Energy and Chemical Energy

Chemistry: The Molecular Nature
of Matter
Seventh Edition
Jespersen; Hyslop
Chapter 6
Energy and Chemical Change
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Chapter in Context (1 of 2)
• Explore the difference between potential and kinetic
energy
• Apply the principle of the conservation of energy
• Examine the connection between energy, heat, and
temperature
• Learn about state functions
• Determine the amount of heat exchanged from the
temperature change of an object
• Understand exo− an endothermic
reactions
2
Copyright ©2014 John Wiley & Sons, Inc.

Chapter in Context (2 of 2)
• Learn and apply the first law of thermodynamics
• Understand the difference between the heat of reaction
under constant pressure versus constant volume
• Explore the utility of and assumptions in
thermochemical equations
• Uses Hess’s law to predict enthalpies of reaction
• Determine and use standard heats of reaction to solve
problems
3

Thermochemistry
• Study of energies given off by or absorbed by reactions
Thermodynamics
• Study of heat transfer or heat flow
Energy (E )
o Ability to do work or to transfer heat
Kinetic Energy (KE)
o Energy of motion
o KE = 2
1
2
mv
4

Potential Energy (PE)
• Stored energy
• Exists in natural attractions
and repulsions
o Gravity
o Positive and negative charges
o Springs
Chemical Energy
o PE possessed by chemicals
o Stored in chemical bonds
o Breaking bonds requires
energy
o Forming bonds releases energy H2 + O2 → H2O
5

Your Turn! 1 (1 of 2)
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
6

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. Answer: A pencil rests on a desk
E. A pencil falls to the floor
7

Factors Affecting Potential Energy (1 of 2)
Increase Potential Energy
• Pull apart objects that attract each
other
o A book is attracted to the earth by
gravity
o North and south poles of magnets
• Push together objects that repel each
other
o Spring compressed
o Same poles on two magnets
o Two like charges
8

Factors Affecting Potential Energy (2 of 2)
Decrease Potential Energy
• Objects that attract each other come together
o Book falls
o North and south poles of two
magnets
• Objects that repel each
other move apart
o North poles on two magnets
o Spring released
o Two like charges
9

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
10

Which of the following represents a decrease in the
A. A book is raised six feet above the floor
B. Answer: A ball rolls downhill
C. Two electrons come close together
D. A spring is stretched completely
E. Two atomic nuclei approach each other
11

Which of the following represents an increase in the
A. +
Na Na + e−
→
B. A periodic table falls off the wall
C. +
Ca2 + 2e Ca
−
→
D. A firecracker explodes
E. Gasoline is burned and creates CO2 and H2O
12

Which of the following represents an increase in the
A. Answer: +
Na Na + e−
→
B. A periodic table falls off the wall
C. +
Ca2 + 2e Ca
−
→
D. A firecracker explodes
E. Gasoline is burned and creates CO2 and H2O
13

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 = Potential Energy + Kinetic Energy
14

Units of Energy (1 of 2)
Joule (J)
• KE possessed by 2 kg object moving at speed of 1 m/s.
( )
2
1 1 m
1 J 2 kg
2 1 s
 
=  
 
2
2
1 kg m
1 J
s
⋅
=
• If calculated value is greater than 1000 J, use
kilojoules (kJ)
• 1 kJ = 1000 J
15

Units of Energy (2 of 2)
A calorie (cal)
• Energy needed to raise the temperature of 1 g H2O by
1 °C
o 1 cal = 4.184 J (exactly)
o 1 kcal = 1000 cal
o 1 kcal = 4.184 kJ
A nutritional Calorie (Cal)
o note capital C
o 1 Cal = 1000 cal = 1 kcal
o 1 kcal = 4.184 kJ
16

Which is a unit of energy?
A. Pascal
B. Newton
C. Joule
D. Watt
E. Ampere
17

Which is a unit of energy?
A. Pascal
B. Newton
C. Answer: Joule
D. Watt
E. Ampere
18

Convert 175.2 kJ into nutritional Calories
A. 3
41.87 10 Cal
×
B. 41.87 cal
C. 733.0 Cal
D. 41.87 Cal
3
733.0 10 Cal
×
E.
19

Convert 175.2 kJ into nutritional Calories
A. 3
41.87 10 Cal
×
B. 41.87 cal
C. 733.0 Cal
D. Answer: 41.87 Cal
3
733.0 10 Cal
×
E.
175.2 kJ 1 kcal
×
4.184 kJ
1 Cal
1 kcal
× = 41.87 Cal
20

Convert 57.9 kJ into calories
A. 13800 Cal
B. 13.8 cal
C. 242 kcal
D. 242 Cal
E. 13.8 kcal
21

Convert 57.9 kJ into calories
A. 13800 Cal
B. 13.8 cal
C. 242 kcal
D. 242 Cal
E. Answer: 13.8 kcal
57.9 kJ 1000 J
×
1 kJ
1 cal
4.184 J
× =
13800 cal
kcal
1000 cal
× = 13.8 kcal
22

Temperature vs. Heat
Temperature
• Proportional to average kinetic energy of object’s particles
• Higher average kinetic energy means
o Higher temperature
o 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
23

Heat Transfer
• Hot and cold objects placed in contact
o Molecules in hot object moving faster
• KE transfers from hotter to colder object
o A decrease in average KE of hotter object
o An increase in average KE of colder object
• Over time
o Average KEs of both objects becomes the same
o Temperature of both becomes the same
hot cold
24

Heat
• Pour hot coffee into cold cup
o Heat flows from hot coffee to cold cup
o Faster coffee molecules bump into wall of cup
o Transfer kinetic energy
o Eventually, the cup and the coffee reach the same temperature
Thermal Equilibrium
o When both cup and coffee reach same average kinetic energy
and same temperature
• Energy transferred through heat comes from object’s internal
energy
25

Internal Energy (E)
• Sum of energies of all particles in system
o E = total energy of system
o E = potential + kinetic = PE + KE
Change in Internal Energy
o ΔE = Efinal – Einitial
o Δ means change
o final – initial
o What we can actually measure
• Want to know change in E associated with given process
26

