unit 13: thermochemistry

Unit 13: Unit 13: ThermochemistryThermochemistry

Chapter 17Chapter 17

By: Jennie BordersBy: Jennie Borders

Section 17.1 – The Flow of Section 17.1 – The Flow of EnergyEnergy

EnergyEnergy is the capacity to do is the capacity to do workwork or or supply supply heatheat..

Energy has no Energy has no mass or volumemass or volume.. Chemical potential energyChemical potential energy is energy is energy

stored in stored in chemicalschemicals.. The kinds of The kinds of atomsatoms and the and the arrangement arrangement

of the atoms in a substance determine of the atoms in a substance determine the amount of the amount of energyenergy stored in the stored in the substance.substance.

HeatHeat HeatHeat is a form of is a form of energyenergy that always that always flowsflows

from a from a warmerwarmer object to a object to a coolercooler object. object. Heat is represented by Heat is represented by qq..

ThermochemistryThermochemistry

ThermochemistryThermochemistry is the is the study of the study of the heatheat changes changes that occur during that occur during chemical reactionschemical reactions and and physical changesphysical changes of state. of state.

The The law of conservation law of conservation of energyof energy states that in states that in any chemical or physical any chemical or physical process, process, energy is neither energy is neither created nor destroyedcreated nor destroyed..

The Great The Great DebateDebate

1. Exothermic reactions lose heat.1. Exothermic reactions lose heat.

2. Endothermic reactions absorb heat.2. Endothermic reactions absorb heat.

Exothermic and Exothermic and EndothermicEndothermic

ThermochemistryThermochemistry is concerned with the is concerned with the flow of heatflow of heat between a between a chemical system chemical system (reaction) and its surroundings(reaction) and its surroundings..

A A systemsystem is the specific part of the is the specific part of the universe on which you focus your universe on which you focus your attention.attention.

The The surroundingssurroundings include everything include everything outside the system.outside the system.

The The systemsystem and the and the surroundingssurroundings constitute the constitute the universeuniverse..

Exothermic and Exothermic and EndothermicEndothermic

In thermochemical calculations In thermochemical calculations the direction of the the direction of the heat flowheat flow is is given from the point of view of given from the point of view of the the systemsystem..

A process that A process that absorbsabsorbs heat heat from the surroundings is called from the surroundings is called an an endothermic processendothermic process..

A process that A process that losesloses heat to the heat to the surroundings is called an surroundings is called an exothermic processexothermic process..

Exothermic and Endothermic

Units of HeatUnits of Heat A A caloriecalorie is the quantity of is the quantity of

heat that raises the heat that raises the temperature of temperature of 1 gram1 gram of of pure water pure water 11ooCC..

A A CalorieCalorie, or dietary Calorie, , or dietary Calorie, is equal to is equal to 1000 calories1000 calories..

A A JouleJoule is the SI unit of is the SI unit of heatheat and and energyenergy..

1 Calorie = 1000 cal = 1 kcal 1 Calorie = 1000 cal = 1 kcal = 4184 J= 4184 J

1 cal = 4.184 J1 cal = 4.184 J

Practice ProblemsPractice ProblemsMake the following conversions.Make the following conversions.

1.1. 444 calories to Joules444 calories to Joules

2.2. 850 Joules to calories850 Joules to calories

444 cal x 444 cal x 4.184 J4.184 J = = 1857.7 1857.7 J J 1 cal1 cal

850 J x 850 J x 1 cal 1 cal = = 203.2 cal203.2 cal 4.184 J4.184 J

Heat CapacityHeat Capacity The The heat capacityheat capacity of an object is the amount of an object is the amount

of of heatheat it takes to change an object’s it takes to change an object’s temperature by exactly temperature by exactly 11ooCC..

The greater the The greater the massmass of an object, the of an object, the greater the greater the heat capacityheat capacity..

The The heat capacityheat capacity of an object also depends of an object also depends on its on its chemical compositionchemical composition..

Specific HeatSpecific Heat The The specific heat specific heat

capacitycapacity of a of a substance is the substance is the amount of amount of heatheat it it takes to raise the takes to raise the temperature of temperature of 1 gram1 gram of the substance of the substance 11ooCC..

