV and qP
 Difference between qV and qP can be significant
 Reactions involving large volume changes,
 Consumption or production of gas
 Consider gas phase reaction in cylinder
immersed in bucket of water
 Reaction vessel is cylinder topped by piston
 Piston can be locked in place with pin
 Cylinder immersed in insulated bucket containing
weighed amount of water
 Calorimeter consists of piston, cylinder, bucket, and
water
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Comparing qV and qP
 Heat capacity of calorimeter = 8.101 kJ/°C
 Reaction run twice, identical amounts of reactants
Run 1: qV - Constant Volume
 Same reaction run once at constant
volume and once at constant pressure
 Pin locked;
 ti = 24.00 C; tf = 28.91 C
qCal = Ct
= 8.101 J/C  (28.91 – 24.00)C = 39.8 kJ
qV = – qCal = –39.8 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Comparing qV and qP
Run 2: qP
 Run at atmospheric pressure
 Pin unlocked
 ti = 27.32 C; tf = 31.54 C
 Heat absorbed by calorimeter is
qCal = Ct
= 8.101 J/C  (31.54  27.32)C
= 34.2 kJ
qP = – qCal = –34.2 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
63
Comparing qV and qP
 qV = -39.8 kJ
 qP = -34.2 kJ
 System (reacting mixture) expands, pushes
against atmosphere, does work
 Uses up some energy that would otherwise be heat
 Work = (–39.8 kJ) – (–34.2 kJ) = –5.6 kJ
 Expansion work or pressure volume work
 Minus sign means energy leaving system
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
64
Work Convention
Work = –P × V
 P = opposing pressure against which piston
pushes
 V = change in volume of gas during expansion
 V = Vfinal – Vinitial
 For Expansion
 Since Vfinal > Vinitial
 V must be positive
 So expansion work is negative
 Work done by system
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
65
Your Turn!
Calculate the work associated with the expansion
of a gas from 152.0 L to 189.0 L at a constant
pressure of 17.0 atm.
A. 629 L atm
B. –629 L atm
C. –315 L atm
Work = –P × V
D.
E.
w = –17.0 atm × 37.0 L
171 L atm
315 L atm
Jespersen/Brady/Hyslop
V = 189.0 L – 152.0 L
Chemistry: The Molecular Nature of Matter, 6E
66
Your Turn!
A chemical reaction took place in a 6 liter
cylindrical enclosure fitted with a piston. Over the
course of the reaction, the system underwent a
volume change from 0.400 liters to 3.20 liters.
Which statement below is always true?
A.Work was performed on the system.
B.Work was performed by the system.
C.The internal energy of the system increased.
D.The internal energy of the system decreased.
E.The internal energy of the system remained
unchanged.
Jespersen/Brady/Hyslop
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First Law of Thermodynamics
 In an isolated system, the change in internal
energy (E) is constant:
 E = Ef – E i = 0
 Can’t measure internal energy of anything
 Can measure changes in energy
E is state function
E = heat + work
E = q + w
E = heat input + work input
Jespersen/Brady/Hyslop
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68
First Law of Thermodynamics
 Energy of system may be transferred as heat or
work, but not lost or gained.
 If we monitor heat transfers (q) of all materials
involved and all work processes, can predict
that their sum will be zero
 Some energy transfers will be positive, gain in
energy
 Some energy transfers will be negative, a loss in
energy
 By monitoring surroundings, we can predict
what is happening to system
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
69
First Law of Thermodynamics
 E = q + w
q is (+)
q is (–)
w is (+)
w is (–)
Heat absorbed by system (IN)
Heat released by system (OUT)
Work done on system (IN)
Work done by system (OUT)
Endothermic reaction
 E = +
Exothermic reaction
 E = –
Jespersen/Brady/Hyslop
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E is Independent of Path
q and w
 NOT path
independent
 NOT state
functions
 Depend on
how change
takes place
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
71
Discharge of Car Battery
Path a
 Short out with wrench
 All energy converted to heat, no work
 E = q
(w = 0)
Path b
 Run motor
 Energy converted to work and little heat
 E = w + q
(w >> q)
 E is same for each path
 Partitioning between two paths differs
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
72
Your Turn!
A gas releases 3.0 J of heat and then performs
12.2 J of work. What is the change in internal
energy of the gas?
