chapter #10 energy. there are things that do not have mass and volume. these things fall into a...
TRANSCRIPT
Chapter #10Energy
Energy• 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 Factors1 calorie (cal) = 4.184 joules (J)1 Calorie (Cal) = 1000 calories (cal)
1 kilowatt-hour (kWh) = 3.60 x 106 joules (J)
Energy Use
Unit
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
joule (J) 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
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
Pot
enti
al e
nerg
y
Reactants
Products
Surroundings
reaction
Amount of energy released
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
Pot
enti
al e
nerg
y
Products
Reactants
Surroundings
reaction
Amount of energy absorbed
Thermochemical EquationsWhen 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 EquationsWhen 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) 3CO2(g) + 3 H2O (g) ΔH = -1790 kjExothermic
H2O (l) H2O (g) ΔH = 44.01 kjEndothermic
Thermochemical ConversionsHow many kj of heat are released when 709 g of C3H6O are burned?
Thermochemical ConversionsHow many kj of heat are released when 709 g of C3H6O are burned?
709 g C3H6O
C3H6O (l ) 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj
58.1 g C3H6O
mole C3H6O
Thermochemical ConversionsHow many kj of heat are released when 709 g of C3H6O are burned?
709 g C3H6O
C3H6O (l ) 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj
58.1 g C3H6O
mole C3H6O
Thermochemical ConversionsHow many kj of heat are released when 709 g of C3H6O are burned?
709 g C3H6O
C3H6O (l ) 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj
58.1 g C3H6O
mole C3H6Omole C3H6O1790 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.
26
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.
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 CapacitiesSubstance 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 jg- °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 jg- °C
7.40 g
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 jg- °C
7.40 g 20 °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 jg- °C
7.40 g 20 °C= 620 j
ThermodynamicsFirst 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 systemw is the amount of work on the systemThe 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? YesHow 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? YesHow 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? YesHow 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 ProcessesSome process proceed without constant outside intervention.
Spontaneous ProcessesSome 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 ProcessesSome 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 ProcessesSome 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 ProcessesSome 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 ProcessesSome 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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 SpontaneitySpontaneous 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
kj2000 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
kj2000 kj
= 886 g water
Practice5.53 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
25.19 j
mole-°C
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
1.20 kg
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
1.20 kg
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C= 3138 j
Evaporating 10.0 mL of water
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C= 3138 j
Evaporating 10.0 mL of water
2257 jg
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C= 3138 j
Evaporating 10.0 mL of water
2257 jg
10.0 g
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C= 3138 j
Evaporating 10.0 mL of water
2257 jg
10.0 g =22570 j
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
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 jg-°C
10.0g 75 °C= 3138 j
Evaporating 10.0 mL of water
2257 jg
10.0 g =22570 j
Now Combine 3138 j + 22570 j = 25708j
Sample Problem Solution
25.19 j
mole-°C
mole
55.85 g
103 g
kg
1.20 kg
25708 j
= 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.
1. If a reaction is reversed, H sign changes. N2(g) + O2(g) 2NO(g) H = 180 kJ 2NO(g) N2(g) + O2(g) H = 180 kJ
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
Calculations via Hess’s Law
H2(g) + 1/2O2(g) H2O(l) -285.8 kJC2H4(g) + 3O2(g) 2H2O(l) + 2CO2(g) -1411 kJC2H6(g) + 7/2O2(g) 3H2O(l) + 2CO2(g) -1560 kJ
Calculate the enthalpy change for C2H4(g) + H2(g) C2H6(g) using the following data.
Example
H2(g) + 1/2O2(g) H2O(l) -285.8 kJC2H4(g) + 3O2(g) 2H2O(l) + 2CO2(g) -1411 kJ
Calculate the enthalpy change for C2H4(g) + H2(g) C2H6(g) using the following data.
Example
3H2O(l) + 2CO2(g) 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