Module 3Module 3

Heat and Energy in Chemical Reactions.

Introduction: ThermochemistryIntroduction: Thermochemistry

Chemical reactions often involve changes in temperature, and in the heat energy that causes temperature to change. This branch of chemistry that studies heat changes is called THERMOCHEMISTRY.

Some Quick DefinitionsSome Quick Definitions

EnergyEnergy: : The ability to (a) do work or (b) The ability to (a) do work or (b) to supply heat.to supply heat.

Kinetic EnergyKinetic Energy: : energy associated with energy associated with movement of objects or particles. movement of objects or particles.

Potential EnergyPotential Energy: : energy that has not energy that has not produced heat or motion yet. Potential produced heat or motion yet. Potential energy may be associated with energy may be associated with gravitation and with chemical bondsgravitation and with chemical bonds..

Background InformationBackground Information(notes optional)(notes optional)

Most energy on earth comes from the Most energy on earth comes from the sun, but a small amount comes from the sun, but a small amount comes from the radioactive decay of certain atoms, and radioactive decay of certain atoms, and an even smaller amount from the tidal an even smaller amount from the tidal effects of the moon.effects of the moon.

Most of the energy that we experience is Most of the energy that we experience is thermal energy, that is, one of several thermal energy, that is, one of several forms of energy related to heatforms of energy related to heat

(Background Info: Notes optional)(Background Info: Notes optional)

Thermal energy can be transferred in Thermal energy can be transferred in many ways:many ways:

Conduction: heat transferred between touching Conduction: heat transferred between touching objects. This form of heat transfer is the most objects. This form of heat transfer is the most important in chemistry.important in chemistry.

Convection: heat transferred to air molecules, Convection: heat transferred to air molecules, which then may expand and may rise due to their which then may expand and may rise due to their reduced density. This is important in meteorologyreduced density. This is important in meteorology

Radiant heat: A hot object, like a heat lamp, may Radiant heat: A hot object, like a heat lamp, may directly radiate heat in the form of infrared directly radiate heat in the form of infrared radiation. This is important in physicsradiation. This is important in physics

Heat and TemperatureHeat and Temperature

HeatHeat is a form of is a form of thermal energy thermal energy that is that is transferred when two systems with transferred when two systems with different temperatures come into contact different temperatures come into contact with each otherwith each other..

Heat “flows” from one object to anotherHeat “flows” from one object to another..

TemperatureTemperature is a measure of the agitation is a measure of the agitation of atoms or molecules in a system. of atoms or molecules in a system.

Temperature is directly related to the average Temperature is directly related to the average kinetic energy of the moleculeskinetic energy of the molecules..

Heat vs. WorkHeat vs. Work

When energy transfer is in the form of “work” When energy transfer is in the form of “work” the transfer is orderly. A whole object or the transfer is orderly. A whole object or group of molecules move in an orderly group of molecules move in an orderly fashion in one direction.fashion in one direction.

When energy transfer is in the form of “heat” When energy transfer is in the form of “heat” the transfer is more random. The molecules the transfer is more random. The molecules of the heated object move more quickly in of the heated object move more quickly in random directions. random directions.

P. 128-130 Questions p. 130P. 128-130 Questions p. 130

Law of Conservation of Law of Conservation of EnergyEnergy

Energy can be neither created nor Energy can be neither created nor destroyed in any chemical process.destroyed in any chemical process.

However:However: Energy can be transferred from one object Energy can be transferred from one object

to another. to another. Eg. Collisions transfer energy between objects.Eg. Collisions transfer energy between objects.

Energy can be transformed from one type Energy can be transformed from one type of energy to another. of energy to another.

Eg. Potential energy can be transformed into Eg. Potential energy can be transformed into kinetic energy, and vice versa.kinetic energy, and vice versa.

Nuclear reactions can change matter into Nuclear reactions can change matter into energy or vice versaenergy or vice versa

Systems and Their Systems and Their SurroundingsSurroundings

SystemSystem: the part of the universe on : the part of the universe on which we are focusing our attention which we are focusing our attention (usually a beaker, flask or test tube)(usually a beaker, flask or test tube)

SurroundingsSurroundings: Everything else in : Everything else in the universe.the universe.

A system may be:A system may be: OpenOpen: : matter & energy easily exchanged matter & energy easily exchanged

between the system and its surroundingsbetween the system and its surroundings ClosedClosed: matter cannot be exchanged, but : matter cannot be exchanged, but

some energy can escape or enter the systemsome energy can escape or enter the system IsolatedIsolated: neither matter nor energy can be : neither matter nor energy can be

exchanged between system and exchanged between system and surroundings.surroundings.

CalorimetersCalorimeters

A calorimeter is an instrument used to measure A calorimeter is an instrument used to measure the amount of heat released or absorbed during the amount of heat released or absorbed during a change.a change.

A calorimeter is a closed system containing a A calorimeter is a closed system containing a fixed amount of water that can absorb heatfixed amount of water that can absorb heat

The The ΔΔH can be calculated by measuring the H can be calculated by measuring the water temperature before and after the reactionwater temperature before and after the reaction

Calorimeters can be as complicated as the Calorimeters can be as complicated as the “bomb” calorimeter illustrated on p. 132 or as “bomb” calorimeter illustrated on p. 132 or as simple as a Styrofoam cupsimple as a Styrofoam cup

The Calorimeter FormulaThe Calorimeter Formula

Q=mcΔTQ=mcΔTWhere Q=Heat Energy

m = mass of the water

c = specific heat capacity of the water

ΔT = the change in temperature

As You Remember!As You Remember! Specific Heat capacitySpecific Heat capacity (c) is the (c) is the

amount of energy required to raise amount of energy required to raise the temperature of one gram of a the temperature of one gram of a substance by one degree Celsius.substance by one degree Celsius.

For water, c is 4.19 J / gFor water, c is 4.19 J / g°C°C

Simple examples: copy and complete*Simple examples: copy and complete*

Mass of Water

Initial Temp

Final Temp

Change ΔT

HeatQ

250 g 22°C 47 °C

200 g 20 °C 18 °C

150 g 30 °C +12 °C

100 g 25 °C -16 °C

22 °C +30 °C 12000 J

*Assume measurements are precise, c = 4.19J/g° for water and that no heat is lost in the calorimeter.

26187.5 J+25°C

-2°C -1676 J

7542 J

-6704 J

18°C

9°C

52°C95.5g

Molar heat of ReactionMolar heat of Reaction

The molar heat of a reaction is the The molar heat of a reaction is the amount of heat released / absorbed amount of heat released / absorbed by the reaction of one mole of the by the reaction of one mole of the materialmaterial

ΔΔH H (Molar)(Molar) = Q / n = Q / n A.K.A. molar heat of combustion, A.K.A. molar heat of combustion,

molar heat of solution, etc.molar heat of solution, etc.

Example: copy and try thisExample: copy and try this

The following data was recorded for the partial The following data was recorded for the partial combustion of wax (Ccombustion of wax (C2525HH5252) in a calorimeter. Use ) in a calorimeter. Use this data to calculate the molar heat of this data to calculate the molar heat of combustion of wax.combustion of wax.

Initial mass of waxInitial mass of wax 22.35 g22.35 g Final mass of waxFinal mass of wax 12.08 g12.08 g Volume of water in calorimeterVolume of water in calorimeter 352.5 mL352.5 mL Initial temperature of waterInitial temperature of water 12.6 12.6 °C°C Final temperature of waterFinal temperature of water 43.5 43.5 °C°C

Solution: part 1- Solution: part 1- heat capturedheat captured

Mass of water: Mass of water: m=352.5 g m=352.5 g (water)(water)

Change of temp:Change of temp: ΔΔT= 30.9 T= 30.9 °C°C Specific heat:Specific heat: c= 4.19 J/g °Cc= 4.19 J/g °C

Q=mc Q=mc ΔΔTT = 352.5= 352.5××4.19 4.19 ×× 30.9 30.9 QQ = = 45638.5275 Joules45638.5275 Joules But… that is not for a whole mole.But… that is not for a whole mole.