A. –15.2 J
B. 15.2 J
C. –9.2 J
D. 9.2 J
E. 3.0 J
Jespersen/Brady/Hyslop
E= q+w
E = – 3.0 J + (–12.2 J)
Chemistry: The Molecular Nature of Matter, 6E
73
Your Turn!
Which of the following is not an expression for the
First Law of Thermodynamics?
A. Energy is conserved
B. Energy is neither created nor destroyed
C. The energy of the universe is constant
D.Energy can be converted from work to heat
E. The energy of the universe is increasing
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
74
Bomb Calorimeter (Constant V)
 Apparatus for
measuring E in
reactions at constant
volume
 Vessel in center with
rigid walls
 Heavily insulated vat
 Water bath
 No heat escapes
 E = qv
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
75
Ex. 4 Calorimeter Problem
When 1.000 g of olive oil is completely burned in pure
oxygen in a bomb calorimeter, the temperature of the
water bath increases from 22.000 ˚C to 26.049 ˚C.
a) How many Calories are in olive oil, per gram? The
heat capacity of the calorimeter is 9.032 kJ/˚C.
t = 26.049 ˚ C – 22.000 ˚C = 4.049 ˚C
qabsorbed by calorimeter = Ct = 9.032 kJ/°C × 4.049
˚C
= 36.57 kJ
qreleased by oil = – qcalorimeter = – 36.57 kJ
q oil
 36.57 kJ
1 kcal
1 Cal
(in cal/g) 


1.000 g
4.184 kJ 1 kcal
–8.740 Cal/g oil
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
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Ex. 4 Calorimeter Problem (cont)
b) Olive oil is almost pure glyceryl trioleate,
C57H104O6. The equation for its combustion is
C57H104O6(l) + 80O2(g)  57CO2(g) + 52H2O
What is E for the combustion of one mole of
glyceryl trioleate (MM = 885.4 g/mol)? Assume the
olive oil burned in part a) was pure glyceryl
trioleate.
885.4 g C 57H104O6
 36.57 kJ

1.000 g C 57H104O6
1 mol C 57H104O6
E = qV = –3.238 × 104 kJ/mol oil
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
77
Your Turn!
A bomb calorimeter has a heat capacity of 2.47 kJ/K.
When a 3.74×10–3 mol sample of ethylene was
burned in this calorimeter, the temperature increased
by 2.14 K. Calculate the energy of combustion for one
mole of ethylene.
qcal = Ct
A. –5.29 kJ/mol
B.
5.29 kJ/mol
C.
–148 kJ/mol
D. –1410 kJ/mol
E.
1410 kJ/mol
Jespersen/Brady/Hyslop
= 2.47 kJ/K × 2.14 K = 5.286 kJ
qethylene = – qcal = – 5.286 kJ
DE ethylene =
-5.286 kJ
-3
3.74 ´ 10 mol
Chemistry: The Molecular Nature of Matter, 6E
78
Enthalpy (H)
 Heat of reaction at constant Pressure (qP)
H = E + PV
 Similar to E, but for systems at constant P
 Now have PV work + heat transfer
 H = state function
 At constant pressure
H = E + PV = (qP + w) + PV
If only work is P–V work, w = – P V
H = (qP + w) – w = qP
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
79
Enthalpy Change (H)
H is a state function
 H = Hfinal – Hinitial
 H = Hproducts – Hreactants
 Significance of sign of H
Endothermic reaction
 System absorbs energy from surroundings
 H positive
Exothermic reaction
 System loses energy to surroundings
 H negative
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
80
Coffee Cup Calorimeter
 Simple
 Measures qP
 Open to atmosphere
 Constant P
 Let heat be exchanged
between reaction and water,
and measure change in
temperature
 Very little heat lost
 Calculate heat of reaction
 qP = Ct
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
81
Ex. 5 Coffee Cup Calorimetry
NaOH and HCl undergo rapid and exothermic reaction
when you mix 50.0 mL of 1.00 M HCl and 50.0 mL of
1.00 M NaOH. The initial t = 25.5 °C and final t =
32.2 °C. What is H in kJ/mole of HCl? Assume for
these solutions s = 4.184 J g–1°C–1. Density: 1.00 M
HCl = 1.02 g mL–1; 1.00 M NaOH = 1.04 g mL–1.