ΔE, Change in Internal Energy (2 of 2)
• For reaction: reactants → products
• ΔE = Eproducts – Ereactants
o Can use to do something useful
• Work
• Heat
• If system absorbs energy during reaction
o Energy coming into system has a positive sign (+)
o Final energy > initial energy
Example: Photosynthesis or charging battery
o As system absorbs energy
• Increase potential energy
• Available for later use
27

Kinetic Molecular Theory
• Kinetic Molecular Theory tells us
Temperature
o Related to average kinetic energy of particles in object
Internal energy
o Related to average total molecular kinetic energy
o Includes molecular potential energy
Average kinetic energy
o Implies distribution of kinetic energies among molecules in
object
28

Temperature and Average Kinetic Energy
In a large collection of gas molecules
o Wide distribution of kinetic energy (KE)
o Small number with KE = 0
• Collisions can momentarily stop a molecule’s motion
o Very small number with very high KE
o Most molecules intermediate KEs
o Collisions tend to average kinetic energies
• Result is a distribution of energies
29

Distribution of Kinetic Energy (1 of 2)
1: lower temperature
2: higher temperature
At higher temperature, distribution shifts to higher kinetic energy
30

Distribution of Kinetic Energy (2 of 2)
Temperature
o Average KE of all atoms and molecules in object
o Average speed of particles
o Kelvin temperature of sample
• ( )
K 2
1
2
Avg KE avg
T mv
∝ =
o 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 so v = 0
31

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
o Same average kinetic energy
o But very different potential energy
32

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.
33

Which statement about kinetic energy (KE) is true?
A. Answer: 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.
34

ΔE, Change in Internal Energy (1 of 2)
• ΔE = Eproducts – Ereactants
o Energy change can appear entirely as heat
o Can measure heat
o Can’t measure Eproduct or Ereactant
o Importantly, we can measure ΔE
o 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
35

State of Object or System
• Complete list of properties that specify object’s current condition
For Chemistry
o Defined by physical properties
• Chemical composition
• Substances
• Number of moles
• Pressure
• Temperature
• Volume
36

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 :
o Internal energy E
o Pressure P
o Temperature t
o Volume V
37

State of an object
• If tc = 25 °C, tells us all we need to know
o 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:
o Δt = tfinal – tinitial
o Don’t need to know how this occurred, just need to know initial and
final values
• What does Δt tell us?
o Change in average KE of particles in object
o Change in object’s total KE
o Heat energy
38

Defining the System
System
• What we are interested in studying
o Reaction in beaker
Surroundings
• Everything else
o Room in which reaction is run
Boundary
• Separation between system and surroundings
o Visible Example: Walls of beaker
o Invisible Example: Line separating warm and cold fronts
39

Three Types of Systems (1 of 2)
Open System
o Open to atmosphere
o Gain or lose mass and energy across
boundary
o Most reactions done in open systems
Open system
Closed System
o Not open to atmosphere
o Energy can cross boundary, but
mass cannot
Closed system
40

Three Types of Systems (2 of 2)
Isolated System
o No energy or matter can cross boundary
o Energy and mass are constant
Example: Thermos bottle
Isolated system
41

Adiabatic Process
Adiabatic Process
• Process that occurs in isolated system
• Process where neither energy nor matter crosses the
system/surrounding boundary
42

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
43

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. Answer: absorb or lose energy, but not mass
44

Heat (q)
• Cannot measure heat directly
• Heat (q) gained or lost by an object
o Directly proportional to temperature change (∆t) it undergoes
o Adding heat, increases temperature
o Removing heat, decreases temperature
• Measure changes in temperature to quantify amount of heat
transferred
q = C × Δt
• C = heat capacity
45

Heat Capacity (C )
• Amount of heat (q) required to raise temperature of object by 1 °C
Heat Exchanged = Heat Capacity × ∆t
q = C × ∆t
o Units for C 1
J / C or J C−
= ° ⋅ °
• Extensive property
• Depends on two factors
1. Sample size or amount (mass)
• Doubling amount doubles heat capacity
2. Identity of substance
• Water versus iron
46

Learning Check: Heat Capacity (1 of 2)
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 t
= ×V
( )
J
C
720 23.5 19.2 C
q °
= × − °
J
C
720
q °
= 4.3 C
× °
( )
3
3.1 10 J
q
= ×
47

Learning Check: Heat Capacity (2 of 2)
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
=
D
1.00 g water o
4.184 J
1.00 C
C = = o
4.18 J / C
48

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. 837 J/ °C
D. 3
1.26 10 J / °C
×
E. 3
2.51 10 J / °C
×
49

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. 837 J/ °C
D. Answer: 3
1.26 10 J / °C
×
E. 3
2.51 10 J / °C
×
object
2510 J
2.00 C
C
= =
°
1255 J/ °C
50

A copper mug has a heat capacity of 77.5 J/ °C. After adding hot water
to the mug, the temperature of the mug changed from 77.0 °F to 185 °F.
How much heat did the mug absorb from the water?
A. 3
4.65 10 J
×
B. 1.29 J
C. 3
8.37 10 J
×
D. 3
5.97 10 J
×
E. 3
1.43 10 J
×
51

A copper mug has a heat capacity of 77.5 J/ °C. After adding hot water
to the mug, the temperature of the mug changed from 77.0 °F to 185 °F.
How much heat did the mug absorb from the water?
A. Answer: 3
4.65 10 J
×
B. 1.29 J
C. 3
8.37 10 J
×
D. 3
5.97 10 J
×
E. 3
1.43 10 J
×
77.5 J / C
q C t t
= × = ×
D ° D
Note units: temps. must be in °C
ti = 25.0 °C, tf = 85.0 °C,
So ∆t = 85.0 °C – 25.0 °C = 60.0 °C
77.5 J
C
q =
°
60.0 C
× °
3
4.65 10 J
= ×
52