Specific heat is Specific heat is represented by represented by CC..

The The unitsunits of specific of specific heat are heat are J/gJ/gooCC..

WaterWater has a has a higherhigher specific heat than specific heat than most substances.most substances.

HeatHeat

Heat = mass x specific heat x change in Heat = mass x specific heat x change in temptemp

q = mq = m..CC..TT

Mass is in gramsMass is in gramsSpecific heat is in J/gSpecific heat is in J/gooCCChange in temp is in Change in temp is in ooCC

Sample Problem Sample Problem

The temperature of a 95.4g piece of The temperature of a 95.4g piece of copper increases form 25copper increases form 25ooC to 48C to 48ooC C when the copper absorbs 849J of when the copper absorbs 849J of heat. What is the specific heat of heat. What is the specific heat of copper?copper?

q = m.c.T c = q T = 48oC – 25oC = 23oC m.T

c = 849J = 0.39 J/goC 95.4g.23oC

Practice ProblemsPractice Problems

1. When 435J of heat is added to 3.4g 1. When 435J of heat is added to 3.4g of olive oil at 21of olive oil at 21ooC, the temperature C, the temperature increases to 85increases to 85ooC. What is the specific C. What is the specific heat of the olive oil?heat of the olive oil?

q = m.c.T c = q T = 85oC – 21oC = 64oC m.T

c = 435J = 1.99 J/goC 3.4g.64oC


2. How much heat is required to raise 2. How much heat is required to raise the temperature of 250g of mercury the temperature of 250g of mercury 5252ooC? (specific heat of mercury = 0.14 C? (specific heat of mercury = 0.14 J/gJ/gooC)C)

q = m.c.T

q = 250g (0.14J/goC) (52oC) = 1820J

Section 17.1 ReviewSection 17.1 Review1.1. In what direction does heat flow between In what direction does heat flow between

two objects?two objects?

2.2. How do endothermic processes differ How do endothermic processes differ from exothermic processes?from exothermic processes?

3.3. On what factors does the heat capacity On what factors does the heat capacity of an object depend?of an object depend?

4.4. How many kilojoules of heat are How many kilojoules of heat are absorbed when 1000g of water is heated absorbed when 1000g of water is heated from 18from 18ooC to 85C to 85ooC? (specific heat of C? (specific heat of water = 4.184 J/gwater = 4.184 J/gooC)C)q = 1000g (4.184J/goC) (67oC) = 280328J

280328J x 1kJ = 280.328kJ 1000 J

Section 17.1 ReviewSection 17.1 Review

5.5. Using Using caloriescalories, calculate how much heat , calculate how much heat 32.0g of water absorbs when it is heated 32.0g of water absorbs when it is heated from 25from 25ooC to 80C to 80ooC. How many joules is C. How many joules is this? (specific heat of water = 4.184 this? (specific heat of water = 4.184 J/gJ/gooC)C)

q = 32g (4.184 J/goC) (55oC) = 7363.8J

7363.8J x 1 cal = 1759.9 cal 4.184 J

Section 17.2 – Measuring and Section 17.2 – Measuring and Expressing Enthalpy ChangesExpressing Enthalpy Changes

CalorimetryCalorimetry is the accurate and is the accurate and precise measurement of heat change precise measurement of heat change for for chemical and physical processeschemical and physical processes..

CalorimetersCalorimeters are devices used to are devices used to measure the amount of measure the amount of heat absorbed heat absorbed or releasedor released during chemical and during chemical and physical processes.physical processes.

EnthalpyEnthalpy is the is the heat contentheat content of a of a system at system at constant pressureconstant pressure..

Enthalpy is represented by Enthalpy is represented by HH..

CalorimeterCalorimeterq = q = H = m H = m . . C C . . TT

Heat Change Sign Heat Change Sign ConventionConvention

Direction of Direction of Heat FlowHeat Flow SignSign Reaction TypeReaction Type

Heat Flows Heat Flows Out of the Out of the

SystemSystem--HH ExothermicExothermic

Heat Flows Heat Flows Into the Into the SystemSystem

++HH EndothermicEndothermic

Thermochemical EquationsThermochemical Equations

An An equationequation that included the that included the heat heat changechange is a is a thermochemical equationthermochemical equation..