Solution: part 2 – Solution: part 2 – number of number of molesmoles

Mass Mass (wax)(wax) m=10.27 g m=10.27 g Molar mass Molar mass (wax)(wax) M=352.61 g/molM=352.61 g/mol Moles of wax:Moles of wax: n= m n= m ÷M÷M n=n= 10.27 ÷ 10.27 ÷ 352.61352.61 n=n= 0.0291 mol0.0291 mol The molar heat of combustion is the The molar heat of combustion is the

heat produced per mole of waxheat produced per mole of wax

Solution: part 3 - molar Solution: part 3 - molar heatheat

Q Q ÷n ÷n = 45638.5275 ÷ = 45638.5275 ÷ 0.02910.0291 Solution= 1568334.278Solution= 1568334.278

Round to 3 significant digitsRound to 3 significant digits

ANSWER:ANSWER: The molar heat of combustion of wax The molar heat of combustion of wax

is about 1570000 Joulesis about 1570000 Joules//mole mole or 1.57x10or 1.57x1066 J J//molmol or 1570 kJor 1570 kJ//molmol

Heat TransferHeat Transfer

When two bodies (or substances) of different When two bodies (or substances) of different temperatures come into contact, heat is temperatures come into contact, heat is transferred from the warmer body (or transferred from the warmer body (or substance) to the cooler body, until substance) to the cooler body, until equilibrium is reached, when the equilibrium is reached, when the temperature is the same.temperature is the same.

At equilibrium, the heat lost (Q-loss) by the At equilibrium, the heat lost (Q-loss) by the substance that was warmer, must equal the substance that was warmer, must equal the heat gained (Q-gain) by the substance which heat gained (Q-gain) by the substance which was cooler. was cooler.

Heat TransferHeat Transfer

Hot Body Cool BodyEquilibrium

Heat moves from Hot to Cold

Until they are the same temperature

Q loss Q gain=

-Q1 Q2=

-Q1 = Q2Then

and

If

Q=mcΔT

-m1c1ΔT1 = m2c2

ΔT2

Example: Copy and try this!Example: Copy and try this!

A calorimeter contains 230.0g of A calorimeter contains 230.0g of water at 25.0water at 25.0°C. A 200.0g sample of °C. A 200.0g sample of copper at 47.0 °C is placed inside the copper at 47.0 °C is placed inside the calorimeter. Calculate the final calorimeter. Calculate the final temperature of this system. (the temperature of this system. (the specific heat capacity of copper is specific heat capacity of copper is 0.390 J/g °C)0.390 J/g °C)

SolutionSolution Q=mcQ=mcΔΔTT For water: QFor water: Qgaingain=230 =230 ×× 4.19 4.19 ×× ΔΔTTwaterwater

For copper: QFor copper: Qlossloss=200 =200 ×× 0.39 0.39 ×× ΔΔTTCuCu Since -QSince -Q11 = Q = Q22 this means that: this means that: -(200 -(200 ×× 0.39 0.39 ×× ΔΔTTCuCu)) = 230x4.19x = 230x4.19x ΔΔTTwaterwater Let Let xx represent the final temperature represent the final temperature So So ΔΔTTwaterwater = ( = (xx––2525 ) ) °C°C And And ΔΔTTCuCu = ( = (xx– – 4747)) °C °C We can substitute: We can substitute:

-200 -200 gg ×× 0.39 0.39 j/g°Cj/g°C × × ((xx-47)-47)°C°C = 230 = 230 gg ×× 4.19 4.19 j/g°Cj/g°C×× ( (xx--25) 25) °C°C

-78 -78 g•j/g°Cg•j/g°C ((x x - 47) - 47) °C °C = 963.7 = 963.7 g•j/g°Cg•j/g°C((xx-25) -25) -78-78x x + 3666 = 963.7+ 3666 = 963.7x - x - 24092.524092.5

-963.7-963.7xx - 78 - 78xx = -3666 - 24092.5 = -3666 - 24092.5

1041.7 1041.7 xx = 27758.5 = 27758.5 26.64726.647 (round to 3 sig. dig.)(round to 3 sig. dig.)

Answer: Answer: 26.626.6°C°C

ΔT (water)

ΔT (copper) m (water) c (water)c (copper)m(copper)

This is true when liquids are This is true when liquids are mixed too!mixed too!

When liquids are mixed, the heat lost When liquids are mixed, the heat lost by the warm liquid must be the same by the warm liquid must be the same as the heat gained by the cooler one.as the heat gained by the cooler one.

In this case, not only do the In this case, not only do the substances transfer heat to each substances transfer heat to each other, but they will mix as well.other, but they will mix as well.

Example: copy and tryExample: copy and try A mixture is made of 100.0 mL of A mixture is made of 100.0 mL of

water at 90.0water at 90.0°C and 100.0 mL of °C and 100.0 mL of water at 25.0°C. What will the water at 25.0°C. What will the final temperature be? final temperature be?

M(hot)=100g m(cold)=100g

C(hot)=4.19 j/g c(cold)=4.19 j/g

Ti(hot)=90 Ti(cold)=25

Tf=x Tf=x

-ΔT(hot) =Tf –Ti(hot)ΔTcold =Tf –Ti(cold)

The same

Data:

Solution: part 1Solution: part 1 Q=mcQ=mcΔΔT so… T so… QQgaingain=m=mcold cold x cx ccold cold x x ΔΔTTcoldcold

-Q-Qlossloss = m = mhothot x c x chothot x x ΔΔTThothot

For cold water, For cold water, QQgaingain=100 =100 ×× 4.19 4.19 ×× ΔΔTTcoldcold

For hot water For hot water QQlossloss=100 =100 ×× 4.19 4.19 ×× ΔΔTThothot Since –QSince –Q(lost)(lost) = Q = Q(gained)(gained) this means thatthis means that -100 -100 ×× 4.19 4.19 ×× ΔΔTThothot = 100 = 100 ×× 4.19 4.19 ×× ΔΔTTcoldcold

ΔΔTTcoldcold = ( = (xx--25)25) let let xx= final temp= final temp --ΔΔTThothot= (= (xx--9090)) Note: since Note: since ΔΔTTcoldcold = = ΔΔTThot hot there is a short cut there is a short cut

you can use, we’ll look at that later.you can use, we’ll look at that later.

Solution: part 2Solution: part 2

-100 -100 ×× 4.19 ( 4.19 (xx--9090)) =100 =100 ×× 4.19 ( 4.19 (xx--25)25) ––((xx – – 9090) == x x -25-25 xx-25=--25=-xx+90 +90 x x ++ x x == 90 + 2590 + 25 2 2 xx = 115 = 115 xx = 57.5 = 57.5 Answer: the final temperature is 57.5 Answer: the final temperature is 57.5 °C°C Of course, in this case you could have done the Of course, in this case you could have done the

problem more simply, since you knew you had equal problem more simply, since you knew you had equal amounts of the same substance (ie. same mass & amounts of the same substance (ie. same mass & specific heat capacity) , you could have just averaged specific heat capacity) , you could have just averaged the temperature!the temperature!

