11

Chapter 5Chapter 5

EnergyEnergy

ThermodynamicsThermodynamics(rev. 0910)(rev. 0910)

22

DefinitionDefinition Thermodynamics- Thermodynamics-

is the study of energy transformations.

33

Chemical ReactionsChemical Reactions Chemical reactions involve not just

the conversion of reactants into products, but also involve an energy change in the form of heat—heat released as the result of a reaction, or heat absorbed as a reaction proceeds.

Energy changes accompany all chemical reactions and are due to rearranging of chemical bonding.

44

Making BondsMaking Bonds Addition of energy is always a

requirement for the breaking of bonds but the breaking of bonds in and of itself does not release energy.

Energy release occurs when new bonds are formed.

55

Bond EnergyBond Energy If more energy is released when new

bonds form than was required to break existing bonds, then the difference will result in an overall release of energy.

If, on the other hand, more energy is required to break existing bonds than is released when new bonds form, the difference will result overall in energy being absorbed.

66

Overall Reaction EnergyOverall Reaction Energy Whether or not an overall reaction

releases or requires energy depends upon the final balance between the breaking and forming of chemical bonds.

77

Energy is...Energy is... the ability to do work or produce the ability to do work or produce

heat.heat. conserved.conserved. made of heat and work.made of heat and work. a state function. a state function. ((Energy is a property that

is determined by specifying the condition or “state” (e.g., temperature, pressure, etc.) of a system or substance.)

independent of the path, or how you independent of the path, or how you get from point A to B.get from point A to B.

88

EnergyEnergy While the total internal energy of a

system (E) cannot be determined, changes in internal energy (E) can be determined.

The change in internal energy will be the amount of energy exchanged between a system and its surroundings during a physical or chemical change.

Δ E = E final - E initial

99

DefinitionsDefinitions

Work is a force acting over a Work is a force acting over a distance.distance.

Heat is energy transferred Heat is energy transferred between objects because of between objects because of temperature difference. temperature difference. ((Heat is not a property of a system or substance and is not a state function. Heat is a process—the transfer of energy from a warm to a cold object.)

1010

System vs SurroundingsSystem vs Surroundings The Universe is divided into two The Universe is divided into two

halves.halves.• system and the surroundings.

In a chemistry setting, a system includes all substances undergoing a physical or chemical change.

The surroundings would include everything else that is not part of the system.

1111

HeatHeat

Most commonly, energy is exchanged between a system and its surroundings in the form of heat.

Heat will be transferred between objects at different temperatures.

Thermochemistry is the study of thermal energy changes.

1212

Exo vs EndoExo vs Endo

Exothermic reactions release Exothermic reactions release energy to the surroundings.energy to the surroundings.

Endothermic reactions absorb Endothermic reactions absorb energy from the surroundings.energy from the surroundings.

1313

CH + 2O CO + 2H O + Heat4 2 2 2

CH + 2O 4 2

CO + 2 H O 2 2

Pote

nti

al en

erg

y

Heat

1414

N + O2 2

Pote

nti

al en

erg

y

Heat

2NO

N + O 2NO2 2 + heat

1515

Three PartsThree PartsEvery energy measurement has Every energy measurement has three parts:three parts:

1.1. A unit ( Joules of calories).A unit ( Joules of calories).

2.2. A number.A number.

3.3. a sign to tell direction.a sign to tell direction.negative - exothermic

positive- endothermicpositive- endothermic

1616

System

Surroundings

Energy

E <0

1717

System

Surroundings

Energy

E >0

1818

Same rules for heat and workSame rules for heat and work

Heat given off is “negative”.Heat given off is “negative”. Heat absorbed is “positive”.Heat absorbed is “positive”.

Work done Work done by the systemby the system on the on the surroundings is surroundings is negativenegative..

Work done Work done on the systemon the system by the by the surroundings is surroundings is positivepositive..

1919

First Law of ThermodynamicsFirst Law of Thermodynamics The energy of the universe is constant.The energy of the universe is constant. It is also called the: It is also called the: Law of conservation of energy.Law of conservation of energy.

q = heatq = heat w = workw = work In a chemical system, the energy exchanged

between a system and its surroundings can beaccounted for by heat (q) and work (ww).