A A heat of reactionheat of reaction is the heat change for is the heat change for the equation the equation exactly as writtenexactly as written..

Ex:Ex:

CaOCaO(s)(s) + H + H22OO(l)(l) Ca(OH) Ca(OH)2(s)2(s) H = -65.2 KJH = -65.2 KJ

2NaHCO2NaHCO3(s)3(s) Na Na22COCO3(s)3(s) +H +H22OO(g)(g) +CO +CO2(g) 2(g) H = +129 KJH = +129 KJ

Sample ProblemSample Problem

Calculate the amount of heat (in kJ) Calculate the amount of heat (in kJ) required to decompose 2.24 moles of required to decompose 2.24 moles of NaHCONaHCO33..

2NaHCO2NaHCO33 Na Na22COCO33 + H + H22O + COO + CO22

H = 129kJH = 129kJ

2.24 mol NaHCO3 x 129kJ = 144.48kJ 2 mol NaHCO3


1.1. Calculate the amount of heat (in kJ) Calculate the amount of heat (in kJ) absorbed when 5.66g of carbon absorbed when 5.66g of carbon disulfide is formed.disulfide is formed.

C + 2S C + 2S CS CS22 H = 89.3kJH = 89.3kJ

5.66g CS2 x 1 mol CS2 x 89.3kJ = 6.65kJ 76g CS2 1 mol CS2


2. How many kilojoules of heat are 2. How many kilojoules of heat are produced when 3.40 mole Feproduced when 3.40 mole Fe22OO33 reacts reacts with an excess of CO?with an excess of CO?

FeFe22OO33 + 3CO + 3CO 2Fe + 3CO 2Fe + 3CO22

H = -26.33 kJH = -26.33 kJ

3.40 mol Fe2O3 x -26.3kJ = -89.42kJ 1 mol Fe2O3


1.1. When 2 mol of solid magnesium When 2 mol of solid magnesium combines with 1 mol of oxygen gas, combines with 1 mol of oxygen gas, 2 mol of solid magnesium oxide is 2 mol of solid magnesium oxide is formed and 1204kJ of heat is formed and 1204kJ of heat is release. Write the thermochemical release. Write the thermochemical equation for this combustion equation for this combustion reaction.reaction.

2Mg(s) + O2(g) 2MgO H = -1204kJ


2. How much heat is released when 2. How much heat is released when 12.5g of ethanol burns?12.5g of ethanol burns?CC22HH55OH + 3OOH + 3O22 2CO 2CO22 + 3H + 3H22O O H = -H = -

1368kJ1368kJ

12.5g x 1mol x -1368kJ = -371.74kJ 46g 1 mol

Section 17.3 – Heat in Changes Section 17.3 – Heat in Changes of Stateof State

The The heat of combustionheat of combustion is the heat of reaction is the heat of reaction for the for the complete burning complete burning of one mole of a of one mole of a substancesubstance..

H (fusion and solidification)H (fusion and solidification) The The heat absorbedheat absorbed by one mole of a by one mole of a

substance substance melting from a solid to a melting from a solid to a liquidliquid at constant temperature is the at constant temperature is the molar heat of fusionmolar heat of fusion..

The The heat lostheat lost when one mole of a when one mole of a liquid changes to a solidliquid changes to a solid at a at a constant temperature is the constant temperature is the molar molar heat of solidificationheat of solidification..

HHfusfus = = -- HHsolidsolid

H (vaporization and H (vaporization and condensation)condensation)

The The heat absorbedheat absorbed by one mole of a by one mole of a substance changing from substance changing from a liquid to a a liquid to a vaporvapor is the is the molar heat of molar heat of vaporizationvaporization..

The The heat releasedheat released by one mole of a by one mole of a substance changing from substance changing from a vapor to a a vapor to a liquidliquid is the is the molar heat of molar heat of condensationcondensation..

HHvapvap = = -- HHcondcond

Practice ProblemsPractice Problems1.1. What is the What is the

HHvapvap of of acetone?acetone?

