starlanter, profile picture
Uploaded bystarlanter
PPT, PDF285 views

Thermochem

Thermodynamics is the study of heat and work, and state functions. Energy exists in many forms including heat, light, electrical, kinetic, and potential. Heat is the transfer of energy between objects due to a temperature difference. Chemical reactions are driven by thermodynamics and kinetics. Reactions that transfer energy to the surroundings are product-favored. The first law of thermodynamics states that the change in energy of a system equals heat transferred plus work done. Enthalpy represents the heat transferred at constant pressure for a chemical reaction.

Embed presentation

Download to read offline
1THERMOCHEMISTRYTHERMOCHEMISTRY ThermodynamicsThermodynamics The study of Heat and WorkThe study of Heat and Work and State Functionsand State Functions
2 Energy & ChemistryEnergy & ChemistryEnergy & ChemistryEnergy & Chemistry ENERGYENERGY is the capacity tois the capacity to do work or transfer heat.do work or transfer heat. HEATHEAT is the form of energyis the form of energy that flows between 2that flows between 2 objects because of theirobjects because of their difference in temperature.difference in temperature. Other forms of energy —Other forms of energy — • lightlight • electricalelectrical • kinetic and potentialkinetic and potential
3 Energy & ChemistryEnergy & Chemistry • Burning peanutspeanuts supply sufficientsupply sufficient energy to boil a cupenergy to boil a cup of water.of water. • Burning sugarBurning sugar (sugar reacts with(sugar reacts with KClOKClO33, a strong, a strong oxidizing agent)oxidizing agent)
4 Energy & ChemistryEnergy & Chemistry • These reactions areThese reactions are PRODUCTPRODUCT FAVOREDFAVORED • They proceed almost completelyThey proceed almost completely from reactants to products, perhapsfrom reactants to products, perhaps with some outside assistance.with some outside assistance.
5 Energy & ChemistryEnergy & Chemistry 2 H2 H22(g) + O(g) + O22(g) -->(g) --> 2 H2 H22O(g) + heat and lightO(g) + heat and light This can be set up to provideThis can be set up to provide ELECTRIC ENERGYELECTRIC ENERGY in ain a fuel cellfuel cell.. Oxidation:Oxidation: 2 H2 H22 ---> 4 H---> 4 H++ + 4 e+ 4 e-- Reduction:Reduction: 4 e4 e-- + O+ O22 + 2 H+ 2 H22O ---> 4 OHO ---> 4 OH-- CCR, page 845
6 Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy PotentialPotential energyenergy —— energy aenergy a motionlessmotionless body has bybody has by virtue of itsvirtue of its position.position.
7 • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy Potential EnergyPotential Energy on the Atomic Scaleon the Atomic Scale NaCl — composed ofNaCl — composed of NaNa++ and Cland Cl-- ions.ions.
8 • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy Potential EnergyPotential Energy on the Atomic Scaleon the Atomic Scale
9 Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy Kinetic energyKinetic energy — energy of— energy of motionmotion •• TranslationTranslation
10 Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy Kinetic energyKinetic energy — energy of— energy of motion.motion. translate rotate vibrate translate rotate vibrate
11 Internal Energy (E)Internal Energy (E) • PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U) • Int. E of a chemical systemInt. E of a chemical system depends ondepends on • number of particlesnumber of particles • type of particlestype of particles • temperaturetemperature
12 Internal Energy (E)Internal Energy (E) • PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U) QuickTime™ and a Graphics decompressor are needed to see this picture.
13 Internal Energy (E)Internal Energy (E) • The higher the TThe higher the T the higher thethe higher the internal energyinternal energy • So, use changesSo, use changes in T (∆T) toin T (∆T) to monitor changesmonitor changes in E (∆E).in E (∆E).
14 ThermodynamicsThermodynamics • Thermodynamics is the science of heat (energy) transfer. Heat energy is associatedHeat energy is associated with molecular motions.with molecular motions. Heat transfers until thermal equilibrium is established.
15 Directionality of Heat TransferDirectionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes downT(system) goes down T(surr) goes upT(surr) goes up
16 Directionality of Heat TransferDirectionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes upT(system) goes up T (surr) goes downT (surr) goes down
17 Energy & ChemistryEnergy & Chemistry All of thermodynamics dependsAll of thermodynamics depends on the law ofon the law of CONSERVATION OF ENERGYCONSERVATION OF ENERGY.. • The total energy is unchangedThe total energy is unchanged in a chemical reaction.in a chemical reaction. • If PE of products is less thanIf PE of products is less than reactants, the difference mustreactants, the difference must be released as KE.be released as KE.
18 Energy Change inEnergy Change in Chemical ProcessesChemical Processes Reactants Products Kinetic Energy PE PE of system dropped. KE increased. Therefore,PE of system dropped. KE increased. Therefore, you often feel a T increase.you often feel a T increase.
19 UNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGY 1 calorie = heat required to1 calorie = heat required to raise temp. of 1.00 g ofraise temp. of 1.00 g of HH22O by 1.0O by 1.0 oo C.C. 1000 cal = 1 kilocalorie = 11000 cal = 1 kilocalorie = 1 kcalkcal 1 kcal = 1 Calorie (a food1 kcal = 1 Calorie (a food “calorie”)“calorie”) But we use the unit calledBut we use the unit called thethe JOULEJOULE 1 cal = 4.184 joules1 cal = 4.184 joules James JouleJames Joule 1818-18891818-1889
20 HEAT CAPACITYHEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?Which has the larger heat capacity?
21 SpecificSpecific Heat CapacityHeat CapacitySpecificSpecific Heat CapacityHeat Capacity How much energy isHow much energy is transferred due to Ttransferred due to T difference?difference? The heatThe heat (q)(q) “lost” or “gained”“lost” or “gained” is related tois related to a)a) sample masssample mass b)b) change in T andchange in T and c)c) specific heat capacityspecific heat capacity Specific heat capacity= heat lost or gained by substance (J) (mass, g)(T change,K)
22 Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K) HH22OO 4.1844.184 Ethylene glycolEthylene glycol 2.392.39 AlAl 0.8970.897 glassglass 0.840.84 AluminumAluminum
23 Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity If 25.0 g of Al cool fromIf 25.0 g of Al cool from 310310 oo C to 37C to 37 oo C, howC, how many joules of heatmany joules of heat energy are lost byenergy are lost by the Al?the Al? Specific heat capacity= heat lost or gained by substance (J) (mass, g)(T change,K)
24 Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity If 25.0 g of Al cool from 310If 25.0 g of Al cool from 310 oo C to 37C to 37 oo C, how manyC, how many joules of heat energy are lost by the Al?joules of heat energy are lost by the Al? heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = Twhere ∆T = Tfinalfinal - T- Tinitialinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 Jq = - 6120 J Notice that the negative sign on q signalsNotice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al. Notice that the negative sign on q signalsNotice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.
