Enthalpy of fusion - Wikipedia

The enthalpy of fusion of a substance, also known as (latent) heat of fusion, is the change ... water from 20 °C to 0 °C. Silicon has a heat of fusion of 50.21 kJ/mol. Enthalpyoffusion FromWikipedia,thefreeencyclopedia Jumptonavigation Jumptosearch Enthalpychangewhenasubstancemelts Enthalpiesofmeltingandboilingforpureelementsversustemperaturesoftransition,demonstratingTrouton'srule Theenthalpyoffusionofasubstance,alsoknownas(latent)heatoffusion,isthechangeinitsenthalpyresultingfromprovidingenergy,typicallyheat,toaspecificquantityofthesubstancetochangeitsstatefromasolidtoaliquid,atconstantpressure.Forexample,whenmelting1 kgofice(at0 °Cunderawiderangeofpressures),333.55kJofenergyisabsorbedwithnotemperaturechange.Theheatofsolidification(whenasubstancechangesfromliquidtosolid)isequalandopposite. Thisenergyincludesthecontributionrequiredtomakeroomforanyassociatedchangeinvolumebydisplacingitsenvironmentagainstambientpressure.Thetemperatureatwhichthephasetransitionoccursisthemeltingpointorthefreezingpoint,accordingtocontext.Byconvention,thepressureisassumedtobe1 atm(101.325 kPa)unlessotherwisespecified. Contents 1Overview 2Examples 3Solubilityprediction 3.1Proof 4Seealso 5Notes 6References Overview[edit] The'enthalpy'offusionisalatentheat,because,whilemelting,theheatenergyneededtochangethesubstancefromsolidtoliquidatatmosphericpressureislatentheatoffusion,asthetemperatureremainsconstantduringtheprocess.Thelatentheatoffusionistheenthalpychangeofanyamountofsubstancewhenitmelts.Whentheheatoffusionisreferencedtoaunitofmass,itisusuallycalledthespecificheatoffusion,whilethemolarheatoffusionreferstotheenthalpychangeperamountofsubstanceinmoles. Theliquidphasehasahigherinternalenergythanthesolidphase.Thismeansenergymustbesuppliedtoasolidinordertomeltitandenergyisreleasedfromaliquidwhenitfreezes,becausethemoleculesintheliquidexperienceweakerintermolecularforcesandsohaveahigherpotentialenergy(akindofbond-dissociationenergyforintermolecularforces). Whenliquidwateriscooled,itstemperaturefallssteadilyuntilitdropsjustbelowthelineoffreezingpointat0 °C.Thetemperaturethenremainsconstantatthefreezingpointwhilethewatercrystallizes.Oncethewateriscompletelyfrozen,itstemperaturecontinuestofall. Theenthalpyoffusionisalmostalwaysapositivequantity;heliumistheonlyknownexception.[1]Helium-3hasanegativeenthalpyoffusionattemperaturesbelow0.3K.Helium-4alsohasaveryslightlynegativeenthalpyoffusionbelow0.77 K(−272.380 °C).Thismeansthat,atappropriateconstantpressures,thesesubstancesfreezewiththeadditionofheat.[2]Inthecaseof4He,thispressurerangeisbetween24.992and25.00 atm(2,533 kPa).[3] StandardenthalpychangeoffusionofperiodthreeStandardenthalpychangeoffusionofperiodtwooftheperiodictableofelements Substance Heatoffusion (cal/g) (J/g) water 79.72 333.55 methane 13.96 58.99 propane 19.11 79.96 glycerol 47.95 200.62 formicacid 66.05 276.35 aceticacid 45.90 192.09 acetone 23.42 97.99 benzene 30.45 127.40 myristicacid 47.49 198.70 palmiticacid 39.18 163.93 sodiumacetate 63–69 264–289[4] stearicacid 47.54 198.91 gallium 19.2 80.4 paraffinwax(C25H52) 47.8–52.6 200–220 ThesevaluesaremostlyfromtheCRCHandbookofChemistryandPhysics,62ndedition.Theconversionbetweencal/gandJ/gintheabovetableusesthethermochemicalcalorie(calth)=4.184joulesratherthantheInternationalSteamTablecalorie(calINT)=4.1868joules. Examples[edit] Toheat1 