Chapter #10
Energy
Energy
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There are things that do not have mass and volume.
These things fall into a category we call energy.
Energy is anything that has the capacity to do work.
Work if force times distance
Work units are Joules = kgm2/s2
Force is a push and has units of Newtons (kgm/s2)
Although chemistry is the study of matter, matter is
effected by energy.
– It can cause physical and/or chemical changes in
matter.
Law of Conservation of Energy
• “Energy can neither be created nor destroyed.”
• The total amount of energy in the universe is
constant. There is no process that can increase
or decrease that amount.
• However, we can transfer energy from one
place in the universe to another, and we can
change its form.
Matter Possesses Energy
• When a piece of matter possesses
energy, it can give some or all of it
to another object.
– It can do work on the other object.
• All chemical and physical changes
result in the matter changing
energy.
Kinetic and Potential Energy
• Potential energy is energy that is
stored.
– Water flows because gravity pulls it
downstream.
– However, the dam won’t allow it to
move, so it has to store that energy.
• Kinetic energy is energy of motion,
or energy that is being transferred
from one object to another.
– When the water flows over the
dam, some of its potential energy
is converted to kinetic energy of
motion.
Forms of Energy
• Electrical
– Kinetic energy associated with the flow of electrical charge.
• Heat or Thermal Energy
– Kinetic energy associated with molecular motion.
• Light or Radiant Energy
– Kinetic energy associated with energy transitions in an atom.
• Nuclear
– Potential energy in the nucleus of atoms.
• Chemical
– Potential energy in the attachment of atoms or because of their
position.
Converting Forms of Energy
• When water flows over the dam, some of its
potential energy is converted to kinetic energy.
– Some of the energy is stored in the water
because it is at a higher elevation than the
surroundings.
• The movement of the water is kinetic energy.
• Along the way, some of that energy can be used
to push a turbine to generate electricity.
– Electricity is one form of kinetic energy.
• The electricity can then be used in your home.
For example, you can use it to heat cake batter
you mixed, causing it to change chemically and
storing some of the energy in the new molecules
that are made.
Using Energy
• We use energy to accomplish all kinds of
processes, but according to the Law of
Conservation of Energy we don’t really use it up!
• When we use energy we are changing it from one
form to another.
– For example, converting the chemical energy
in gasoline into mechanical energy to make
your car move.
“Losing” Energy
• If a process was 100% efficient, we could
theoretically get all the energy transformed into
a useful form.
• Unfortunately we cannot get a 100% efficient
process.
• The energy “lost” in the process is energy
transformed into a form we cannot use.
There’s No Such Thing as a Free Ride
• When you drive your car, some of the chemical
potential energy stored in the gasoline is released.
• Most of the energy released in the combustion of
gasoline is transformed into sound or heat energy
that adds energy to the air rather than move your car
down the road.
Units of Energy
• Calorie (cal) is the amount of energy needed to raise
one gram of water by 1 °C.
– kcal = energy needed to raise 1000 g of water 1 °C.
– food calories = kcals.
Energy Conversion Factors
1 calorie (cal)
1 Calorie (Cal)
1 kilowatt-hour (kWh)
=
=
=
4.184 joules (J)
1000 calories (cal)
3.60 x 106 joules (J)
Energy Use
Energy Required to
Raise Temperature
of 1 g of Water by
1°C
Energy
Required to
Light 100-W
Bulb for 1
Hour
Energy Used
by Average
U.S. Citizen
in 1 Day
4.18
3.6 x 105
9.0 x 108
calorie (cal)
1.00
8.60 x 104
2.2 x 108
Calorie (Cal)
1.00 x 10-3
86.0
2.2 x 105
kWh
1.1 x 10-6
0.100
2.50 x 102
Unit
joule (J)
Chemical Potential Energy
• The amount of energy stored in a material is its
chemical potential energy.
• The stored energy arises mainly from the
attachments between atoms in the molecules and
the attractive forces between molecules.
• When materials undergo a physical change, the
attractions between molecules change as their
position changes, resulting in a change in the
amount of chemical potential energy.