25 Heat TransferHeat Transfer No Change in StateNo Change in State q transferred = (sp. ht.)(mass)(∆T)
26Heat Transfer withHeat Transfer with Change of StateChange of State Heat Transfer withHeat Transfer with Change of StateChange of State Changes of state involve energyChanges of state involve energy (at constant T)(at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid waterIce + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass)q = (heat of fusion)(mass)
27 Heat Transfer andHeat Transfer and Changes of StateChanges of State Heat Transfer andHeat Transfer and Changes of StateChanges of State Requires energyRequires energy (heat).(heat). This is the reasonThis is the reason a)a) you cool down afteryou cool down after swimmingswimming b)b) you use water to putyou use water to put out a fire.out a fire. + energy Liquid ---> VaporLiquid ---> Vapor
28 Heat waterHeat water Evaporate waterEvaporate water Melt iceMelt ice Heating/Cooling Curve for WaterHeating/Cooling Curve for Water Note that T isNote that T is constant as ice meltsconstant as ice melts Note that T isNote that T is constant as ice meltsconstant as ice melts
29 Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g What quantity of heat is required toWhat quantity of heat is required to melt 500. g ofmelt 500. g of iceice and heat theand heat the water towater to steamsteam at 100at 100 oo C?C? Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State +333 J/g+333 J/g +2260 J/g+2260 J/g
30 How much heat is required to melt 500. g ofHow much heat is required to melt 500. g of ice and heat the water to steam at 100ice and heat the water to steam at 100 oo C?C? 1.1. To melt iceTo melt ice q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 1055 JJ 2.2. To raise water from 0To raise water from 0 oo C to 100C to 100 oo CC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 xq = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 101055 JJ 3.3. To evaporate water at 100To evaporate water at 100 oo CC q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 JJ 4.4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 1066 J =J = Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State
31 ChemicalChemical ReactivityReactivity What drives chemical reactions? How do theyWhat drives chemical reactions? How do they occur?occur? The first is answered byThe first is answered by THERMODYNAMICSTHERMODYNAMICS and the second byand the second by KINETICSKINETICS.. Have already seen a number of “drivingHave already seen a number of “driving forces” for reactions that areforces” for reactions that are PRODUCT-PRODUCT- FAVOREDFAVORED.. •• formation of a precipitateformation of a precipitate •• gas formationgas formation •• HH22O formation (acid-base reaction)O formation (acid-base reaction) •• electron transfer in a batteryelectron transfer in a battery
32 ChemicalChemical ReactivityReactivity But energy transfer also allows us to predictBut energy transfer also allows us to predict reactivity.reactivity. In general, reactions that transfer energyIn general, reactions that transfer energy to their surroundings are product-to their surroundings are product- favored.favored. So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.
33 Heat Energy Transfer inHeat Energy Transfer in a Physical Processa Physical Process COCO22 (s, -78(s, -78 oo C) ---> COC) ---> CO22 (g, -78(g, -78 oo C)C) Heat transfers from surroundings to system in endothermic process.
34 Heat Energy Transfer inHeat Energy Transfer in a Physical Processa Physical Process • COCO22 (s, -78(s, -78 oo C) --->C) ---> COCO22 (g, -78(g, -78 oo C)C) • A regular array ofA regular array of molecules in amolecules in a solidsolid -----> gas phase-----> gas phase molecules.molecules. • Gas moleculesGas molecules have higher kinetichave higher kinetic energy.energy.
35 Energy Level DiagramEnergy Level Diagram for Heat Energyfor Heat Energy TransferTransfer ∆E = E(final) - E(initial) = E(gas) - E(solid) COCO22 solidsolid COCO22 gasgas
36 Heat Energy Transfer inHeat Energy Transfer in Physical ChangePhysical Change • Gas molecules have higherGas molecules have higher kinetic energy.kinetic energy. • Also,Also, WORKWORK is done by theis done by the system in pushing aside thesystem in pushing aside the atmosphere.atmosphere. COCO22 (s, -78(s, -78 oo C) ---> COC) ---> CO22 (g, -78(g, -78 oo C)C) Two things have happened!Two things have happened!
37 FIRST LAW OFFIRST LAW OF THERMODYNAMICSTHERMODYNAMICS ∆∆E = q + wE = q + w heat energy transferredheat energy transferred energyenergy changechange work donework done by theby the systemsystem Energy is conserved!Energy is conserved!
38 heat transfer outheat transfer out (exothermic), -q(exothermic), -q heat transfer inheat transfer in (endothermic), +q(endothermic), +q SYSTEMSYSTEM ∆E = q + w w transfer inw transfer in (+w)(+w) w transfer outw transfer out (-w)(-w)
39 ENTHALPYENTHALPY Most chemical reactions occur at constant P, soMost chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small) ∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E ∆∆H = change inH = change in heat contentheat content of the systemof the system ∆∆H = HH = Hfinalfinal - H- Hinitialinitial and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small) ∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E ∆∆H = change inH = change in heat contentheat content of the systemof the system ∆∆H = HH = Hfinalfinal - H- Hinitialinitial Heat transferred at constant P = qHeat transferred at constant P = qpp qqpp == H∆H∆ wherewhere H = enthalpyH = enthalpy Heat transferred at constant P = qHeat transferred at constant P = qpp qqpp == H∆H∆ wherewhere H = enthalpyH = enthalpy
40 IfIf HHfinalfinal < H< Hinitialinitial then H is negative∆then H is negative∆ Process isProcess is EXOTHERMICEXOTHERMIC IfIf HHfinalfinal < H< Hinitialinitial then H is negative∆then H is negative∆ Process isProcess is EXOTHERMICEXOTHERMIC IfIf HHfinalfinal > H> Hinitialinitial then H is positive∆then H is positive∆ Process isProcess is ENDOTHERMICENDOTHERMIC IfIf HHfinalfinal > H> Hinitialinitial then H is positive∆then H is positive∆ Process isProcess is ENDOTHERMICENDOTHERMIC ENTHALPYENTHALPY ∆∆H = HH = Hfinalfinal - H- Hinitialinitial
41 Consider the formation of waterConsider the formation of water HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) +O(g) + 241.8241.8 kJkJ USING ENTHALPYUSING ENTHALPY Exothermic reaction — heat is a “product”Exothermic reaction — heat is a “product” and H = – 241.8 kJ∆and H = – 241.8 kJ∆
42 MakingMaking liquidliquid HH22O from HO from H22 + O+ O22 involvesinvolves twotwo exoexothermic steps.thermic steps. USING ENTHALPYUSING ENTHALPY H2 + O2 gas Liquid H2OH2O vapor
