THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6 Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that

THERMOCHEMISTRY OR THERMODYNAMICS

THERMOCHEMISTRY OR THERMODYNAMICS

ChapterChapter 66

Energy and ChemistryEnergy 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 nuclearnuclear

kinetic kinetic potentialpotential

Temperature v. HeatTemperature 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

Law of Conservation of Energy

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)

Kinetic and Potential EnergyKinetic and Potential EnergyKinetic 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 EnergyKinetic and Potential EnergyKinetic and Potential Energy

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

Units of EnergyUnits of EnergyUnits 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 kcal

1 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 JOULE

1 cal = 4.184 joules1 cal = 4.184 joules

James JouleJames Joule1818-18891818-1889

Extensive & Intensive Properties

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)

System and SurroundingsSystem 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 EndothermicExo and Endothermic

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

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

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

Endo- and ExothermicEndo- and ExothermicEndo- and ExothermicEndo- and Exothermic

SurroundingsSurroundings

SystemSystem

heatheat

SurroundingsSurroundings

SystemSystem

heatheat

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

EnthalpyEnthalpyEnthalpyEnthalpy

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

Upon adding potassium hydroxide to water

the following reaction takes place

NaOH(S) NaOH(aq)for this reaction at constant pressure, ∆H= -43 kj/mol1 .Is the reaction exo- or

endothermic2. Does the water get warmer?

3. What is the enthalpy change for the solution if 14 g of NaOH is

added?

Upon adding potassium hydroxide to water

the following reaction takes place

NaOH(S) NaOH(aq)for this reaction at constant pressure, ∆H= -43 kj/mol1 .Is the reaction exo- or

endothermic2. Does the water get warmer?

3. What is the enthalpy change for the solution if 14 g of NaOH is

added?

ExerciseExerciseConsider the combustion of propane:Consider the combustion of propane:

CC33HH88((gg) + 5O) + 5O22((gg) ) → 3CO→ 3CO22((gg) + 4H) + 4H22O(O(ll))

ΔΔHH = –2221 kJ = –2221 kJ

Assume that all of the heat comes from the Assume that all of the heat comes from the combustion of propane. Calculate combustion of propane. Calculate ΔΔHH in which in which 5.00 g of propane is burned in excess oxygen at 5.00 g of propane is burned in excess oxygen at constant pressure.constant pressure.

––252 kJ252 kJ

Consider the reactionConsider the reaction

HH22((gg) + O) + O22((gg) ) H H22O(O(ll) ) HH° = –286 kJ° = –286 kJ

Which of the following is true?Which of the following is true?a)a) The reaction is exothermic.The reaction is exothermic.b)b) The reaction is endothermic.The reaction is endothermic.c)c) The enthalpy of the products is less than that of The enthalpy of the products is less than that of

the reactants.the reactants.d)d) Heat is absorbed by the system.Heat is absorbed by the system.e) Both A and C are true. ACe) Both A and C are true. AC

Consider the reaction:Consider the reaction:

When a 24.8-g sample of ethyl alcohol When a 24.8-g sample of ethyl alcohol (molar mass = 46.07 g/mol) is burned, (molar mass = 46.07 g/mol) is burned, how much energy is released as heat? how much energy is released as heat?

cc

32 5 2 2 2C H OH( ) + 3O ( ) 2CO ( ) + 3H O( ), 1.37 × kJ10l g g l H

The total volume of hydrogen gas needed to fill The total volume of hydrogen gas needed to fill the Hindenburg was 2.01x10the Hindenburg was 2.01x1088 L at 1.00 atm L at 1.00 atm and 24.7°C.and 24.7°C.

How much energy was evolved when it How much energy was evolved when it burned? burned?

2H2H22((gg) + O) + O22((gg) ) 2H 2H22O(O(ll) ) HH° = –286 kJ ° = –286 kJ

7.37 7.37 10 1022 kJ kJ

2.35 109 kJ

06_74

Styrofoamcups

Stirrer

Styrofoamcover

Thermometer

q = msq = 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))

Some Heat Exchange TermsSome 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

Heat CapacityHeat Capacity

C = heat absorbed

increase in temperature =

JC

or JK

Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity

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

HH22OO 4.1844.184

AlAl 0.9020.902

glassglass 0.840.84

AluminumAluminum


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?