Homework Questions:Homework Questions:

What will the final temperature be if you mix What will the final temperature be if you mix 100.0 g. of water at 20.0100.0 g. of water at 20.0°C and 40.0 g. of water °C and 40.0 g. of water at 80.0 °C?at 80.0 °C?

What is the molar heat of combustion of methane What is the molar heat of combustion of methane (CH(CH44)if burning 1.00 gram of it in a calorimeter )if burning 1.00 gram of it in a calorimeter raises the temperature of 800.0g of water from raises the temperature of 800.0g of water from 27.0 °C to 42.0 °C?27.0 °C to 42.0 °C?

What will the final temperature be if 50.0g of What will the final temperature be if 50.0g of copper (c=0.39 J/g °) at 80.0 °C is dropped into copper (c=0.39 J/g °) at 80.0 °C is dropped into 200.0 mL of 20.0°C water?200.0 mL of 20.0°C water?

Review ConceptsReview Concepts

There are three types of ChangeThere are three types of Change Physical changePhysical change: does not change the : does not change the

composition for example a change of statecomposition for example a change of state Chemical changeChemical change: Changes the composition, : Changes the composition,

examples: effervescence, decomposition, change of examples: effervescence, decomposition, change of colour, precipitation, combustion.colour, precipitation, combustion.

Nuclear change:Nuclear change: Changes in the atom Changes in the atom nucleii, nucleii, examples: formation of isotopes, examples: formation of isotopes, radioactive decay, nuclear fission, nuclear fusionradioactive decay, nuclear fission, nuclear fusion

Remember:Remember:Kinetic Energy of particles:Kinetic Energy of particles:

Particles (ie. Molecules) can have 3 Particles (ie. Molecules) can have 3 types of motion, giving them kinetic types of motion, giving them kinetic energyenergy Vibrational kinetic energyVibrational kinetic energy Rotational kinetic energyRotational kinetic energy Translational kinetic energyTranslational kinetic energy

State & Kinetic EnergyState & Kinetic Energy Solids exhibit only vibrational energySolids exhibit only vibrational energy

Virtually none of their kinetic energy comes from Virtually none of their kinetic energy comes from translation or rotation.translation or rotation.

Liquids exhibit mostly rotational energyLiquids exhibit mostly rotational energy A small portion of their kinetic energy can come A small portion of their kinetic energy can come

from vibration or translation.from vibration or translation. Gases exhibit mostly translational energyGases exhibit mostly translational energy

a tiny portion of their kinetic energy can come from a tiny portion of their kinetic energy can come from vibration or rotation. vibration or rotation.

Melting a Pure Solid andMelting a Pure Solid andBoiling a Pure LiquidBoiling a Pure Liquid

As you slowly add thermal energy to a pure As you slowly add thermal energy to a pure solid, its temperature will rise as the solid, its temperature will rise as the molecular vibration increases.molecular vibration increases.

There will come a point, however, where There will come a point, however, where heating the solid does not increase the heating the solid does not increase the temperature. Instead, the increased energy temperature. Instead, the increased energy is used to change the type of motion, is used to change the type of motion, making the molecules tumble or rotate. making the molecules tumble or rotate. This is called the melting point.This is called the melting point.

The solid becomes a liquid, and the The solid becomes a liquid, and the temperature again rises with increased temperature again rises with increased thermal energy, until the liquid begins to thermal energy, until the liquid begins to boil or evaporate.boil or evaporate.

Phase DiagramPhase DiagramT

em

per

atu

re (

°C o

r K

)

Thermal Energy Added (joules or kilojoules)

Solid

Vibrational

Liquid

Mostly R

otation

Gas

Translatio

n

Melting

Boiling

Vibration

&

Rotation

Rotation

Translation

Info from a phase diagramInfo from a phase diagram(diagram shown in (diagram shown in endothermicendothermic direction, increasing absorption of direction, increasing absorption of

heat)heat)

Te

mp

era

ture

°C

Energy Added (Joules / gram) 10 20 30 40 50 60 70 80 90

100

80

60

20

40

0

-20

-40

Melting Point (18°C)

Boiling Point (56°C)

Heat of Fusion (ΔHfus=10 J/g)Heat of Fusion (ΔHfus=10 J/g)Heat of vaporization (ΔH vap=15 J/g)Heat of vaporization (ΔH vap=15 J/g)

Specific Heat capacity (c) of the solid is the

inverse of the slope, ie:

Run over Rise

30÷60 = 0.5 J/g°

Specific Heat capacity (c) of the solid is the

inverse of the slope, ie:

Run over Rise

30÷60 = 0.5 J/g°

Info from a phase diagramInfo from a phase diagram(diagram shown in (diagram shown in ExothermicExothermic direction, increasing release of direction, increasing release of

heat)heat)

T

em

per

atu

re °

C

Energy Released (Joules / gram) 10 20 30 40 50 60 70 80 90

100

80

60

20

40

0

-20

-40

Freezing Point(18°C)

Condensation Point (56°C)

Heat of condensation (ΔHcond=15 J/g)Heat of condensation (ΔHcond=15 J/g)

Heat of solidification (ΔH=10 J/g)Heat of solidification (ΔH=10 J/g)

Heat Diagrams for MixturesHeat Diagrams for MixturesSo far we have only shown heat diagrams for So far we have only shown heat diagrams for pure pure substancessubstances. For . For mixturesmixtures the “plateaus” are less the “plateaus” are less clear, and the melting and boiling points less clearly clear, and the melting and boiling points less clearly defined.defined.

Melt

ing

Ran

ge

Boilin

g R

an

ge

solid

liquid

slushy

foamy

gas

First fractional distillation pointSecond fractional distillation point

Third fractional distillation point

Sometimes the difference in boiling points for different substances in a liquid mixture can be used to separate them by fractional distillation.

AssignmentsAssignments

Textbook Reading pp. 125-136Textbook Reading pp. 125-136 Textbook Questions pp. 145, #1-19Textbook Questions pp. 145, #1-19

Module 3, Lesson 2Module 3, Lesson 2

Endothermic and Exothermic Reactions

Exothermic ReactionsExothermic Reactions

ExoExo=outer, =outer, ThermTherm=heat=heat Exothermic reactions are chemical Exothermic reactions are chemical

reactions which release heat energy.reactions which release heat energy. You can recognize exothermic You can recognize exothermic

reactions because the products are reactions because the products are hotter than the reactants.hotter than the reactants.

Example: burning wood is an Example: burning wood is an exothermic reactionexothermic reaction..

Endothermic ReactionsEndothermic Reactions

EndoEndo = inner; = inner; ThermTherm = heat = heat Endothermic reactions absorb heat Endothermic reactions absorb heat

from their surroundings.from their surroundings. You can recognize endothermic You can recognize endothermic

reactions because the products reactions because the products become colder than the reactants become colder than the reactants were.were.

Examples: instant “ice packs”, Examples: instant “ice packs”,

More About EnergyMore About Energy Potential Energy (EPotential Energy (EPP) is stored energy. In ) is stored energy. In

chemistry it is usually stored as chemical chemistry it is usually stored as chemical bonds.bonds.

Kinetic Energy (EKinetic Energy (EKK) is energy of motion. In ) is energy of motion. In chemistry it usually revealed by chemistry it usually revealed by temperaturetemperature, caused by moving , caused by moving molecules.molecules.

Heat Energy (Q) is energy transferred from Heat Energy (Q) is energy transferred from one body to another due to a difference in one body to another due to a difference in temperature between the bodiestemperature between the bodies

Types of Heat Change in ChemistryTypes of Heat Change in Chemistry

Enthalpy (H):Enthalpy (H): The amount of total energy in a The amount of total energy in a substance, most of it is in the form of substance, most of it is in the form of potentialpotential or “hidden” heat energy. or “hidden” heat energy. (Detailed (Detailed

discussion will follow)discussion will follow) Heat of Reaction (Heat of Reaction (ΔΔH):H): amount of energy amount of energy

absorbed or released during a reaction. It absorbed or released during a reaction. It represents an amount that the represents an amount that the enthalpy has enthalpy has changed. changed.