43 Making HMaking H22O from HO from H22 involves two steps.involves two steps. HH22(g) + 1/2 O(g) + 1/2 O22(g) ---> H(g) ---> H22O(g) + 242 kJO(g) + 242 kJ HH22O(g) ---> HO(g) ---> H22O(liq) + 44 kJO(liq) + 44 kJ ------------------------------------------------------------------ ----- HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liq) + 286 kJO(liq) + 286 kJ Example ofExample of HESS’S LAWHESS’S LAW—— If a rxn. is the sum of 2 or moreIf a rxn. is the sum of 2 or more others, the net ∆H is the sum of theothers, the net ∆H is the sum of the ∆H’s of the other rxns.∆H’s of the other rxns. USING ENTHALPYUSING ENTHALPY
44 Hess’s LawHess’s Law & Energy Level Diagrams& Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
45 Hess’s LawHess’s Law & Energy Level Diagrams& Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
46 • This equation is valid becauseThis equation is valid because ∆H is a∆H is a STATE FUNCTIONSTATE FUNCTION • These depend only on the stateThese depend only on the state of the system andof the system and notnot on howon how the system got there.the system got there. • V, T, P, energy — and your bankV, T, P, energy — and your bank account!account! • Unlike V, T, and P, one cannotUnlike V, T, and P, one cannot measure absolute H. Can onlymeasure absolute H. Can only measure ∆H.measure ∆H. ΣΣ ∆∆H along one path =H along one path = ΣΣ ∆∆H along another pathH along another path ΣΣ ∆∆H along one path =H along one path = ΣΣ ∆∆H along another pathH along another path
47 Standard Enthalpy ValuesStandard Enthalpy Values Most H values are labeled∆Most H values are labeled∆ H∆H∆ oo Measured underMeasured under standard conditionsstandard conditions P = 1 bar = 10P = 1 bar = 1055 Pa = 1 atm /Pa = 1 atm / 1.013251.01325 Concentration = 1Concentration = 1 mol/Lmol/L T = usually 25T = usually 25 oo CC with all species in standard stateswith all species in standard states e.g., C = graphite and Oe.g., C = graphite and O22 = gas= gas
48 Enthalpy ValuesEnthalpy Values HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g) ∆∆H˚ = -242 kJH˚ = -242 kJ 2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g)O(g) ∆∆H˚ = -484 kJH˚ = -484 kJ HH22O(g) ---> HO(g) ---> H22(g) + 1/2 O(g) + 1/2 O22(g)(g) ∆∆H˚ = +242 kJH˚ = +242 kJ HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liquid)O(liquid) Depend onDepend on how the reaction is writtenhow the reaction is written and on phasesand on phases of reactants and productsof reactants and products Depend onDepend on how the reaction is writtenhow the reaction is written and on phasesand on phases of reactants and productsof reactants and products
49 Standard Enthalpy ValuesStandard Enthalpy Values NIST (Nat’l Institute for Standards andNIST (Nat’l Institute for Standards and Technology) gives values ofTechnology) gives values of ∆∆HHff oo = standard molar enthalpy of= standard molar enthalpy of formationformation —— the enthalpy change when 1 mol ofthe enthalpy change when 1 mol of compound is formed from elementscompound is formed from elements under standard conditions.under standard conditions. See Table 6.2See Table 6.2
50 ∆∆HHff oo , standard molar, standard molar enthalpy of formationenthalpy of formation Enthalpy change when 1 mol ofEnthalpy change when 1 mol of compound is formed from thecompound is formed from the corresponding elements undercorresponding elements under standard conditionsstandard conditions HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g) ∆∆HHff oo (H(H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol By definition,By definition, ∆∆HHff oo = 0 for elements in their standard= 0 for elements in their standard states.states.
51 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values Use H ’s to calculate∆ ˚Use H ’s to calculate∆ ˚ enthalpy changeenthalpy change forfor HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) (product is called “(product is called “water gaswater gas”)”)
52 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) From reference books we findFrom reference books we find • HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) H∆O(g) H∆ ff = - 242˚ = - 242˚ kJ/molkJ/mol • C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) H∆H∆ ff = - 111˚ = - 111˚ kJ/molkJ/mol
53 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) H∆(g) H∆ oo = +242= +242 kJkJ C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) H∆H∆ oo = -111= -111 kJkJ ----------------------------------------------------------------------- --------- To convert 1 mol of water to 1 mol eachTo convert 1 mol of water to 1 mol each of Hof H22 and COand CO requiresrequires 131 kJ of energy.131 kJ of energy. The “water gas” reaction isThe “water gas” reaction is ENDOENDOthermic.thermic. HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) ∆∆HHoo netnet = +131 kJ= +131 kJ HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) ∆∆HHoo netnet = +131 kJ= +131 kJ
54 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values In general, whenIn general, when ALLALL enthalpies of formation areenthalpies of formation are known:known: Calculate ∆H ofCalculate ∆H of reaction?reaction? ∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (products) -(products) - ΣΣ H∆H∆ ff oo (reactants)(reactants)∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (products) -(products) - ΣΣ H∆H∆ ff oo (reactants)(reactants) Remember that ∆ always = final – initial
55 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values Calculate the heat of combustion ofCalculate the heat of combustion of methanol, i.e., ∆Hmethanol, i.e., ∆Hoo rxnrxn forfor CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g) ∆∆HHoo rxnrxn == ΣΣ ∆H∆Hff oo (prod) -(prod) - ΣΣ ∆H∆Hff oo (react)(react)
56 Using Standard Enthalpy ValuesUsing Standard Enthalpy Values ∆∆HHoo rxnrxn = ∆H= ∆Hff oo (CO(CO22) + 2 ∆H) + 2 ∆Hff oo (H(H22O)O) - {3/2 ∆H- {3/2 ∆Hff oo (O(O22) + ∆H) + ∆Hff oo (CH(CH33OH)}OH)} = (-393.5 kJ) + 2 (-241.8 kJ)= (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)} ∆∆HHoo rxnrxn = -675.6 kJ per mol of methanol= -675.6 kJ per mol of methanol CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g) ∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (prod) -(prod) - ΣΣ H∆H∆ ff oo (react)(react)
57 Measuring Heats of Reaction CALORIMETRYCALORIMETRY Constant Volume “Bomb” Calorimeter • Burn combustible sample. • Measure heat evolved in a reaction. • Derive ∆E for reaction.
58 CalorimetryCalorimetry Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb
59 Calculate heat of combustion of octane.Calculate heat of combustion of octane. CC88HH1818 + 25/2 O+ 25/2 O22 --> 8 CO--> 8 CO22 + 9 H+ 9 H22OO •• Burn 1.00 g of octaneBurn 1.00 g of octane • Temp rises from 25.00 to 33.20Temp rises from 25.00 to 33.20 oo CC • Calorimeter contains 1200 g waterCalorimeter contains 1200 g water • Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY
60 Step 1Step 1 Calc. heat transferred from reaction toCalc. heat transferred from reaction to water.water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2Step 2 Calc. heat transferred from reaction toCalc. heat transferred from reaction to bomb.bomb. q = (bomb heat capacity)( T)∆q = (bomb heat capacity)( T)∆ = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J Step 3Step 3 Total heat evolvedTotal heat evolved 41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY

More Related Content

PPT
Heat in changes of state
byCurrituck County High School
 
PPTX
Heat phenomenon
byAndrei Matias
 
PPT
3.1 thermal concepts
bygavin40
 
PPTX
Tp 13 specific heat capacity (shared)
byLThistlewood
 
PPTX
Thisispartoneentalphy 121121135040-phpapp02
byDr Robert Craig PhD
 
PPT
Ccp intro to thermal energy fall 2010
bysbarkanic
 
PPTX
Heat
byYhanzieCapilitan
 
PPTX
Wk 5 p1 wk 6-p2_12.1-12.2_thermal properties of materials
bychris lembalemba
 
Heat in changes of state
byCurrituck County High School
 
Heat phenomenon
byAndrei Matias
 
3.1 thermal concepts
bygavin40
 
Tp 13 specific heat capacity (shared)
byLThistlewood
 
Thisispartoneentalphy 121121135040-phpapp02
byDr Robert Craig PhD
 
Ccp intro to thermal energy fall 2010
bysbarkanic
 
Heat
byYhanzieCapilitan
 
Wk 5 p1 wk 6-p2_12.1-12.2_thermal properties of materials
bychris lembalemba
 

What's hot

PPTX
Causes of change
byAbhishek Gunasekaran
 
PPT
Thermal energy
byDuluth Middle
 
PPT
Lecture 11 heat and phase changes
byAlbania Energy Association
 
PPTX
F4.4.3 heat
bySyiera Rahman
 
PPTX
Heat transfer & heat exchangers
byMohamed Alsalihi
 
PPT
3.2
bykrause004anj
 
PPT
2 latent heat
byMissingWaldo
 
PPT
Heat Capacity
bymeenng
 
PPTX
Heat & Thermodynamics
byitutor
 
PPTX
Heat Capacity and Specific Heat Capacity
bystemegypt.edu.eg
 
PPTX
Thermochemistry
byMak Tutorials
 
PPT
Thermochemistry
byLumen Learning
 
PPT
chemistry-enthalpy power point
byShmiley3000
 
PPTX
Tp 14 specific latent heat (shared)
byLThistlewood
 
PPTX
Energy ch 16
byAkhil Reddy
 
PPT
Chemical reactions
byPriyavrat Bhardwaj
 
PPT
Heat Energy
byfurrs
 
PPT
Change of Phase and Latent Heat
byLyn Agustin
 
PPSX
Capter 10 for 9th grade physics
byPhysics Amal Sweis
 
Causes of change
byAbhishek Gunasekaran
 
Thermal energy
byDuluth Middle
 
Lecture 11 heat and phase changes
byAlbania Energy Association
 
F4.4.3 heat
bySyiera Rahman
 
Heat transfer & heat exchangers
byMohamed Alsalihi
 
3.2
bykrause004anj
 
2 latent heat
byMissingWaldo
 
Heat Capacity
bymeenng
 
Heat & Thermodynamics
byitutor
 
Heat Capacity and Specific Heat Capacity
bystemegypt.edu.eg
 
Thermochemistry
byMak Tutorials
 
Thermochemistry
byLumen Learning
 
chemistry-enthalpy power point
byShmiley3000
 
Tp 14 specific latent heat (shared)
byLThistlewood
 
Energy ch 16
byAkhil Reddy
 
Chemical reactions
byPriyavrat Bhardwaj
 
Heat Energy
byfurrs
 
Change of Phase and Latent Heat
byLyn Agustin
 
Capter 10 for 9th grade physics
byPhysics Amal Sweis
 

Similar to Thermochem

PPT
Chemical Thermodynamics
bysuresh gdvm
 
PPTX
Unit 11 thermochemistry
bycpasilla
 
PDF
Science chemistry thermochemistryfinal_version2.pdf
bykitbal471
 
PPTX
fnd605 prof 2024 lecture.pptx. Food analy
byMarilynGoreh
 
PPTX
Thermochemistry
byGeorge Pescasio Jr.
 
PPT
005 lecture f
byEng Mahamed
 
PPT
Chemical reactions
bySyed Shah
 
PPT
Thermochemistry ch 16
bytanzmanj
 
PDF
Chapter 20 The First Law of Thermodynamics.pdf
byBelnSevilla2
 
PPTX
Exothermic and Endothermic Processes.pptx
byRaellenRegalado
 
PPTX
Heat and Thermodynamics ro4 class 11 .pptx
byDipendarThakur
 
PPTX
Thermodynamicrererererereerererrrer.pptx
bysalimaabajja1
 
PDF
TOPIC 8 Temperature and Heat.pdf physics
byssuserddc89b
 
PPTX
6.Thermochemistry111111111111111111.pptx
byHappyFrando
 
PPSX
Notes for Unit 17 of AP Chemistry (Thermodynamics)
bynoahawad
 
PPTX
IB Chemistry on Energetics experiment and Thermodynamics
byLawrence kok
 
PPT
Chapter 4
byhansdg1
 
PDF
IB Chemistry on Energetics experiment, Thermodynamics and Hess's Law
byLawrence kok
 
PPT
Thermodynamics
bycit-cit
 
PPT
Energy
byPaul Schumann
 
Chemical Thermodynamics
bysuresh gdvm
 
Unit 11 thermochemistry
bycpasilla
 
Science chemistry thermochemistryfinal_version2.pdf
bykitbal471
 
fnd605 prof 2024 lecture.pptx. Food analy
byMarilynGoreh
 
Thermochemistry
byGeorge Pescasio Jr.
 
005 lecture f
byEng Mahamed
 
Chemical reactions
bySyed Shah
 
Thermochemistry ch 16
bytanzmanj
 
Chapter 20 The First Law of Thermodynamics.pdf
byBelnSevilla2
 
Exothermic and Endothermic Processes.pptx
byRaellenRegalado
 
Heat and Thermodynamics ro4 class 11 .pptx
byDipendarThakur
 
Thermodynamicrererererereerererrrer.pptx
bysalimaabajja1
 
TOPIC 8 Temperature and Heat.pdf physics
byssuserddc89b
 
6.Thermochemistry111111111111111111.pptx
byHappyFrando
 
Notes for Unit 17 of AP Chemistry (Thermodynamics)
bynoahawad
 
IB Chemistry on Energetics experiment and Thermodynamics
byLawrence kok
 
Chapter 4
byhansdg1
 
IB Chemistry on Energetics experiment, Thermodynamics and Hess's Law
byLawrence kok
 