E = q + wE = q + w Take the system’s point of viewTake the system’s point of view to decide to decide

signs.signs.

2020

Conservation of EnergyConservation of Energy

Energy exchanged between a system and its surroundings can be considered to off set one another.

The same amount of energy leaving a system will enter the surroundings (or vice versa), so the total amount of energy remains constant.

2121

Metric UnitsMetric Units

The SI (Metric System) unit for all forms of energy is the joule (J).

2222

Heat and WorkHeat and Work E = q + wE = q + w - q is exothermic - q is exothermic -q = -∆H-q = -∆H +q is endothermic+q is endothermic -w is done “by” the system-w is done “by” the system +w is done “on” the system+w is done “on” the system

Note: Note: ∆∆H stands for H stands for enthalpyenthalpy which is the which is the

heat of reactionheat of reaction

2323

Practice ProblemPractice Problem A gas absorbs 28.5 J of heat and

then performs 15.2 J of work. The change in internal

energy of the gas is:(a) 13.3 J(b) - 13.3 J(c) 43.7 J(d) - 43.7 J(e) none of the above

2424

AnswerAnswer

(b) E = q + w 28.5 J - 15.2 J = + 13.3 J

2525

Practice ProblemPractice ProblemWhich of the following statements correctly describes the signs of q and w for the following exothermic process at 1 atmosphere pressure and 370 Kelvin?

H2O(g) → H2O(l)

(a) q and w are both negative(b) q is positive and w is negative(c) q is negative and w is positive(d) q and w are both positive(e) q and w are both zero

2626

AnswerAnswer

(c). An exothermic indicates q is negative and the gas is condensing to a liquid so it is exerting less pressure on its surroundings indicating w is positive.

2727

What is work?What is work? Work is a force acting over a distance.Work is a force acting over a distance.

w= F x w= F x ddP = F/ areaP = F/ aread = V/aread = V/areaw= (P x area) x w= (P x area) x (V/area)= P (V/area)= PVV

Work can be calculated by multiplying Work can be calculated by multiplying pressure by the change in volume at pressure by the change in volume at constant pressure.constant pressure.

Use units of liter•atm or L•atmUse units of liter•atm or L•atm

2828

Pressure and Volume WorkPressure and Volume Work

Work refers to a force that moves an object over a distance.

Only pressure/volume work (i.e., the expansion/contraction of a gas) is of significance in chemical systems and only when there is an increase or decrease in the amount of gas present.

2929

Work needs a signWork needs a sign If the volume of a gas increases, If the volume of a gas increases,

the system has done work on the the system has done work on the surroundings.surroundings.

work is negativework is negative w = - Pw = - PVV Expanding work is negative.Expanding work is negative. Contracting, surroundings do work Contracting, surroundings do work

on the system W is positive.on the system W is positive. 1 L•atm = 101.3 J1 L•atm = 101.3 J

3030

Example When, in a chemical reaction,

there are more moles of product gas compared toreactant gas, the system can be thought of as performing work on its surroundings (making w < 0) because it is “pushing back,” or moving back the atmosphere to make room for the expanding gas. When the reverse is true, w > 0.

3131

Compressing and Expanding GasesCompressing and Expanding Gases

Compressing gasCompressing gas– Work on the system is positive– Work is going into the system

Expanding gasExpanding gas– Work on the surroundings is negative– Work is leaving the system

3232

Clarification InfoClarification Info If the reaction is performed in a rigid

container, there may be a change in pressure,but if there is no change in volume, the atmosphere outside the container didn’t “move” and without movement, no work is done by or on the system.

If there is no change in volume (V= 0), then no work is done by or on the system (w= 0) and the change in internal energy will be entirely be due to the heat involved ( ΔE = q).

3333

ExamplesExamples What amount of work is done when What amount of work is done when

15 L of gas is expanded to 25 L at 15 L of gas is expanded to 25 L at 2.4 atm pressure?2.4 atm pressure?

If 2.36 J of heat are absorbed by the If 2.36 J of heat are absorbed by the gas above. what is the change in gas above. what is the change in energy?energy?

