Name_________________________
TeamName______________________
CHM111Lab–EnthalpyofHydrationofSodiumAcetate–GradingRubric
Criteria
Pointspossible
Pointsearned
LabPerformance
Printedlabhandoutandrubricwasbroughttolab
Safetyandproperwastedisposalproceduresobserved
Followedprocedurecorrectlywithoutdependingtoomuch
oninstructororlabpartner
Workspaceandglasswarewascleanedup
3
2
3
1
1
2
1
1
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2
2
20
LabReport
Datarecordedclearlywithproperunits
Calculationsforqshownclearlywithcorrectsigns
Question1
Question2
Question3–Hess’sLawcalculationshownandexplained
clearly
Question4
Question5
Total
Subjecttoadditionalpenaltiesaspertheinstructor
EnthalpyofHydrationofSodiumAcetate
Goal
Todeterminetheenthalpy(ΔH)forthefollowingprocess:
NaC2H3O2(s)+3H2O(l)àNaC2H3O2.3H2O(s)
Introduction
Mostchemicalreactionsareaccompaniedbyachangeinenergy(heat).Thereactionswilleithergiveoffheator
absorb heat. Thermochemistry is the study of heat exchanged during reactions. In thermochemistry, the atoms and
molecules doing the reacting are referred to as the system and everything else in the universe is referred to as the
surroundings. Reactions that release heat to the surroundings are exothermic reactions – these reactions cause the
surroundings to increase in temperature. A reaction that absorbs heat is an endothermic reaction and it causes the
surroundings to decrease in temperature. The sign for the change in enthalpy of a reaction (ΔH) is negative for
exothermicreactionsandpositiveforendothermicreactions.Considerthefollowingexothermicreaction:
(1)
NaOH(s)àNa+(aq)+OH-(aq) ΔH=-44.4kJ/mol
Whenonemoleofsodiumhydroxideisdissolvedinwater,thereaction(thesystem)releases44.4kJofenergy.
Thisenergyistransferredtothesurroundingsandthetemperatureofthesurroundingswillincrease.ΔHexperiments
areoftendoneinaqueoussolutionininsulatedcoffeecups.Thecoffeecupensuresthatvirtuallyallofenergyinthe
systemistransferredtothesolution,nottothelabbenchortheair.
Calculatingqreaction
Thefirstlawofthermodynamicsstatesthatthenetchangeinenergyintheuniverseisequaltozero:
(2)
ΔEuniverse=qsystem+qsurroundings=0
Rearrangingequation(2)andnotingthatinthiscaseqsystemisequivalenttoqreaction,andqsurroundingsisequivalenttoqsolution,
weget:
(3)
qsystem=-qsurroundings or
qreaction=-qsolution
Heatforthesystemandsurroundingshavethesamemagnitude,butoppositesign.JustlikeΔH,whenheatislost,qis
negative,whenheatisgained,qispositive.
qreactioncannotbedetermineddirectly.However,qsolutioncanbedeterminedusingthespecificheatequation(4).
misthetotalmassofthesolution,Csisthespecificheatofthesolution(usuallysimilartothespecificheatofwaterfor
dilutesolutions)andTfandTiarethefinalandinitaltemperaturesofthesolution.
(4)
qsolution=mCs(Tfinal–Tinitial)
Notethataccordingtoequation(4),whenthetemperatureofthesolutionincreases,thesignofqsolutionwillbepositive,
whenthetemperatureofthesolutiondecreases,thesignofqsolutionwillbenegative.
CalculatingΔH
ΔHistheheatreleasedwhentheamountswritteninthebalancedequationarereacted.Forequation(1),that
wouldbe1moleofNaOH(s)reacting.However,reactinganentiremoleiswastefulandunnecessary.TofindΔH,we
reactamuchsmalleramount,findtheheatreleasedorabsorbed(qreaction),thenextrapolatehowmuchheatwouldhave
beenreleasedifwereactedonemole.ΔHiscalculatedasfollows:
(5)
ΔH=qreaction/mol
ForanexampleofhowtheΔHinequation(1)mighthavebeendetermined,if0.013molesofNaOHaredissolved,-572J
ofheatwillbereleased.Pluggingintoequation(5)ΔH=-572J/0.013mol=-44,000J=44kJ.Inthiswayweareableto
findΔHusingmuchsmallerquantitiesofreagents.