Specific Heat (s)
• Amount of heat energy needed to raise temperature of 1 g
substance by 1 °C
C = s × m or
C
s
m
=
• Intensive property
• Ratio of two extensive properties
• Units
• ( ) or 1 1
J/ g C J g C
− −
° °
• Unique to each substance
• Large specific heat means substance releases large amount of heat
as it cools
53

Learning Check 4
• Calculate the specific heat of water if the heat capacity of 100. g
of water is 418 J/°C.
C
s
m
=
418 J/°C
100. g
=
s 4.18 J/g°C
• 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
1.00 g
=
s 4.18 J/g°C
• Thus, heat capacity is independent of amount of substance
54

The specific heat of silver 1 1
0.235 J g C .
− −
° 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
D. 235 J/°C
E. 3
2.35 10 / C
× °
55

The specific heat of silver 1 1
0.235 J g C .
− −
° What is the
heat capacity of a 100. g sample of silver?
A. 0.235 J/°C
B. 2.35 J/°C
C. Answer: 23.5 J/°C
D. 235 J/°C
E. 3
2.35 10 / C
× °
C = s × m
0.235
=
J
C
g
100.
×
°
g
C
23.5
=
°
J
C
C
56

Specific Heats of Some Substances
Table 6.1 Specific Heats
57

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
o Important to body (~60% water)
• Used to cushion temperature changes
o Why coastal temperatures are different from inland temperatures
58

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 t
= × × ∆
235 J
32.91 g 2.35 C
q
s
m t
= =
× ∆ × °
J
2.82
g C
s =
°
59

Your Turn! (1 of 2)
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
B. 4
1.37 10 J
×
C. 4
3.47 10 J
×
D. 34.7 J
E. 4
4.74 10 J
×
60

Your Turn! (2 of 2)
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
B. 4
1.37 10 J
×
C. Answer: 4
3.47 10 J
×
D. 34.7 J
E. 4
4.74 10 J
×
q m s t
= × ×∆
(88.5 25.0) C
t
∆
= − °
1420 g
q =
J
0.385
g
×
C
°
63.5 C
× °
4
3.47 10 J
q
= ×
61

Direction of Heat Flow
• Heat is the energy transferred between two objects
o Heat lost by one object has the same magnitude as heat gained by
other object
• Sign of q indicates direction of heat flow
o Heat is gained, q is positive (+)
o Heat is lost, q is negative (–)
q1 = −q2
Example: 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
62

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 has the same sign as
the heat transfer for the water
63

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. Answer: The heat transfer for the skillet has the same
sign as the heat transfer for the water
64

Example 1: Using Heat Capacity (1 of 2)
A ball bearing at 260.0 °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
3
12.87 10 J
= ×
65

Example 1: Using Heat Capacity (2 of 2)
A ball bearing at 260.0 °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
1. Determine how much heat ball bearing lost
qball bearing = – qwater 3
12.87 10 J
= ×
2. Determine T change of ball bearing
∆tball bearing = (37.3 °C – 260.0 °C) = –222.7 °C
3. Calculate C of ball bearing
3
–12.87 10 J
222.7 C
×
= =
∆ − °
q
C
t
= 57.8 J/°C
66

Example 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 37.0 °C? (Assume
density of coffee = 1.00 g/mL, scoffee = swater = 4.18 J/g°C)
q = s × m × Δt
Δt = 37.0 − 65.0 ℃ = − 28.0 ℃
4.18 J/ g
=
q C
° 250.mL
× 1.00 g
× / mL × 28.0 C
− °
( )
( )
3
29.3 10 J
=
− ×
q = −29.3 kJ
67

Example 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°C.
Need to find tfinal ∆t = tf – ti
First use q = s × m × ∆t to calculate ∆t
q
t
s m
∆ =
×
297 J
0.129 J
=
/ g C 38.6 g
° ×
= 59.6 °C
Next calculate tfinal
59.6 °C = tf – 24.5 °C
tf = 59.6 °C + 24.5 °C = 84.1 °C
68

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 25.0 °C in the
container and the final temperature is 61.0 °C?
A. 35 J/°C
B. 3
4.12 10 J/ C
× °
C. 21 J/°C
D. 3
4.53 10 J/ C
× °
E. 50. J/°C
69

A. Answer: 35 J/°C
B. 3
4.12 10 J/ C
× °
C. 21 J/°C
D. 3
4.53 10 J/ C
× °
E. 50. J/°C
70

Your Turn! − Solution
qlost by hot water = m × ∆t × s
=(100. g )(61.0 °C 100.°C
− )(4.18 J/ g°C ) = 1.63 ? 104 J
= (100.g
−
gained by cold water
q )(61.0 °C 25.0 °C
− )(4.18J/ g °C )
= 1.50 × 104 J
4 4 3
= 1.50 10 J + ( 1.63 10 J) = 1.3 10 J
× − × − ×
lost by system
q
= − 3
container lost by system = +1.3×10 J
q q
3
1.3 10
(61.0 25.0)
q J
C
t C
×
= =
∆ −  = 36 J/°C
71

Chemical Bonds and Energy
Chemical bond
• Attractive forces that bind
o Atoms to each other in molecules, or
o Ions to each other in ionic compounds
o Give rise to compound’s potential energy
Chemical energy
o Potential energy stored in chemical bonds
Chemical reactions
• Generally involve both breaking and making chemical bonds
72

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
73

Exothermic Reaction
• Reaction where products have less chemical energy than reactants
o Some chemical heat energy converted to kinetic energy
o Reaction releases heat energy to surroundings
o Heat leaves the system; q is negative ( − )
o Heat energy is a product
o Reaction gets warmer, temperature increases
Example:
4 2 2 2
CH ( ) 2O ( ) CO ( ) 2H O( ) heat
g g g g
+ → + +
74

Endothermic Reaction
• Reaction where products have more chemical energy than reactants
o Some kinetic energy converted to chemical energy
o Reaction absorbs heat from surroundings
o Heat added to system; q is positive (+)
o Heat energy is a reactant
o Reaction becomes colder,
temperature decreases
Example: Photosynthesis
6CO2(g) + 6H2O(g) + solar energy → C6H12O6(s) + 6O2(g)
75