Spec of Al=0.902Spec of Al=0.902

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 C, how many joules of heat energy are lost by the Al?joules 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)q = (0.902 J/g•K)(25.0 g)(37 - 310)q = - 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, how C, how many joules of heat energy are lost by the many joules of heat energy are lost by the Al? Al?

q = - 6160 Jq = - 6160 J

Notice that the negative sign on q signals heat Notice that the negative sign on q signals heat “lost by” or transferred out of Al.bv “lost by” or transferred out of Al.bv 233hhbn 233hhbn

Copper has a specific hear Copper has a specific hear of .382j/gof .382j/gooC. If 2.51 g of C. If 2.51 g of cooper absorbs 2.75 j of heat , cooper absorbs 2.75 j of heat , what is the change in temp ?what is the change in temp ?

Cooper has a specific heat of .382j/g/ Cooper has a specific heat of .382j/g/ ooC. the temperature of an C. the temperature of an unknown mass of cooper unknown mass of cooper increases by 4.50 increases by 4.50 ooC when it C when it absorbs 3.97J of heat. What is the absorbs 3.97J of heat. What is the mass of the copper?mass of the copper?

Heating curves

REMEMBER!!!REMEMBER!!!

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

In bomb calorimetry, volume is constant.

CalorimetryCalorimetry

Constant volume calorimeter is called a bomb Constant volume calorimeter is called a bomb calorimeter.calorimeter.

Material is put in a container with pure Material is put in a container with pure oxygen. Wires are used to start the oxygen. Wires are used to start the combustion. The container is put into a combustion. The container is put into a container of water.container of water.

The heat capacity of the calorimeter is known The heat capacity of the calorimeter is known and tested.and tested.

Bomb CalorimeterBomb Calorimeter

thermometerthermometer

stirrerstirrer

full of waterfull of water

ignition wireignition wire

Steel bombSteel bomb

samplesample

Suppose we wish to measure the energy of Suppose we wish to measure the energy of combustion of octane (C8H18), a component of combustion of octane (C8H18), a component of gasoline. A 0.5269-g sample of octane is placed gasoline. A 0.5269-g sample of octane is placed in a bomb calorimeter known to have a heat in a bomb calorimeter known to have a heat capacity of 11.3 kJ/ºC. capacity of 11.3 kJ/ºC.

This means that 11.3 kJ of energy is required to This means that 11.3 kJ of energy is required to raise the temperature of the water and other parts raise the temperature of the water and other parts of the calorimeter by 1ºC. The octane is ignited in of the calorimeter by 1ºC. The octane is ignited in the presence of excess oxygen, and the the presence of excess oxygen, and the temperature increase of the calorimeter is 2.25ºC. temperature increase of the calorimeter is 2.25ºC.

The amount of energy released is The amount of energy released is calculated as follows:calculated as follows:

Energy released by the reactionEnergy released by the reaction

= ΔT x heat capacity of calorimeter= ΔT x heat capacity of calorimeter

A bomb calorimeter has a heat capacity of A bomb calorimeter has a heat capacity of 9.47 kJ/K. When a 2.01-g sample of 9.47 kJ/K. When a 2.01-g sample of (C3H6) was burned in this calorimeter, the (C3H6) was burned in this calorimeter, the temperature increased by 4.26 K. temperature increased by 4.26 K. Calculate the energy of combustion for the Calculate the energy of combustion for the sample.sample.

A bomb calorimeter has a heat capacity of A bomb calorimeter has a heat capacity of 9.47 kJ/K. When a 2.01-g sample of 9.47 kJ/K. When a 2.01-g sample of (C3H6) was burned in this calorimeter, the (C3H6) was burned in this calorimeter, the temperature increased by 4.26 K. temperature increased by 4.26 K. Calculate the energy of combustion for the Calculate the energy of combustion for the sample.sample.

A 2.200-g sample of quinine (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854kj/ºC. the temperature of the calorimeter plus the contents increased from 23.44ºC to 30.57ºC.