2.2. What is the What is the HHcondcond of water? of water?

3.3. What is the What is the HHfusfus of rubbing of rubbing alcohol?alcohol?

4.4. What is the What is the HHsolidsolid of diethyl of diethyl ether?ether?

29 kJ/mol

-41 kJ/mol

6 kJ/mol

-7 kJ/mol

H (solution)H (solution)

The The heat changeheat change caused by caused by dissolutiondissolution of of one mole of a substance is the one mole of a substance is the molar heat molar heat of solutionof solution..

Ex.Ex.

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

(aq)(aq)

HHsolnsoln = -445.1 KJ = -445.1 KJ

Heating Curve for WaterHeating Curve for Water


How much heat (in kJ) is absorbed How much heat (in kJ) is absorbed when 24.8g Hwhen 24.8g H22OO(l) (l) at 100at 100ooC and C and 101.3kPa is converted to steam at 101.3kPa is converted to steam at 100100ooC. C.

24.8g H2O x 1 mol H2O x 40.7kJ = 56.08kJ 18g H2O 1 mol


1. How much heat is absorbed when 1. How much heat is absorbed when 63.7g H63.7g H22OO(l) (l) at 100at 100ooC and 101.3kPa is C and 101.3kPa is converted to steam at 100converted to steam at 100ooC?C?

63.7g H2O x 1 mol H2O x 40.7kJ = 144.03kJ 18g H2O 1 mol H2O


2. How many kilojoules of heat are 2. How many kilojoules of heat are absorbed when 0.46g of chloroethane absorbed when 0.46g of chloroethane (C(C22HH55Cl) vaporizes at its boiling point? Cl) vaporizes at its boiling point? (The molar heat of vaporization for (The molar heat of vaporization for chloroethane is 26.4 kJ/mol.)chloroethane is 26.4 kJ/mol.)

0.46g C2H5Cl x 1 mol C2H5Cl x 26.4kJ = 0.19kJ 64g C2H5Cl 1 mol


3. How much heat (in kJ) is released 3. How much heat (in kJ) is released when 2.5 mol of NaOH is dissolved in when 2.5 mol of NaOH is dissolved in water? (The molar heat of solution is -water? (The molar heat of solution is -445.1 kJ/mol.)445.1 kJ/mol.)

2.5 mol NaOH x -445.1kJ = -1112.75kJ 1 mol


4. How many moles of NH4. How many moles of NH44NONO33 must be must be dissolved in water so that 88kJ of heat dissolved in water so that 88kJ of heat is absorbed from the water? (The is absorbed from the water? (The molar heat of solution 25.7 kJ/mol.)molar heat of solution 25.7 kJ/mol.)

88kJ x 1 mol = 3.42mol NH4NO3

25.7kJ

Heating Curve ProblemsHeating Curve Problems

Sometimes a Sometimes a questionquestion might ask you might ask you about a about a physical changephysical change that that involves a involves a temperaturetemperature change and a change and a phasephase change. change.

You will have to do these problems in You will have to do these problems in multiple stepsmultiple steps. .

You have to You have to workwork your way through your way through the the heating curveheating curve..

Heating CurveHeating Curve

Sample ProblemSample Problem Calculate the amount of heat Calculate the amount of heat

required to change 50g of water at required to change 50g of water at 7575ooC to steam.C to steam.q = m.c.T q = 50g (4.184 J/goC) (25oC) = 5230J

50g H2O x 1 mol H2O x 40.7kJ = 113.06kJ 18g H2O 1 mol

113.06kJ = 113060J

113060J+ 5230J118290J

Practice ProblemsPractice Problems1. Calculate the amount of heat needed 1. Calculate the amount of heat needed to change 175g of ice at -20to change 175g of ice at -20ooC to water at C to water at 1515ooC.C.q = 175g (2.1 J/goC) (20oC) = 7350J

175g H2O x 1 mol H2O x 6.01kJ = 58.43kJ 18g H2O 1 mol

q = 175g (4.184 J/goC) (15oC) = 10983J

58.43kJ = 58430J

10983J58430J+7350J76763J

Practice ProblemsPractice Problems2. Calculate the amount of heat released 2. Calculate the amount of heat released when 100g of steam at 130when 100g of steam at 130ooC is cooled to C is cooled to water at 60water at 60ooC.C.

