1 © 2006 Brooks/Cole - Thomson Thermochemistry and Thermodynamics Chemistry: The central Science AP...

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© 2006 Brooks/Cole - Thomson

Thermochemistry and Thermodynamics

Chemistry: The central Science AP version (10th edition, chapter 5) Additional References and Resources

Powerpoint presentation from Melissa Brophy’s website (modified for our needs)

Chemtour Animations and Vocab flashcards from W.W. Norton site:

http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch05/studyplan.asp

(a) State functions and path functions (b) Internal Energy (c) PV work (d) Calorimetry (e) Heating Curves

POGILS:

Thermochemistry and Calorimetry

Hess’s Law: Enthalpy is a state function

Internal Energy and Enthalpy

Thermochemistry package: vocab, diagrams and equations, practice problems (M. Brophy)

Lab experiments: (1) Heat capacity of a metal

(2) Heat of neutralization (HCl + NaOH)

Activity (sign conventions for q and w) from chap 5 online resources and movies (thermite; AlBr3; NI3). Online practice quiz

Chemguy videos for introduction to energy and thermochemistry or thermodynamics

http://www.youtube.com/watch?v=DIpGoOu-rt8&p=93017BDB7EDFE758&index=1&feature=BF

Chem guy videos – thermodynamics 1-5 – one video clip runs into the next one for more advanced aspect of thermodynamics

For thermodynamics 1: spontaneous process, enthalpy, entropy (system, surroundings, universe)http://www.youtube.com/watch?v=94p9ZfTjeKc&p=2FDA79D443854B33&index=2&playnext=2

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Chemistry and Chemical Reactivity 6th Edition

John C. Kotz Paul M. Treichel

Gabriela C. Weaver

CHAPTER 6

Principles of Reactivity: Energy and Chemical Reactions

© 2006 Brooks/Cole Thomson

Lectures written by John Kotz

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Thermodynamics and Thermochemistry

Thermodynamics• Study of energy and its

transformations (or Interconversions)

Thermochemistry• That part of

thermodynamics that deals with the relationships between chemical reactions and energy changes involving heat

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Energy & ChemistryEnergy & ChemistryENERGYENERGY is the capacity to do work or transfer heat. is the capacity to do work or transfer heat.

work work is the energy used to cause an object with mass is the energy used to cause an object with mass to move (w= F x d). When chemical reactions to move (w= F x d). When chemical reactions involve gases, the work done may involve involve gases, the work done may involve compression or expansion of gases. compression or expansion of gases.

(w = -P∆V where P = pressure and ∆V = V(w = -P∆V where P = pressure and ∆V = Vfinalfinal-V-Vinitialinitial))

heatheat is the form of energy that flows between 2 is the form of energy that flows between 2 objects because of their difference in temperature.objects because of their difference in temperature.

Other forms of energy -Other forms of energy -LightLight, , electricalelectrical, , kinetic and kinetic and potentialpotential

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Internal Energy (E)Internal Energy (E)Internal Energy (E)Internal Energy (E)

• PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)

• Internal E of a chemical system Internal E of a chemical system depends ondepends on

• number of particlesnumber of particles

• type of particlestype of particles

• temperaturetemperature

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GO TO chemtours animation on Internal Energy

• http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch05/studyplan.asp

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Changes in Internal Energy

• When a system undergoes any chemical or physical change, the change in internal energy is

• ∆E = q + w [ ∆ E = Efinal- Einitial ]

• q = heat added or liberated from the system

• w = work done on or by the system

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heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEMSYSTEMSYSTEM

∆E = q + w∆E = q + w

w transfer inw transfer in(+w)(+w)

w transfer outw transfer out(-w)(-w)

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Go To TB animations and to Chem Tours

• Activity (sign conventions for q and w) from chap 5 online resources and movies (thermite; AlBr3; NI3).