• When materials undergo a chemical change, the
structures of the molecules change, resulting in a
change in the amount of chemical potential
energy.
Energy Changes in Reactions
• Chemical reactions happen most readily when
energy is released during the reaction.
• Molecules with lots of chemical potential
energy are less stable than ones with less
chemical potential energy.
• Energy will be released when the reactants
have more chemical potential energy than the
products.
Exothermic Processes
• When a change results in the release of energy it is
called an exothermic process.
• An exothermic chemical reaction occurs when the
reactants have more chemical potential energy than the
products.
• The excess energy is released into the surrounding
materials, adding energy to them.
– Often the surrounding materials get hotter from the
energy released by the reaction.
An Exothermic Reaction
Surroundings
reaction
Potential energy
Reactants
Amount
of energy
released
Products
Endothermic Processes
• When a change requires the absorption of energy it
is called an endothermic process.
• An endothermic chemical reaction occurs when the
products have more chemical potential energy than
the reactants.
• The required energy is absorbed from the
surrounding materials, taking energy from them.
– Often the surrounding materials get colder due to
the energy being removed by the reaction.
An Endothermic Reaction
Surroundings
reaction
Potential energy
Products
Amount
of energy
absorbed
Reactants
Thermochemical Equations
When a chemical or physical change takes place
energy is either lost of gained. A Thermochemical
equation describes this change. Equations gaining
energy are called endothermic and equations losing
energy are called exothermic.
Thermochemical Equations
When a chemical or physical change takes place
energy is either lost of gained. A Thermochemical
equation describes this change. Equations gaining
energy are called endothermic and equations losing
energy are called exothermic.
Examples:
C3H6O (l ) 4O2 (g)
H2O (l)
3CO2(g) + 3 H2O (g)
Exothermic
H2O (g)
ΔH = 44.01 kj
Endothermic
ΔH = -1790 kj
Thermochemical Conversions
How many kj of heat are released when 709 g of C3H6O
are burned?
Thermochemical Conversions
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) 4O2 (g)
709 g C3H6O
3CO2(g) + 3 H2O (g)
mole C3H6O
58.1 g C3H6O
ΔH = -1790 kj
Thermochemical Conversions
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) 4O2 (g)
709 g C3H6O
3CO2(g) + 3 H2O (g)
mole C3H6O
58.1 g C3H6O
ΔH = -1790 kj
Thermochemical Conversions
How many kj of heat are released when 709 g of C3H6O
are burned?
C3H6O (l ) 4O2 (g)
709 g C3H6O
3CO2(g) + 3 H2O (g)
mole C3H6O
58.1 g C3H6O
1790 kj
mole C3H6O
ΔH = -1790 kj
= 21800 kj
Energy and the Temperature of Matter
• The amount the temperature of an object
increases depends on the amount of heat energy
added (q).
– If you double the added heat energy the
temperature will increase twice as much.
• The amount the temperature of an object
increases depending on its mass.
– If you double the mass, it will take twice as
much heat energy to raise the temperature the
same amount.
Heat Capacity
• Heat capacity is the amount of heat a substance
must absorb to raise its temperature by 1 °C.
– cal/°C or J/°C.
– Metals have low heat capacities; insulators
have high heat capacities.
• Specific heat = heat capacity of 1 gram of the
substance.
– cal/g°C or J/g°C.
– Water’s specific heat = 4.184 J/g°C for liquid.
• Or 1.000 cal/g°C.
• It is less for ice and steam.
26
Specific Heat Capacity
• Specific heat is the amount of energy required to
raise the temperature of one gram of a substance by 1
°C.
• The larger a material’s specific heat is, the more
energy it takes to raise its temperature a given
amount.
• Like density, specific heat is a property of the type of
matter.
– It doesn’t matter how much material you have.
– It can be used to identify the type of matter.
• Water’s high specific heat is the reason it is such a
good cooling agent.
– It absorbs a lot of heat for a relatively small mass.
