THERMOCHEMISTRY OR THERMODYNAMICS

THERMOCHEMISTRY OR THERMODYNAMICS

ChapterChapter 66

Chemical ReactivityChemical Reactivity

What drives chemical reactions? How do they occur?What drives chemical reactions? How do they occur?The first is answered by The first is answered by THERMODYNAMICSTHERMODYNAMICS and and

the second by the second by KINETICSKINETICS..We have already seen a number of “driving forces” for We have already seen a number of “driving forces” for

reactions that are 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

Energy and Chemistry

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

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

Other forms of energy —Other forms of energy —lightlight electricalelectrical nuclearnuclearkinetic kinetic potentialpotential

Law of Conservation of Energy

Energy can be converted from one form to Energy can be converted from one form to another but can neither be created nor another but can neither be created nor destroyed.destroyed.

((EEuniverseuniverse is constant) is constant)

The Two Types of Energy

Potential: Potential: due to position or composition - can be due to position or composition - can be converted to workconverted to work

PE = mghPE = mgh(m = mass, g = acceleration of gravity, and h = height)(m = mass, g = acceleration of gravity, and h = height)

Kinetic: Kinetic: due to motion of the objectdue to motion of the object

KE = KE = 11/2 /2 mvmv22

((mm = mass, = mass, vv = velocity) = velocity)

Kinetic and Potential EnergyKinetic and Potential Energy

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

Kinetic and Potential EnergyKinetic and Potential Energy

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

• • TranslationTranslation• • RotationRotation• • VibrationVibration

Units of EnergyUnits of Energy

1 calorie = heat required to raise temp. 1 calorie = heat required to raise temp. of 1.00 g of Hof 1.00 g of H22O by 1.0 O by 1.0 ooC.C.

1000 cal = 1 kilocalorie = 1 kcal1000 cal = 1 kilocalorie = 1 kcal1 kcal = 1 Calorie (a food “calorie”)1 kcal = 1 Calorie (a food “calorie”)

But we use the unit called the JOULEBut we use the unit called the JOULE1 cal = 4.184 joules1 cal = 4.184 joules

James JouleJames Joule1818-18891818-1889

Temperature v. Heat

TemperatureTemperature reflects reflects random motions random motions of of particles, therefore related to kinetic particles, therefore related to kinetic energy of the system.energy of the system.

HeatHeat involves a involves a transfer of energy transfer of energy betweenbetween 2 objects due to a temperature difference2 objects due to a temperature difference

Extensive & Intensive Properties

Extensive properties Extensive properties depends directly on the depends directly on the amount of substance amount of substance present.present.

•massmass•volumevolume•heatheat•heat capacity (C)heat capacity (C)

Intensive properties is Intensive properties is not related to the amount not related to the amount of substance.of substance.

•temperaturetemperature•concentrationconcentration•pressurepressure•specific heat (s)specific heat (s)

State Function

Depends only on the present state of the Depends only on the present state of the system - not how it arrived there.system - not how it arrived there.

It is It is independent of pathwayindependent of pathway..

Energy changeEnergy change is is independentindependent of the pathway of the pathway (and, therefore, a state function), while (and, therefore, a state function), while work work and heatand heat are are dependentdependent on the pathway. on the pathway.

System and Surroundings

SystemSystem: That on which we focus attention: That on which we focus attention

SurroundingsSurroundings: Everything else in the universe: Everything else in the universe

Universe = System + SurroundingsUniverse = System + Surroundings

Exo and Endothermic

Heat exchange accompanies chemical Heat exchange accompanies chemical reactions.reactions.

ExothermicExothermic: Heat flows : Heat flows outout of the system (to of the system (to the surroundings).the surroundings).

EndothermicEndothermic: Heat flows : Heat flows intointo the system the system (from the surroundings).(from the surroundings).