How much heat would it take to How much heat would it take to change the gas without changing change the gas without changing the internal energy of the gas? the internal energy of the gas?

3434

EnthalpyEnthalpy The symbol for Enthalpy is HThe symbol for Enthalpy is H H = E + PV (that’s the definition)H = E + PV (that’s the definition) at constant pressure.at constant pressure. H = H = E + PE + PVV

the heat at constant pressure qthe heat at constant pressure qpp can can

be calculated frombe calculated from

E = qE = qpp + w = q + w = qpp - P - PVV

qqpp = = E + P E + P V = V = HH

3535

H = H = E + PE + PVV

Using Using H = H = E + PE + PVV

the heat at constant pressure qthe heat at constant pressure qpp can can

be calculated from:be calculated from:

E = qE = qpp + w (if w = - P + w (if w = - PV then…) V then…)

E = qE = qpp – P – PV (now rearrange)V (now rearrange)

qqpp = = E + P E + P V = V = HH

3636

Examples of Enthapy Changes

KOH(s) → K(aq) + OH-1 (aq) ΔHsolution = - 57.8 kJ mol1

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHcombustion = -2221kJ/mol

H2O(s) → H2O(l) ΔHfusion = 6.0 kJ/mol

Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s) ΔHreaction - 852 kJ/mol

Ca(s) + O2(g) H2(g) → Ca(OH)2(s) ΔHformation - 986 kJ/mol1

3737

3 Methods3 Methods There are a variety of methods

for calculating overall enthalpy changes that you should be familiar with.

The three most common are the:

1. the use of Heats of Formation 2. Hess’s Law 3. the use of Bond Energies

3838

Heat of ReactionHeat of Reaction To compare heats of reaction for

different reactions, it is necessary to know the temperatures at which heats of reaction are measured and the physical states of the reactants and products.

Look in the Appendix of the textbook to find Standard Enthapy tables.

3939

Standard Enthalpy of FormationStandard Enthalpy of Formation Measurements have been made and tables

constructed of Standard Enthalpies of Formation with reactants in their “standard states”.

Use the symbol Use the symbol HºHºff

Standard state is the most stable physical state of reactants at:

1 atmosphere pressurespecified temperature—usually 25 °C 1 M solutions

For solids which exist in more than one allotropic form, a specific allotrope must be specified.

4040

ΔH°formation

It is important to recognize that the ΔH°formation (abbreviated as ΔH°f) is really just the heat of reaction for a chemical change involving the formation of a compound from its elements intheir standard states.

4141

Standard Enthalpies of FormationStandard Enthalpies of Formation The standard heat of formation is The standard heat of formation is

the amount of heat needed to form the amount of heat needed to form 1 mole of a compound from its 1 mole of a compound from its elements in their standard states.elements in their standard states.

See the table in the Appendix See the table in the Appendix Remember: For an element the Remember: For an element the

value is 0value is 0

4242

Equation PracticeEquation Practice You need to be able to write the You need to be able to write the

equation correctly before solving equation correctly before solving the problem.the problem.

Try…Try… What is the equation for the What is the equation for the

formation of NOformation of NO22 ? Try writing the ? Try writing the

equation.equation.

4343

Practice AnswerPractice Answer

½N½N2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) (g)

You must make You must make one mole one mole to meet to meet the definition.the definition.

4444

Since we can manipulate the equationsSince we can manipulate the equations

We can use heats of formation to We can use heats of formation to figure out the heat of reaction.figure out the heat of reaction.

Lets do it with this equation.Lets do it with this equation. CC22HH55OH +3OOH +3O22(g) (g) 2CO 2CO22 + 3H + 3H22OO

which leads us to this rule.which leads us to this rule.

( H products) - ( H reactants) = Hfo

fo o

4545

Hess’s LawHess’s Law Definition:Definition: When a reaction may be expressed as

the algebraic sum of other reactions, the enthalpy change of the reaction is the algebraic sum of the enthalpy changes for the combined reactions.