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(aq)
qabsorbed by solution = mass  s  t
massHCl = 50.0 mL  1.02 g/mL = 51.0 g
massNaOH = 50.0 mL  1.04 g/mL = 52.0 g
massfinal solution = 51.0 g + 52.0 g = 103.0 g
t = (32.2 – 25.5) °C = 6.7 °C
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
82
Ex. 5 Coffee Cup Calorimetry
qcal = 103.0 g  4.184 J g–1 °C–1  6.7 °C = 2890 J
Rounds to qcal = 2.9  103 J = 2.9 kJ
qrxn = –qcalorimeter = –2.9 kJ
1 mol HCl
0.0500 L HCl soln 
1 L HCl soln
= 0.0500 mol HCl
Heat evolved per mol HCl =
-2.9 kJ
DH =
= -58 kJ/mol
0.0500 mol HCl
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
83
Ex. 5: Coffee Cup Calorimetry
When 50.0 mL of 0.987 M H2SO4 is added to 25.0 mL
of 2.00 M NaOH at 25.0 °C in a calorimeter, the
temperature of the aqueous solution increases to
33.9 °C. Calculate H in kJ/mole of limiting
reactant. Assume: specific heat of the solution is
4.184 J/g°C, density is 1.00 g/mL, and the
calorimeter absorbs a negligible amount of heat.
Write balanced equation
2NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2H2O(aq)
Determine heat absorbed by calorimeter
masssoln = (25.0 mL + 50.0 mL) × 1.00 g/mL = 75.0 g
qsoln = 75.0 g × (33.9 – 25.0)°C × 4.184 J/g°C =
2.8×103 J Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
84
Ex. 6 Determine Limiting Reagent
1 L H2SO 4
0.987 mol H2 SO 4
50.0 mL H2 SO 4 

1000 mL H2SO 4
1 L H2SO 4
= 0.04935 mol H2SO4 present
2 mol NaOH
= 0.0987 mol NaOH
0.04935 mol H2SO4 ´
1 mol H2SO4 needed
1 L NaOH
2 mol NaOH
25.0 mL NaOH 

1000 mL NaOH
1 L NaOH
= 0.0500 mol NaOH present
NaOH is limiting
 2.8  103 J
1 kJ
H 

= –56 kJ/mol
0.0500 mol NaOH 1000 J
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
Your Turn!
A 43.29 g sample of solid is transferred from boiling
water (t = 99.8 ˚ C) to 152 g water at 22.5 ˚C in a
coffee cup. The temperature of the water rose to 24.3
˚C. Calculate the specific heat of the solid.
A. –1.1 × 103 J g–1 ˚C–1
q = m × s × t
B. 1.1 × 103 J g–1 ˚C–1
C. 1.0 J
g–1
˚C–1
D. 0.35 J g–1 ˚C–1
(
)
4.184 J
qwater = 152 g ´
´ 24.3 - 22.5 °C
g °C
E. 0.25 J g–1 ˚C–1
= 1.1 × 103 J
qsample = – qwater = – 1.1 × 103 J
s 
Jespersen/Brady/Hyslop
 1.1  10 3 J
43.29 g  (24.3 – 99.8) C
Chemistry: The Molecular Nature of Matter, 6E
86
Enthalpy Changes in Chemical
Reactions
 Focus on systems
 Endothermic
 Reactants + heat  products
 Exothermic
 Reactants  products + heat
 Want convenient way to use enthalpies to
calculate reaction enthalpies
 Need way to tabulate enthalpies of reactions
Jespersen/Brady/Hyslop
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87
The Standard State
 A standard state specifies all the necessary
parameters to describe a system. Generally this
includes the pressure, temperature, and amount
and state of the substances involved.
 Standard state in thermochemistry
 Pressure = 1 atmosphere
 Temperature = 25 °C = 298 K
 Amount of substance = 1 mol (for formation
reactions and phase transitions)
 Amount of substance = moles in an equation
(balanced with the smallest whole number
coefficients)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
88
Thermodynamic Quantities
 E and H are state functions and are also
extensive properties
 E and H are measurable changes but still
extensive properties.