Specific Heat Capacities
Substance
Specific Heat
J/g°C
Aluminum
0.903
Carbon (dia)
0.508
Carbon (gra)
0.708
Copper
0.385
Gold
0.128
Iron
0.449
Lead
0.128
Silver
0.235
Ethanol
2.42
Water (l)
4.184
Water (s)
2.03
Water (g)
2.02
Heat Gain or Loss by an Object
• The amount of heat energy gained or lost by an
object depends on 3 factors: how much material
there is, what the material is, and how much the
temperature changed.
Practice—Calculate the Amount of Heat Released
When 7.40 g of Water Cools from 49° to 29 °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g 20 °C
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g 20 °C
g- °C
= 620 j
Thermodynamics
First Law of thermodynamics: Energy cannot be
created nor destroyed.
Mathematical Statement: ΔE = q + w
ΔE is the change in internal energy of matter.
q is the amount of heat into the system
w is the amount of work on the system
The system is the test tube, beaker or flask
The Second Law of Thermodynamics: Entropy of the
universe is always increasing.
Entropy
Entropy is a word used to describe the spreading of
matter. For example consider the Universe, is it
expanding?
Entropy
Entropy is a word used to describe the spreading of
matter. For example consider the Universe, is it
expanding? Yes
How about where you live, does matter spread
there?
Entropy
Entropy is a word used to describe the spreading of
matter. For example consider the Universe, is it
expanding? Yes
How about where you live, does matter spread
there? Yes
How about our natural resources, are they being
spread about? Yes
Entropy
Entropy is a word used to describe the spreading of
matter. For example consider the Universe, is it
expanding? Yes
How about where you live, does matter spread
there? Yes
How about our natural resources, are they being
spread about?
The symbol for entropy is S and the change in
entropy is ΔS. ΔS>0, means the spreading of
matter.
House cleaning would have ΔS<0 (more order)
Spontaneous Processes
Some process proceed without constant outside
intervention.
Spontaneous Processes
Some process proceed with constant outside
intervention.
For example air escaping out of a tire with a hole in it.
Have you ever observed air flowing into a tire with a
hole in it?
How about aging? Have you ever observed someone
getting younger?
Spontaneous Processes
Some process proceed with constant outside
intervention.
For example air escaping out of a tire with a hole in it.
Have you ever observed air flowing into a tire with a
hole in it?
How about aging? Have you ever observed someone
getting younger?
How about shuffling a deck of cards? Does it ever
become organized like it came from the factory?
Spontaneous Processes
Some process proceed with constant outside
intervention.
For example air escaping out of a tire with a hole in it.
Have you ever observed air flowing into a tire with a
hole in it?
How about aging? Have you ever observed someone
getting younger?
How about shuffling a deck of cards? Does it ever
become organized like it came from the factory?
Theoretically, it is possible to shuffle a deck of cards
until it has the same order as a new deck of cards.
Spontaneous Processes
Some process proceed with constant outside
intervention.
For example air escaping out of a tire with a hole in it.
Have you ever observed air flowing into a tire with a
hole in it?
How about aging? Have you ever observed someone
getting younger?
How about shuffling a deck of cards? Does it ever
become organized like it came from the factory?
Theoretically, it is possible to shuffle a deck of cards
until it has the same order as a new deck of cards.
How many shuffles until a deck has a new order?
Spontaneous Processes
Some process proceed with constant outside
intervention.
For example air escaping out of a tire with a hole in it.
Have you ever observed air flowing into a tire with a
hole in it?
How about aging? Have you ever observed someone
getting younger?
How about shuffling a deck of cards? Does it ever
become organized like it came from the factory?
Theoretically, it is possible to shuffle a deck of cards
until it has the same order as a new deck of cards.
How many shuffles until a deck has a new order? 1064
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous?
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous? Yes What is being
spread, energy or matter?
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous? Yes What is being
spread, energy or matter? Matter, right?
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous? Yes What is being
spread, energy or matter? Matter, right? Is burning
of gasoline spontaneous?