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
o Weak bonds require less energy to break than strong bonds
• Key to understanding reaction energies
Example: If reaction has
o Weak bonds in reactants and
o Stronger bonds in products
o Heat released
76

Why Fuels Release Heat
• Methane and oxygen have weaker bonds
• Water and carbon dioxide have stronger bonds
77

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
78

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. Answer: the potential energy which resides in
chemical bonds
E. the energy living plants receive from solar radiation
79

Which of the following is an endothermic process?
A. Na+ + e− → Na
B. wood burning
C. CH4(g) → C(g) + 4H(g)
D. a bomb exploding
E. water condensing
80

Which of the following is an endothermic process?
A. Na+ + e− → Na
B. wood burning
C. Answer: CH4(g) → C(g) + 4H(g)
D. a bomb exploding
E. water condensing
Attraction of charges releases heat
Produces heat (i.e., heat released)
Breaking bonds requires heat
Produces heat (i.e., heat released)
Heat must be released
81

Heat of Reaction
• Amount of heat absorbed or released in chemical reaction
• Determined by measuring temperature change they cause in
surroundings
Calorimeter
o Instrument used to measure temperature changes
o Container of known heat capacity
o Use results to calculate heat of reaction
Calorimetry
o Science of using calorimeter to determine heats of reaction
82

Heats of Reaction
• Calorimeter design not standard
o Depends on
• Type of reaction
• Precision desired
• Usually measure heat of reaction under one of two sets of
conditions
o Constant volume, qV
• Closed, rigid container
o Constant pressure, qP
• Open to atmosphere
83

What is Pressure?
• Amount of force acting on unit area
force
Pressure
area
=
Atmospheric Pressure
o Pressure exerted by Earth’s atmosphere by virtue of its weight.
o 2
14.7 Ib /in
:
• Container open to atmosphere
o Under constant P conditions
o 2
P 14.7 Ib /in 1 atm 1 bar
: : :
84

Comparing qV and qP (1 of 4)
• Difference between qV and qP can be significant
• Reactions involving large volume changes,
o Consumption or production of gas
• Consider gas phase reaction in cylinder immersed in bucket of
water
o Reaction vessel is cylinder topped by piston
o Piston can be locked in place with pin
o Cylinder immersed in insulated bucket containing weighed
amount of water
o Calorimeter consists of piston, cylinder, bucket, and water
85

• Heat capacity of calorimeter = 8.101 kJ/°C
• Reaction run twice, identical amounts of reactants
Run 1: qV − Constant Volume
o Same reaction run once at constant
volume and once at constant pressure
o Pin locked
o ti = 24.00 °C; tf = 28.91 °C
qCal = CΔt
8.101 J/ C
= ° ( )
28.91 24.00 C
× − ° 39.8 kJ
=
qV = − qCal = −39.8 kJ
86

Run 2: qP
• Run at atmospheric pressure
• Pin unlocked
• ti = 27.32 °C; tf = 31.54 °C
• Heat absorbed by calorimeter is
qCal = CΔt
8.101 / C
J
= ° 31.54 27.3
( 2 C
)
− °
= 34.2 kJ
qP = – qCal = –34.2 kJ
87

• qV = −39.8 kJ
• qP = −34.2 kJ
• System (reacting mixture) expands, pushes against atmosphere,
does work
o Uses up some energy that would otherwise be heat
o Work = (−39.8 kJ) − (−34.2 kJ) = −5.6 kJ
• Expansion work or pressure volume work
o Minus sign means energy leaving system
88

Work Convention
Work = −P × ΔV
• P = opposing pressure against which piston pushes
• ΔV = change in volume of gas during expansion
• ΔV = Vfinal − Vinitial
• For expansion
o Since Vfinal > Vinitial
o ΔV must be positive
o So expansion work is negative
o Work done by system
89

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
D. 171 L atm
E. 315 L atm
90

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. Answer: −629 L atm
C. −315 L atm
D. 171 L atm
E. 315 L atm
Work = −P × ΔV
ΔV = 189.0 L − 152.0 L = 37.0 L
w = −17.0 atm × 37.0 L
91

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.
92

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. Answer: 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.
93

First Law of Thermodynamics (1 of 3)
• In an isolated system, the change in internal energy (ΔE) is
constant:
ΔE = Ef − Ei = 0
• Can’t measure internal energy of anything
• Can measure changes in energy
E is state function
ΔE = heat + work
ΔE = q + w
ΔE = heat input + work input
94

• 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
o Some energy transfers will be positive, gain in energy
o Some energy transfers will be negative, a loss in energy
• By monitoring surroundings, we can predict what is happening to
system
95

• ΔE = q + w
q is (+) Heat absorbed by system (IN)
q is (−) Heat released by system (OUT)
w is (+) Work done on system (IN)
w is (−) Work done by system (OUT)
Endothermic reaction
• ΔE = +
Exothermic reaction
• ΔE = –
96

ΔE is Independent of Path
q and w
• NOT path
independent
• NOT state
functions
• Depend on
how change
takes place
97

Discharge of Car Battery
Path a
• Short out with wrench
• All energy converted to heat, no work
o ∆E = q (w = 0)
Path b
• Run motor
• Energy converted to work and little heat
o ∆E = w + q (w >> q)
• ∆E is same for each path
o Partitioning between two paths differs
98

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
99

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. Answer: −15.2 J
B. 15.2 J
C. −9.2 J
D. 9.2 J
E. 3.0 J
E = q + w
E = − 3.0 J + (−12.2 J)
100

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
101

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. Answer: The energy of the universe is increasing
102

A reaction contracts by 1.534 L under a constant pressure of 2.134
atm while releasing 200.7 J of heat to the surrounding. What is the
change in internal energy of the system? 1.000 Latm = 101.3 J
A.131.0 J
B. 532.4 J
C. −131.0 J
D. −532.4 J
E. 331.6 J
103

