1 thermochemistry thermodynamics the study of heat and work and state functions

1THERMOCHEMISTRYTHERMOCHEMISTRYThermodynamicsThermodynamics

The study of Heat and The study of Heat and Work and State FunctionsWork and State Functions

THERMOCHEMISTRYTHERMOCHEMISTRYThermodynamicsThermodynamics

The study of Heat and The study of Heat and Work and State FunctionsWork and State Functions

Energy & ChemistryEnergy & ChemistryEnergy & ChemistryEnergy & Chemistry

ENERGYENERGY is the capacity to do is the capacity to do work or transfer heat.work or transfer heat.

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

Other forms of energy —Other forms of energy —

• lightlight

• electricalelectrical

• kinetic and potentialkinetic and potential

Potential & Kinetic Potential & Kinetic EnergyEnergy

Potential energy Potential energy — energy a — energy a motionless motionless body has by body has by virtue of its virtue of its position.position.

• 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

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

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

• Two atoms can bond

• As the particles attract they have a lower potential energy

ScaleScale

Kinetic energy Kinetic energy — energy of — energy of motionmotion

• • TranslationTranslation

7Potential & Kinetic Potential & Kinetic EnergyEnergy

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

translate

rotate

vibratetranslate

rotate

vibrate

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)

• Int. E of a chemical system Int. E of a chemical system depends ondepends on

• number of particlesnumber of particles

• type of particlestype of particles

• temperaturetemperature

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.

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.

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

Chemical ProcessesChemical Processes

Reactants

Products

Kinetic Energy

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.

Heat TransferHeat TransferNo Change in StateNo Change in State

q transferred = (sp. ht.)(mass)(∆T)

13Heat Transfer with Heat Transfer with Change of StateChange of State

Heat Transfer with Heat Transfer with Change of StateChange 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)

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

Heat waterHeat water

Evaporate waterEvaporate water

Melt iceMelt ice

Heating/Cooling Curve for Water

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

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

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

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.

Heat Energy Transfer Heat Energy Transfer in a Physical Processin a Physical Process• COCO2 2 (s, -78 (s, -78 ooC) ---> C) --->

COCO2 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 higher kinetic energy.energy.

20Energy 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

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!

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!

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

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

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

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

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

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

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 ∆H = – 241.8 kJand ∆H = – 241.8 kJ

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

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

28Hess’s Law Hess’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.

Figure 6.18, page 227

29Hess’s Law Hess’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.

Figure 6.18, page 227

• 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

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 conditionsP = 1 atmP = 1 atmConcentration = 1 mol/LConcentration = 1 mol/LT = 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

32Enthalpy 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

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 Appendix A.2See Table Appendix A.2

∆∆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.

35Using Standard Enthalpy Using Standard Enthalpy ValuesValues

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”)”)

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

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

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

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)

∆∆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)

CalorimetryCalorimetry

Some 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

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

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

Measuring Heats of ReactionMeasuring Heats of ReactionCALORIMETRYCALORIMETRY

1 thermochemistry thermodynamics the study of heat and work and state functions

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