ENERGY DIAGRAMSENERGY DIAGRAMS

ExothermicExothermic

EndothermicEndothermic

Endo- and ExothermicEndo- and Exothermic

SurroundingsSurroundingsSystemSystem

qqsystemsystem > 0 > 0 w > 0w > 0

heatheat

ENDOTHERMICENDOTHERMICE goes upE goes up

Endo- and ExothermicEndo- and Exothermic


qqsystemsystem > 0 > 0 w > 0w > 0

heatheat


qqsystemsystem < 0 < 0 w < 0w < 0

heatheat

ENDOTHERMICENDOTHERMIC EXOTHERMICEXOTHERMICE(system) goes upE(system) goes up E(system) goes downE(system) goes down

First Law

First Law of ThermodynamicsFirst Law of Thermodynamics: : The energy of the universe is The energy of the universe is constant.constant.

EnthalpyEnthalpy

H = HH = Hfinalfinal - H - Hinitialinitial

If HIf Hfinalfinal > H > Hinitialinitial then then H is positiveH is positive

Process is Process is ENDOTHERMICENDOTHERMIC

If HIf Hfinalfinal < H < Hinitialinitial then then H is negativeH is negative

Process is Process is EXOTHERMICEXOTHERMIC

First Law

EE = = qq + + ww

EE = change in system’s internal energy = change in system’s internal energyqq = heat = heatww = work = work

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P = FA

Initialstate

P = FA

Finalstate

h h

Area = A A

V

(a) (b)

Piston moving a distance against Piston moving a distance against a pressure does work.a pressure does work.

Work

w & w & V must have opposite signs, since work is being done by the system to V must have opposite signs, since work is being done by the system to expand the gas.expand the gas.wwsystemsystem = -P = -P VV

1 Latm = 101.3 J1 Latm = 101.3 J

1 J = kgm1 J = kgm22/s/s22

P = F/AP = F/AF = PAF = PAw = F w = F hhw = PA w = PA hhw = P w = P VV

Enthalpy

Enthalpy = Enthalpy = HH = = EE + + PVPVEE = = HH PPVVHH = = EE + + PPVV

At constant pressure,At constant pressure,qqPP = = EE + + PPVV, ,

where where qqPP = = HH at constant pressure at constant pressure

HH = energy flow as heat (at constant pressure) = energy flow as heat (at constant pressure)

Some Heat Exchange Terms

specific heat capacity (s)specific heat capacity (s)heat capacity per gram = J/°C g or J/K gheat capacity per gram = J/°C g or J/K g

molar heat capacity (s)molar heat capacity (s)heat capacity per mole = J/°C mol or J/K molheat capacity per mole = J/°C mol or J/K mol

Specific Heat CapacitySpecific Heat Capacity

SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K)HH22OO 4.1844.184

AlAl 0.9020.902glassglass 0.840.84

AluminumAluminum

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Styrofoamcups

Stirrer

Styrofoamcover

Thermometer

HHoo = - q = - qpp qqpp = ms = mstt

Simple CalorimeterSimple Calorimeterq = heat (J)q = heat (J)m = mass (g)m = mass (g)s = specific heat (j/gCs = specific heat (j/gCoo))t = “change” in temperature (Ct = “change” in temperature (Coo))HHoo = “change in” enthalpy (kJ) = “change in” enthalpy (kJ)


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 C, how many joules of heat energy are lost by the Al?many joules of heat energy are lost by the Al?

where where T = TT = Tfinalfinal - T - Tinitialinitial

heat gain/lost = q = m s T


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 C, how many joules of heat energy are lost by the Al?of heat energy are lost by the Al?

where where T = TT = Tfinalfinal - T - Tinitialinitial

q = (0.902 J/g•K)(25.0 g)(37 - 310)Kq = (0.902 J/g•K)(25.0 g)(37 - 310)Kq = - 6160 Jq = - 6160 J

heat gain/lost = q = m s T


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, C, how many joules of heat energy are lost how many joules of heat energy are lost by the Al?by the Al?

q = - 6160 Jq = - 6160 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.

