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HONPrinciples of Chemistry II

by: Gavin Harvey

HONPrinciples of Chemistry II CHEM 112

Marketplace > University of Idaho > Chemistry > CHEM 112 > HONPrinciples of Chemistry II
Gavin Harvey
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This 41 page Class Notes was uploaded by Gavin Harvey on Friday October 23, 2015. The Class Notes belongs to CHEM 112 at University of Idaho taught by Staff in Fall. Since its upload, it has received 76 views. For similar materials see /class/227960/chem-112-university-of-idaho in Chemistry at University of Idaho.


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Date Created: 10/23/15
Chapter 6 Thermochemistry Thermodynamics is the science of the relationship between heat and other forms of energy Chapter 1920 Chem 112 Thermochemistry is the study of the quantity of heat absorbed or evolved by chemical reactions A subset of thermodynamics Chapter 6 this course Law of Conservation of Energy Energy is neither created nor destroyed Energy can be converted from one form to another Energy can be transferred from one system to another 0 2 ways energy can be transferred from one system to another Work 1w energy exchange that results when force F moves an object through a distance d w forcedistance w pressurevol change PAV Consider Nas H20 gt 2 H2Cg NaOHaq ngre 6 9 Pressurewo ume Work um 53575 kl Ilefare Alhr Heat jg energy that flows out of the system because of a difference in temperature between the system and the surroundings Internal Energy U sum of the kinetic and potential energies of the particles making up the system AU q W Enthalpy H is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction Defined as H U PV or AH AU PAV constant P Remember that W PAV amp AU q W Plug these definitions into AH AU PAV We have AH q PAV PAV AH 11 const P At constant pressure enthalyv is heat All of considerations in Chem 1 I I will be at constant pressure Sign Convention of q and AH Sign of q is Sign of q is Sign convention of enthalpy changes AH and heat q in chemical reactions at constant pressure Reactants Product lt H enthalpy O gt AH 0 AH v Product m EXOtheI IIIiC RXH Endotherniic Rxn Thermochemical Eguations are the chemical equations for reactions including phase labels in which the equation is given a molar interpretation and the enthalpy of reaction for these molar amounts is written directly after the equation N2g 3H2g a 2NH3g AH 918 kJ The minus sign indicates that this is an exothermic rxn N2g 3H2g gt 2NH3g AH 9l8 M M is kilojoules Where joule kg mZs2 note that l calorie 4184 J Stoichiometg and Thennochemical Eguations N2g 3H2g gt 2NH3g AH 9l8 kJ Calculate the amount of heat q released by the combination of 350 g N2g and 221 g of H2g 350gN2 moleN2 9l8k115k 2802gN2 moleN2 moldI2 9l8kJ 2016gH2 3m0leH2 221gII2 335k Answer q ll5 kJ Endothermic chemical rxns have a positive AH C02g 2 H201gt CH4g 2 02g AH 890 kJ Rules for manipulation of chemical rxns and AH 1 If we reverse this rxn then we must reverse the sign of AH CH4g 2 02g gt C02g 2 H200 AH 890 Id 2 If we multiply the coefficients of the rxn by a constant we must multiply AH by that constant CH4g 2 02g gt C02g 2 H200 AH 890 H 12 CH4g 02g gt 12 C02g H200 AH 445 H 3 If we add two rxns together in order to obtain a third rxn we must add their AH s rxn l Cs 12 02g gt COgAH1 1105 M rxn 2 COGI 12 032 gt C0201 AH 2830 N rxn l 2 Cs 12 02g GG gH 12 02g gt GG g C02g Cs 02g gt C02g AH3 AH1 AHZ 3935 M Collectively these rules fonn the basis of Hess Law Hess s law of heat summation states that for a chemical equation that can be w1itten as the sum of two or more steps the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps In class Given the following two rxns Ss02g gt S02g AH 297 kJ 2503g a 2502g 02g AH 198 kJ Calculate AH for 255 3 02g gt 2503g AH 7 Answer 792 kJ Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data 