A reaction contracts by 1.534 L under a constant pressure of 2.134
atm while releasing 200.7 J of heat to the surrounding. What is the
change in internal energy of the system? 1.000 Latm = 101.3 J
A.Answer: 131.0 J
B. 532.4 J
C. −131.0 J
D. −532.4 J
E. 331.6 J
E = q + w = −200.7 J + w
Work = −P ×ΔV
= −2.134 atm × 1.534 L = 3.274 Latm
3.274 Latm
=
101.3 J
×
1.000 Latm
331.7 J
=
E = q + w = −200.7 J + 331.7
= 131.0 J
104

Bomb Calorimeter (Constant V)
• Apparatus for measuring
ΔE in reactions at constant
volume
• Vessel in center with rigid
walls
• No change in volume, so
∆V = 0 so P∆V = 0
• Heavily insulated vat
o Water bath
o No heat escapes
• ∆E = q − P∆V
• ∆E = q + 0 = qv
• subscript ‘v’ emphasizes
constant volume
105

Example 4: Calorimeter Problem (1 of 2)
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.
Δt = 26.049 °C – 22.000 °C = 4.049 °C
absorbed by calorimeter = = 9.032 kJ / C
∆ °
q C t 4.049 C
× °
= 36.57 kJ
qreleased by oil = − qcalorimeter = − 36.57 kJ
36.57 kJ
(in cal/g)
−
=
oil
q
1 kcal
1.000 g
×
4.184 kJ
1 Cal
1 kcal
×
−8.740 Cal/g oil
106

Example 4: Calorimeter Problem (2 of 2)
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.
57 104 6
36.57 kJ
1.000 g C H O
− 57 104 6
885.4 g C H O
×
57 104 6
1 mol C H O
− 4
= 3.238×10 kJ /mol oil
V
E q
Δ =
107

A bomb calorimeter has a heat capacity of 2.47 kJ/K.
When a –3
3.74 10 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.
A. −5.29 kJ/mol
B. 5.29 kJ/mol
C. −148 kJ/mol
D. −1410 kJ/mol
E. 1410 kJ/mol
108

A bomb calorimeter has a heat capacity of 2.47 kJ/K.
When a –3
3.74 10 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.
A. −5.29 kJ/mol
B. 5.29 kJ/mol
C. −148 kJ/mol
D. Answer: −1410 kJ/mol
E. 1410 kJ/mol
qcal = C∆t
2.47 kJ/ K
= 2.14 K
× 5.286 kJ
=
qethylene = – qcal = – 5.286 kJ
ethylene 3
5.286 kJ
3.74 10 mol
E −
−
∆ =
×
ethylene
E 1410 kJ/mol
∆ =
−
109

Enthalpy (H)
• Heat of reaction at constant pressure (qP)
H = E + PV
• Similar to E, but for systems at constant P
• Now have PΔV work + heat transfer
• H = state function
• At constant pressure
ΔH = ΔE + PΔV = (qP + w) + PΔV
If only work is P−V work, w = − P ΔV
ΔH = (qP + w) − w = qP
110

Enthalpy Change (ΔH)
H is a state function
o ΔH = Hfinal – Hinitial
o ΔH = Hproducts – Hreactants
• Significance of sign of ΔH
Endothermic reaction
o System absorbs energy from surroundings
o ΔH positive
Exothermic reaction
o System loses energy to surroundings
o ΔH negative
111

Coffee Cup Calorimeter
• Simple
• Measures qP
• Open to atmosphere
o Constant P
• Let heat be exchanged between
reaction and water, and measure
change in temperature
o Very little heat lost
• Calculate heat of reaction
o qP = CΔt
112

Example 5: Coffee Cup Calorimetry (1 of 2)
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 ΔH in kJ/mole of
HCl? Assume for these solutions –1 –1
4.184 J g C .
= 
s
Density: 1.00 M –1 –1
HCl 1.02 gmL ; 1.00 NaOH 1.04 gmL .
M
= =
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(aq)
qabsorbed by solution = mass × s × Δt
HCl
mass 50.0 mL
= 1.02 g/ mL
× 51.0 g
=
NaOH
mass 50.0 mL
= 1.04 g/ mL
× 52.0 g
=
massfinal solution = 51.0 g + 52.0 g = 103.0 g
Δt = (32.2 − 25.5) °C = 6.7 °C
113

Example 5: Coffee Cup Calorimetry (2 of 2)
cal 103.0 g
q = –1
4.184J g
× C
°
–1
6.7 C
× ° 2890 J
=
3
cal
Rounds to 2.9 10 J 2.9 kJ
q = × =
qrxn = −qcalorimeter = −2.9 kJ
0.0500 L HCl soln
1 mol HCl
1 L HCl soln
×
= 0.0500 mol HCl
Heat evolved per mol HCl =
2.9 kJ
0.0500 mol HCl
−
∆ =
H = −58 kJ/mol
114

Example 6: 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 ΔH 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
soln
mass 25.0 mL
= + 50.0 mL
( ) 1.00 g/mL
× 75.0 g
=
soln 75.0 g
q = ( )
33.9 – 25.0 C
× ° 4.184 J/ g
× C
° = 3
2.8×10 J
115

Example 6: Determine Limiting Reagent
2 4
50.0 mL H SO
2 4
1 L H SO
×
2 4
1000 mL H SO
2 4
2 4
0.987 mol H SO
1 L H SO
×
= 0.04935 mol H2SO4 present
2 4
0.04935 mol H SO
2 4
2 mol NaOH
1 mol H SO
× = 0.0987 mol NaOH
needed
25.0 mL NaOH
1 L NaOH
×
1000 mL NaOH
2 mol NaOH
1 L NaOH
×
= 0.0500 mol NaOH present
NaOH is limiting
3
2.8 10 J
H
− ×
∆ =
1 kJ
0.0500 mol NaOH 1000 J
× = −56 kJ/mol
116

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. 3 –1 –1
–1.1 10 J g C
× 
B. 3 –1 –1
1.1 10 J g C
× 
C. –1 –1
1.0 J g C

D. –1 –1
0.35 J g C

E. –1 –1
0.25 J g C

117

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. 3 –1 –1
–1.1 10 J g C
× 
B. 3 –1 –1
1.1 10 J g C
× 
C. –1 –1
1.0 J g C

