1 All Bold Numbered Problems. 2 Chapter 6 Outline Energy -vs- HeatEnergy -vs- Heat Specific...

All Bold Numbered ProblemsAll Bold Numbered Problems

Chapter 6 OutlineChapter 6 Outline

• Energy -vs- HeatEnergy -vs- Heat

• Specific HeatSpecific Heat

• First Law of ThermodynamicsFirst Law of Thermodynamics

• q (Heat)q (Heat)

• Hess’s LawHess’s Law

THERMOCHEMISTRYTHERMOCHEMISTRYoror

ThermodynamicsThermodynamics

Energy & ChemistryEnergy & Chemistry

• Burning peanuts Burning peanuts supply sufficient supply sufficient energy to boil a cup energy to boil a cup of water.of water.

• Burning sugar Burning sugar (sugar reacts with (sugar reacts with KClOKClO33, a strong , a strong oxidizing agent)oxidizing agent)

Energy and ChemistryEnergy and Chemistry

2 H2 H22(g) + O(g) + O22(g) --> 2 H(g) --> 2 H22O(g) + heat and lightO(g) + heat and light

This can be set up to provide This can be set up to provide

ELECTRIC ENERGYELECTRIC ENERGY in a fuel cell. in a fuel cell.

Oxidation:Oxidation: 2 H 2 H22 ---> 4 H ---> 4 H++ + 4 e + 4 e--

Reduction: 4 eReduction: 4 e-- + O + O22 + 2 H + 2 H22O ---> 4 OHO ---> 4 OH--

Energy and Chemistry

ENERGYENERGY is the capacity to do work is the capacity to do work or transfer heat.or 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 types of energy Other types of energy

• lightlight

• electricalelectrical

• kinetic and potentialkinetic and potential

Kinetic and Potential EnergyKinetic and Potential Energy

Potential energy: Potential energy: energy stored in energy stored in chemical due to chemical due to its structure.its structure.

The energy of a The energy of a motionless body motionless body due to its position.due to its position.

Kinetic and Potential EnergyKinetic and Potential Energy

Kinetic energy: Kinetic energy: energy of motion.energy of motion.

• TranslationalTranslational

• RotationalRotational

• Vibrational Vibrational (IR)(IR)

Thermodynamics

Thermodynamics is the science of heat (energy) transfer.

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

Energy and Chemistry

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

THE CONSERVATION THE CONSERVATION OF ENERGYOF ENERGY..

• The The total energytotal energy of a of a

system is system is constantconstant..

UNITS OF ENERGY

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

James JouleJames Joule1818-18891818-1889

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”)

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”)

The S.I sytem uses The S.I sytem uses the unit called the the unit called the JOULEJOULE1 cal = 4.18 Joules1 cal = 4.18 Joules

Specific heat capacity =

heat lost or gained by substance (J)

(mass, g)(T change, K)

Specific Heat Capacity

Thermochemistry is the science of Thermochemistry is the science of heat (energy) flow. A difference in heat (energy) flow. A difference in temperature leads to energy transfer.temperature leads to energy transfer.

The heat “lost” or “gained” is related to The heat “lost” or “gained” is related to a)a) sample masssample massb) b) change in T change in T c) c) specific heat capacity specific heat capacity

Specific Heat CapacitySpecific Heat Capacity

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

HH22OO 4.184.18

AlAl 0.9020.902

glassglass 0.840.84

AluminumAluminum

CHEMICAL REACTIVITYCHEMICAL REACTIVITYWhat drives chemical reactions? What drives chemical reactions?

How do they occur?How do they 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

CHEMICAL REACTIVITYCHEMICAL REACTIVITYBut ENERGY TRANSFER also allows us to But ENERGY TRANSFER also allows us to

predict reactivity.predict reactivity.