In addition to chemical reactions, energy can In addition to chemical reactions, energy can be absorbed or released in other changes. The be absorbed or released in other changes. The symbol symbol ΔΔH H can also be used for these.can also be used for these.

Variations on Variations on ΔΔHH(enthalpy can change in many types of reaction)(enthalpy can change in many types of reaction)

Heat of FormationHeat of Formation ( (ΔΔHHFF): the amount of energy ): the amount of energy absorbed/released when compound is made from its absorbed/released when compound is made from its elements.elements.

Heat of DissolutionHeat of Dissolution ( (ΔΔHHdd): The amount of energy ): The amount of energy absorbed/released when a solute dissolves. absorbed/released when a solute dissolves.

Heat of NeutralizationHeat of Neutralization ( (ΔΔHHnn): The amount of energy ): The amount of energy absorbed/released when a solute dissolves. absorbed/released when a solute dissolves.

Heat of CombustionHeat of Combustion ( (ΔΔHHcombustioncombustion): amount of energy ): amount of energy released when a material burns.released when a material burns.

Heat of Fusion (Heat of Fusion (meltingmelting) or Solidification () or Solidification (freezingfreezing)) ((ΔΔHHff)=-)=-((ΔΔHHss) : amount of energy absorbed when a ) : amount of energy absorbed when a solid melts or released when a liquid freezessolid melts or released when a liquid freezes

Heat of Vaporization or CondensationHeat of Vaporization or Condensation ( (ΔΔHHvv)=-)=-((ΔΔHHcondensationcondensation) : amount of energy absorbed when a ) : amount of energy absorbed when a liquid evaporates or released when a gas condensesliquid evaporates or released when a gas condenses

Recapping EnthalpyRecapping Enthalpy Enthalpy is the Enthalpy is the total heat total heat content of a content of a

substance, including the energy that substance, including the energy that was stored in the bonds of the was stored in the bonds of the substance during its formation.substance during its formation.

Enthalpy is mostly Enthalpy is mostly potential energypotential energy.. Enthalpy cannot be measured directly, Enthalpy cannot be measured directly,

but it can be calculated by the amount but it can be calculated by the amount of energy released/absorbed during of energy released/absorbed during reactions.reactions.

The The Heat of ReactionHeat of Reaction ( (ΔΔH) or H) or enthalpy enthalpy changechange is the difference between the is the difference between the Heats of FormationHeats of Formation ( (ΔΔHHFF) of the products ) of the products and the reactantsand the reactants Next

Enthalpy Diagram: Exothermic

Po

ten

tial E

ner

gy (

Ent

hal

py)

Progress of Reaction (Time)

Reactants

Products

ΔH is Negative

(ΔH <0)

Enthalpy Diagram: Enthalpy Diagram: EndothermicEndothermic

Po

ten

tial E

ner

gy (

Ent

hal

py)

Progress of Reaction (Time)

Reactants

Products

ΔH is Positive(ΔH >0)

Some books show Enthalpy Graphs like this:

ΔH

ΔH

Exothermic

Endothermic

It means the same as the ones before!

Another Type of Enthalpy GraphThis is the type shows what happens during a reaction.

We will examine this type in more detail in the next module.

Reaction progress (time)

En

thalp

y

ΔH

Activation Energy

Trick Questions

• Some physical processes are tricky to classify as exothermic or endothermic. The following are guides:

EXOTHERMICCondensing (gasliquid)

Freezing (liquidsolid)

Solidifying (gassolid)

ΔH is negative

ENDOTHERMICEvaporating (liquidgas)

Sublimating (solidgas)

Boiling (liquidgas)

Melting ΔH positive

Dissolving can be EXOTHERMIC or ENDOTHERMIC depending on the solute.

Example 1:

Which of the followings Statements about enthalpy is true? A) enthalpy always increases during a chemical

change B) enthalpy always decreases during a chemical

change C) enthalpy remains unchanged when chemical

bonds form D) enthalpy always decreases during a

exothermic reaction.

Example 2:

Which of the following processes releases more energy than it absorbs: A) Water cools down when NH4Cl is dissolved in it. B) Solid margarine melts on a hot plate C) Gasoline is burned in an automobile engine D) River water evaporates

Example 3:

The molar heat of formation of a compound is: A) The heat required to atomize one mole of the

compound. B) The heat of reaction for making one mole of

that compound from its elements. C) The heat of reaction for formation of one mole

of the compound from the reaction of two other compounds.

D) The heat liberated during bond formation for one mole of the compound.

Example 4:

ΔH

What type of reaction is this? Exothermic

References and practice

Textbook Reading: p. 653-671 Study guide reading: 3-5 to 3-6 Extra Practice: Student study guide

pages 3-6 to 3-10 # 1-14 Correct these yourself using the answer

key.

Module 3, Lesson 3Module 3, Lesson 3

Calorimeters and Heat Transfer

Extra Practice

Student Study Guide pp. 3-14 to 3-16 #1-12 Do these on your own and correct

yourself. Keep your answers in your folder.

Lab Activity

Heat Exchange

Lab Activity: Heat Exchange

• Purpose: – (a) to investigate heat exchange when mixing liquids

of different temperatures and verify the formula:

Qloss = Qgain, or mcΔT(hot)=mcΔT(cold).

– (b) to determine the specific heat capacity of an unknown metal, and to use that information to identify the metal.

• Materials:– Styrofoam cups, Graduated cylinder, Thermometer,

hot water (from a water bath), cold water (from tap).

Never leave a thermometer unattended!

Procedure (part 1-4)

• Place a measured amount of cold water in a Styrofoam cup (measure to 1/10 mL).

• Place a measured amount of hot water into water a Styrofoam cup.

• Take the temperature of both cups (to 1/10 °C).• Pour the cold water into the hot water.• Stir and take the temperature of the mixture.• Repeat this with four different volume

combinations (see suggested volumes)

Rewrite these instructions in procedure format

Procedure (part 5)

• Put about 50 ml (measure an exact amount) of cold water into a Styrofoam cup and measure its temperature.

• Take a piece of metal out of boiling water and add it to the cold water (boiling water should be 100°C, but use a thermometer to be sure).

• Use the beaker tag to find the mass of metal.• Find the temperature of the water after the metal

has been added.

Sample Beaker Tagfound at the corner stations for step 5

Metal #5

Mass = 23.58g ± 0.03g

H2O use ≈ 60mL

Note: use both pieces!

Suggested amount of water to use

Mass of metalMetal #

Other instructions

Suggested volumes(do not try to get these exactly. Come close but don’t pour

back and forth. Measure to 3 significant digits)

Run Mass*water

(cold)

Temp (cold)

Mass* (hot)

Temp (hot)

Final Temp

1 ≈100 g ≈ 8 °C ≈100 g ≈ 55 °C ? °C

2 ≈ 50 g °C ≈100 g °C ? °C

3 ≈100 g ≈ 8 °C ≈50 g ≈ 55 °C ? °C

4 ≈100 g °C ≈75 g ° C

5 50-70 g(see beaker tag for suggestion)

≈ 8 °C metal g(see beaker tag for

mass of metal)

≈100 °C ? °C

*for water, mass in grams = volume in mL

If time allows

Specific heat capacities of common metals.