Thermodynamics
bycit-cit
 
Energy
byPaul Schumann
 

More from starlanter

PDF
Chemical equilibrium
bystarlanter
 
PDF
Electrochemistry
bystarlanter
 
PDF
Reproductive system
bystarlanter
 
PPTX
Mahayana reporting
bystarlanter
 
DOCX
Summary of-xhtml-basics
bystarlanter
 
DOCX
3rd grading-reviewer-pc-assembly-and-networking
bystarlanter
 
PPT
Spontaneity entropy___free_energy
bystarlanter
 
PPTX
Equilibrium constant-presentation
bystarlanter
 
Chemical equilibrium
bystarlanter
 
Electrochemistry
bystarlanter
 
Reproductive system
bystarlanter
 
Mahayana reporting
bystarlanter
 
Summary of-xhtml-basics
bystarlanter
 
3rd grading-reviewer-pc-assembly-and-networking
bystarlanter
 
Spontaneity entropy___free_energy
bystarlanter
 
Equilibrium constant-presentation
bystarlanter
 

Recently uploaded

PDF
International Men's Day Event: Breaking the Silence: Embracing Vulnerability ...
byAssociation for Project Management
 
PPTX
REVISED DEFENSE MECHANISM / ADJUSTMENT MECHANISM-FINAL
byShimla
 
PPTX
PURPOSIVE SAMPLING IN EDUCATIONAL RESEARCH RACHITHRA RK.pptx
byssuser047b711
 
PPTX
ATTENTION -PART 2.pptx Shilpa Hotakar for I semester BSc students
bySHILPA HOTAKAR
 
PPTX
Introduction-to-Anatomy-and-Physiology.pptx
byAshish Umale
 
PPTX
Details of Epithelial and Connective Tissue.pptx
byAshish Umale
 
PPTX
4 G8_Q3_L4 (Evaluating opinion editorials for textual evidence and quality).pptx
byNorafeSahagun
 
PPTX
TAMIS & TEMS - HOW, WHY and THE STEPS IN PROCTOLOGY
byJohn Thanakumar
 
PPTX
Semester 6 Unit 2 Club foot (talipes).pptx
bysachin7989
 
PPTX
How to Configure Push & Pull Rule in Odoo 18 Inventory
byCeline George
 
PPTX
PARENTAL ROUTES OF DRUGS ADMINISTRATION .pptx
byAneetaSharma15
 
PDF
DHA/HAAD/MOH/DOH OPTOMETRY MCQ PYQ. .pdf
byAnmol Singh
 
PDF
Risk management in Moroccan public hospitals_ a literature review
byMan_Ebook
 
PPTX
Steve Hale - Developing Elite Young Goalkeepers 2024.pptx
bypjmfd
 
PPTX
Partial Correlation of Coefficient - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂
bySundar B N
 
PDF
Scalable-MADDPG-Based Cooperative Target Invasion for a Multi-USV System.pdf
byMan_Ebook
 
PPTX
3 G8_Q3_L3_ (Cartoon as Representation in Opinion Editorial Article).pptx
byNorafeSahagun
 
PPTX
How to Create Lead_Opportunity From Odoo 18 Website
byCeline George
 
PPTX
AN EXTREMELY BORING GENERAL QUIZ FOR UG.pptx
bySriramG47
 
PPTX
Access partner report in Odoo 18.2 _ Odoo 19
byCeline George
 
International Men's Day Event: Breaking the Silence: Embracing Vulnerability ...
byAssociation for Project Management
 
REVISED DEFENSE MECHANISM / ADJUSTMENT MECHANISM-FINAL
byShimla
 
PURPOSIVE SAMPLING IN EDUCATIONAL RESEARCH RACHITHRA RK.pptx
byssuser047b711
 
ATTENTION -PART 2.pptx Shilpa Hotakar for I semester BSc students
bySHILPA HOTAKAR
 
Introduction-to-Anatomy-and-Physiology.pptx
byAshish Umale
 
Details of Epithelial and Connective Tissue.pptx
byAshish Umale
 
4 G8_Q3_L4 (Evaluating opinion editorials for textual evidence and quality).pptx
byNorafeSahagun
 
TAMIS & TEMS - HOW, WHY and THE STEPS IN PROCTOLOGY
byJohn Thanakumar
 
Semester 6 Unit 2 Club foot (talipes).pptx
bysachin7989
 
How to Configure Push & Pull Rule in Odoo 18 Inventory
byCeline George
 
PARENTAL ROUTES OF DRUGS ADMINISTRATION .pptx
byAneetaSharma15
 
DHA/HAAD/MOH/DOH OPTOMETRY MCQ PYQ. .pdf
byAnmol Singh
 
Risk management in Moroccan public hospitals_ a literature review
byMan_Ebook
 
Steve Hale - Developing Elite Young Goalkeepers 2024.pptx
bypjmfd
 
Partial Correlation of Coefficient - Values of r₁₂.₃, r₂₃.₁ & r₁₃.₂
bySundar B N
 
Scalable-MADDPG-Based Cooperative Target Invasion for a Multi-USV System.pdf
byMan_Ebook
 