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

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

Heat Energy Transfer in Heat Energy Transfer in Chemical ProcessesChemical Processes

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

FIRST LAW OF THERMODYNAMICS

E = q + wE = q + w

heat energy transferred heat energy transferred (to the system)(to the system)

energyenergychangechange

work donework done(by the (by the surroundings)surroundings)

Energy is conserved!Energy is conserved!

18 The First Law of Thermodynamics is

the law of conservation of energy. E = q + wE = q + w

E is the change in Energy, q is heat, w is workE is the change in Energy, q is heat, w is work

• Enthalpy (H), defined H = E + PV.

H = H = E + PE + PVV at constant pressure (only V and T change). • If w = - PV (because V does not change in a typical lab experiment,

only T changes) and H = q + w + PV

H = qH = qpp (for chemist in a lab which means (for chemist in a lab which means only T changes)only T changes)

ENTHALPYENTHALPY

Most chemical reactions occur at Most chemical reactions occur at constant P. constant P.

qqpp = = HH

H = change in heat content of the systemH = change in heat content of the system

H = HH = Hfinalfinal - H - Hinitialinitial

How do we measure q in the lab?

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

Specific heat capacity Specific heat capacity == heat lost or gained by substance (J)heat lost or gained by substance (J)

(mass,(mass, g)(T change,g)(T change, K)K)

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

HH22OO 4.184.18

AlAl 0.9020.902

glassglass 0.840.84

( )( )( )q C m T C = Specific Heat, units of J/(gK)C = Specific Heat, units of J/(gK)

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

heat gain/lost = q = (C)(mass)(heat gain/lost = q = (C)(mass)(T)T)

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.)K

q = - 6160 Jq = - 6160 J

Heat flows FROM the Heat flows FROM the SYSTEMSYSTEM into the into the

SURROUNDINGSSURROUNDINGS isis an an EXOTHERMICEXOTHERMIC process.process.

q is “-”q is “-”

The reactionThe reactionfeels hot.feels hot.

Energy givenEnergy givenOff!Off!

Heat flows INTO the Heat flows INTO the SYSTEMSYSTEM from the from the

SURROUNDINGSSURROUNDINGS is an is an ENDOTHERMICENDOTHERMIC process.process.

q is “+”q is “+”

The reactionThe reactionfeels cold.feels cold.

Energy Added!Energy Added!

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?

q = - 6160 Jq = - 6160 J

q = - 6.16 kJq = - 6.16 kJ

Notice that the Notice that the negative signnegative sign on q on q

signals heat signals heat “lost by”“lost by” or transferred or transferred

out of Al. Is this Exo or Endothermic?out of Al. Is this Exo or Endothermic?

A 500.0 g piece of metal is heated to 75.0 A 500.0 g piece of metal is heated to 75.0 00C. It C. It is placed into 300. mL of water at 22.0 is placed into 300. mL of water at 22.0 00C and C and the final temperature is 31.0 the final temperature is 31.0 00C. Calculate the C. Calculate the specific heat of the metal.specific heat of the metal.

qqwaterwater + q + qmetal metal = 0= 0

qqwaterwater = -q = -qmetal metal

(300. g)(4.18 J/gK)(9.0(300. g)(4.18 J/gK)(9.0ooC) = -(500.0 g)(c)(-44.0C) = -(500.0 g)(c)(-44.0ooC)C)

c = 0.51 J/gKc = 0.51 J/gK

qqwaterwater + q + qmetal metal = 0= 0

qqwaterwater = -q = -qmetal metal

(300. g)(4.18 J/gK)(9.0(300. g)(4.18 J/gK)(9.0ooC) = -(500.0 g)(c)(-44.0C) = -(500.0 g)(c)(-44.0ooC)C)

c = 0.51 J/gKc = 0.51 J/gK

To 150 g of water at 25oC is added 45 g of Al at 115oC, what will be the final temperature?