Aluminum 0.91 Magnesium 1.05

Antimony 0.21** Mercury 0.14*

Copper 0.39 Nickel 0.59

Gold 0.13* Platinum 0.13*

Iron 0.46 Tin 0.21

Lead 0.13* Zinc 0.39

•*Gold, lead, platinum and mercury have similar specific heats, but are easy to tell apart by other properties

** Antimony and Tin are similar, but we don’t use antimony.

Module 3, Lesson 4Module 3, Lesson 4

Thermo-chemical Equations

Thermo-chemical EquationsThermo-chemical Equations

An equation that includes the heat change An equation that includes the heat change is a thermo-chemical equation.is a thermo-chemical equation.

Thermo-chemical equations are just Thermo-chemical equations are just ordinary chemical equations, but they also ordinary chemical equations, but they also show how much heat (or energy) was show how much heat (or energy) was given off or absorbed during a reaction.given off or absorbed during a reaction.

Unless stated otherwise, the amount of Unless stated otherwise, the amount of energy is in kilojoules per mole.energy is in kilojoules per mole.

2 Types of Thermo-chemical Equation2 Types of Thermo-chemical Equation

Thermo-chemical equations can be Thermo-chemical equations can be expressed using two methods:expressed using two methods:

First method: First method: By adding the heat to the equation directly, orBy adding the heat to the equation directly, or

Second method: Second method: By writing the By writing the ΔΔH value at the end of the H value at the end of the

equation.equation.*Preferred Method

Method 1: ExamplesMethod 1: Examples

Add the heat factor to either side:Add the heat factor to either side: For exothermic reactions:For exothermic reactions:

NaOHNaOH(s)(s) Na Na+ + (aq)(aq) + OH + OH-- (aq)(aq) + 44.51 kJ/mol + 44.51 kJ/mol

For endothermic reactions:For endothermic reactions: NHNH44NONO33(s)(s) + 25.7 kJ/mol + 25.7 kJ/mol NH NH44

++ + NO + NO33--

This method is simple, but it is not widely used.This method is simple, but it is not widely used. Remember, in exothermic reactions the heat is listed Remember, in exothermic reactions the heat is listed

afterafter the arrow, in endothermic reactions the heat is the arrow, in endothermic reactions the heat is listed listed beforebefore the arrow. the arrow.

Method 2: ExamplesMethod 2: Examples

Add the Add the ΔΔH value to the end of the H value to the end of the equation. In this case…equation. In this case… Negative means exothermicNegative means exothermic Positive means endothermicPositive means endothermic

ExothermicExothermic NaOHNaOH(s)(s) Na Na+ +

(aq)(aq) + OH + OH-- (aq)(aq) ΔΔH =-44.51 kJ/molH =-44.51 kJ/mol

EndothermicEndothermic NHNH44NONO33(s)(s) NH NH44

++ + NO + NO33- - ΔΔHH =+=+ 25.7 kJ/mol 25.7 kJ/mol

Examples:

Are the following equations endothermic or exothermic?

C2H5OH + 3O2 2CO2 + 3H2O ΔH=-1235 kJ

2NaHCO3+129kJ Na2CO3+H2O+CO2

C+O2CO2 + 349 kJ

NH4NO3 + H2O NH4++NO3

-+H2O ΔH=25.7kJ

Exothermic

Exothermic

Endothermic

Endothermic

Convert the following “type 1” thermochemical equations into “type 2” thermochemical equations.

NaOH(s) Na+(aq) + OH-

(aq)+ 44.5kJ/mol C2H5OH + 3O2 2CO2 + 3H2O + 1234kJ/mol

C(s) + S2(s) + 89kJ/mol CS2(l)

C2H5Cl(l) + 26.4kJ/mol C2H5Cl(g)

H2O(g) H2O(l) + 40.7kJ/molNaOH(s) Na+(aq) + OH-

(aq) ΔH=–44.5kJ/mol

C2H5OH + 3O2 2CO2 + 3H2O ΔH=–1234kJ/mol

C + S2(s) CS2(l) ΔH=+89kJ/mol

C2H5Cl(l) C2H5Cl(g) ΔH=+26.4kJ/mol

H2O(g) H2O(l) ΔH=–40.7kJ/mol

Module 3, Lesson 5Module 3, Lesson 5

Hess’ Law

Hess’s LawHess’s Law

If two or more thermo-chemical If two or more thermo-chemical equations are added together to give equations are added together to give a final equation, then the enthalpy a final equation, then the enthalpy changes can be added to give the changes can be added to give the enthalpy change for the final enthalpy change for the final equation.equation.

Example of Adding Example of Adding EquationsEquations

Find the equation for:Find the equation for: NN22 + 2O + 2O2 2 2 NO 2 NO22 ΔΔH=?H=?

By combining the equations:By combining the equations: NN22+O+O22 2 NO 2 NO ΔΔH=+180.6 H=+180.6

kJ/molkJ/mol 2 NO+O2 NO+O22 2 NO 2 NO22 ΔΔH=-144.4 kJ/molH=-144.4 kJ/mol

Adding EquationsAdding Equations

ReactantsReactants ProductsProducts Enthalpy (Enthalpy (ΔΔH)H)

NN22 ++OO22 2 NO2 NO +180.6kJ+180.6kJ

2 NO2 NO++OO22 2 NO2 NO22 - 144.4 kJ- 144.4 kJ

NN22++2NO2NO++2O2O

22

2 NO2 NO + + 2NO2NO22 + 36.2 kJ+ 36.2 kJ

NN22 + 2O + 2O22 2 NO2 NO22 + 36.2 kJ+ 36.2 kJ

In the example, we cancelled the like In the example, we cancelled the like substances at the end of the problem. substances at the end of the problem. You may also cancel them earlier, as long You may also cancel them earlier, as long

as you are careful to only cancel a as you are careful to only cancel a substance in the reactant column with an substance in the reactant column with an identical substance in the product column.identical substance in the product column.

You may multiply an entire equation, You may multiply an entire equation, including the including the ΔΔH value by a simple number H value by a simple number to make the coefficients match. to make the coefficients match.

You can only cancel if coefficents they match You can only cancel if coefficents they match exactly!exactly!

You may switch the substances in the You may switch the substances in the reactant column with the substances in the reactant column with the substances in the product column, but you then have to product column, but you then have to change the sign of the change the sign of the ΔΔH value.H value.

Example 2: Example 2: ((see study guide page 47 question 2)see study guide page 47 question 2)

Find the heat of combustion of methanol Find the heat of combustion of methanol (CH(CH33OH) from the following reactions:OH) from the following reactions: CC(s)(s) + O + O22(g)(g) CO CO22(g)(g) + 394 kJ + 394 kJ HH22(g)(g) + ½ O + ½ O22(g)(g) H H22OO(l)(l) + 242 kJ + 242 kJ CC(s)(s) + 2H + 2H22(g)(g) + ½ O + ½ O22(g)(g) CH CH33OHOH(l)(l)+239 kJ+239 kJ

Note: for simplicity, the “phase markers” ie.Note: for simplicity, the “phase markers” ie.(s) (l) (s) (l)

(g) (aq) (g) (aq) will be left out of the solution this time. Be will be left out of the solution this time. Be aware that they aware that they are importantare important if a material’s if a material’s changes statechanges state before or after it reacts. before or after it reacts.