kgofliquidwaterfrom0 °Cto20 °Crequires83.6 kJ(seebelow).However,heating0 °Ciceto20 °Crequiresadditionalenergytomelttheice.Wecantreatthesetwoprocessesindependently;thus,toheat1 kgoficefrom273.15 Ktowaterat293.15 K(0 °Cto20 °C)requires: (1)333.55 J/g(heatoffusionofice)=333.55 kJ/kg=333.55 kJfor1 kgoficetomelt,plus (2)4.18 J/(g⋅K)×20 K=4.18 kJ/(kg⋅K)×20 K=83.6 kJfor1 kgofwatertoincreaseintemperatureby20 K (1+2)333.55 kJ+83.6 kJ=417.15 kJfor1 kgoficetoincreaseintemperatureby20 K Fromthesefiguresitcanbeseenthatoneparticeat0 °Cwillcoolalmostexactly4partswaterfrom20 °Cto0 °C.Siliconhasaheatoffusionof50.21 kJ/mol.50 kWofpowercansupplytheenergyrequiredtomeltabout100 kgofsiliconinonehour: 50 kW=50kJ/s=180000kJ/h 180000kJ/h×(1 molSi)/50.21kJ×28gSi/(molSi)×1kgSi/1000gSi=100.4kg/h Solubilityprediction[edit] Theheatoffusioncanalsobeusedtopredictsolubilityforsolidsinliquids.Providedanidealsolutionisobtainedthemolefraction ( x 2 ) {\displaystyle(x_{2})} ofsoluteatsaturationisafunctionoftheheatoffusion,themeltingpointofthesolid ( T fus ) {\displaystyle(T_{\text{fus}})} andthetemperature ( T ) {\displaystyle(T)} ofthesolution: ln ⁡ x 2 = − Δ H fus ∘ R ( 1 T − 1 T fus ) {\displaystyle\lnx_{2}=-{\frac{\DeltaH_{\text{fus}}^{\circ}}{R}}\left({\frac{1}{T}}-{\frac{1}{T_{\text{fus}}}}\right)} Here, R {\displaystyleR} isthegasconstant.Forexample,thesolubilityofparacetamolinwaterat298Kispredictedtobe: x 2 = exp ⁡ [ − 28100   Jmol − 1 8.314   JK − 1   mol − 1 ( 1 298   K − 1 442   K ) ] = 0.0248 {\displaystylex_{2}=\exp{\left[-{\frac{28100~{\text{Jmol}}^{-1}}{8.314~{\text{JK}}^{-1}~{\text{mol}}^{-1}}}\left({\frac{1}{298~{\text{K}}}}-{\frac{1}{442~{\text{K}}}}\right)\right]}=0.0248} Sincethemolarmassofwaterandparacetamolare18.0153gmol−1and151.17gmol−1andthedensityofthesolutionis1000gL−1,anestimateofthesolubilityingramsperliteris: 0.0248 × 1000   gL − 1 18.0153   gmol − 1 1 − 0.0248 × 151.17   gmol − 1 = 213.4   gL − 1 {\displaystyle{\frac{0.0248\times{\frac{1000~{\text{gL}}^{-1}}{18.0153~{\text{gmol}}^{-1}}}}{1-0.0248}}\times151.17~{\text{gmol}}^{-1}=213.4~{\text{gL}}^{-1}} whichisadeviationfromtherealsolubility(240 g/L)of11%.Thiserrorcanbereducedwhenanadditionalheatcapacityparameteristakenintoaccount.[5] Proof[edit] Atequilibriumthechemicalpotentialsforthepuresolventandpuresolidareidentical: μ solid ∘ = μ solution ∘ {\displaystyle\mu_{\text{solid}}^{\circ}=\mu_{\text{solution}}^{\circ}\,} or μ solid ∘ = μ liquid ∘ + R T ln ⁡ X 2 {\displaystyle\mu_{\text{solid}}^{\circ}=\mu_{\text{liquid}}^{\circ}+RT\lnX_{2}\,} with R {\displaystyleR\,} thegasconstantand T {\displaystyleT\,} thetemperature. Rearranginggives: R T ln ⁡ X 2 = − ( μ liquid ∘ − μ solid ∘ ) {\displaystyleRT\lnX_{2}=-\left(\mu_{\text{liquid}}^{\circ}-\mu_{\text{solid}}^{\circ}\right)\,} andsince Δ G fus ∘ = μ liquid ∘ − μ solid ∘ {\displaystyle\DeltaG_{\text{fus}}^{\circ}=\mu_{\text{liquid}}^{\circ}-\mu_{\text{solid}}^{\circ}\,} theheatoffusionbeingthedifferenceinchemicalpotentialbetweenthepureliquidandthepuresolid,itfollowsthat R T ln ⁡ X 2 = − ( Δ G fus ∘ ) {\displaystyleRT\lnX_{2}=-\left(\DeltaG_{\text{fus}}^{\circ}\right)\,} ApplicationoftheGibbs–Helmholtzequation: ( ∂ ( Δ G fus ∘ T ) ∂ T ) p = − Δ H fus ∘ T 2 {\displaystyle\left({\frac{\partial\left({\frac{\DeltaG_{\text{fus}}^{\circ}}{T}}\right)}{\partialT}}\right)_{p\,}=-{\frac{\DeltaH_{\text{fus}}^{\circ}}{T^{2}}}} ultimatelygives: ( ∂ ( ln ⁡ X 2 ) ∂ T ) = Δ H fus ∘ R T 2 {\displaystyle\left({\frac{\partial\left(\lnX_{2}\right)}{\partialT}}\right)={\frac{\DeltaH_{\text{fus}}^{\circ}}{RT^{2}}}} or: ∂ ln ⁡ X 2 = Δ H fus ∘ R T 2 × δ T {\displaystyle\partial\lnX_{2}={\frac{\DeltaH_{\text{fus}}^{\circ}}{RT^{2}}}\times\deltaT} andwithintegration: ∫ X 2 = 1 X 2 = x 2 δ ln ⁡ X 2 = ln ⁡ x 2 = ∫ T fus T Δ H fus ∘ R T 2 × Δ T {\displaystyle\int_{X_{2}=1}^{X_{2}=x_{2}}\delta\lnX_{2}=\lnx_{2}=\int_{T_{\text{fus}}}^{T}{\frac{\DeltaH_{\text{fus}}^{\circ}}{RT^{2}}}\times\DeltaT} theendresultisobtained: ln ⁡ x 2 = − Δ H fus ∘ R ( 1 T − 1 T fus ) {\displaystyle\lnx_{2}=-{\frac{\DeltaH_{\text{fus}}^{\circ}}{R}}\left({\frac{1}{T}}-{\frac{1}{T_{\text{fus}}}}\right)} Seealso[edit] Heatofvaporization Heatcapacity Thermodynamicdatabasesforpuresubstances Jobackmethod(Estimationoftheheatoffusionfrommolecularstructure) Latentheat Latticeenergy Heatofdilution Notes[edit] ^Atkins&Jones2008,p. 236. ^Ott&Boerio-Goates2000,pp. 92–93. ^Hoffer,J.K.;Gardner,W.R.;Waterfield,C.G.;Phillips,N.E.(April1976)."Thermodynamicpropertiesof4He.II.ThebccphaseandtheP-TandVTphasediagramsbelow2K".JournalofLowTemperaturePhysics.23(1):63–102.Bibcode:1976JLTP...23...63H.doi:10.1007/BF00117245.S2CID 120473493. ^IbrahimDincerandMarcA.Rosen.ThermalEnergyStorage:SystemsandApplications,page155 ^MeasurementandPredictionofSolubilityofParacetamolinWater-IsopropanolSolution.Part2.PredictionH.HojjatiandS.RohaniOrg.ProcessRes.Dev.;2006;10(6)pp1110–1118;(Article)doi:10.1021/op060074g References[edit] Atkins,Peter;Jones,Loretta(2008),ChemicalPrinciples:TheQuestforInsight(4th ed.),W.H.FreemanandCompany,p. 236,ISBN 978-0-7167-7355-9 Ott,BJ.Bevan;Boerio-Goates,Juliana(2000),ChemicalThermodynamics:AdvancedApplications,AcademicPress,ISBN 0-12-530985-6 vteStatesofmatter(list)State Solid Liquid Gas/Vapor Plasma Lowenergy Bose–Einsteincondensate Fermioniccondensate Degeneratematter QuantumHall Rydbergmatter Rydbergpolaron Strangematter Superfluid Supersolid Photonicmolecule Highenergy QCDmatter LatticeQCD Quark–gluonplasma Color-glasscondensate Supercriticalfluid Otherstates Colloid Glass Crystal Liquidcrystal Timecrystal Quantumspinliquid Exoticmatter Programmablematter Darkmatter Antimatter Magneticallyordered Antiferromagnet Ferrimagnet Ferromagnet String-netliquid Superglass Transitions Boiling Boilingpoint Condensation Criticalline Criticalpoint Crystallization Deposition Evaporation Flashevaporation Freezing Chemicalionization Ionization Lambdapoint Melting Meltingpoint Recombination Regelation Saturatedfluid Sublimation Supercooling Triplepoint Vaporization Vitrification Quantities Enthalpyoffusion Enthalpyofsublimation Enthalpyofvaporization Latentheat Latentinternalenergy Trouton'srule Volatility Concepts Baryonicmatter Binodal Compressedfluid Coolingcurve Equationofstate Leidenfrosteffect Macroscopicquantumphenomena Mpembaeffect Orderanddisorder(physics) Spinodal Superconductivity Superheatedvapor Superheating 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