3 G8_Q3_L3_ (Cartoon as Representation in Opinion Editorial Article).pptx
byNorafeSahagun
 
How to Create Lead_Opportunity From Odoo 18 Website
byCeline George
 
AN EXTREMELY BORING GENERAL QUIZ FOR UG.pptx
bySriramG47
 
Access partner report in Odoo 18.2 _ Odoo 19
byCeline George
 

Thermochem

  • 1.
    1THERMOCHEMISTRYTHERMOCHEMISTRY ThermodynamicsThermodynamics The study ofHeat and WorkThe study of Heat and Work and State Functionsand State Functions
  • 2.
    2 Energy & ChemistryEnergy& ChemistryEnergy & ChemistryEnergy & Chemistry ENERGYENERGY is the capacity tois the capacity to do work or transfer heat.do work or transfer heat. HEATHEAT is the form of energyis the form of energy that flows between 2that flows between 2 objects because of theirobjects because of their difference in temperature.difference in temperature. Other forms of energy —Other forms of energy — • lightlight • electricalelectrical • kinetic and potentialkinetic and potential
  • 3.
    3 Energy & ChemistryEnergy& Chemistry • Burning peanutspeanuts supply sufficientsupply sufficient energy to boil a cupenergy to boil a cup of water.of water. • Burning sugarBurning sugar (sugar reacts with(sugar reacts with KClOKClO33, a strong, a strong oxidizing agent)oxidizing agent)
  • 4.
    4 Energy & ChemistryEnergy& Chemistry • These reactions areThese reactions are PRODUCTPRODUCT FAVOREDFAVORED • They proceed almost completelyThey proceed almost completely from reactants to products, perhapsfrom reactants to products, perhaps with some outside assistance.with some outside assistance.
  • 5.
    5 Energy & ChemistryEnergy& Chemistry 2 H2 H22(g) + O(g) + O22(g) -->(g) --> 2 H2 H22O(g) + heat and lightO(g) + heat and light This can be set up to provideThis can be set up to provide ELECTRIC ENERGYELECTRIC ENERGY in ain a fuel cellfuel cell.. Oxidation:Oxidation: 2 H2 H22 ---> 4 H---> 4 H++ + 4 e+ 4 e-- Reduction:Reduction: 4 e4 e-- + O+ O22 + 2 H+ 2 H22O ---> 4 OHO ---> 4 OH-- CCR, page 845
  • 6.
    6 Potential & KineticEnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy PotentialPotential energyenergy —— energy aenergy a motionlessmotionless body has bybody has by virtue of itsvirtue of its position.position.
  • 7.
    7 • Positive and negativeparticles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy Potential EnergyPotential Energy on the Atomic Scaleon the Atomic Scale NaCl — composed ofNaCl — composed of NaNa++ and Cland Cl-- ions.ions.
  • 8.
    8 • Positive and negativeparticles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy Potential EnergyPotential Energy on the Atomic Scaleon the Atomic Scale
  • 9.
    9 Potential & KineticEnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy Kinetic energyKinetic energy — energy of— energy of motionmotion •• TranslationTranslation
  • 10.
    10 Potential & KineticEnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy Kinetic energyKinetic energy — energy of— energy of motion.motion. translate rotate vibrate translate rotate vibrate
  • 11.
    11 Internal Energy (E)InternalEnergy (E) • PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U) • Int. E of a chemical systemInt. E of a chemical system depends ondepends on • number of particlesnumber of particles • type of particlestype of particles • temperaturetemperature
  • 12.
    12 Internal Energy (E)InternalEnergy (E) • PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U) QuickTime™ and a Graphics decompressor are needed to see this picture.
  • 13.
    13 Internal Energy (E)InternalEnergy (E) • The higher the TThe higher the T the higher thethe higher the internal energyinternal energy • So, use changesSo, use changes in T (∆T) toin T (∆T) to monitor changesmonitor changes in E (∆E).in E (∆E).
  • 14.
    14 ThermodynamicsThermodynamics • Thermodynamics isthe science of heat (energy) transfer. Heat energy is associatedHeat energy is associated with molecular motions.with molecular motions. Heat transfers until thermal equilibrium is established.
  • 15.
    15 Directionality of HeatTransferDirectionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes downT(system) goes down T(surr) goes upT(surr) goes up
  • 16.
    16 Directionality of HeatTransferDirectionality of Heat Transfer • Heat always transfer from hotter object to cooler one. • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes upT(system) goes up T (surr) goes downT (surr) goes down
  • 17.
    17 Energy & ChemistryEnergy& Chemistry All of thermodynamics dependsAll of thermodynamics depends on the law ofon the law of CONSERVATION OF ENERGYCONSERVATION OF ENERGY.. • The total energy is unchangedThe total energy is unchanged in a chemical reaction.in a chemical reaction. • If PE of products is less thanIf PE of products is less than reactants, the difference mustreactants, the difference must be released as KE.be released as KE.
  • 18.
    18 Energy Change inEnergyChange in Chemical ProcessesChemical Processes Reactants Products Kinetic Energy PE PE of system dropped. KE increased. Therefore,PE of system dropped. KE increased. Therefore, you often feel a T increase.you often feel a T increase.
  • 19.
    19 UNITS OF ENERGYUNITSOF ENERGYUNITS OF ENERGYUNITS OF ENERGY 1 calorie = heat required to1 calorie = heat required to raise temp. of 1.00 g ofraise temp. of 1.00 g of HH22O by 1.0O by 1.0 oo C.C. 1000 cal = 1 kilocalorie = 11000 cal = 1 kilocalorie = 1 kcalkcal 1 kcal = 1 Calorie (a food1 kcal = 1 Calorie (a food “calorie”)“calorie”) But we use the unit calledBut we use the unit called thethe JOULEJOULE 1 cal = 4.184 joules1 cal = 4.184 joules James JouleJames Joule 1818-18891818-1889
  • 20.
    20 HEAT CAPACITYHEAT CAPACITY Theheat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity?Which has the larger heat capacity?
  • 21.
    21 SpecificSpecific Heat CapacityHeatCapacitySpecificSpecific Heat CapacityHeat Capacity How much energy isHow much energy is transferred due to Ttransferred due to T difference?difference? The heatThe heat (q)(q) “lost” or “gained”“lost” or “gained” is related tois related to a)a) sample masssample mass b)b) change in T andchange in T and c)c) specific heat capacityspecific heat capacity Specific heat capacity= heat lost or gained by substance (J) (mass, g)(T change,K)
  • 22.
    22 Specific Heat CapacitySpecificHeat CapacitySpecific Heat CapacitySpecific Heat Capacity SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K) HH22OO 4.1844.184 Ethylene glycolEthylene glycol 2.392.39 AlAl 0.8970.897 glassglass 0.840.84 AluminumAluminum
  • 23.
    23 Specific Heat CapacitySpecificHeat CapacitySpecific Heat CapacitySpecific Heat Capacity If 25.0 g of Al cool fromIf 25.0 g of Al cool from 310310 oo C to 37C to 37 oo C, howC, how many joules of heatmany joules of heat energy are lost byenergy are lost by the Al?the Al? Specific heat capacity= heat lost or gained by substance (J) (mass, g)(T change,K)
  • 24.
    24 Specific Heat CapacitySpecificHeat CapacitySpecific Heat CapacitySpecific Heat Capacity If 25.0 g of Al cool from 310If 25.0 g of Al cool from 310 oo C to 37C to 37 oo C, how manyC, how many joules of heat energy are lost by the Al?joules of heat energy are lost by the Al? heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = Twhere ∆T = Tfinalfinal - T- Tinitialinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 Jq = - 6120 J Notice that the negative sign on q signalsNotice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al. Notice that the negative sign on q signalsNotice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.