1 atmosphere pressure and the specified temperature usually 25 OC Note that the standard state condition is not to be confused with STP standard state 1 atm amp 25 0C 298 K STP 1 atm amp 0 0C 273 K The enthalpy change for a reaction in which reactants are in their standard states is denoted AH delta H zero or delta H naught The standard enthalpy of formation of a substance denoted AHfO is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state Note that the standard enthalpy of formation for a pure element in its standard state is zero The law of summation of heats of formation states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants AH0 Z nAH f products Z mAH f reactants 2 is the mathematical symbol meaning the sum of and m and n are the coefficients of the substances in the chemical equation Example Calculate AHO for Cs V2 02g gt C02g AH C0 AHfO 02g 0 AEWXMQ65kamwh62 aD AW35M mmmm35m In class Calculate AHO for 4NH3 g 502g gt 4N0g 6H20g Using data from table 62 p260 in your text see next page of these lecture notes Answer AHO 906 kJ Note Be careful of arithmetic signs as they are a likely source of mistakes ilk E SW u u n HI NW at Limaquot SH mm lm nii Mmll fig 393 g2 25 E sil H IT H m u mun 4M Hun SHEA HEW 39 l l HamII El EEMITJ EEJEI mg n I 139 Nn gl 61193 Mil Iiiquot HEW5amp3 39 HEPth 35 Hip li l39u39i EHETHEI39J EMI39 Hum u im H J IJ WI l3 l Hn lf yin 1W1 i l E55 quotBilTERI quotHEME 39JT Emma El amp 1 H3I Sitar Elli1 I SEE Mill539 Mi ELI35 SiF g IMH mayme ti l W finhm lm 1 Emil TIE Ml 139J El graph ll a J HE J 391 did I 9 i39iinmn Hg X I LE M5 4733 g ag5 Haul I Hmfmqla EampIJ Hktr l Hui EFLng HF TIM E1 mamd Fm EH Huhnil FluInd JIM m Hl lfgaqj 4113 HERE 11 my W3 E1 39 45 EMZE HEB11mg YIDEIJEI Elli J HEM 1 9313 5433 H mm 143 Elalm H j lling m9 EE 1111 HELEN 34 E Brill I IIIglilm 35 MM 5 E Ilg39 PM 513 I ll I39m 55395I 51m 1E E1121 1 mm D lm ngmnnndhn 2 Ag I39m I FEB 11 SID ELng 925553 M mw I111 H23ij 32 hair I FluHEM 31 II39jI3 113 13 ESE wkquot 1quot39 Fquot I m 33539quot hi nj FILE ill mg qu l i 39E39nm mul I39m I ilhl 12 Measuring Heats of Reaction To see how heats of reactions are measured we must look at the heat required to raise the temperature of a substance because a thermochemical measurement is based on the relationship between heat and temperature change The heat required to raise the temperature of a substance is its heat capacity The heat cagaci C of a sample of substance is the quantity of heat required to raise the temperature of the sample of substance one degree Celsius Changing the temperature of the sample requires heat equal to q CAT note units of C JOC Suppose a piece of iron requires 670 J of heat to raise its temperature by one degree Celsius The quantity of heat required to raise the temperature of the piece of iron from 250 0C to 350 0C is q CAT 670 J0 Cgtlt350 0C 250 0C q 670J Heat capacities are also compared for one gram amounts of substances The specific heat capacity or specific heat is the heat required to raise the temperature of one gram of a substance by one degree Celsius To find the heat required you must multiply the specific heat s of the substance times its mass in grams m and the temperature change AT q SmAT Molar heat capacity is the heat required to raise the temperature of one mole of a substance by one degree Celsius Each substance has a unique specific heat and molar heat capacity this is a physical property From Table 61 p 251 Specific Heat Molar Heat Capacity JgOC JmolOC Als 0901 243 Cus 03 84 244 Fes 0449 25 1 H20l 418 753 Calculate the heat absorbed when the temperature of 150 grams of water is raised from 200 0C to 500 0C The speci c heat of water is 4184 Jg39OC q 4184 i x 150ggtlt 500 2000C q s x m x AT q188gtlt103J A calorimeter is a device used to measure the heat absorbed or evolved during a physical or chemical change see Figure 611 The heat