The Reactions we know:The Reactions we know: CC(s)(s) + O + O22(g)(g) CO CO22(g)(g) + 394 kJ + 394 kJ HH22(g)(g) + ½ O + ½ O22(g)(g) H H22OO(l)(l) + 242 kJ + 242 kJ CC(s)(s) + 2H + 2H22(g)(g) + ½ O + ½ O22(g)(g) CH CH33OHOH(l)(l)+239 kJ+239 kJ

The reaction we are trying to get will contain The reaction we are trying to get will contain the following:the following:

CHCH33OHOH(l)(l)+ O+ O22(g)(g) CO CO22 + H + H22OO We can balance it to find the coefficients, but I’m We can balance it to find the coefficients, but I’m

not going to do that yet, since all I need to know for not going to do that yet, since all I need to know for now is what side of the arrow each substance is on. now is what side of the arrow each substance is on.

SolutionSolution

ReactantsReactants ProductsProducts ΔΔHH

C + OC + O22 COCO22 -394 kJ-394 kJ

HH2 2 +½ O+½ O22 HH22OO -242 kJ-242 kJ

-239 kJ-239 kJC+2HC+2H22+½ O+½ O22 CHCH33OHOH

Step 1: copy the reactionsStep 2: CH3OH is on wrong sideReverse the 3rd equation

++

2H2H2 2 + O+ O22 2H 2H22OO -484 kJ-484 kJ

and change the signStep 3: double everything in equation 2

CHCH33OH+ 2OOH+ 2O22COCO22+2H+2H22O+ ½ OO+ ½ O22

CHCH33OH+ 1½ OOH+ 1½ O22 CO CO22+2H+2H22OO

-639 kJ-639 kJ

-639 kJ-639 kJ

flip

x2

Step 4: simplify by cancellation

Assignments• Read Textbook pages 672 to 673• Practice problems: p. 674 #9-10

– (correct yourself from answer key in back, keep answers in your assignment folder)

• Assigned problems: p. 681 # 31-32– (put the answers in your assignment folder, I will

correct later)

• Student Study Guide: Read pp. 3-17 to 3-18• Study Guide Extra practice: 3-18 to 3-20 #1-6

– (correct yourself but keep answers in folder)

Module 3, Lesson 6Module 3, Lesson 6

Heat of Formation

Heat of Formation Tables

Standard Heat of FormationStandard Heat of Formation

The standard heat of formation of a The standard heat of formation of a compound is the change in enthalpy compound is the change in enthalpy that accompanies the formation of that accompanies the formation of one mole of the compound from its one mole of the compound from its elements, with all substances in their elements, with all substances in their standard states at 25standard states at 25°°CC

The symbol for heat of formation is The symbol for heat of formation is ΔΔHHff

In other words:In other words:

It’s how much heat was given off (or It’s how much heat was given off (or absorbed) when you combined the absorbed) when you combined the elements into a compoundelements into a compound

For example:For example: When you burn hydrogen and oxygen When you burn hydrogen and oxygen

together you make water (steam together you make water (steam actually) and a lot of heat is given off actually) and a lot of heat is given off (usually a flaming explosion!)(usually a flaming explosion!)

The Example of WaterThe Example of Water

Hydrogen + Oxygen Hydrogen + Oxygen water water Balanced Equation:Balanced Equation:

2 H2 H22 + O + O22 2 H 2 H22O, O, ΔΔH = -570 kJ (for 2 mol)H = -570 kJ (for 2 mol) But we want it per mole, so But we want it per mole, so ÷2÷2

HH22 + ½ O + ½ O22 H H22O, O, ΔΔH = -285 kJ/molH = -285 kJ/mol So the standard heat of formation for So the standard heat of formation for

water from hydrogen & oxygen is -285 water from hydrogen & oxygen is -285 kJ/molkJ/mol

Some background info:Some background info:

When calculating the standard heat When calculating the standard heat of formation we arbitrarily set the of formation we arbitrarily set the enthalpy of the most common form of enthalpy of the most common form of the elements to zero. This gives us a the elements to zero. This gives us a negative heat of formation if creating negative heat of formation if creating the compound was an exothermic the compound was an exothermic reaction, or positive if making the reaction, or positive if making the compound was endothermiccompound was endothermic

Heat of FormationHeat of Formation

ΔH

ΔH

Exothermic

Endothermic

elementselements

compounds

compounds

0 0

Heat of Formation TablesHeat of Formation Tables

Table 8.4 on page 418 of your text Table 8.4 on page 418 of your text book is an example of a Heat of book is an example of a Heat of Formation Table.Formation Table.

It gives the heat of formation of It gives the heat of formation of several common organic compounds.several common organic compounds.

Using Heat of FormationUsing Heat of Formation

The heat of reaction is equal to the The heat of reaction is equal to the difference between the heat of difference between the heat of formation of the reactants and formation of the reactants and products.products.

ΔΔHHReactionReaction = = ΔΔHHF(Products)F(Products) – – ΔΔHHF(Reactants)F(Reactants)

Example: Find the heat of reaction for Example: Find the heat of reaction for the formation of hydrogen peroxide:the formation of hydrogen peroxide:

2 H2 H22OO(l)(l) + O + O22(g)(g) 2 H 2 H22OO22(l)(l)

Step 1. look up heats of formation in tables:Step 1. look up heats of formation in tables: HH22OO(l)(l) = -285.8 kJ/mol = -285.8 kJ/mol (table 8.4 p. 418)(table 8.4 p. 418) OO22(g)(g) = 0.00 kJ/mol= 0.00 kJ/mol (element)(element) HH22OO22 = -187.8 kJ/mol= -187.8 kJ/mol (table 8.4 p. 418)(table 8.4 p. 418)

Step 2. Find the total Step 2. Find the total ΔΔHHff of the reactants: of the reactants: 2 (-285.8) + 1(0) = -571.6 kJ 2 (-285.8) + 1(0) = -571.6 kJ (just kJ, no longer kJ/mol)(just kJ, no longer kJ/mol)

Step 3. Find the total of the product:Step 3. Find the total of the product: 2 (-187.8) = -375.6 kJ 2 (-187.8) = -375.6 kJ

Step 4. Subtract: Step 4. Subtract: ΔΔHHff (products)(products) – – ΔΔHHff ((reactants)reactants) -375.6 – (-571.6) = -375.6+571.6 = +196 kJ-375.6 – (-571.6) = -375.6+571.6 = +196 kJ

Answer: The heat of reaction is:Answer: The heat of reaction is: Δ ΔH =+H =+196 kJ196 kJ (for 2 moles of peroxide). (for 2 moles of peroxide). The molar heat” of reaction is The molar heat” of reaction is ΔΔH°= 98 kJ/molH°= 98 kJ/mol) )

Assignments

• Read pp. 674 -676 in textbook

• Practice: question 11 and 12 on p. 677 (check your own answers in answer key)

• Assigned: #33 on page 681

Heat of Reaction/Hess’ LawLaboratory Activity

• In this lab you will be calculating the heat energy given off by dissolving and neutralizing sodium hydroxide

• You should also be able to use these calculations to demonstrate the accuracy of Hess’ Law.NaOH(s) NaOH(aq)

NaOH(aq) + HCl(aq) NaCl(aq) +H2O(l)

NaOH(s) + HCl(aq) NaCl(aq) +H2O(l)

Heat of Reaction/Hess’ LawHeat of Reaction/Hess’ LawLaboratory ActivityLaboratory Activity

Pointers for this lab:Pointers for this lab: Be careful with the thermometers! Be careful with the thermometers! Keep Keep

them in the casethem in the case when not using them!!!! when not using them!!!! DON’TDON’T put contaminated NaOH back into put contaminated NaOH back into

the vial. Throw excess down the sink. the vial. Throw excess down the sink. MeasureMeasure close to, close to, but not abovebut not above the the

required measurements, for example required measurements, for example don’t try to get exacty 2.00 g, just very don’t try to get exacty 2.00 g, just very close.close.