  • 25.
    25 Heat TransferHeat Transfer NoChange in StateNo Change in State q transferred = (sp. ht.)(mass)(∆T)
  • 26.
    26Heat Transfer withHeatTransfer with Change of StateChange of State Heat Transfer withHeat Transfer with Change of StateChange of State Changes of state involve energyChanges of state involve energy (at constant T)(at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid waterIce + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass)q = (heat of fusion)(mass)
  • 27.
    27 Heat Transfer andHeatTransfer and Changes of StateChanges of State Heat Transfer andHeat Transfer and Changes of StateChanges of State Requires energyRequires energy (heat).(heat). This is the reasonThis is the reason a)a) you cool down afteryou cool down after swimmingswimming b)b) you use water to putyou use water to put out a fire.out a fire. + energy Liquid ---> VaporLiquid ---> Vapor
  • 28.
    28 Heat waterHeat water EvaporatewaterEvaporate water Melt iceMelt ice Heating/Cooling Curve for WaterHeating/Cooling Curve for Water Note that T isNote that T is constant as ice meltsconstant as ice melts Note that T isNote that T is constant as ice meltsconstant as ice melts
  • 29.
    29 Heat of fusionof ice = 333 J/gHeat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g What quantity of heat is required toWhat quantity of heat is required to melt 500. g ofmelt 500. g of iceice and heat theand heat the water towater to steamsteam at 100at 100 oo C?C? Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State +333 J/g+333 J/g +2260 J/g+2260 J/g
  • 30.
    30 How much heatis required to melt 500. g ofHow much heat is required to melt 500. g of ice and heat the water to steam at 100ice and heat the water to steam at 100 oo C?C? 1.1. To melt iceTo melt ice q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 1055 JJ 2.2. To raise water from 0To raise water from 0 oo C to 100C to 100 oo CC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 xq = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 101055 JJ 3.3. To evaporate water at 100To evaporate water at 100 oo CC q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 JJ 4.4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 1066 J =J = Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State
  • 31.
    31 ChemicalChemical ReactivityReactivity What driveschemical reactions? How do theyWhat drives chemical reactions? How do they occur?occur? The first is answered byThe first is answered by THERMODYNAMICSTHERMODYNAMICS and the second byand the second by KINETICSKINETICS.. Have already seen a number of “drivingHave already seen a number of “driving forces” for reactions that areforces” for reactions that are PRODUCT-PRODUCT- FAVOREDFAVORED.. •• formation of a precipitateformation of a precipitate •• gas formationgas formation •• HH22O formation (acid-base reaction)O formation (acid-base reaction) •• electron transfer in a batteryelectron transfer in a battery
  • 32.
    32 ChemicalChemical ReactivityReactivity But energytransfer also allows us to predictBut energy transfer also allows us to predict reactivity.reactivity. In general, reactions that transfer energyIn general, reactions that transfer energy to their surroundings are product-to their surroundings are product- favored.favored. So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.
  • 33.
    33 Heat Energy TransferinHeat Energy Transfer in a Physical Processa Physical Process COCO22 (s, -78(s, -78 oo C) ---> COC) ---> CO22 (g, -78(g, -78 oo C)C) Heat transfers from surroundings to system in endothermic process.
  • 34.
    34 Heat Energy TransferinHeat Energy Transfer in a Physical Processa Physical Process • COCO22 (s, -78(s, -78 oo C) --->C) ---> COCO22 (g, -78(g, -78 oo C)C) • A regular array ofA regular array of molecules in amolecules in a solidsolid -----> gas phase-----> gas phase molecules.molecules. • Gas moleculesGas molecules have higher kinetichave higher kinetic energy.energy.
  • 35.
    35 Energy Level DiagramEnergyLevel Diagram for Heat Energyfor Heat Energy TransferTransfer ∆E = E(final) - E(initial) = E(gas) - E(solid) COCO22 solidsolid COCO22 gasgas
  • 36.
    36 Heat Energy TransferinHeat Energy Transfer in Physical ChangePhysical Change • Gas molecules have higherGas molecules have higher kinetic energy.kinetic energy. • Also,Also, WORKWORK is done by theis done by the system in pushing aside thesystem in pushing aside the atmosphere.atmosphere. COCO22 (s, -78(s, -78 oo C) ---> COC) ---> CO22 (g, -78(g, -78 oo C)C) Two things have happened!Two things have happened!
  • 37.
    37 FIRST LAW OFFIRSTLAW OF THERMODYNAMICSTHERMODYNAMICS ∆∆E = q + wE = q + w heat energy transferredheat energy transferred energyenergy changechange work donework done by theby the systemsystem Energy is conserved!Energy is conserved!
  • 38.
    38 heat transfer outheattransfer out (exothermic), -q(exothermic), -q heat transfer inheat transfer in (endothermic), +q(endothermic), +q SYSTEMSYSTEM ∆E = q + w w transfer inw transfer in (+w)(+w) w transfer outw transfer out (-w)(-w)
  • 39.
    39 ENTHALPYENTHALPY Most chemical reactionsoccur at constant P, soMost chemical reactions occur at constant P, so and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small) ∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E ∆∆H = change inH = change in heat contentheat content of the systemof the system ∆∆H = HH = Hfinalfinal - H- Hinitialinitial and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small) ∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E ∆∆H = change inH = change in heat contentheat content of the systemof the system ∆∆H = HH = Hfinalfinal - H- Hinitialinitial Heat transferred at constant P = qHeat transferred at constant P = qpp qqpp == H∆H∆ wherewhere H = enthalpyH = enthalpy Heat transferred at constant P = qHeat transferred at constant P = qpp qqpp == H∆H∆ wherewhere H = enthalpyH = enthalpy
  • 40.
    40 IfIf HHfinalfinal <H< Hinitialinitial then H is negative∆then H is negative∆ Process isProcess is EXOTHERMICEXOTHERMIC IfIf HHfinalfinal < H< Hinitialinitial then H is negative∆then H is negative∆ Process isProcess is EXOTHERMICEXOTHERMIC IfIf HHfinalfinal > H> Hinitialinitial then H is positive∆then H is positive∆ Process isProcess is ENDOTHERMICENDOTHERMIC IfIf HHfinalfinal > H> Hinitialinitial then H is positive∆then H is positive∆ Process isProcess is ENDOTHERMICENDOTHERMIC ENTHALPYENTHALPY ∆∆H = HH = Hfinalfinal - H- Hinitialinitial
  • 41.
    41 Consider the formationof waterConsider the formation of water HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) +O(g) + 241.8241.8 kJkJ USING ENTHALPYUSING ENTHALPY Exothermic reaction — heat is a “product”Exothermic reaction — heat is a “product” and H = – 241.8 kJ∆and H = – 241.8 kJ∆
  • 42.
    42 MakingMaking liquidliquid HH22Ofrom HO from H22 + O+ O22 involvesinvolves twotwo exoexothermic steps.thermic steps. USING ENTHALPYUSING ENTHALPY H2 + O2 gas Liquid H2OH2O vapor
  • 43.
    43 Making HMaking H22Ofrom HO from H22 involves two steps.involves two steps. HH22(g) + 1/2 O(g) + 1/2 O22(g) ---> H(g) ---> H22O(g) + 242 kJO(g) + 242 kJ HH22O(g) ---> HO(g) ---> H22O(liq) + 44 kJO(liq) + 44 kJ ------------------------------------------------------------------ ----- HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liq) + 286 kJO(liq) + 286 kJ Example ofExample of HESS’S LAWHESS’S LAW—— If a rxn. is the sum of 2 or moreIf a rxn. is the sum of 2 or more others, the net ∆H is the sum of theothers, the net ∆H is the sum of the ∆H’s of the other rxns.∆H’s of the other rxns. USING ENTHALPYUSING ENTHALPY