absorbed by the calorimeter and its contents are the negative of the heat of reaction qcalorimeter qrxn F1g 61184612 CaCl2sH20CaCl2 aq When 236 grams of calcium chloride CaClZ was dissolved in water in a calorimeter the temperature rose from 250 0C to 387 0C If the heat capacity of the solution and the calorimeter is 1258 JOC what is the enthalpy change per mole of calcium chloride Note that we must know the heat capacity of the solution and the calorimeter in order to do this problem qwl CAT 1258 J gtlt387 Jo 250 0C 0C qcal 172x104J Now we must calculate the heat per mole of calcium chloride 236 CaCl XM 0212 mol CaCl g 2 2 1111g AH 0 3917392 1 811kJmol m0 CaCl2 0212 mol In Class 0500 L of 0200 M NaClaq was added to 0500 L of 0200 M AgN03aq in a calorimeter The temperature rose from 25000 0C to 26423 OC The heat capacity of the solution and the calorimeter is 460103 JOC What was the complete balanced chemical equation for the reaction What is AH per mole of this reaction Answer AH 655 kJmol Chapter 11 States of Matter 0 Gases are compressible uids Their molecules are Widely separated o Liquids are relatively incompressible uids Their molecules are more tightly pac e o Solids are nearly incompressible and rigid Their molecules or ions are in close contact and do not moVe Intermolecular Forces Explaining Liguid Properties 0 Many of the physical properties of liquids and certain solids can be explained in terms of intermolecular forces the forces of attraction between molecules 0 Three types of forces are known to exist between neutral molecules 1Dipoledipole forces aka van der Waals 2London or dispersion forces aka van der Waals 3Hydrogen bonding DipoleDipole Forces 0 Polar molecules can attract one another through dipoledipole forces 0 The dipoledipole force is an attractive intermolecular force resulting from the tendency of polar molecules to align themselves positive end to negative end 0 Use VSEPR rules Chapter 10 to obtain the 3D structure The First Chem 112 7 Viscosity Week of September 9 o Viscosity is the resistance to ow exhibited by all liquids and gases 0 Viscosity can be illustmted by measuring the time required for a steel ball to fall through a column of the liquid see Figures 1119 and 1120 c In comparisons ofsubstances as intermolecular forces increase viscosity usually increases London is ersion Forces 0 London forces are the weak attractive forces resulting from instantaneous dipoles that occur due to the distortion of the electron cloud surrounding a molecule Figure 1122 0 London forces increase with molecular weight The larger a molecule the more easily it can be distorted to give an instantaneous dipole 0 Higher MW larger electron cloud 0 All covalent molecules exhibit some London force Example at Room Temperature and Pressure STP conditions I is a liquid F2 is a gas Why Hydrogen Bonding 0 Hydrogen bonding is a force that exists between a hydrogen atom covalently bonded to a very electronegative atom X and a lone pair of electrons on a very electronegative atom Y 0 To exhibit hydrogen bonding one of the following three structures must be present H N H O H F o On N O and F are electrone ative enough to leave the hydrogen nucleus exposed How is Hbonding different from dipoledipole attractions 0 A hydrogen atom bonded to an electronegative atom appears to be special 0 The electrons in the 0H bond are drawn to the O atom leaving the dense positive charge of the hydrogen nucleus exposed 0 It s the strong attraction of this exposed nucleus for the lone pair on an adjacent molecule that accounw for the strong attraction o A similar mechanism explains the attractions in HF and NH3 Hbonding is a signi cant attractive force in liquid Water and solid Water ice Fig 1125 Hyumgen bond Stmct39ure ofice is based on Hbonding between molecules Hydrogenbonding is a signi cant force in biological m olecules Stmctures of proteins