For reaction#3 use the colour coded For reaction#3 use the colour coded graduated cylinders from the drawersgraduated cylinders from the drawers

Yellow cylinderYellow cylinder for for NaOHNaOH solution solution Red cylinderRed cylinder for for HClHCl solution solution

Heat of Reaction/Hess’ LawHeat of Reaction/Hess’ LawLaboratory ActivityLaboratory Activity

These slides will show some of the These slides will show some of the calculations required to complete the calculations required to complete the Lab.Lab.

Note: The computer program will Note: The computer program will calculate the class results, and will use calculate the class results, and will use adjustments to improve the accuracy adjustments to improve the accuracy of individual results, but you should do of individual results, but you should do your own calculations as well.your own calculations as well.

Data for Reaction #1Data for Reaction #1NaOH (s) NaOH (s) NaOH (aq) NaOH (aq)

Temperature of WaterTemperature of Water =_____=_____°°CC TTii

Volume of the WaterVolume of the Water =_____ mL=_____ mL mmww

Final temperatureFinal temperature =_____ =_____ °°CC TTff

Mass of CupMass of Cup =__ =__ (used to calculate m(used to calculate mNaOHNaOH))

Mass of Cup+NaOHMass of Cup+NaOH =__ =__ (used to calculate m(used to calculate mNaOHNaOH))

Mass of NaOHMass of NaOH =_____g=_____g mmNaOHNaOH

Symbol

Calculations for Reaction #1Calculations for Reaction #1

Find Find ΔΔ T: T: ΔΔT= TT= Tff – T – Tii

Find Find MMNaOHNaOH (use periodic table)(use periodic table) Find Q Find Q (calorimeter)(calorimeter)Q=mQ=mwaterwaterccwaterwater ΔΔTT Find n Find n (#mols NaOH)(#mols NaOH) n=mn=mNaOHNaOH ÷÷MMNaOHNaOH

Find Find ΔΔH H ΔΔH= Q ÷ nH= Q ÷ n Where Where ΔΔT=temperature change, T=temperature change, MMNaOHNaOH =molar mass, n= moles of NaOH =molar mass, n= moles of NaOH Q = overall heat of reaction, Q = overall heat of reaction, ΔΔH =H =molar heat of reactionmolar heat of reaction

Data for Reaction #2Data for Reaction #2NaOHNaOH(s)(s) + HCl + HCl(aq)(aq) NaCl NaCl(aq)(aq)+ +

HH22OO Temperature of HClTemperature of HCl =_____=_____°°CC TTii

Volume of the HClVolume of the HCl =_____ mL=_____ mL mmSS

Final temperatureFinal temperature =_____ =_____ °°CC TTff

Mass of CupMass of Cup (used to calculate mass of NaOH)(used to calculate mass of NaOH)

Mass of Cup+NaOHMass of Cup+NaOH (used to calculate mass of NaOH)(used to calculate mass of NaOH)

Mass of NaOHMass of NaOH =_____g=_____g mmNaOHNaOH

Calculations for Reaction #2Calculations for Reaction #2

Find Find ΔΔ T: T: ΔΔT= TT= Tff – T – Tii

Find Find MMNaOHNaOH (use periodic table)(use periodic table) Find Q Find Q (calorimeter)(calorimeter)Q=mQ=mSSc c ΔΔTT Find n Find n (#mols NaOH)(#mols NaOH) n=mn=mNaOHNaOH ÷÷MMNaOHNaOH

Find Find ΔΔH H ΔΔH= Q ÷ nH= Q ÷ n Assume that HCl solution has same specific heat Assume that HCl solution has same specific heat

capacity and density as water. This is not capacity and density as water. This is not exactly true, but it is close enough for your own exactly true, but it is close enough for your own calculation. The computer analysis will be more calculation. The computer analysis will be more accurate.accurate.

Data for Reaction #3Data for Reaction #3NaOHNaOH(aq)(aq) + HCl + HCl(aq)(aq) NaCl NaCl(aq)(aq)+ +

HH22OO Temperature of HClTemperature of HCl =_____=_____°°CC TT11

Volume of the HClVolume of the HCl =_____ mL=_____ mLmm11

Temperature of NaOHTemperature of NaOH =_____ =_____ °°CC TT22

Volume NaOHVolume NaOH =_____ mL =_____ mL m m22

Final TemperatureFinal Temperature =_____ =_____ °°C C T Tff

Calculations for Reaction#3Calculations for Reaction#3

Find combined solution mass:Find combined solution mass: mmss = m = m11 + + mm22 Average Initial Temp:Average Initial Temp: TTi i = = TT11+ T+ T22 ) )

(note: we are using a simple average here)(note: we are using a simple average here) 2 2 Find Find ΔΔT: T: ΔΔT= TT= Tff – T – Tii Find Q Find Q (calorimeter)(calorimeter)Q=Q=mmSSc c ΔΔTT Find n Find n (#mols NaOH)(#mols NaOH) ((vv22 / / 1000)1000)LL * 1.00 * 1.00 molmol//LL

Divide volume of NaOH Solution by 1000 to get LitresDivide volume of NaOH Solution by 1000 to get Litres Multiply by the concentration (mol/L)… in this case just 1Multiply by the concentration (mol/L)… in this case just 1

Find Find ΔΔHH ΔΔH= QH= Q÷÷nn Notes: To more accurately compare with other Notes: To more accurately compare with other

sources, you may wish to change your H values to sources, you may wish to change your H values to kilojoules/mol instead of joules/mol (divide them by kilojoules/mol instead of joules/mol (divide them by 1000) 1000)

m2=v2

Submit Your Data SheetSubmit Your Data Sheet

One partner from each pair should One partner from each pair should submit a data collection sheet to the submit a data collection sheet to the teacher at the end of the lab class.teacher at the end of the lab class.

The other partner should keep a copy The other partner should keep a copy of the data collection sheet in a safe of the data collection sheet in a safe place (ie. Their lab folder)place (ie. Their lab folder)

The submitted sheet will be used for The submitted sheet will be used for computer analysis of your results.computer analysis of your results.

Lab Write UpLab Write Up

Wait until the class results are back Wait until the class results are back from computer analysis before from computer analysis before finishing your good copy.finishing your good copy.

Do and record all your calculations in Do and record all your calculations in your rough copy, so that you will be your rough copy, so that you will be ready to finish your good copy and ready to finish your good copy and compare your calculations to the compare your calculations to the computer analysiscomputer analysis

Individual & Class ResultsIndividual & Class Results

ReactionReaction #1#1

ReactionReaction #3#3

ReactionReaction #1+ #1+ #3#3

ReactionReaction #2#2

Our Results Our Results (calculated)(calculated)

Our Results Our Results (computer)(computer)

Our Results Our Results (corrected)(corrected)

Class Results Class Results

(computer)(computer)

Class Results Class Results

(corrected)(corrected)

Expected Expected (from (from

references)references)

4444 5656 100100 100100Results in kJ/mol, with decimal values rounded.

Discussion PointsDiscussion Points Were your results as good as the class Were your results as good as the class

average?average? Were the adjusted or unadjusted results Were the adjusted or unadjusted results

closer to the expected results?closer to the expected results? What were the percentage differences What were the percentage differences

between the calculated and observed between the calculated and observed results?results?

What were the three reactions? Why can What were the three reactions? Why can we say reaction #2 is the sum of we say reaction #2 is the sum of reaction#1 and reaction#3?reaction#1 and reaction#3?

How can this experiment be used to justify How can this experiment be used to justify Hess’ Law?Hess’ Law?