• (To Do as Homework)

• PV work from Chem tours:


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ENTHALPYENTHALPYENTHALPYENTHALPYMost chemical reactions occur at constant P, soMost chemical reactions occur at constant P, so

and so ∆E = ∆H + w (and w is usually small)

∆H = heat transferred at constant P ≈ ∆E

∆H = change in heat content of the system

∆H = Hfinal – Hinitial

More on Enthalpy Later

and so ∆E = ∆H + w (and w is usually small)

∆H = heat transferred at constant P ≈ ∆E

∆H = change in heat content of the system

∆H = Hfinal – Hinitial

More on Enthalpy Later

Heat transferred at constant P = qHeat transferred at constant P = qpp

qqpp = = ∆H ∆H where where H = enthalpyH = enthalpy



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• Recall: PE + KE = Internal energy Recall: PE + KE = Internal energy (E or U)(E or U)

• Internal E of a chemical system Internal E of a chemical system depends ondepends on

• number of particlesnumber of particles

• type of particlestype of particles

• temperaturetemperature

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy

Potential EnergyPotential Energyon the Atomic on the Atomic

ScaleScale


ScaleScale

NaCl — composed of NaCl — composed of NaNa++ and Cl and Cl-- ions. ions.

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• Positive and negative particles (ions) attract one another.

• Two atoms can bond

• As the particles attract they have a lower potential energy


ScaleScale


ScaleScale

+ _

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Electrostatic potential energy (Eel)

• Equation:

• The chemical energy associated with compounds is due to the potential energy stored in the arrangement of atoms in a substance (ex: bond energies)

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Potential & Kinetic Potential & Kinetic EnergyEnergy

Kinetic energy Kinetic energy — — energy of energy of motion.motion.

translate

rotate

vibratetranslate

rotate

vibrate

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• PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)

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• The higher the T The higher the T the higher the the higher the internal energyinternal energy

• So, use changes So, use changes in T (∆T) to in T (∆T) to monitor changes monitor changes in E (∆E).in E (∆E).

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ThermodynamicsThermodynamicsThermodynamicsThermodynamics• Thermodynamics is the science of heat

(energy) transfer.

Heat energy is associated Heat energy is associated with molecular motions.with molecular motions.

Heat transfers until thermal equilibrium is established.

∆T measures energy transferred.

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System and Surroundings

• SYSTEM– The object under study

• SURROUNDINGS– Everything outside the

system

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Heat

The flow of thermal energy from one object to another.

Heat always flows from warmer to cooler objects.

Ice gets warmer while hand gets

cooler

Cup gets cooler while hand gets warmer

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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 downT(surr) goes upT(surr) goes up

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Directionality of Heat TransferDirectionality of Heat Transfer• Heat always transfers from hotter object to

cooler one.

•ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.

T(system) goes upT(system) goes upT (surr) goes downT (surr) goes down

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Go To – chapter 5 – online resources for our TB

• Go to Activity banner and view sign conventions for q and w

• View movies (thermite; AlBr3; NI3)

• Write key points in margins of this page

• Reminder Online practice quiz for chapter 5 from our TB needs to be completed

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Energy & ChemistryEnergy & Chemistry

All of thermodynamics depends on All of thermodynamics depends on the law of the law of

CONSERVATION OF ENERGYCONSERVATION OF ENERGY..

• The total energy is unchanged in The total energy is unchanged in a chemical reaction.a chemical reaction.

• If PE of products is less than If PE of products is less than reactants, the difference must be reactants, the difference must be released as KE.released as KE.

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Energy Change in Energy Change in Chemical ProcessesChemical ProcessesEnergy Change in Energy Change in

Chemical ProcessesChemical Processes

Reactants

Products

Kinetic Energy

PE

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.

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UNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGY1 calorie = heat required to 1 calorie = heat required to

raise temp. of 1.00 g of Hraise temp. of 1.00 g of H22O O by 1.0 by 1.0 ooC.C.