enzymes and nucleic acids DNA Summarx of Intermolecular Attractive Forces Table 1 1 4 Types of Intermolecular and Chemical Bonding Interactions Approximate 1ype of Interaction Energy kJmol Intermolecular Van der Waals dipole dipole London 01 to 10 Hydrogen bonding 10 to 40 Chemical bonding Ionic 100 to 1000 Covalent 100 to 1000 0 Do homework problems 1110 12 13 55 57 Next We Will consider how intermolecular attractive forces affect physical properties of solids and liquids Viscosity upcoming lab Vapor Pressure Boiling Point Freezing Point Changes of State 0 A change of state or phase transition is a change ofa substance from one state to another see Table 111 mile 1 11 Kinds of Eliase Trailsi ons39 Phase Transition Name Examples Solid 4 liquid Solid gt gas Liquid solid Liquid gas Gas gt liquid Gas 8 solid Melting fusion Sublimation Freezing Vaporization Condensation liquefaction Condensation deposition Melting of snow and ice Sublimation of dry ice freezedrying of coffee Freezing of water or a liquid metal Evaporation of water or refrigerant Formation of dew liquefaction of carbon dioxide Formation of frost and snow Vapor Pressure gliguidgas o Liquids are continuously VapoIizing HZOl gt HZOQ vaporization 0 Their Vapors are continuously condensing HZOQ gt HZOl condensation o Ifa liquid is in a closed Vessel with space above it a paltial pressure ofthe Vapor state builds up in this space Example r 4 Tcp of mercury column burure lhc experiment 3 Wulur molecule Water surface o The vapor pressure ofaliquid is the partial pressure of the Vapor oVer the liquid measured at eguilibrium at a given temperature see Figure 114 Equilibrium 7 When the rate of vaporization equals the rate condensation watzr mulecules vapurzzmg jwam mulecules Cundensing new own on 7 7 atgcondansa 1 secund Rme of Equilibrium vapnrizallon mined Rules become equal Ruin Rate of Cnndenszmon I Timta Initial conditions Final conditions Vapor Pressure 0 The Vapor pressure ofa liquid depends on its temperature 0 As the temperature increases the kinetic energy of the molecular motion becomes greater and Vapor pressure increases 0 Liquids and solids With relatively high Vapor pressures at normal temperatures are said to be volatile chloroform Diethyl eurer Vapor pressure mmHg 8 0 Temperature C 0 Explain the VP trend for the graph above 0 Do Homework Problem 1161 62 Boiling Point 0 The temperature at Which the vapor pressure of a liquid equals the pressure exerted on the liquid is called the boiling point 0 As the temperature ofa liquid increases the Vapor pressure increases until it reaches atmospheric pressure 0 At this point stable bubbles of Vapor form Within the liquid This is called boiling o The normal boiling point is the boiling point at 1 atm Almospheric I s r 0 BP are strongly affected by the pressure exerted on their surface Vapor bubble Boiling pnlm C Intermolecular Forces and Bo39 39 g P 39 tTrends 0 Consider the following trends each line represents the series of elements in a column ofthe periodic table 0 First consider the le graph Which include the hydrides of Group VIA elements 0 How can We explain this trend HQ 20 HZSe H35 40 0 20 Molecular wel ghl A 100 HzTe 120 Boiling point C IOU IZU 440 7160 20 44 U 80 00 Molecular welglu B IZO o How can We explain the trends on the righthand graph 0 Do homework problems 1163 64 Freezing Point 0 The temperature at Which a pure liquid changes to a crystalline solid or freezes is called the freezing point 0 The melting point is identical to the freezing point and is de ned as the temperature at Which a solid becomes a liquid FF and BP trends strongly correlate and usually can be explained by the same intermolecular forces Melh39 g nailing Poi n Nam 1y of Solid39 n c Point c Neon Ne Molecular 249 246 Hydrogen sul de H25 Molecular 86 61 Chloroform CHCI Molecular 64 62 Water H20 Molecular 0 100 Acetic acid HC2H302 Molecular 17 