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous? Yes What is being
spread, energy or matter? Matter, right? Is burning
of gasoline spontaneous? Yes, after it starts it
does not stop. What is spread here?
Predicting Spontaneity
Spontaneous process are always accompanied with
spreading of energy or matter; one or the other or
both. Enthalpy is a term used to describe spreading
of energy. When energy is being spread ΔH<0
Is melting of ice spontaneous? Yes What is being
spread, energy or matter? Matter, right? Is burning
of gasoline spontaneous? Yes, after it starts it
does not stop. What is spread here? Both matter
and energy!
Practice
During a strenuous workout, a student generates
2000 kJ of heat energy. What mass of water
would have to evaporate from the student’s skin
to dissipate this much heat?
Practice
During a strenuous workout, a student generates
2000 kJ of heat energy. What mass of water
would have to evaporate from the student’s skin
to dissipate this much heat?
g
2257 j
Practice
During a strenuous workout, a student generates
2000 kJ of heat energy. What mass of water
would have to evaporate from the student’s skin
to dissipate this much heat?
g
2257 j
10 3 j
kj
Practice
During a strenuous workout, a student generates
2000 kJ of heat energy. What mass of water
would have to evaporate from the student’s skin
to dissipate this much heat?
g
2257 j
10 3 j
kj
2000 kj
Practice
During a strenuous workout, a student generates
2000 kJ of heat energy. What mass of water
would have to evaporate from the student’s skin
to dissipate this much heat?
g
2257 j
10 3 j
kj
2000 kj
= 886 g water
Practice
5.53 From Text
Exactly 10 mL of water at 25oC was added to a
hot iron skillet. All of the water was converted
into steam at 100oC. If the mass of the pan was
1.20 kg and the molar heat capacity of iron is
25.19 J/mol•oC, what was the temperature
change of the skillet?
Sample Problem Solution
mole-°C
25.19 j
Sample Problem Solution
mole-°C 55.85 g
25.19 j mole
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
g-°C
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j 10.0 g
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Evaporating 10.0 mL of water
2257 j 10.0 g =22570 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25.19 j mole
103 g 1.20 kg
Now the energy required to heat 10mL of water from 25°C to
100°C and then to vaporize the water is outlined below.
Heating from 25°C to 100°C
4.184 j 10.0g 75 °C
= 3138 j
g-°C
Now Combine
Evaporating 10.0 mL of water 3138 j + 22570 j = 25708j
2257 j 10.0 g =22570 j
g
Sample Problem Solution
mole-°C 55.85 g kg
25708 j
25.19 j mole
103 g 1.20 kg
= 47.5 °C
Hess’s Law
• Hess’s law states that the enthalpy change of a
reaction that is the sum of two or more reactions
is equal to the sum of the enthalpy changes of the
constituent reactions.
Calculations via Hess’s Law
1. If a reaction is reversed, H sign changes.
O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
N2(g) +
2. If the coefficients of a reaction are multiplied by an
integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g)
H = 540 kJ
Example
Calculate the enthalpy change for
C2H4(g) + H2(g)
C2H6(g) using the following data.
H2(g) + 1/2O2(g)
H2O(l)
-285.8 kJ
C2H4(g) + 3O2(g)
2H2O(l) + 2CO2(g) -1411 kJ
C2H6(g) + 7/2O2(g)
3H2O(l) + 2CO2(g) -1560 kJ
Example
Calculate the enthalpy change for
C2H4(g) + H2(g)
C2H6(g) using the following data.
H2(g) + 1/2O2(g)
C2H4(g) + 3O2(g)
3H2O(l) + 2CO2(g)
H2O(l)
-285.8 kJ
2H2O(l) + 2CO2(g) -1411 kJ
C2H6(g) + 7/2O2(g) +1560 kJ
H2(g) +1/2O2(g)+C2H4(g)+3O2(g)+3H2O(l)+2CO2(g)
H2O(l)+ 2H2O(l) + 2CO2(g)+ C2H6(g) + 7/2O2(g) -136.8
simplify
C2H4(g) + H2(g)
C2H6(g) -136.8 kj
The End