Chapter #10 Energy. There are things that do not have mass and volume. These things fall into a...

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

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

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

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.

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?

709 g C3H6O

C3H6O (l ) 4O2 (g) 3CO2(g) + 3 H2O (g) ΔH = -1790 kj

58.1 g C3H6O

mole C3H6O

709 g C3H6O

58.1 g C3H6O

mole C3H6O

709 g C3H6O

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.

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

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

4.184 jg- °C

7.40 g 20 °C

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?

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.

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?

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

Is melting of ice spontaneous?

Is melting of ice spontaneous? Yes What is being spread, energy or matter?

Is melting of ice spontaneous? Yes What is being spread, energy or matter? Matter, right?

Is melting of ice spontaneous? Yes What is being spread, energy or matter? Matter, right? Is burning of gasoline spontaneous?

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?

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?

2257 j

10 3 j

2257 j

10 3 j

kj2000 kj

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

25.19 j

mole-°C

55.85 g

25.19 j

mole-°C

55.85 g

25.19 j

mole-°C

55.85 g

1.20 kg

25.19 j

mole-°C

55.85 g

1.20 kg

25.19 j

mole-°C

55.85 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.

25.19 j

mole-°C

55.85 g

1.20 kg

Heating from 25°C to 100°C

4.184 jg-°C

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C= 3138 j

Evaporating 10.0 mL of water

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C= 3138 j

2257 jg

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C= 3138 j

2257 jg

10.0 g

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C= 3138 j

2257 jg

10.0 g =22570 j

25.19 j

mole-°C

55.85 g

1.20 kg

4.184 jg-°C

10.0g 75 °C= 3138 j

2257 jg

10.0 g =22570 j

Now Combine 3138 j + 22570 j = 25708j

25.19 j

mole-°C

55.85 g

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

Chapter #10 Energy. There are things that do not have mass and volume. These things fall into a...

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