q = 100g (1.7 J/goC) (-30oC) = -5100J

100g H2O x 1 mol H2O x -40.7kJ = -226.11kJ 18g H2O 1 mol

q = 100g (4.184 J/goC) (-40oC) = -16736J

-226.11kJ = -226110J

-226110J-16736J-5100J-247946J

Section 17.3 ReviewSection 17.3 Review1.1. How does the molar heat of fusion How does the molar heat of fusion

of a substance compare to its molar of a substance compare to its molar heat of solidification?heat of solidification?

2.2. How does the molar heat of How does the molar heat of vaporization of a substance vaporization of a substance compare to its molar heat of compare to its molar heat of condensation?condensation?


3. Identify each enthalpy change by 3. Identify each enthalpy change by name and classify each change as name and classify each change as exothermic or endothermic.exothermic or endothermic.

a.a. 1 mol C1 mol C33HH8(l)8(l) 1 mol C 1 mol C33HH8(g)8(g)

b.b. 1 mol Hg1 mol Hg(l)(l) 1 mol Hg 1 mol Hg(s)(s)

c.c. 1 mol NH1 mol NH3(g)3(g) 1 mol NH 1 mol NH3(l)3(l)

d.d. 1 mol NaCl1 mol NaCl(s)(s) + 3.88kJ/mol + 3.88kJ/mol 1 mol 1 mol NaClNaCl(aq)(aq)

e.e. 1 mol NaCl1 mol NaCl(s)(s) 1 mol NaCl 1 mol NaCl(l)(l)

Endothermic

Exothermic

Exothermic

Endothermic

Endothermic

Section 17.4 – Calculating Section 17.4 – Calculating Heats of ReactionHeats of Reaction

Hess’ Law of heat summationHess’ Law of heat summation states that states that if you add two or more if you add two or more thermochemical thermochemical equationsequations to give a final equation, then to give a final equation, then you can also add the you can also add the heat changesheat changes to give to give the the final heat changefinal heat change..

Sample ProblemSample Problem Calculate the enthalpy change for Calculate the enthalpy change for

the following reaction given the the following reaction given the following information:following information:

PbClPbCl22 + Cl + Cl22 PbCl PbCl44 H = ?H = ?

Pb + 2ClPb + 2Cl22 PbCl PbCl44 H = -329.2kJH = -329.2kJ

Pb + ClPb + Cl22 PbCl PbCl22 H = -359.4kJH = -359.4kJ

1st reaction: keep the same H = -329.2kJ2nd reaction: flip the reaction (change sign) +H = 359.4kJ

H = 30.2kJ


1.1. Find the enthalpy change for the Find the enthalpy change for the reaction using the following reaction using the following information.information.

2P + 5Cl2P + 5Cl22 2PCl 2PCl55 H = ?H = ?

PClPCl55 PCl PCl33 + Cl + Cl22 H = 87.9kJH = 87.9kJ

2P + 3Cl2P + 3Cl22 2PCl 2PCl33 H = -574kJH = -574kJ1st reaction: flip reaction (change sign) and x2 H = -175.8kJ2nd reaction: keep the same +H = -574kJ

H = -749.8kJ


2. Calculate the enthalpy change for 2. Calculate the enthalpy change for the reaction using the following the reaction using the following information:information:

NN22 + O + O22 2NO 2NO H = ?H = ?

4NH4NH33 + 3O + 3O22 2N 2N22 + 6H + 6H22OO H = -H = -1530kJ1530kJ

4NH4NH33 + 5O + 5O22 4NO + 6H 4NO + 6H22OO H = -1170kJH = -1170kJ1st reaction: flip reaction (change sign) /2 H = 765kJ2nd reaction: /2 +H = -585kJ H = 180kJ


3. Calculate the enthalpy change for the 3. Calculate the enthalpy change for the reaction using the following information.reaction using the following information.

CC22HH44 + H + H22 C C22HH66 H = ?H = ?