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Water

Insulatingcontainer

StirrerThermometer

Ignitionwires

Reactantsin samplecup

Steelbomb

Bomb CalorimeterBomb Calorimeter

Heat Capacity

C = heat absorbedincrease in temperature

= JC

or JK

E = qE = qv v & q& qvv = -(C = -(C t + ms t + ms t)t)

E = “change in” internal energy (J)E = “change in” internal energy (J)qqvv = heat at constant volume (J) = heat at constant volume (J)C = heat capacity (J/CC = heat capacity (J/Coo))t = “change”in temperature (Ct = “change”in temperature (Coo))

REMEMBER!!!

In regular calorimetry pressure is constant, but the In regular calorimetry pressure is constant, but the volume will change so:volume will change so:

qp = -Hqp = E + p V

In bomb calorimetry, volume is constant so:

qv = Esince since p V = zero. = zero.

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 33.20 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 HHccm = -(Cm = -(Ct + mst + mst) where Ht) where Hc c is heat of combustion.is heat of combustion.

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 JStep 2Step 2 Calc. heat transferred from reaction to bomb.Calc. heat transferred from reaction to bomb.q = C q = C tt = (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 JStep 3Step 3 Total heat evolvedTotal heat evolved 41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 JHeat 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

Hess’s Law

Reactants Reactants Products Products

The change in The change in enthalpy is the same enthalpy is the same whether the reaction takes place in whether the reaction takes place in one one step or a series of stepsstep or a series of steps..

Standard States

CompoundCompound- For a For a gasgas, pressure is exactly , pressure is exactly 1 atmosphere1 atmosphere..- For a For a solutionsolution, concentration is exactly , concentration is exactly 1 molar1 molar..- Pure substance (liquid or solid), it is the pure liquid Pure substance (liquid or solid), it is the pure liquid

or solid.or solid.

ElementElement- The form [NThe form [N22((gg), K(), K(ss)] in which it exists at )] in which it exists at 1 atm 1 atm

and 25°Cand 25°C..

Calculations via Hess’s Law

1.1. If a reaction is If a reaction is reversedreversed, , HH is also reversed. is also reversed.

NN22((gg) + O) + O22((gg) ) 2NO( 2NO(gg) ) HH = 180 kJ = 180 kJ

2NO(2NO(gg) ) N N22((gg) + O) + O22((gg) ) HH = = 180 kJ180 kJ

2.2. If the coefficients of a reaction are multiplied by If the coefficients of a reaction are multiplied by an integer, an integer, H is multiplied by that same integer.H is multiplied by that same integer.

66NO(NO(gg) ) 33NN22((gg) + ) + 33OO22((gg) ) HH = = 540 kJ540 kJ

Using EnthalpyUsing Enthalpy

Consider the decomposition of waterConsider the decomposition of waterHH22O(g) + O(g) + 243 kJ243 kJ ---> H ---> H22(g) + 1/2 O(g) + 1/2 O22(g)(g)

Endothermic reaction — heat is a “reactant”Endothermic reaction — heat is a “reactant”H = + 243 kJH = + 243 kJ

Making HMaking H22 from H from H22O involves two steps.O involves two steps.HH22O(l) + 44 kJ ---> HO(l) + 44 kJ ---> H22O(g)O(g)HH22O(g) + 242 kJ ---> HO(g) + 242 kJ ---> H22(g) + 1/2 O(g) + 1/2 O22(g)(g)----------------------------------------------------------------------------------------------------------------------------------HH22O(l) + 286 kJ --> HO(l) + 286 kJ --> H22(g) + 1/2 O(g) + 1/2 O22(g)(g)Example of Example of HESS’S LAWHESS’S LAW——If a rxn. is the sum of 2 or more others, the net If a rxn. is the sum of 2 or more others, the net H is H is

the sum of the the sum of the H’s of the other rxns.H’s of the other rxns.