4646

Hess’ LawHess’ Law Enthalpy is a state function.Enthalpy is a state function. It is independent of the path.It is independent of the path. We can add equations to come up We can add equations to come up

with the desired final product, and with the desired final product, and add the add the H values.H values.

Two rules to remember:Two rules to remember: If the reaction is reversed the sign If the reaction is reversed the sign

of of H is changedH is changed If the reaction is multiplied, so is If the reaction is multiplied, so is HH

4747

EnthalpyEnthalpy As enthalpy is an extensive property,

the magnitude of an enthalpy change for a chemical reaction depends upon the quantity of material that reacts.

This means: if the amount of reacting

material in an exothermic reaction is doubled, twice the quantity of heat energy will be released.

4848

For the oxidation of sulfur dioxide gas

SO2(g) + ½O2(g) → SO3(g) ΔH° = - 99 kJ/mol

Doubling the reaction results in: 2SO2(g) + O2(g) → 2SO3(g)

ΔH° = - 198 kJ/mol

Notice that if you double the reaction, you must double the ΔH° value.

4949

Sign Change of Sign Change of ΔH 2SO2(g) + O2(g) → 2SO3(g)

ΔH° = - 198 kJ/mol

If the reaction is written as an endothermic reaction:

2SO3(g) → 2SO2(g) + O2(g)

ΔH° = + 198 kJ/mol

5050

Tips for Hess’s Law ProblemsTips for Hess’s Law Problems It is always a good idea to begin by looking

for species that appear as reactants and products in the overall reaction.

This will provide a clue as to whether a reaction needs to be reversed or not.

Second, consider the coefficients of species that appear in the overall reaction.

This will help determine whether a reaction needs to be multiplied before the overall summation.

5151

ExampleExample

C(s) + O2(g) → CO2(g) ΔH°f = - 394 kJ/mol

2H2(g) + O2(g) → 2H2O(l) ΔH°f = - 572 kJ/mol

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol

-----------------------------------------------------

C(s) + 2H2(g) → CH4(g) ΔH°f = - 75 kJ/mol

5252

Hess’s Law ExampleHess’s Law Example

O (g) + H (g) 2OH(g) 2 2 O (g) 2O(g)2 H (g) 2H(g)2

O(g) + H(g) OH(g)

Given

Calculate Hº for this reaction

Hº= +77.9kJHº= +495 kJ

Hº= +435.9kJ

5353

H (g) + 1

2O (g) H (l) 2 2 2O

C(s) + O (g) CO (g) 2 2Hº= -394 kJ

Hº= -286 kJ

C H (g) + 5

2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l

ExampleExample GivenGiven

calculate calculate Hº for this reactionHº for this reaction

Hº= -1300. kJ

2C(s) + H (g) C H (g) 2 2 2

5454

Problems to TryProblems to Try Try 12-3 Practice ProblemsTry 12-3 Practice Problems

5555

N2 2O2

O2 NO2

68 kJ

NO2180 kJ

-112 kJ

H (

kJ)

5656

CalorimetryCalorimetry Calorimetry is the study of the heat

released or absorbed during physical and chemical reactions.

For a certain object, the amount of heat energy lost or gained is proportional to

the temperature change. The initial temperature and the final temperature in the calorimeter

are measured and the temperature difference is used to calculate the heat of reaction.

5757

Equipment: CalorimeterEquipment: Calorimeter There are two kinds of There are two kinds of

calorimeters: calorimeters: – constant pressure

– bomb

5858

CalorimetryCalorimetry Constant pressure calorimeter Constant pressure calorimeter

(called a coffee cup (called a coffee cup calorimeter)calorimeter)

A coffee cup calorimeter measures A coffee cup calorimeter measures H.H.

The calorimeter can be an The calorimeter can be an insulated cup, full of water. insulated cup, full of water.

5959

Definitions: Heat capacity is the amount of energy

required to raise the temperature of an object 1 kelvin or 1 °C.

Specific heat capacity is the heat capacity of 1 gram of a substance.