D. –1 –1
0.35 J g C

E. –1 –1
0.25 J g C

q = m × s × Δt
water 152 g
q =
4.184 J
g
×
°C
( )
24.3 22.5 °C
× −
3
1.1 10 J
= ×
3
– – 1.1 10 J
sample water
q q
= = ×
3
1.1 10 J
43.29 g (24.3 – 99.8) C
− ×
=
× 
s
118

Enthalpy Changes in Chemical Reactions
• Focus on systems
• Endothermic
o Reactants + heat → products
• Exothermic
o Reactants → products + heat
• Want convenient way to use enthalpies to
calculate reaction enthalpies
• Need way to tabulate enthalpies of reactions
119

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
o Pressure = 1 atmosphere
o Temperature = 25 °C = 298 K
o Amount of substance = 1 mol (for formation reactions and
phase transitions)
o Amount of substance = moles in an equation (balanced with the
smallest whole number coefficients)
120

Thermodynamic Quantities
• E and H are state functions and are also
extensive properties
• ΔE and ΔH are measurable changes but still
extensive properties
o Often used where n is not standard, or specified
o ΔE° and ΔH° are standard changes and
intensive properties
o Units of kJ /mol for formation reactions and phase
changes (Example: ΔH°f or ΔH°vap)
o Units of kJ for balanced chemical equations
(ΔH°reaction)
121

ΔH in Chemical Reactions
Standard Conditions for ΔH 's
o 25 °C and 1 atm and 1 mole
Standard Heat of Reaction (ΔH° )
o 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
o When N2 and H2 react to form NH3 at 25 °C and
1 atm 92.38 kJ released
o ΔH = −92.38 kJ
122

Thermochemical Equation (1 of 2)
• Write ΔH immediately after equation
N2(g) + 3H2(g) → 2NH3(g) ΔH = −92.38 kJ
• Must give physical states of products
and reactants
o ΔH different for different states
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l ) ΔH° rxn = −890.5 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH° rxn = −802.3 kJ
o Difference is equal to the energy to vaporize
water
123

Thermochemical Equation (2 of 2)
• Write ∆H immediately after equation
N2(g) + 3H2(g) → 2NH3(g) ∆H= –92.38 kJ
• Assumes coefficients is the number of moles
o 92.38 kJ released when 2 moles of NH3 formed
o If 10 mole of NH3 formed
5N2(g) + 15H2(g) → 10NH3(g) ∆H= –461.9 kJ
• ∆H° = (5 × –92.38 kJ) = – 461.9 kJ
o Can have fractional coefficients
• Fraction of mole, NOT fraction of molecule
½N2(g) + 3/2H2(g) → NH3(g) ∆H°rxn = –46.19 kJ
124

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
125

Learning Check 5
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?
2 2
2511 kJ
2 mol C H
−
2 2
1 mol C H
× = –1,256 kJ
126

Learning Check 6
Given the equation below, how many kJ are required for 44 g
CO2 (MM = 44.01 g/mol) to react with H2O?
6CO2(g) + 6H2O → C6H12O6(s) + 6O2(g)
ΔH°reaction = 2816 kJ
44 g CO
2
2
1 mol CO
×
44.01 g CO 2
2816 kJ
6 mol CO
×
2
= 470 kJ
• If 100. kJ are provided, what mass of CO2 can be converted
to glucose?
100 kJ
2
6 mol CO
×
2816 kJ
2
2
44 g CO
1 mol CO
× = 9.4 g
127

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. 2
3.6 10 kJ
− ×
B. 2
5.2 10 kJ
×
C. 2
4.3 10 kJ
− ×
D. 2
3.6 10 kJ
×
E. 2
5.2 10 kJ
− ×
128

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. 2
3.6 10 kJ
− ×
B. 2
5.2 10 kJ
×
C. 2
4.3 10 kJ
− ×
D. 2
3.6 10 kJ
×
E. Answer: 2
5.2 10 kJ
− ×
434 kJ/ mol
H
∆ =
− 1.2 mol
×
2
520.8 kJ 5.2 10 kJ
H
∆ =
− = ×
129

Rxn. 1: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
Rxn. 2: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
Why does rxn. 2 release more heat than rxn. 1?
A. It shouldn’t, and the ∆H values should be equal
B. More reactants must have been used in rxn. 2
C. In rxn. 1 some of the heat of the reaction is used up
converting liquid water to gas, so less heat can be given off.
D. Liquids always have lower temperatures than gases
130

Rxn. 1: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(g)
Rxn. 2: C3H8(g) + 5O2(g) → 3 CO2(g) + 4 H2O(l )
Why does rxn. 2 release more heat than rxn. 1?
A. It shouldn’t, and the ∆H values should be equal
B. More reactants must have been used in rxn. 2
C. Answer: In rxn. 1 some of the heat of the reaction is used up
converting liquid water to gas, so less heat can be given off.
D. Liquids always have lower temperatures than gases
131

Reversing Thermochemical Equations
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
132

Reversing Thermochemical Equations
Changes sign of ∆H
• Makes sense:
o Get energy out when form products
o Must put energy in to go back to reactants
• Consequence of Law of Conservation of Energy
o Like mathematical equation
o If you know ∆H° for reaction, you also know ∆H° for
the reverse
133

Multiple Paths; Same ∆H°
• Can often get from reactants to products by several different
paths
o Should get same ∆H°
o Enthalpy is state function and path independent
o Let’s see if this is true
134

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
• Chemically and thermochemically, identical results
• True for exothermic and endothermic reactions
135

Ex. 8: Multiple Paths; Same ∆Hrxn
Path a:
N2(g) + 2O2(g) → 2NO2(g) ∆H°rxn = 68 kJ
Path b:
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
136

Enthalpy Diagrams (1 of 2)
• Graphical description of Hess’ Law
o Vertical axis = enthalpy scale
o Horizontal line =various states of reactions
o Higher up = larger enthalpy
o Lower down = smaller enthalpy
137