 Often used where n is not standard, or specified
 E ° and H ° are standard changes and
intensive properties
 Units of kJ /mol for formation reactions and phase
changes (e.g. H °f or H °vap)
 Units of kJ for balanced chemical equations
(H °reaction)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
89
H in Chemical Reactions
Standard Conditions for H 's
 25 °C and 1 atm and 1 mole
Standard Heat of Reaction (H ° )
 Enthalpy change for reaction at 1 atm and 25 °C
Example:
N2(g) + 3H2(g)  2 NH3(g)
1.000 mol
3.000 mol
2.000 mol
 When N2 and H2 react to form NH3 at 25 °C and 1
atm 92.38 kJ released
 H= –92.38 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
90
Thermochemical Equation
 Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g)
H = –92.38 kJ
 Must give physical states of products and
reactants
 H different for different states
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l ) H °
rxn
= –890.5 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H °
rxn
= –802.3 kJ
 Difference is equal to the energy to vaporize
water
Jespersen/Brady/Hyslop
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91
Thermochemical Equation
 Write H immediately after equation
N2(g) + 3H2(g)  2NH3(g)
H= –92.38 kJ
 Assumes coefficients is the number of moles
 92.38 kJ released when 2 moles of NH3 formed
 If 10 mole of NH3 formed
5N2(g) + 15H2(g)  10NH3(g)
H= –461.9 kJ
 H° = (5 × –92.38 kJ) = – 461.9 kJ
 Can have fractional coefficients
 Fraction of mole, NOT fraction of molecule
½N2(g) + 3/2H2(g)  NH3(g)
Jespersen/Brady/Hyslop
H°rxn = –46.19 kJ
Chemistry: The Molecular Nature of Matter, 6E
92
State Matters!
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
ΔH °rxn= –2043 kJ
C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
ΔH °rxn = –2219 kJ
Note: there is difference in energy because
states do not match
If H2O(l ) → H2O(g) ΔH °vap = 44 kJ/mol
4H2O(l ) → 4H2O(g) ΔH °vap = 176 kJ/mol
Or –2219 kJ + 176 kJ = –2043 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
93
Learning Check:
Consider the following reaction:
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
ΔE ° = –2511 kJ
The reactants (acetylene and oxygen) have 2511
kJ more energy than products. How many kJ
are released for 1 mol C2H2?
-2511 kJ
´ 1 mol C2H2 =
2 mol C2H2
Jespersen/Brady/Hyslop
–1,256 kJ
Chemistry: The Molecular Nature of Matter, 6E
94
Learning Check:
Given the equation below, how many kJ are required
for 44 g CO2 (MM = 44.01 g/mol)?
6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g)
ΔH˚reaction = 2816 kJ
1 mol CO2
2816 kJ
44 g CO2 

 470 kJ
44.01 g CO2 6 mol CO2
If 100. kJ are provided, what mass of CO2 can be
converted to glucose?
6 mol CO2 44.0 g CO2
100 kJ 


2816 kJ
1 mol CO2
Jespersen/Brady/Hyslop
9.38 g
Chemistry: The Molecular Nature of Matter, 6E
95
Your Turn!
Based on the reaction
CH4(g) + 4Cl2(g)  CCl4(g) + 4HCl(g)
H˚reaction = – 434 kJ/mol CH4
What energy change occurs when 1.2 moles of
methane reacts?