What is the heat of combustion?

per gram of quinine?

Per mole of Quinine?

.5865g sample of lactic acid HC3H5O3 is burned in a calorimeter whose heat capacity is 4.812kj/ºC. The temperature increases from 23.10ºC to 24.95ºC . Calculate the heat of combustion of lactic acid per gram ?

The heat of combustion of pentane, is -131.64 The heat of combustion of pentane, is -131.64 kJ/g. Combustion of 4.50 g of pentane kJ/g. Combustion of 4.50 g of pentane causes a temperature rise of 2.00°C in a causes a temperature rise of 2.00°C in a certain bomb calorimeter. What is the heat certain bomb calorimeter. What is the heat capacity of this bomb calorimeter?capacity of this bomb calorimeter?

The combustion of 0.1584g benzoic acid increase the Temperature of a bomb calorimeter by 2.54°C. The energy °C. The energy released by the combustion is released by the combustion is 26.42kj/g.Calculate the heat capacity 26.42kj/g.Calculate the heat capacity of the bomb Calorimeter .of the bomb Calorimeter .

Standard StatesStandard 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 1

atm and 25°Catm and 25°C..

Hess’s LawHess’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..

Using EnthalpyUsing Enthalpy

Consider the decomposition of waterConsider the decomposition of waterHH22O(g) + O(g) + 286 kJ286 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 = + 286 kJH = + 286 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 all rxns.H’s of the all rxns.


Calculations via Hess’s LawCalculations 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

Calc. Calc. H for H for

S(s) + 3/2 OS(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

SOSO33(g) --> SO(g) --> SO22(g) + 1/2 O(g) + 1/2 O22(g) +75.2 kJ (g) +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



Determine the heat of reaction for the decomposition of one mole of benzene to acetylene

C6H6(l) 3C2H2(g)

given the following thermo chemical equations:

2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O(g) H = -6271 kJ

2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) H = -2511 kJ

See smart Hess Law See smart Hess Law

Using Standard Enthalpy ValuesUsing Standard Enthalpy ValuesUsing Standard Enthalpy ValuesUsing Standard Enthalpy Values

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

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

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 of the system and not how it got there.got there.





Change in EnthalpyChange 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)

Standard Enthalpy ValuesStandard Enthalpy ValuesStandard 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

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

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

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

HHooff, standard molar enthalpy of , standard molar enthalpy of

formationformation

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.


Use Use HHff’s to calculate enthalpy change ’s to calculate enthalpy change H for H for

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

From reference books we findFrom reference books we find

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

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

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

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


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

HHoorxnrxn for for

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

HHooff (-201.5kj)(-201.5kj) (-393.5kj) (-393.5kj) (-241.8kj) (-241.8kj)

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

hh

2Al(s) + Fe2O3(s)2Al(s) + Fe2O3(s) Al2O3(s)+Fe(s)Al2O3(s)+Fe(s)

FeFe22OO33(s) (s) HHoof =-826kj/molf =-826kj/mol

AlAl22OO33 HHoof = 646kj/molf = 646kj/mol

2KIO3 + 12HCl 2KIO3 + 12HCl 2ICl 2ICl +KCL+6H20+4Cl2+KCL+6H20+4Cl2

KIO3= -501KIO3= -501

HCl= -92HCl= -92

ICl= -24ICl= -24

KCL= -435KCL= -435

6H20= -2866H20= -286

The standard enthalpy of combustion of ethane gas The standard enthalpy of combustion of ethane gas C2H4 is C2H4 is –1411.1 kj/mol–1411.1 kj/mol a 298k a 298k

Given the following enthalpy of formation calculate Given the following enthalpy of formation calculate the enthalpy of formation of ethane gas the enthalpy of formation of ethane gas

COCO22HHoof =-393.5kj/molf =-393.5kj/mol

H20H20HHoof =-285.8kj/molf =-285.8kj/mol

06_77

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

06_1551

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.

06_80

Earth’satmosphere

Infraredradiated bythe earth

Earth

CO2

and 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.

Documents

THERMOCHEMISTRY OR THERMODYNAMICS Chapter 6 Energy and Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that