In this experiment you will be calculating the ΔH for dissolving anhydrous sodium acetate, NaC2H3O2 (s)
(equation6)andΔHfordissolvingsodiumacetatetrihydrate,NaC2H3O2.3H2O(s)(equation7).
(6)
NaC2H3O2(s)àNa+(aq)+C2H3O2-(aq)
ΔHreaction=?(fortheanhydrous)
(7)
NaC2H3O2.3H2O(s)àNa+(aq)+C2H3O2-(aq)+3H2O(l)
ΔHreaction=?(forthehydrate)
ΔHrelationshipsandHess’Law
Ultimatelyinthislab,wewanttocalculatetheΔHforanhydroussodiumacetateconvertingtosodiumacetate
trihydrate(equation8).However,measuringthisdirectlyinthelabisnotfeasible–theprocesscannotbedonein
solution,itwilloccurslowlyoverseveraldaysifanhydroussodiumacetateisexposedtoatmosphericmoisture(leftinan
opencontainer.
(8)
NaC2H3O2(s)+3H2O(l)àNaC2H3O2.3H2O(s)
ΔHreaction=?
anhydrous
hydrate
Since we cannot figure out ΔH for equation (8) directly, we will use Hess’s Law to determine it indirectly. Hess’ Law
worksbecauseenthalpyisastatefunction–itisindependentofhowthesystemispreparedorthepathofthereaction.
Hess’Lawstatesthatthechangeinenthalpyofastepwisereactionisequaltothesumofthechangesinenthalpyfor
theindividualsteps.WemaynotbeabletomeasuretheenthalpyforA+BàDorequation(8)butwecancalculateit
ifweknowtheenthalpiesoftheothersteps.
A+BàC ΔH=-100kJ
CàD
ΔH=+25kJ
A+BàD ΔH=-75kJ
Notethatequations(6)and(7)donotaddtoequation(8)aswritten.Youwillneedtomanipulatethereactions
sothattheyaddtothereactionofinterest.OtherΔHrelationshipsarethatreversingareactionreversesthesignofΔH,
andwhenmultiplyingreactioncoefficientsbyafactor,ΔHismultipliedbythesamefactor.
IfA+BàC ΔH=-100kJ, thenCàA+BΔH=+100kJ
IfCàD
ΔH=+25kJ, then2Cà2DΔH=+50kJ
Youwillusetheseprinciplestomakeequations(6)and(7)addtotheoverallequation(8).Thesumofthetransformed
ΔH’swillthenequaltheΔHoftheoverallreaction(8).
LaboratoryActivity
Equipment
1Largetesttube
Thermometer Chemicals
Anhydroussodiumacetate
Sodiumacetatetrihydrate
A.DissolutionofAnhydrousSodiumAcetate
1. Obtainalargedrytesttubeandathermometer.
2. Measure10.0mLofdeionizedwaterinagraduatedcylinder.Recordthetemperatureofthedeionizedwater.
3. Weighbetween1.20to1.25gramsofanhydroussodiumacetate(NaC2H3O2 (s)).Recordtheexactmass.Closethe
lid of the jar immediately after weighing – anhydrous sodium acetate will absorb atmospheric moisture and
conventtothehydrateifitisexposedtoair–wewanttokeepitanhydrous!
4. Workingquickly,placethesolidintothetesttube.Holdthetesttubeatthetoporinaracktoavoidtransferring
heatfromyourhands.Addthe10.0mLofwatertothetesttube.Besurethattherearenosolidsstickingtothe
sides.Stirgentlywithyourthermometer.Measurethesolution’stemperatureasyoustir,andrecordthehighest
temperaturethatthesolutionreaches.