A. – 3.6 × 102 kJ H = –434 kJ/mol × 1.2 mol
B. +5.2 × 102 kJ
C. – 4.3 × 102 kJ
H = –520.8 kJ = 5.2 × 102 kJ
D. +3.6 × 102 kJ
E. – 5.2 × 102 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
96
Running Thermochemical
Equations in Reverse
Consider
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H°reaction = – 802.3 kJ

Reverse thermochemical equation

Must change sign of H
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H°reaction = 802.3 kJ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
97
Reverse Thermochemical
Equation, Changes sign of H
 Makes sense:
 Get energy out when form products
 Must put energy in to go back to reactants
 Consequence of Law of Conservation of
Energy
 Like mathematical equation
 If you know H ° for reaction, you also know H °
for the reverse
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
98
Multiple Paths; Same H °
 Can often get from reactants to products by
several different paths
Products
Reactants
Intermediate A
Intermediate B
 Should get same H °
 Enthalpy is state function and path independent
 Let’s see if this is true
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
99
Ex. 7 Multiple Paths; Same H °
Path a: Single step
C(s) + O2(g)  CO2(g)
H°rxn = –393.5 kJ
Path b: Two step
Step 1: C(s) + ½O2(g)  CO(g)
H °rxn = –110.5 kJ
Step 2: CO(g) + ½O2(g)  CO2(g) H °rxn = –283.0 kJ
Net Rxn: C(s) + O2(g)  CO2(g) H °rxn = –393.5 kJ
 Chemically and thermochemically, identical results
 True for exothermic reaction or for endothermic
reaction
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
100
Ex. 8 Multiple Paths; Same H °rxn
Path a: N2(g) + 2O2(g)  2NO2(g)
H °rxn = 68 kJ
Path b:
Step 1:
N2(g) + O2(g)  2NO(g)
Step 2: 2NO(g) + O2(g)  2NO2(g)
H °rxn = 180. kJ
H °rxn = –112 kJ
Net rxn: N2(g) + 2O2(g)  2NO2(g) H °rxn =
68 kJ
Hess’s Law of Heat Summation
 For any reaction that can be written into steps, value
of H °rxn for reactions = sum of H °rxn values of
each individual step
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
101
Enthalpy Diagrams
 Graphical description of Hess’ Law
 Vertical axis = enthalpy scale
 Horizontal line =various states of reactions
 Higher up = larger enthalpy
 Lower down = smaller enthalpy
Jespersen/Brady/Hyslop
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102
Enthalpy Diagrams
 Use to measure Hrxn
 Arrow down Hrxn =
negative
 Arrow up Hrxn =
positive
 Calculate cycle
 One step process =
sum of two step
process
Ex. H2O2(l )  H2O(l ) + ½O2(g)
–286 kJ = –188 kJ + Hrxn
Hrxn = –286 kJ – (–188 kJ )
Hrxn = –98 kJ
Jespersen/Brady/Hyslop
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103
Hess’s Law
Hess’s Law of Heat Summation
 Going from reactants to products
 Enthalpy change is same whether reaction takes
place in one step or many
 Chief Use
 Calculation of H °rxn for reaction that can’t be
measured directly
 Thermochemical equations for individual steps
of reaction sequence may be combined to obtain
thermochemical equation of overall reaction
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Rules for Manipulating
Thermochemical Equations
1. When equation is reversed, sign of H°rxn must
also be reversed.
2. If all coefficients of equation are multiplied or
divided by same factor, value of H°rxn must
likewise be multiplied or divided by that factor
3. Formulas canceled from both sides of equation
must be for substance in same physical states
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Strategy for Adding Reactions
Together:
1. Choose most complex compound in equation
for one-step path
2. Choose equation in multi-step path that
contains that compound
3. Write equation down so that compound
 is on appropriate side of equation
 has appropriate coefficient for our reaction
4. Repeat steps 1 – 3 for next most complex
compound, etc.
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Strategy for Adding Reactions (Cont.)
5. Choose equation that allows you to
 cancel intermediates
 multiply by appropriate coefficient
6. Add reactions together and cancel like terms
7. Add energies together, modifying enthalpy
values in same way equation modified
 If reversed equation, change sign on enthalpy
 If doubled equation, double energy
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Ex. 9 Calculate H°rxn for
C (s, graphite)  C (s, diamond)
Given C (s, gr) + O2(g)  CO2(g) H°rxn = –394 kJ
–1[ C (s, dia) + O2(g)  CO2(g)
H°rxn = –396 kJ]
 To get desired equation, must reverse second
equation and add resulting equations
C(s, gr) + O2(g)  CO2(g)
H°rxn = –394 kJ
CO2(g)  C(s, dia) + O2(g)
H°rxn = –(–396 kJ)
C(s, gr) + O2(g) + CO2(g)  C(s, dia) + O2(g) + CO2(g)
H° = –394 kJ + 396 kJ = + 2 kJ
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Learning Check: Ex. 10
Calculate H °rxn for
2 C (s, gr) + H2(g)  C2H2(g)
Given the following:
a. C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l )
H °rxn = –1299.6 kJ
b. C(s, gr) + O2(g)  CO2(g) H °rxn = –393.5 kJ
c. H2(g) + ½O2(g)  H2O(l ) H °rxn = –285.8 kJ
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Ex.10 Calculate for
2C(s, gr) + H2(g)  C2H2(g)
2CO2(g) + H2O(l )  C2H2(g) + 5/2O2(g)
H°rxn = – (–1299.6 kJ) = +1299.6
kJ
2C(s, gr) + 2O2(g)  2CO2(g)
H°rxn =(2 –393.5 kJ) = –787.0
kJ
H2(g) + ½O2(g)  H2O(l )
kJ
H°rxn = –285.8
2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g)
 C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l )
2C(s, gr) + H2(g)  C2H2(g)
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H°rxn = +226.8 kJ
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Your Turn!