1000 cal = 1 kilocalorie = 1 kcal1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food 1 kcal = 1 Calorie (a food

“calorie”)“calorie”)

But we use the unit called the But we use the unit called the JOULEJOULE

1 cal = 4.184 joules1 cal = 4.184 joules James JouleJames Joule1818-18891818-1889

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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?

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Specific Heat Specific Heat CapacityCapacity


How much energy is transferred due to T difference?How much energy is transferred due to T difference?

The heat The heat (q)(q) “lost” or “gained” is related to “lost” or “gained” is 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)

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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

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If 25.0 g of Al cool from If 25.0 g of Al cool from 310 310 ooC to 37 C to 37 ooC, how C, how many joules of heat many joules of heat energy are lost by energy are lost by the Al?the Al?

Specific heat capacity =

heat lost or gained by substance (J)

(mass, g)(T change, K)

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If 25.0 g of Al cool from 310 If 25.0 g of Al cool from 310 ooC to 37 C to 37 ooC, how many joules of heat energy are lost by the Al?C, how many 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 signals Notice 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 signals Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.

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Heat/Energy TransferHeat/Energy TransferNo Change in StateNo Change in State

q transferred = (sp. ht.)(mass)(∆T)q transferred = (sp. ht.)(mass)(∆T)

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Heat Heat TransferTransferHeat Heat TransferTransfer

• Use heat transfer as a way to find specific heat capacity, Cp

• 55.0 g Fe at 99.8 ˚C

• Drop into 225 g water at 21.0 ˚C

• Water and metal come to 23.1 ˚C

• What is the specific heat capacity of the metal? (Lab Activity)

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Because of conservation of energy,Because of conservation of energy,

q(Fe) = –q(Hq(Fe) = –q(H22O) (heat out of Fe = heat into HO) (heat out of Fe = heat into H22O)O)

or or q(Fe) + q(Hq(Fe) + q(H22O) = 0O) = 0

q(Fe) = (55.0 g)(q(Fe) = (55.0 g)(CCpp)(23.1 ˚C – 99.8 ˚C))(23.1 ˚C – 99.8 ˚C)

q(Fe) = –4219 • q(Fe) = –4219 • CCpp

q(Hq(H22O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C)O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C)

q(Hq(H22O) = 1977 JO) = 1977 J

q(Fe) + q(Hq(Fe) + q(H22O) = –4219 O) = –4219 CCpp + 1977 = 0 + 1977 = 0

CCpp = 0.469 J/K•g = 0.469 J/K•g

Heat TransferHeat TransferHeat TransferHeat Transfer

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Heat Transfer Heat Transfer withwith Change of State Change of State

Heat Transfer Heat Transfer withwith Change of State Change of State

Changes of state involve energy Changes 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)

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Heat Transfer and Heat Transfer and Changes of StateChanges of State

Heat Transfer and Heat Transfer and Changes of StateChanges of State

Requires energy (heat).Requires energy (heat).

This is the reasonThis is the reason

a)a) you cool down you cool down after swimming after swimming

b)b) you use water to you use water to put out a fire.put out a fire.

+ energy

Liquid ---> VaporLiquid ---> Vapor

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GO TO chemtours animation on Heating Curves


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Heating/Cooling Curve for Water

Heating/Cooling Curve for Water

Note that T is Note that T is constant as ice meltsconstant as ice melts

Note that T is Note that T is constant as ice meltsconstant as ice melts

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Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/gSpecific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•KHeat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

What quantity of heat is required to melt What quantity of heat is required to melt 500. g of ice and heat the water to steam 500. g of ice and heat the water to steam at 100 at 100 ooC?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

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What quantity of heat is required to melt 500. g of ice and What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 heat the water to steam at 100 ooC?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 J J

2.2. To raise water from 0 To raise water from 0 ooC to 100 C to 100 ooCC

q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 10q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 1055 J J

3.3. To evaporate water at 100 To evaporate water at 100 ooCC

q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 1066 J J

4. 4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 1066 J = 1510 kJ J = 1510 kJ

Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State

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ChemicalChemical Reactivity ReactivityChemicalChemical Reactivity ReactivityWhat drives chemical reactions? How do they What drives chemical reactions? How do they

occur?occur?