Molar heat capacity is the heat capacity of1 mole of a substance.

6060

Heat CapacityHeat Capacity Heat capacity is an extensive property,

meaning it depends on the amount present—a large amount of a substance would require more heat to raise the temperature 1 K than a small amount of the same substance.

6161

Specific Heat CapacitySpecific Heat Capacity

Definition: The specific heat capacity of each

substance is an intensive property which relates the heat capacity to the mass of the substance.

6262

Specific Heat Capacity “c”Specific Heat Capacity “c” q = mass x c x q = mass x c x TT Or written as Or written as

q = mcq = mcTT

This is the main equation for This is the main equation for calorimetry calculationscalorimetry calculations

mass will be in gramsmass will be in grams Units for “c” are J/g K or J/g °CUnits for “c” are J/g K or J/g °C The specific heat of water is 1 cal/g ºCThe specific heat of water is 1 cal/g ºC

6363

Molar Heat CapacityMolar Heat Capacity Molar Heat Capacity is the heat capacity Molar Heat Capacity is the heat capacity

of 1 mole of a substance.of 1 mole of a substance.

molar heat capacity = c/molesmolar heat capacity = c/moles

heat = molar heat x moles x heat = molar heat x moles x TT

Remember that heat is shown as ∆HRemember that heat is shown as ∆H Make the units work and you’ve done Make the units work and you’ve done

the problem right.the problem right.

6464

Sign of qSign of q• If a process results in the sample losing heat energy, the loss in heat is designated as q is negative.The temperature of the surroundings will increase during this exothermic process.

• If the sample gains heat during the process, then q is positive. The temperature of thesurroundings will decrease during an endothermic process.

The amount of heat that an object gains or loses is directly proportional to the change intemperature.

6565

ExamplesExamples The specific heat of graphite is The specific heat of graphite is

0.71 J/gºC. 0.71 J/gºC. Calculate the energy needed to Calculate the energy needed to

raise the temperature of 75 kg of raise the temperature of 75 kg of graphite from 294 K to 348 K.graphite from 294 K to 348 K.

6666

Extra ProblemExtra Problem A 46.2 g sample of copper is A 46.2 g sample of copper is

heated to 95.4ºC and then placed heated to 95.4ºC and then placed in a calorimeter containing 75.0 g in a calorimeter containing 75.0 g of water at 19.6ºC. The final of water at 19.6ºC. The final temperature of both the water and temperature of both the water and the copper is 21.8ºC. the copper is 21.8ºC.

What is the specific heat of What is the specific heat of copper?copper?

6767

CalorimetryCalorimetry Constant volume calorimeter is Constant volume calorimeter is

called a bomb calorimeter.called a bomb calorimeter. Material is put in a container with Material is put in a container with

pure oxygen. Wires are used to pure oxygen. Wires are used to start the combustion. The container start the combustion. The container is put into a container of water.is put into a container of water.

The heat capacity of the The heat capacity of the calorimeter is known and tested.calorimeter is known and tested.

Since Since V = 0, PV = 0, PV = 0, V = 0, E = q E = q

6868

Bomb CalorimeterBomb Calorimeter thermometerthermometer

stirrerstirrer

full of waterfull of water

ignition wireignition wire

Steel bombSteel bomb

samplesample

6969

PropertiesProperties intensive properties are not related intensive properties are not related

to the amount of substance.to the amount of substance. Examples: density, specific heat, Examples: density, specific heat,

temperature.temperature.

Extensive property - does depend Extensive property - does depend on the amount of substance.on the amount of substance.

Examples: Heat capacity, mass, Examples: Heat capacity, mass, heat from a reaction.heat from a reaction.

7070

Collegeboard. (2007-2008). Professional Development workshop Collegeboard. (2007-2008). Professional Development workshop materials: Special focus thermochemistry. materials: Special focus thermochemistry. http://apcentral.collegeboard.com/apc/public/repository/5886-http://apcentral.collegeboard.com/apc/public/repository/5886-3_Chemistry_pp.ii-88.pdf3_Chemistry_pp.ii-88.pdf