Enthalpy Diagrams (2 of 2)
• Use to measure ∆Hrxn
o Arrow down ∆Hrxn = negative
o Arrow up ∆Hrxn = positive
• Calculate cycle
o One step process = sum of two step process
Example: H2O2(l ) → H2O(l ) + ½O2(g)
–286 kJ = –188 kJ + ∆Hrxn
∆Hrxn = –286 kJ – (–188 kJ)
∆Hrxn = –98 kJ
138

What is your change in altitude if you climb up a 7,005 ft.
mountain, drive a car from that spot down 2,003 ft, then parachute
down another 11,508 ft. into a canyon?
Draw a diagram that reflects your changes and your calculation.
A. −6506 ft.
B. 20516 ft.
C. 16310 ft.
D. −20516 ft.
E. No change
139

What is your change in altitude if you climb up a 7,005 ft.
mountain, drive a car from that spot down 2,003 ft, then parachute
down another 11,508 ft. into a canyon?
Draw a diagram that reflects your changes and your calculation.
A. Answer: −6506 ft.
B. 20516 ft.
C. 16310 ft.
D. −20516 ft.
E. No change
Note: Change is negative because altitude decreased Final state is
lower than initial state
140

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
o Calculation of ∆H°rxn for reaction that can’t be measured
directly
o Thermochemical equations for individual steps of reaction
sequence may be combined to obtain thermochemical equation
of overall reaction
141

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
142

Strategy for Adding Reactions
Together: (1 of 2)
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
o is on appropriate side of equation
o has appropriate coefficient for our reaction
4. Repeat steps 1 to 3 for next most complex compound, etc.
143

Strategy for Adding Reactions
Together: (2 of 2)
5. Choose equation that allows you to
o cancel intermediates
o multiply by appropriate coefficient
6. Add reactions together and cancel like terms
7. Add energies together, modifying enthalpy values in same
way equation modified
o If reversed equation, change sign on enthalpy
o If doubled equation, double energy
144

Example 9: Calculate ∆Ho
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
∆H° = –394 kJ + 396 kJ = + 2 kJ
145

Learning Check: Example 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
146

Example 10: Calculate ∆Ho
rxn for
2C(s, gr) + H2(g) → C2H2(g)
147

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.
148

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. Answer: 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.
149

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
D. –226 kJ
E. 620 kJ
150

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. Answer: 226 kJ
B. –1980 kJ
C. –620 kJ
D. –226 kJ
E. 620 kJ
∆Hrxn = +1300. kJ + 2(–394 kJ) + (–286 kJ)
151

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
152

Standard State
• Most stable form and physical state of element at 1 atm
(1 bar) and 25 °C (298 K)
Element Standard state
O O2(g)
C C (s, gr)
H H2(g)
Al Al(s)
Ne Ne(g)
Note: All ∆Hf° of
elements in their
standard states = 0
Forming element from
itself.
• See Appendix in back of textbook and Table 6.2
153

Uses of Standard Enthalpy (Heat) of
Formation, ∆Hf°
1. From definition of ∆Hf°, can write balanced
equations directly
°
2 5
Δ of C H OH( )
f
H l
1
2 2 5
2
2C( , ) + 3H ( ) + O2( ) C H OH( ) = 277.03 kJ/mol
f
s gr g g l H
→ ∆ −

°
2 3
Δ of Fe O ( )
f
H s
3
2 2 3
2
2Fe( ) + O ( ) Fe O ( ) = 822.2 kJ/mol
f
s g s H
→ ∆ −

154

Which reaction corresponds to the standard enthalpy of
formation of NaHCO3(s), ∆Hf° = – 947.7 kJ/mol?
A. 3
1
2 2 3
2 2
Na( ) + H ( ) + O ( ) + C( , ) NaHCO ( )
s g g s gr s
→
B. + + 2 4+
3
Na ( ) + H ( ) + 3O ( ) + C ( ) NaHCO ( )
g g g g s
−
→
C. + + 2 4+
3
Na ( ) + H ( ) + 3O ( ) + C ( ) NaHCO ( )
aq aq aq aq s
−
→
D. 3
1
3 2 2
2 2
NaHCO ( ) Na( ) + H ( ) + O ( ) + C( , )
s s g g s gr
→
E. +
3 3
Na ( ) + HCO ( ) NaHCO ( )
aq aq s
−
→
155

formation of NaHCO3(s), ∆Hf° = – 947.7 kJ/mol?
A. Answer: 3
1
2 2 3
2 2
Na( ) + H ( ) + O ( ) + C( , ) NaHCO ( )
s g g s gr s
→
B. + + 2 4+
3
Na ( ) + H ( ) + 3O ( ) + C ( ) NaHCO ( )
g g g g s
−
→
C. + + 2 4+
3
Na ( ) + H ( ) + 3O ( ) + C ( ) NaHCO ( )
aq aq aq aq s
−
→
D. 3
1
3 2 2
2 2
NaHCO ( ) Na( ) + H ( ) + O ( ) + C( , )
s s g g s gr
→
E. +
3 3
Na ( ) + HCO ( ) NaHCO ( )
aq aq s
−
→
156

formation of C4H3Br2NO2(ℓ)?
A. 4C(s, gr) + 3H(g) + 2Br(s) + N(g) + 2O(g) → C4H3Br2NO2(ℓ)
B. 8C(s, gr) + 3H2(g) + 2Br2(g) + N2(g) + 2O2 (g) → 2C4H3Br2NO2(ℓ)
C. 3 1
2 2 2 2 4 3 2 2
2 2
4C( , ) + H ( ) + Br ( ) + N ( ) + O ( ) C H Br NO ( )
s gr g g g →
 