  • 44.
    44 Hess’s LawHess’s Law &Energy Level Diagrams& Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
  • 45.
    45 Hess’s LawHess’s Law &Energy Level Diagrams& Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.
  • 46.
    46 • This equationis valid becauseThis equation is valid because ∆H is a∆H is a STATE FUNCTIONSTATE FUNCTION • These depend only on the stateThese depend only on the state of the system andof the system and notnot on howon how the system got there.the system got there. • V, T, P, energy — and your bankV, T, P, energy — and your bank account!account! • Unlike V, T, and P, one cannotUnlike V, T, and P, one cannot measure absolute H. Can onlymeasure absolute H. Can only measure ∆H.measure ∆H. ΣΣ ∆∆H along one path =H along one path = ΣΣ ∆∆H along another pathH along another path ΣΣ ∆∆H along one path =H along one path = ΣΣ ∆∆H along another pathH along another path
  • 47.
    47 Standard Enthalpy ValuesStandardEnthalpy Values Most H values are labeled∆Most H values are labeled∆ H∆H∆ oo Measured underMeasured under standard conditionsstandard conditions P = 1 bar = 10P = 1 bar = 1055 Pa = 1 atm /Pa = 1 atm / 1.013251.01325 Concentration = 1Concentration = 1 mol/Lmol/L T = usually 25T = usually 25 oo CC with all species in standard stateswith all species in standard states e.g., C = graphite and Oe.g., C = graphite and O22 = gas= gas
  • 48.
    48 Enthalpy ValuesEnthalpy Values HH22(g)+ 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g) ∆∆H˚ = -242 kJH˚ = -242 kJ 2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g)O(g) ∆∆H˚ = -484 kJH˚ = -484 kJ HH22O(g) ---> HO(g) ---> H22(g) + 1/2 O(g) + 1/2 O22(g)(g) ∆∆H˚ = +242 kJH˚ = +242 kJ HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liquid)O(liquid) Depend onDepend on how the reaction is writtenhow the reaction is written and on phasesand on phases of reactants and productsof reactants and products Depend onDepend on how the reaction is writtenhow the reaction is written and on phasesand on phases of reactants and productsof reactants and products
  • 49.
    49 Standard Enthalpy ValuesStandardEnthalpy Values NIST (Nat’l Institute for Standards andNIST (Nat’l Institute for Standards and Technology) gives values ofTechnology) gives values of ∆∆HHff oo = standard molar enthalpy of= standard molar enthalpy of formationformation —— the enthalpy change when 1 mol ofthe enthalpy change when 1 mol of compound is formed from elementscompound is formed from elements under standard conditions.under standard conditions. See Table 6.2See Table 6.2
  • 50.
    50 ∆∆HHff oo , standard molar,standard molar enthalpy of formationenthalpy of formation Enthalpy change when 1 mol ofEnthalpy change when 1 mol of compound is formed from thecompound is formed from the corresponding elements undercorresponding elements under standard conditionsstandard conditions HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g) ∆∆HHff oo (H(H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol By definition,By definition, ∆∆HHff oo = 0 for elements in their standard= 0 for elements in their standard states.states.
  • 51.
    51 Using Standard EnthalpyValuesUsing Standard Enthalpy Values Use H ’s to calculate∆ ˚Use H ’s to calculate∆ ˚ enthalpy changeenthalpy change forfor HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) (product is called “(product is called “water gaswater gas”)”)
  • 52.
    52 Using Standard EnthalpyValuesUsing Standard Enthalpy Values HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) From reference books we findFrom reference books we find • HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) H∆O(g) H∆ ff = - 242˚ = - 242˚ kJ/molkJ/mol • C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) H∆H∆ ff = - 111˚ = - 111˚ kJ/molkJ/mol
  • 53.
    53 Using Standard EnthalpyValuesUsing Standard Enthalpy Values HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) H∆(g) H∆ oo = +242= +242 kJkJ C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) H∆H∆ oo = -111= -111 kJkJ ----------------------------------------------------------------------- --------- To convert 1 mol of water to 1 mol eachTo convert 1 mol of water to 1 mol each of Hof H22 and COand CO requiresrequires 131 kJ of energy.131 kJ of energy. The “water gas” reaction isThe “water gas” reaction is ENDOENDOthermic.thermic. HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) ∆∆HHoo netnet = +131 kJ= +131 kJ HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g) ∆∆HHoo netnet = +131 kJ= +131 kJ
  • 54.
    54 Using Standard EnthalpyValuesUsing Standard Enthalpy Values In general, whenIn general, when ALLALL enthalpies of formation areenthalpies of formation are known:known: Calculate ∆H ofCalculate ∆H of reaction?reaction? ∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (products) -(products) - ΣΣ H∆H∆ ff oo (reactants)(reactants)∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (products) -(products) - ΣΣ H∆H∆ ff oo (reactants)(reactants) Remember that ∆ always = final – initial
  • 55.
    55 Using Standard EnthalpyValuesUsing Standard Enthalpy Values Calculate the heat of combustion ofCalculate the heat of combustion of methanol, i.e., ∆Hmethanol, i.e., ∆Hoo rxnrxn forfor CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g) ∆∆HHoo rxnrxn == ΣΣ ∆H∆Hff oo (prod) -(prod) - ΣΣ ∆H∆Hff oo (react)(react)
  • 56.
    56 Using Standard EnthalpyValuesUsing Standard Enthalpy Values ∆∆HHoo rxnrxn = ∆H= ∆Hff oo (CO(CO22) + 2 ∆H) + 2 ∆Hff oo (H(H22O)O) - {3/2 ∆H- {3/2 ∆Hff oo (O(O22) + ∆H) + ∆Hff oo (CH(CH33OH)}OH)} = (-393.5 kJ) + 2 (-241.8 kJ)= (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)} ∆∆HHoo rxnrxn = -675.6 kJ per mol of methanol= -675.6 kJ per mol of methanol CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g) ∆∆HHoo rxnrxn == ΣΣ H∆H∆ ff oo (prod) -(prod) - ΣΣ H∆H∆ ff oo (react)(react)
  • 57.
    57 Measuring Heats ofReaction CALORIMETRYCALORIMETRY Constant Volume “Bomb” Calorimeter • Burn combustible sample. • Measure heat evolved in a reaction. • Derive ∆E for reaction.
  • 58.
    58 CalorimetryCalorimetry Some heat fromreaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb
  • 59.
    59 Calculate heat ofcombustion of octane.Calculate heat of combustion of octane. CC88HH1818 + 25/2 O+ 25/2 O22 --> 8 CO--> 8 CO22 + 9 H+ 9 H22OO •• Burn 1.00 g of octaneBurn 1.00 g of octane • Temp rises from 25.00 to 33.20Temp rises from 25.00 to 33.20 oo CC • Calorimeter contains 1200 g waterCalorimeter contains 1200 g water • Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY
  • 60.
    60 Step 1Step 1Calc. heat transferred from reaction toCalc. heat transferred from reaction to water.water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2Step 2 Calc. heat transferred from reaction toCalc. heat transferred from reaction to bomb.bomb. q = (bomb heat capacity)( T)∆q = (bomb heat capacity)( T)∆ = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J Step 3Step 3 Total heat evolvedTotal heat evolved 41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY Measuring Heats of ReactionMeasuring Heats of Reaction CALORIMETRYCALORIMETRY

Editor's Notes

  • #2 To play the movies and simulations included, view the presentation in Slide Show Mode.