118 Solid State 0 A solid is a nearly incompressible state of matter with a wellde ned shape The units making up the solid are in close contact and in xed positions 0 Solids are characterized by the type of force holding the structural units together 0 In some cases these forces are intermolecular but in others they are chemical bonds metallic ionic or covalent Types of Solids o A molecular solid is a solid that consists of atoms or molecules held together by intermolecular forces 0 Many solids are of this type 0 Examples include solid neon solid water ice and solid carbon dioxide dry ice o A metallic solid is a solid that consists of positive cores of atoms held together by a surrounding sea of electrons metallic bonding o In this kind of bonding positively charged atomic cores are surrounded by delocalized electrons 0 Examples include iron copper and silver 0 An ionic solid is a solid that consists of cations and anions held together by electrical attraction of opposite charges ionic bond 0 Examples include cesium chloride sodium chloride and magnesium bromide o A covalent network solid is a solid that consists of atoms held together in large networks or chains by covalent bonds 0 Examples include carbon in its forms as diamond or graphite see Figure 1127 SiOz glass Diamond Graph What is the difference in VSPER and orbital hybridization schemes between diamond and graphite 0 Why is diamond a more durable material for drill bis 0 Why is graphite a lubricant Table 115 summa zes these four types ofsolids fable 1 1 5 lypes of solids Antanive Forces 3am fype of Solid Shuthlrll Units S luduml I ll Examples Vlolcctllar Atoms or mulcculcs Intermolecular forces NCy H10 CD Vlcixllic Atoms positive eoies Metallic bonding Fe Cu Ag surruun y exrrenie deioeniized eleclmn sea bond Ionic Ions Ionic bonding CSCL NaCL ZnS ovaleni neiwotk Moms Covsieni bonding Diamond graphite asbeslns FF and BP Trends are apparent in Table 112 Table 1 1 2 Melting Points and Boiling Points if 1 aim of Several Substances Melting Boillng Poi I Name Npe of Solid39 n C Point nC con Ne Molecular 7249 7246 Hydrogen sul de H25 Molecular 86 761 Clilorcidrm CHCl Molecular 64 62 Water H20 Molecular 0 mo Aeede acid HQH302 Molecular I7 118 Mercury Hg Metallic 739 357 Sodium Na Metallic 93 883 Tungsten W 39 3410 5660 Cesium chloride CsCI Ionic 645 290 odium chlon39de Na 1 ionic 801 1413 Magnesium oxide Mgo Ionic 2800 3600 Quanz Sio2 Covalent network 16 10 2230 Diamond c Cov em network 3 50 4327 39Typcs of solids are discussed in Seclinn 115 Do Homework Problems 1165 67 69 71 73 Heats of Phase Transitions MeltingFusion To melt a pure substance at its melting point requires an extra boost of energy to overcome lattice energies o The heat needed to melt 1 mol of a pure substance is called the heat of melting or fusion and denoted AI39Imelt o For ice the heat of fusion is 601 kJmol H20s gt H20l AHfus 601 kJ o The heat of freezing for ice is 601 kJmol AHfreezing 39 AHmelt 39 AHfusion 20 BoilingCondensation To boil a pure substance at its boiling point requires an extra boost of energy to overcome intermolecular forces 0 The heat needed to boil 1 mol of a pure substance is called the heat of vaporization and denoted AHvap see Figure 119 o For ice the heat of vaporization is 4066 kJmol H20l gt H20g AHvap 4066 kJ 0 AHcondensa on 4066 kJmol 21 Heating curve for water Temperature C Water and steam Time a Heal added at constant rate Example Suppose We are interested in the amount of heat required to take 346 g ofwater from 0 C to steam at 182 C Important constanw We will have to look up Speci c heat ofwater Cs1 4184 Jg C Speci c heat ofsteam ng 199 Jg C Step 1 346 g H201 0 C gt 346 g H201 100 C AH1 145 kJ Step 2 vaporize 346 g of water to steam AH2 781 kJ Step 3 346 g steam 100 0C to 182 0C AH3 565 kJ AHtotal Homework 1137 amp 39 23


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