D. 3 1
2 2 2 2 4 3 2 2
2 2
4C( , ) + H ( ) + Br ( ) + N ( ) + O ( ) C H Br NO ( )
s gr g s g g → 
157

formation of C4H3Br2NO2(ℓ)?
A. 4C(s, gr) + 3H(g) + 2Br(s) + N(g) + 2O(g) → C4H3Br2NO2(ℓ)
B. 8C(s, gr) + 3H2(g) + 2Br2(g) + N2(g) + 2O2 (g) → 2C4H3Br2NO2(ℓ)
C. Answer: 3 1
2 2 2 2
2 2
4 3 2 2
4C( , ) + H ( ) + Br ( ) + N ( ) + O ( )
C H Br NO ( )
s gr g g g →


D. 3 1
2 2 2 2 4 3 2 2
2 2
4C( , ) + H ( ) + Br ( ) + N ( ) + O ( ) C H Br NO ( )
s gr g s g g → 
158

The standard enthalpy of formation of sulfur dioxide is
−296.9 kJ. What is DH for the formation of 16.03 g of
sulfur dioxide in its standard state from its elements in
their standard states?
A. 148.4 kJ
B. −296.9 kJ
C. −4,759 kJ
D. −148.4 kJ
E. 593.6 kJ
159

The standard enthalpy of formation of sulfur dioxide is
−296.9 kJ. What is DH for the formation of 16.03 g of
sulfur dioxide in its standard state from its elements in
their standard states?
A. 148.4 kJ
B. −296.9 kJ
C. −4,759 kJ
D. Answer: −148.4 kJ
E. 593.6 kJ
2
16.03 g SO
2
mol SO
×
2
32.07 g SO 2
296.9 kJ
mol SO
−
×
= −148.4 kJ
160

Using ∆Hf°
2. Way to apply Hess’s Law without needing to manipulate thermochemical
equations
reaction
Sum of all Sum of all
= of all of of all of
the products the reactants
f f
H H H
   
   
∆ ∆ − ∆
   
   
   
  
Consider the reaction:
aA + bB → cC + dD
reaction f f
f f
= c × ( ) + d × ( )
{a× ( ) + b× ( )}
H H C H D
H A H B
∆ ∆ ∆
− ∆ ∆
  
 
• ∆H°rxn has units of kJ because
• Coefficients × heats of formation have units of mol × kJ/mol
( ) ( )
( ) ( )
o o
rxn f
o
f
products moles of product
reactants moles of reactant
H H
H
 
∆ = ∆ × −
 
 
∆ ×
 
∑
∑
∆H°rxn has units of kJ ∆H°f has units of kJ/mol
161

Ex. 11: Calculate ∆Ho
rxn Using ∆Hf°
Calculate ∆H°rxn using ∆Hf° data for the reaction
1
3 2 2
2
SO ( ) SO ( ) + O ( )
g g 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 o
rxn f
o
f
H H
H
 
∆ = ∆ × −
 
 
∆ ×
 
∑
∑
∆H°rxn has units of kJ
∆H°f has units of kJ/mol
1
( ) ( ) ( )
rxn f 2 f 2 f 3
2
(SO ) (O ) (SO )
g g g
H H H H
∆ = ∆ + ∆ − ∆
   
1
rxn 2
297 kJ/mol (0 kJ/mol) ( 396 kJ/mol)
H
∆ =− + − −

∆H°rxn = 99 kJ
162

Learning Check 1
Calculate ∆H°rxn using ∆Hf° for the reaction
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l )
2 2 3 2
rxn f NO ( ) f H O( ) f NH ( ) f O ( )
4 6 4 7
g l g g
H H H H H
∆ = ∆ + ∆ − ∆ − ∆
    
rxn 4 mol
H
∆ =

(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
163

Check Using Hess’s Law
2 2 3 2
rxn f NO ( ) f H O( ) f NH ( ) f O ( )
4 6 4 7
g l g g
H H H H H
∆ = ∆ + ∆ − ∆ − ∆
    
Same as before
164

Example 12: Other Calculations (1 of 2)
• Don’t always want to know ∆H°rxn
• Can use Hess’s Law and ∆H°rxn to calculate ∆Hf° for compound
where not known
Example: 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) ∆Ho
f = –239.7 kJ/mol
NaC2H3O2•3H2O(s) ∆Ho
f = 710.4 kJ/mol
H2O(l) ∆Ho
f = 285.9 kJ/mol
165

Example 12: Other Calculations (2 of 2)
∆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
166

Learning Check 2
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 )
rxn 2 mol
H°
∆ = ( 696.5 kJ/ mol
× − ) 1 mol
− + 3 0 2 0
6 mol
× − ×
− ( 285.8 kJ/ mol
× − )
∆H°rxn = –1393 kJ + 1714.8 kJ
∆H°rxn = 321.8 kJ
167

Learning Check 3
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 )
rxn 2 0 1 mol
H°
∆ = × + ( 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
168

Calculate ΔHf° for FeO(s) using the information below. ΔHf°
values are shown below each substance.
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ∆Ho = 21.9 kJ
−1120.9 kJ −110.5 kJ ?? – 393.5 kJ
A. 272.0 kJ
B. −816.0 kJ
C. −272.0 kJ
D. 26.00 J
E. −38.60 kJ
169

Calculate ΔHf° for FeO(s) using the information below. ΔHf°
values are shown below each substance.
Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ∆Ho = 21.9 kJ
−1120.9 kJ −110.5 kJ ?? – 393.5 kJ
A. 272.0 kJ
B. −816.0 kJ
C. Answer: −272.0 kJ
D. 26.00 J
E. −38.60 kJ
170

Your Turn! sol’n
( ) ( )
( ) ( )
o o
rxn f
o
f
H H
H
 
∆ = ∆ × −
 
 
∆ ×
 
∑
∑
ΔH°rxn = [3ΔHf° (FeO, s) + ΔHf°(O2, g)]
– [ΔHf°(Fe3O4, s) + ΔHf°(CO, g)]
Important
+21.9 kJ = [3ΔHf° (FeO, s) + −393.5 kJ)]
– [−1120.9 kJ + − 110.5 kJ)]
+21.9 kJ = [3ΔHf° (FeO, s) + 837.9 kJ]
−816.0 kJ = 3ΔHf° (FeO, s)
−272.0 kJ = ΔHf° (FeO, s)
171

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from the use of the information contained herein.
172

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