AssignmentAssignment All questions on page 680-681 that All questions on page 680-681 that

were not previously assigned,were not previously assigned, Module 3 (chapter 27) Test first class Module 3 (chapter 27) Test first class

next week:next week:

Answers to page 680

• 13. Kinetic energy is active energy of motion, such as heat energy. Potential energy is hidden or stored energy, such as enthalpy

• 14. Not all of gasoline’s energy is used to give the car motion. Some of its energy is converted into heat that is “wasted” in the radiator. Other energy is used to overcome mechanical friction or air resistance or to brake the car.

• 15. Two components of internal energy change are work (w) and heat flow (q)

• 16. The processes are classified as:– A) exothermic (the battery gives off energy)– B) endothermic (plants absorb sun energy)– C) exothermic (burning releases heat)– D) endothermic (the potato absorbs heat)

• 17. In this case q is positive and w is negative, so… – -24.6kJ + 14.6kJ = 10.0 kJ

• 18. A calorimeter measures the amount of heat absorbed or released in a physical or chemical process.

• 19. A Styrofoam cup has several problems as a calorimeter:

• heat escapes through the top, • vapour or gases formed during reactions escape• . It is difficult to calculate how much heat passes

into the Styrofoam.

• 20. There are two ways to do this:• 1st method: Q=mcΔT where m=100g. (mass of

both liquids), c=4.19 (assume liquids are mostly water) and ΔT= 26.3-24.4 = 1.9°C, so Q=100 x 4.19 x 1.9 = 791.6J

• 2nd method uses the data about the calorimeter in the problem335 J/°C x 1.9°C= 636.5J

• The two methods give different results. In this case the 2nd method is better, since the first method used several assumptions.

• 21. q=ΔH if the pressure is constant

• 22. The work must be negligible (≈0)

• 23. if the reaction of 1 mol Fe2O3 produces 26.3 kJ then the reaction of 3.4 mol would produce 3.4 x 26.3 = 89.42 kJ

• 24. – The formation of 1 mol of Al2O3 would have an

enthalpy change of 3352 kJ / 2 or 1676 kJ ie ΔH=+1676 kJ

– The reaction would be endothermic.

• 25. (a) heat of vaporization, (b) heat of solution, (c ) heat of fusion, (d) heat of condensation, (e) heat of solidification

• 26. (a) endo, (b) endo, (c ) endo, (d) exo, (e) exo

• 27. The heat that flows into the system is used to melt the ice (change vibrations into rotation) instead of raising the temperature

#1: hot solid and cold water1. Qloss = Qgain

mcΔT (hot) = mcΔT(cold)

let t represent the final temperature…

2.05g • 0.519 j/g˚ (74.21˚C - t ) = 26.05 g• 4.19 j/g ( t - 27.2˚C)

1.064 (74.21 – t) = 109.1 ( t – 27.2)

78.95-1.064t = 109.1t - 2969

3048 = 110.2t

so..

t= 3048 / 110.2

t=27.66˚C 27.66˚C is the final temperature.

a) ΔT (water) = 0.46˚C

ΔT (solid) = 46.55˚C

b) yes, this makes sense because: i) the final temperature is between the hot and cold temperature, ii)the specific heat capacity of water is much greater than that of the solid

Data

m(hot)=2.05g

c(hot)=0.519 J/g˚C

Ti(hot)=74.21˚C

m(cold)=26.05g

c(cold)=4.19 J/g ˚C

Ti(cold)=27.2 ˚C

2. Calorimeter Problem (Q = mcΔT )

Data

mc=1.56 kJ/K

ΔT= 3.2 K

m(quinine) =0.1964g

M(quinine) = 108 g/mol

n(quinine) = m / M

=0.1964/108

= 0.001818 mol

Formula:

Q= mc ΔT

Q=1.56 • 3.2

Q=4.992 kJ

molar heat formula:

ΔH=Q / n

ΔH= 4.992 kJ / 0.001818 mol

ΔH= 2745 kJ/mol(heat of combustion per mole)

Heat per gram=Q/m

= 4.992 kJ / 0.1964 g

= 25.4 kJ/g (heat of combustion per gram)

#3. (Hot metal in cold ethanol)

Qloss = Qgain

mcΔThot =mcΔTcold

15.6 •0.146 (160-t) = 125 • 2.45 (t-20)

2.27(160-t) = 306.25 (t-20)

364.4 – 2.27 t =306.3 t – 6125

6489.4 = 308.6 t

So… t = 21.03

The final temperature is 21.03˚C

Data

Halfnium=hot

Ethanol = cold

mHot=15.6g

cHot = 0.146

Ti Hot=160˚C

mcold=125g

Ccold=2.45

Ti cold = 20˚C

#4

Data:

Q=1430 J

ΔT=1.93˚C

Heat capacity =mc

Formula: Q=mcΔT

1430 = mc (1.93)

1430 / 1.93 =mc

So:

mc =740.9 J/K

The heat capacity is 741 J/K

#5

Data:

mc=741 j/K

ΔT=3.46˚C

Formula: Q=mcΔT

Q=741 x 3.46

Q=2563.86

Answer:

The heat evolved was 2560 J

Or 2.56 kJ

#6

(a) N2H4 + 2Cl2 4HCl + N2 ΔH=-420.0kJIf one mol of hydrazine produces -420.0 kJ

then 0.794 mol will produce:0.794 mol • -420.0 kJ/mol

= - 333.5 kJ

Data:

m(hydrazine)=25.4g

M(hydrazine)= 2(14)+4(1)

= 32 g/molFor part (a):

n (hydrazine) =0.794mol

Formula (a): n=m/M

n=25.4 g / 32 g/mol

n=0.7937 mol

b) N2H4 + 2Cl2 4HCl + N2 H=-420.0kJ1 2 4 1

1.45moln mol

For part (b): n (hydrazine) =(1.45)(1) / 4 = 0.3625 mol

So: 0.3625mol • -420 kJ/mol = -152 kJ

#7 (Use Hess’s Law)

CO2 + SiO SiO2 + CO H=-520.9

8CO2 + Si3N4 3SiO2 +2N2O+8CO H=+461.05

Looking for: 5CO2 + Si3N43SiO + 2N2O + 5 CO

Rx3: 3SiO2 + 3CO 3CO2 + 3SiO H=+1562.7

5CO2 + Si3N4 3SiO +2N2O +5CO H=+2023.75

5/ /

5

Answer: the enthalpy change for the reaction is… ΔH=+2024 kJ

#8ΔHf

C3H8(g)=-103.9 kj/mol

CO2(g)=-393.5 kj/mol

H2O(g) = -241 kJ/mol

O2 = 0 kJ/mol

Lesson 1 Summary

• Thermochemistry – studies heat changes in chemical reactions

• Molecules have three types of motion– Vibrational (mostly in solids)– Rotational (mostly in liquids)– Translational (mostly in gases)

• A phase diagram shows the transition of states– Use a phase diagram to find melting point,

boiling point, heat of fusion, heat of evaporation, specific heat capacity.

Lesson 2 Summary

• Exothermic reactions release heat• Endothermic reactions absorb heat• “Hidden heat” is called enthalpy (H)

– It is a form of potential energy– It cannot be measured directly, but can be calculated

by experiment.

• In an exothermic reaction, enthalpy decreases• In and endothermic reaction, enthalpy increases.• The enthalpy change or “heat of reaction” is the

amount of heat absorbed or released (ΔH)– Is negative for exothermic reactions, positive for

endothermic ones.

Lesson 3 summary

• A Calorimeter is a device used to find the amount of heat absorbed or lost in a reaction.

• The calorimeter formula is Q=mcΔT

• Molar heat of reaction is the heat released or absorbed by one mole of a reactant: H=Q/n