Which of the following is a statement of Hess's Law?
A.H for a reaction in the forward direction is equal to H
for the reaction in the reverse direction.
B.H for a reaction depends on the physical states of the
reactants and products.
C.If a reaction takes place in steps, H for the reaction will
be the sum of Hs for the individual steps.
D.If you multiply a reaction by a number, you multiply H
by the same number.
E.H for a reaction in the forward direction is equal in
magnitude and opposite in sign to H for the reaction in
the reverse direction.
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Your Turn!
Given the following data:
C2H2(g) + O2(g)  2CO2(g) + H2O(l ) H rxn= –1300. kJ
C(s) + O2(g)  CO2(g)
Hrxn = –394 kJ
H2(g) + O2(g)  H2O(l )
Hrxn = –286 kJ
Calculate for the reaction
2C(s) + H2(g)  C2H2(g)
A.
226 kJ
B. –1980 kJ
C.
–620 kJ
Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)
D. –226 kJ
E.
620 kJ
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Tabulating H° values
 Need to Tabulate H° values
 Major problem is vast number of reactions
 Define standard reaction and tabulate these
 Use Hess’s Law to calculate H° for any other
reaction
Standard Enthalpy of Formation, Hf°
 Amount of heat absorbed or evolved when one
mole of substance is formed at 1 atm (1 bar) and
25 °C (298 K) from elements in their standard
states
 Standard Heat of Formation
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Standard State
 Most stable form and physical state of
element at 1 atm (1 bar) and 25 °C (298 K)
Element
O
C
Standard
state
O2(g)
C (s, gr)
H
Al
Ne
H2(g)
Al(s)
Ne(g)
Note: All Hf° of
elements in their
standard states = 0
Forming element from
itself.
 See Appendix C in back of textbook and Table 7.2
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Uses of Standard Enthalpy (Heat)
of Formation, Hf°
1. From definition of Hf°, can write balanced
equations directly
Hf° of C2H5OH(l )
2C(s, gr) + 3H2(g) + ½O2(g)  C2H5OH(l )
Hf° = –277.03
kJ/mol
Hf° of Fe2O3(s)
2Fe(s) + 3/2O2(g)  Fe2O3(s) Hf° = –822.2
kJ/mol
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Your Turn!
What is the reaction that corresponds to the standard
enthalpy of formation of NaHCO3(s ), Hf ° = – 947.7
kJ/mol?
A. Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr)  NaHCO3(s)
B. Na+(g) + H+(g) + 3O2–(g) + C4+(g)  NaHCO3(s)
C. Na+(aq) + H+(aq) + 3O2–(aq) + C4+(aq)  NaHCO3(s)
D. NaHCO3(s)  Na(s) + ½H2(g) + 3/2O2(g) + C(s, gr)
E. Na+(aq) + HCO3–(aq)  NaHCO3(s)
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Using Hf°
2. Way to apply Hess’s Law without needing to manipulate
thermochemical equations
Sum of all H°f
Sum of all
H°reaction = of all of the
– H°f of all of
products
the reactants
Consider the reaction:
aA + bB  cC + dD
H°reaction = c × H°f(C) + d × H°f(D)
– {a×H°f(A) + b×H°f(B)}
 H°rxn has units of kJ because
 Coefficients  heats of formation have units of mol  kJ/mol
(
) (
)
o
DH rxn
= åéëDH fo products ´ moles of product ùû éDH o reactants ´ moles of reactant ù
åë f
û
(
) (
H°rxn has units of kJ
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)