2H2H22 + O + O22 2H 2H22OO H = -572kJH = -572kJ

CC22HH44 + 3O + 3O22 2H 2H22O + 2COO + 2CO22 H = -H = -1401kJ1401kJ

2C2C22HH66 + 7O + 7O22 6H 6H22O + 4COO + 4CO22 H = -3100kJH = -3100kJ1st reaction: /2 H = -286kJ2nd reaction: keep the same H = -1401kJ3rd reaction: flip reaction (change sign) /2 +H = 1550kJ H = -137kJ

Standard Heat of FormationStandard Heat of Formation

The The standard heat of formationstandard heat of formation of a of a compound is the change in enthalpy that compound is the change in enthalpy that accompanies the accompanies the formation of one moleformation of one mole of of the compound from its element with all the compound from its element with all substances in their substances in their standard states at 25standard states at 25ooCC..

The The HHffoo of a of a free elementfree element in its standard in its standard

state is state is zerozero..

HHoo = = HHffoo (products) – (products) – HHff

oo (reactants) (reactants)


Calculate the standard heat of Calculate the standard heat of formation for the following reaction.formation for the following reaction.

2CO2CO(g) (g) + O+ O2(g) 2(g) 2CO 2CO2(g)2(g)

Ho = Hfo (products) – Hf

o (reactants)

Ho = (-787kJ) – (-221kJ + 0kJ) = -566kJ

CO2(g) = 2mol x -110.5 kJ/mol = -221kJO2(g) = 1 mol x 0 kJ/mol = 0kJCO2(g) = 2 mol x -393.5 kJ/mol = -787kJ


1.1. Calculate the standard heat of Calculate the standard heat of formation for the following reaction.formation for the following reaction.

CaCOCaCO3(s) 3(s) CaO CaO(s) (s) + CO+ CO2(g) 2(g)

CaCO3(s) = 1 mol x -1207 kJ/mol = -1207kJCaO(s) = 1 mol x -635.1 kJ/mol = -635.1kJCO2(g) = 1 mol x -393.5 kJ/mol = -393.5kJ


o (reactants)

Ho = (-635.1kJ + -393.5kJ) – (-1207kJ) = 178.4kJ


2. Calculate the standard heat of 2. Calculate the standard heat of formation for the following reaction.formation for the following reaction.

2NO2NO(g) (g) + O+ O2(g) 2(g) 2NO 2NO2(g) 2(g)

NO(g) = 2 mol x 90.37 kJ/mol = 180.74kJO2(g) = 1 mol x 0 kJ/mol = 0kJNO2(g) = 2 mol x 33.85 kJ/mol = 67.7kJ


o (reactants)

Ho = (67.7kJ) – (180.74kJ + 0kJ) = -113.04kJ


1.1. Calculate the enthalpy change in kJ Calculate the enthalpy change in kJ for the following reaction.for the following reaction.

2Al + Fe2Al + Fe22OO33 2Fe + Al 2Fe + Al22OO33

Use the enthalpy changes for the Use the enthalpy changes for the combustion of aluminum and iron:combustion of aluminum and iron:

2Al + 3/2O2Al + 3/2O22 Al Al22OO33 H = -1669.8kJH = -1669.8kJ

2Fe + 3/2O2Fe + 3/2O22 Fe Fe22OO33 H = -824.2kJH = -824.2kJ

1st reaction: keep the same H = -1669.8kJ2nd reaction: flip reaction (change sign) +H = 824.2kJ

H = -845.6kJ


2. What is the standard heat of 2. What is the standard heat of reaction for the decomposition of reaction for the decomposition of hydrogen peroxide?hydrogen peroxide?

2H2H22OO2(l)2(l) 2H 2H22OO(l)(l) + O + O2(g)2(g)

H2O2(l) = 2 mol x -187.8kJ/mol = -375.6kJH2O(l) = 2 mol x -285.8kJ/mol = -571.6kJO2(g) = 1 mol x 0kJ/mol = 0kJ


o (reactants)

Ho = (-571.6kJ + 0kJ) – (-375.6kJ) = -196kJ

THE ENDTHE END

unit 13: thermochemistry

Documents

j of heat

flow of heat

heat flow

heat changes

gramsspecific heat

quantity of heat

supply heat

specific heat of copper