5. Pourthesolutiondownthesinkandwashdrythetesttube.
B.DissolutionofSodiumAcetateTrihydrate
6. Measureanother10.0mLofwater.Recordthetemperatureofthedeionizedwater.
7. Weighoutbetween1.96to2.04gramsofsodiumacetatetrihydrate(NaC2H3O2.3H2O(s)).Recordtheexactmass.
8. Place the solid in the washed test tube and dried test tube. Add the 10.0 mL of water, and begin to stir gently.
Again,besurethatnosolidisstucktothesidesofthetesttube.Measurethesolution’stemperatureasyoustir,and
recordthelowesttemperaturethatthesolutiongetsto.
9. Pourthesolutiondownthesinkandwashthetesttube.
C.Calculations
10. ForbothPartIandPartII,calculatetheheatgainedorlostbythesolutionusingequation(3).Assumethespecific
heatofthesolutionisthesameaspurewater(4.18J/goC)andthatthesolutionhasthesamedensityaspurewater
(1.00g/mL).
Theheatgainedorlostbythereactionshouldbeofthesamemagnitudeasheatgainedorlostbythereaction,but
havetheoppositesign.Useequation(2)tofindqreaction.
FindΔHforequations(5)and(6)bydividingqreactionbymolesofreactant(equation4)
Finally,manipulateequations(5)and(6)sothattheyaddtoequation(7)andapplyHess’sLawtodetermineitsΔH.
EnthalpyofHydration:DataSheet
Showallyourworkclearlyandincludeappropriateunits.
A:DissolutionofAnhydrousSodiumAcetate
Massofwater
MassofNaC2H3O2
Massofsolution
RoomTemperature
FinalTemperature
ΔT
Specificheatcapacityofthesolution
4.18J/goC
Calculationsforqsolution:
qsolution=________________ qreaction=________________
B.DissolutionofSodiumAcetateTrihydrate
Massofwater
.
MassofNaC2H3O2 3H2O
Massofsolution
RoomTemperature
FinalTemperature
ΔT
Specificheatcapacityofthesolution
4.18J/goC
Calculationsforqsolution:
qsolution=________________ qreaction=_______________
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EnthalpyofHydration:PostLabQuestions
Name________________________
Showallyourworkandincludeappropriateunits.
Q1.CalculatemolesofNaC2H3O2(s),thenuseqreactiontocalculateΔHreactionforanhydroussodiumacetateinkilojoules.
NaC2H3O2(s)àNa+(aq)+C2H3O2-(aq)
Q2.CalculatemolesofNaC2H3O2.3H2O(s),thenuseqreactiontocalculateΔHreactionforsodiumacetatetrihydratein
kilojoules.
NaC2H3O2.3H2O(s)àNa+(aq)+C2H3O2-(aq)+3H2O(l)
Q3.UseHess’Lawandtheresultsfromquestions1and2tofindtheenthalpyofhydration(ΔHreaction)foranhydrous
sodiumacetateinkJ.Showyourworkclearly–rewritetransformedreactionstepsfromquestions1and2above.
NaC2H3O2(s)+3H2O(l)àNaC2H3O2.3H2O(s)
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EnthalpyofHydration:PostLabQuestions
Name________________________
Q4.AnotherwaytocalculateΔHvaluesthatcannotbemeasuredexperimentallyiswithstandardheatsofformation.
a)UsingtheΔHfovaluesprovided,calculateΔHrxnforNaC2H3O2(s)+3H2O(l)àNaC2H3O2.3H2O(s)withtheequation:
ΔHrxno=ΣnpΔHfo(products)–ΣnrΔHfo(reactants)
ΔHfo(kJ/mol)
NaC2H3O2(s)
–708.8
NaC2H3O2.3H2O(s)
–1604
H2O(l)
–285.8
b)Compareyourvaluefrom4a)toyourexperimentallydeterminedvalueinQ3.Determinethepercenterror.
Q5.UsethestandardreactionenthalpiesgivenbelowtodetermineΔH°rxnforthefollowingreaction.Showyourworkin
detail.
Find:
2CH4+2O2àCH2CO+3H2OΔH°rxn=?
Given:
CH2CO+2O2à2CO2+H2O
ΔH°rxn=-981.1kJ
CH4+2O2àCO2+2H2O
ΔH°rxn=-802.3kJ
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