The first is answered by The first is answered by THERMODYNAMICSTHERMODYNAMICS and the second by and the second by KINETICSKINETICS..

Have already seen a number of “driving forces” Have already seen a number of “driving forces” for reactions that are for reactions that are PRODUCT-FAVOREDPRODUCT-FAVORED..

•• 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

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ChemicalChemical Reactivity ReactivityBut energy transfer also allows us to predict But energy transfer also allows us to predict

reactivity.reactivity.

In general, reactions that transfer In general, reactions that transfer energy to their surroundings are energy to their surroundings are product-favored.product-favored.

So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.

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Heat Energy Transfer Heat Energy Transfer in a Physical Processin a Physical ProcessHeat Energy Transfer Heat Energy Transfer in a Physical Processin a Physical Process

COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2 (g, -78 (g, -78 ooC)C)

Heat transfers from surroundings to system in endothermic process.

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Heat Energy Transfer in Heat Energy Transfer in a Physical Processa Physical Process

• COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2

(g, -78 (g, -78 ooC)C)

• A regular array of A regular array of molecules in a solid molecules in a solid -----> gas phase -----> gas phase molecules. molecules.

• Gas molecules have Gas molecules have higher kinetic energy.higher kinetic energy.

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Energy Level Energy Level Diagram for Heat Diagram for Heat Energy Energy TransferTransfer

Energy Level Energy Level Diagram for Heat Diagram for Heat Energy Energy TransferTransfer

∆E = E(final) - E(initial) = E(gas) - E(solid)

COCO22 solid solid

COCO22 gas gas

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Heat Energy Transfer in Heat Energy Transfer in Physical ChangePhysical Change

Heat Energy Transfer in Heat Energy Transfer in Physical ChangePhysical Change

• Gas molecules have higher kinetic Gas molecules have higher kinetic energy.energy.

• Also, Also, WORKWORK is done by the system is done by the system in pushing aside the atmosphere.in pushing aside the atmosphere.

COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2 (g, -78 (g, -78 ooC)C)

Two things have happened!Two things have happened!

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FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS

FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS

∆∆E = q + wE = q + w

heat energy transferredheat energy transferred

energyenergychangechange

work donework doneby the by the systemsystem

Energy is conserved!Energy is conserved!

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heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEMSYSTEMSYSTEM

∆E = q + w∆E = q + w

w transfer inw transfer in(+w)(+w)

w transfer outw transfer out(-w)(-w)

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ENTHALPYENTHALPYENTHALPYENTHALPYMost 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 in H = change in heat content heat 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 in H = change in heat content heat content of the systemof the system

∆∆H = HH = Hfinalfinal - H - Hinitialinitial





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If If HHfinalfinal < H < Hinitialinitial then ∆H is negative then ∆H is negative

Process is Process is EXOTHERMICEXOTHERMIC

If If HHfinalfinal < H < Hinitialinitial then ∆H is negative then ∆H is negative

Process is Process is EXOTHERMICEXOTHERMIC

If If HHfinalfinal > H > Hinitialinitial then ∆H is positive then ∆H is positive

Process is Process is ENDOTHERMICENDOTHERMIC

If If HHfinalfinal > H > Hinitialinitial then ∆H is positive then ∆H is positive

Process is Process is ENDOTHERMICENDOTHERMIC

ENTHALPYENTHALPYENTHALPYENTHALPY∆∆H = HH = Hfinalfinal - H - Hinitialinitial

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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) + 241.8 kJ241.8 kJ

USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY

Exothermic reaction — heat is a “product” Exothermic reaction — heat is a “product” and and ∆H = – 241.8 kJ∆H = – 241.8 kJ

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Making Making liquidliquid H H22O from HO from H22 + +

OO22 involves involves twotwo exoexothermic thermic

steps. steps.

USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY

H2 + O2 gas

Liquid H2OH2O vapor

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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 kJ O(liq) + 44 kJ

-----------------------------------------------------------------------

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(liq) + 286 kJO(liq) + 286 kJ

Example of Example of HESS’S LAWHESS’S LAW——

If a rxn. is the sum of 2 or more If a rxn. is the sum of 2 or more others, the net ∆H is the sum of others, the net ∆H is the sum of the ∆H’s of the other rxns.the ∆H’s of the other rxns.

USING ENTHALPYUSING ENTHALPY

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ChemToursGO TO: http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch05/studyplan.asp

View animation on

(a)State functions and path functions

Write key points on this page

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Hess’s Law Hess’s Law & Energy Level & Energy Level DiagramsDiagrams

Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Active Figure 6.18

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Hess’s Law Hess’s Law & Energy Level & Energy Level DiagramsDiagrams

Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Active Figure 6.18

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• This equation is valid because This equation is valid because ∆H is a ∆H is a STATE FUNCTIONSTATE FUNCTION

• These depend only on the state These depend only on the state of the system and not how it got of the system and not how it got there.there.

• V, T, P, energy — and your bank V, T, P, energy — and your bank account!account!

• Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot measure absolute H. Can only measure absolute H. Can only measure ∆H.measure ∆H.

∆∆H along one path =H along one path =

∆∆H along another pathH along another path

∆∆H along one path =H along one path =

∆∆H along another pathH along another path

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Standard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesMost ∆H values are labeled Most ∆H values are labeled ∆H∆Hoo

Measured under Measured under standard conditionsstandard conditions

P = 1 atm = 101.3 kPa = 760 mmHg P = 1 atm = 101.3 kPa = 760 mmHg

Concentration = 1 mol/LConcentration = 1 mol/L

T = usually 25 T = usually 25 ooCC

with all species in standard stateswith all species in standard states

e.g., C = graphite and Oe.g., C = graphite and O22 = gas = gas

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Enthalpy ValuesEnthalpy Values 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)

∆∆H˚ = -286 kJH˚ = -286 kJ

Depend on how the reaction is written and on phases Depend on how the reaction is written and on phases of reactants and productsof reactants and productsDepend on how the reaction is written and on phases Depend on how the reaction is written and on phases of reactants and productsof reactants and products

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Standard Enthalpy ValuesStandard Enthalpy Values

NIST (Nat’l Institute for Standards and NIST (Nat’l Institute for Standards and Technology) gives values ofTechnology) gives values of

∆∆HHffoo = standard molar enthalpy of = standard molar enthalpy of

formationformation

— — the enthalpy change when 1 mol of the enthalpy change when 1 mol of compound is formed from elements under compound is formed from elements under standard conditions.standard conditions.

See Table 5.3 and Appendix C of OUR TB and AP reference See Table 5.3 and Appendix C of OUR TB and AP reference TablesTables

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Reference Tables for Thermochemistry

• Text Book: Table 5.3 Standard enthalpies of formation and Appendix C: Thermodynamic Quantities for Selected Substances at 298.15 K (25 °C) Use these for problems from our textbook.

• AP test reference tables

• Regents Reference tables

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∆∆HHffoo, standard molar , standard molar

enthalpy of formationenthalpy of formation

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g)O(g)

∆∆HHffoo (H (H22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol

By definition, By definition,

∆∆HHffoo

= 0 for elements in their = 0 for elements in their

standard states.standard states.