H°f has units of kJ/mol
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Ex. 11 Calculate H°rxn Using Hf°
Calculate H°rxn using Hf° data for the reaction
SO3(g)  SO2(g) + ½O2(g)
1. Multiply each Hf° (in kJ/mol) by the number of moles in the
equation
2. Add the Hf° (in kJ/mol) multiplied by the number of moles in the
equation of each product
3. Subtract the Hf° (in kJ/mol) multiplied by the number of moles in
the equation of each reactant
o
DH rxn
= åéëDH fo products ´ moles of product ùû éDH o reactants ´ moles of reactant ù
åë f
û
(
) (
H°rxn has units of kJ
H°f has units of kJ/mol

H rxn
 H f (SO 2 ( g )) 
1
(
)
) (
)



H
(O
(
g
)
)


H
f
2
f (SO 3 ( g ))
2
DHrxn = -297 kJ/mol + 1 2 (0 kJ/mol) - ( - 396 kJ/mol)
H°rxn = 99 kJ
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Learning Check
Calculate H°rxn using Hf° for the reaction
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(l )
DHrxn = 4DH f NO (g ) + 6DH f H O(l ) - 4DH f NH ( g ) - 7DH f O ( g )
2
2
3
2
DH rxn = 4 mol(34 kJ/mol) + 6 mol( - 285.9 kJ/mol)
- 4 mol( - 46.0 kJ/mol) - 7 mol(0 kJ/mol)
H°rxn = [136 – 1715.4 + 184] kJ
H°rxn = –1395 kJ
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Check Using Hess’s Law
4 ×[NH3(g)  ½N2(g) + 3/2H2(g)] – 4 × Hf°(NH3,
g)
–7 × Hf°(O2,
7 ×[ O2(g)  O2(g) ]
g)
4 ×[ O2(g) + ½ N2(g)  NO2(g)]
g)
4 × Hf°(NO2,
6 ×[ H2(g) + ½ O2(g)  H2O(l ) ] 6 × Hf°(H2O,
DHl)rxn = 4DH f NO (g ) + 6DH f H O(l ) - 4DH f NH ( g ) - 7DH f O ( g )
2
2
3
2
4NH
+ before
7O2(g)  4NO2(g) + 6H2O(l )
3(g)as
Same
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Other Calculations
 Don’t always want to know H°rxn
 Can use Hess’s Law and H°rxn to calculate
Hf° for compound where not known
Ex. Given the following data, what is the value of
Hf°(C2H3O2–, aq)?
Na+(aq) + C2H3O2–(aq) + 3H2O(l )  NaC2H3O2·3H2O(s)
H°rxn = –19.7 kJ/mol
Na aq)
H
f
= –239.7 kJ/mol
NaC2H3O2•3H2O(s)
H
f
= 710.4 kJ/mol
H2O(l)
H
f
= 285.9 kJ/mol
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Ex. 13 cont.
H°rxn = Hf° (NaC2H3O2·3H2O, s) – Hf° (Na+,
aq) – Hf° (C2H3O2–, aq) – 3Hf° (H2O, l )
Rearranging
Hf°(C2H3O2–, aq) = Hf°(NaC2H3O2·3H2O, s) –
Hf°(Na+, aq) – H°rxn – 3Hf° (H2O, l)
Hf°(C2H3O2–, aq) =
–710.4 kJ/mol – (–239.7kJ/mol) – (–19.7
kJ/mol) – 3(–285.9 kJ/mol)
= +406.7 kJ/mol
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Learning Check
Calculate H for this reaction using Hf° data.
2Fe(s) + 6H2O(l)  2Fe(OH)3(s) + 3H2(g)
Hf° 0
–285.8
–696.5
0
H°rxn = 2×Hf°(Fe(OH)3, s) + 3×Hf°(H2, g)
– 2× Hf°(Fe, s) – 6×Hf°(H2O, l )
H°rxn = 2 mol× (– 696.5 kJ/mol) + 3×0 – 2×0
– 6 mol× (–285.8 kJ/mol)
H°rxn = –1393 kJ + 1714.8 kJ
H°rxn = 321.8 kJ
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Learning Check
Calculate H°rxn for this reaction using Hf° data.
CO2(g) + 2H2O(l )  2O2(g) + CH4(g)
Hf°
–393.5
–285.8
0
– 74.8
H°rxn = 2×Hf°(O2, g) + Hf°(CH4, g)
–Hf°(CO2, g) – 2× Hf°(H2O, l )
H°rxn = 2 × 0 + 1 mol × (–74.8 kJ/mol) – 1 mol
× (–393.5 kJ/mol) – 2 mol × (–285.8 kJ/mol)
H°rxn = –74.8 kJ + 393.5 kJ + 571.6 kJ
H°rxn = 890.3 kJ
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