Calc. Calc. H for S(s) + 3/2 OH for S(s) + 3/2 O22(g) --> SO(g) --> SO33(g)(g)

S(s) + OS(s) + O22(g) --> SO(g) --> SO22(g) (g) -320.5 kJ-320.5 kJ

SOSO22(g) + 1/2 O(g) + 1/2 O22(g) --> SO(g) --> SO33(g) (g) -75.2 kJ -75.2 kJ

______________________________________________________________________________

S(s) + 3/2 OS(s) + 3/2 O22(g) --> SO(g) --> SO33(g)(g) -395.7 kJ -395.7 kJ


S solid

SO3 gas

SO2 gas

direct path

+ 3/2 O2

H = -395.7 kJ

energy

+O2H 1 = -320.5 kJ

+ 1/2 O2H 2 = -75.2 kJ

H along one path =H along one path = H along another pathH along another path

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

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 measure H.H.

H along one path =H along one path = H along another pathH along another path

Standard Enthalpy ValuesStandard Enthalpy Values

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

HHooff = standard molar enthalpy of formation = standard molar enthalpy of formation

This is the enthalpy change when 1 mol of compound is This is the enthalpy change when 1 mol of compound is formedformed from elements under standard conditions. from elements under standard conditions. HHoo

ff is always stated in terms of moles of product is always stated in terms of moles of product formed.formed.

See Appendix A21-A24.See Appendix A21-A24.

HHooff, standard molar enthalpy of , standard molar enthalpy of

formationformation

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

HHooff = -241.8 kJ/mol = -241.8 kJ/mol

By definition, By definition, HHoof f = 0 for elements = 0 for elements

in their standard states.in their standard states.

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

Use Use HH’s to calculate enthalpy change for ’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”)”)


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) 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)

HH ff of CO = - 111 kJ/mol of CO = - 111 kJ/mol


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

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

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

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

HHoonetnet = +131 kJ = +131 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.

Change in Enthalpy

Can be calculated from enthalpies of formation of Can be calculated from enthalpies of formation of reactantsreactants and and productsproducts..

HHrxnrxn° = ° = nnppHHff((productsproducts) ) nnrrHHff((reactantsreactants))

H is an extensive property--kJ/molH is an extensive property--kJ/mol

For the reaction: 2HFor the reaction: 2H22 (g) (g) + O+ O2 (g)2 (g) ---> 2H ---> 2H22OO(g)(g)

Enthalpy would be twice the Enthalpy would be twice the H value for the combustion of hydrogen.H value for the combustion of hydrogen.


Calculate the heat of combustion of Calculate the heat of combustion of methanol, i.e., methanol, i.e., HHoo

rxnrxn for for

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

HH22O(g)O(g)

HHoorxnrxn = = HHoo

f f (prod) - (prod) - HHoof f (react)(react)


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

HHoorxnrxn = = HHoo

f f (prod) - (prod) - HHoof f (react)(react)

HHoorxnrxn = = HHoo

f f (CO(CO22) + 2 ) + 2 HHoof f (H(H22O) O)

- {3/2 - {3/2 HHoof f (O(O22) + ) + HHoo

f f (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 = -675.6 kJ per molper mol of methanol of methanolHHoo

rxn rxn is always in terms of moles of reactant.is always in terms of moles of reactant.

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CH4(g)

C(s)

CO2(g)

2H2(g)

2H2O(l)

2O2(g) 2O2(g)

Reactants Elements Products

(a)

(b)

(d)

(c)

Pathway for the Combustion of MethanePathway for the Combustion of Methane

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2O2(g)

CH4(g) CO2(g) 2H2O(l)

- 394 kJ75 kJ

0 kJ

- 572 kJ

2O2(g)C(s)

2H2(g)

Reactants Elements

Products

= Products = Elements = Reactants

Energy

Schematic diagram of the energy changes for the Schematic diagram of the energy changes for the combustion of methane.combustion of methane.

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Earth’satmosphere

Infraredradiated bythe earth

Earth

CO2and H2Omolecules

Visible lightfrom the sun

Greenhouse EffectGreenhouse Effect

Greenhouse Gases:Greenhouse Gases:

COCO22 HH22OO

CHCH44 NN22OO

-- a warming effect exerted by the earth’s atmosphere due to-- a warming effect exerted by the earth’s atmosphere due to thermal energy retained by absorption of infrared radiation.thermal energy retained by absorption of infrared radiation.

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