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Using Standard Enthalpy Using Standard Enthalpy ValuesValues


Use ∆H˚’s to calculate enthalpy change for Use ∆H˚’s to calculate enthalpy change for

HH22O(g) + C(graphite) --> HO(g) + C(graphite) --> H22(g) + CO(g)(g) + CO(g)

(product is called “(product is called “water gaswater gas”)”)

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From reference books we findFrom reference books we find

• HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) O(g)

∆ ∆HHff˚ of H˚ of H22O vapor = - 242 kJ/molO vapor = - 242 kJ/mol

• C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g)

∆ ∆HHff˚ of CO = - 111 kJ/mol˚ of CO = - 111 kJ/mol

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HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) ∆H(g) ∆Hoo = +242 kJ = +242 kJ

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) ∆H∆Hoo = -111 kJ = -111 kJ

--------------------------------------------------------------------------------

To convert 1 mol of water to 1 mol each of HTo convert 1 mol of water to 1 mol each of H22

and CO and CO requiresrequires 131 kJ of energy. 131 kJ of energy.

The “water gas” reaction is The “water gas” reaction is ENDOENDOthermic.thermic.


∆∆HHoonetnet = +131 kJ = +131 kJ


∆∆HHoonetnet = +131 kJ = +131 kJ

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In general, when In general, when ALLALL

enthalpies of formation are enthalpies of formation are

known, known,

Calculate ∆H of Calculate ∆H of reaction?reaction?

∆∆HHoorxnrxn = =

∆ ∆HHffoo (products) (products)

- - ∆H ∆Hffoo (reactants)(reactants)

∆∆HHoorxnrxn = =

∆ ∆HHffoo (products) (products)

- - ∆H ∆Hffoo (reactants)(reactants)

Remember that ∆ always = final – initial

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Calculate the heat of combustion of Calculate the heat of combustion of

methanol, i.e., ∆Hmethanol, i.e., ∆Hoorxnrxn for for

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

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∆∆HHoorxnrxn = ∆H = ∆Hff

oo (CO(CO22) + 2 ∆H) + 2 ∆Hff

oo (H(H22O) O)

- {3/2 ∆H- {3/2 ∆Hffoo

(O(O22) + ∆H) + ∆Hffoo

(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)}

∆∆HHoorxnrxn = -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)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)

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Calorimetry ChemToursGO TO: http://www.wwnorton.com/college/chemistry/gilbert2/contents/ch05/studyplan.asp

View animation on calorimetry and Write key points in margins of this page

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Measuring Heats of ReactionCALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

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Measuring Heats of ReactionCALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

Constant Volume “Bomb” Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for reaction.

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CalorimetryCalorimetrySome heat from reaction warms waterqwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms “bomb”qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

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Calculate heat of combustion of Calculate heat of combustion of octane. octane. CC88HH1818 + 25/2 O + 25/2 O22 --> -->

8 CO8 CO22 + 9 H + 9 H22OO

•• Burn 1.00 g of octaneBurn 1.00 g of octane• Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20 ooCC• 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 ReactionCALORIMETRYCALORIMETRY


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Step 1Step 1 Calc. heat transferred from reaction to water.Calc. heat transferred from reaction to 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 to bomb.Calc. heat transferred from reaction to 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



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Activities and Problem set 8 (due date_______)

TextBook ch. 5 – all sections required for regents (in part), SAT II and AP exams

POGIL activities on Thermochemistry and Calorimetry, Hess’s Law, Internal Energy and Enthalpy

Vocab from Norton flash cards

Lab activities: (specific heat capacity and heat of neutralization)

Online practice quiz ch 5 due by_____

• Ch 5 Problems TO DO: write out

questions and answers & show work

• Study (don’t write out) the sample

exercises

• Do all in-chapter practice exercises

• Do all GIST and Visualizing concepts

problems

• Do end of chapter 5 exercises: 9,13, 18,

25, 28 , 29, 39, 40,42 , 43, 49, 51, 53, 54,

57, 60, 61, 63, 65, 67, 71 a,c, 74, 77, 85, 89,

90.

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