Al 0.902Al 0.902

Heat Transfer and Heat Transfer and Changes of StateChanges of State

Changes of state involve energyChanges of state involve energy

Ice -----> WaterIce -----> Water 333 J/g (heat of fusion)333 J/g (heat of fusion)

+ energy+ energy

Exo or Endo? Sign of q?Exo or Endo? Sign of q?

Heat Transfer and Heat Transfer and Changes of StateChanges of State

Liquid ---> VaporLiquid ---> Vapor

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

This is the reasonThis is the reason

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

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

+ energy+ energy

What quantity of heat is required to change What quantity of heat is required to change 10.0 g of ice at -50.0 10.0 g of ice at -50.0 ooC to steam at 200.0 C to steam at 200.0 ooC? C? Endo or Exo? Sign of q?Endo or Exo? Sign of q?

Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g

Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g

Heat and Changes of StateHeat and Changes of StateHeat and Changes of StateHeat and Changes of State

+333 J/g+333 J/g +2260 J/g+2260 J/g

Heating/Cooling Curve for WaterHeating/Cooling Curve for WaterHeating/Cooling Curve for WaterHeating/Cooling Curve for Water

1.1. To heat iceTo heat ice

q = (10.0 g)(2.09 J/g•K)(50.0K) = 1050Jq = (10.0 g)(2.09 J/g•K)(50.0K) = 1050J

2.2. To melt iceTo melt ice q = (10.0 g)(333 J/g) = 3330J q = (10.0 g)(333 J/g) = 3330J 3.3. To heat waterTo heat water q = (10.0 g)(4.18 J/g•K)(100.K) = 4180Jq = (10.0 g)(4.18 J/g•K)(100.K) = 4180J4.4. To evaporate waterTo evaporate water q = (10.0 g)(2260 J/g) = 22600Jq = (10.0 g)(2260 J/g) = 22600J5.5. To heat steamTo heat steam q = (10.0 g)(2.03 J/g•K)(100.0K) = 2030Jq = (10.0 g)(2.03 J/g•K)(100.0K) = 2030J

Total heat energy = 33190.J = 33.190kJTotal heat energy = 33190.J = 33.190kJ

Calculate the amount of heat energy necessary Calculate the amount of heat energy necessary to change 25.0 g of copper solid at 925to change 25.0 g of copper solid at 925o o C to C to liquid at 1083liquid at 1083o o C. Melting point = 1083C. Melting point = 1083o o C, C,

ccsolidsolid = 0.382 = 0.382 J/gJ/g..K, K, Heat of fusion = 205 Heat of fusion = 205 J/g.J/g.

qq11 = (25.0 g)(.382 J/gK)(1083 = (25.0 g)(.382 J/gK)(1083ooC-925C-925ooC)= 1510 JC)= 1510 J

qq2 2 = (25.0 g)(205 J/g) = 5120 J= (25.0 g)(205 J/g) = 5120 J

qqTT = q = q1 1 + q+ q22 = 6630 J = 6630 J

Endo- and ExothermicEndo- and Exothermic

qqsystemsystem > 0 > 0

ENDOTHERMICENDOTHERMICHeat goes in the systemHeat goes in the system

T(system) goes upT(system) goes up

qqsystemsystem < 0 < 0

EXOTHERMICEXOTHERMICHeat leavesHeat leavesthe systemthe system

T(system) goes downT(system) goes down

Consider the decomposition of waterConsider the decomposition of water

HH22O(g) + O(g) + 242 kJ242 kJ ---> H ---> H22(g) + 1/2 O(g) + 1/2 O22(g)(g)

Endo or Exo?Endo or Exo?

Endothermic reaction Endothermic reaction

Heat is a “reactant”Heat is a “reactant”

H = + 242 kJH = + 242 kJ

USING ENTHALPYUSING ENTHALPY

Making HMaking H22 from H from H22O involves two steps.O involves two steps.

Each step requires energy.Each step requires energy.

Liquid HLiquid H22OO

HH22 + O + O22 gas gas

HH22O vaporO vapor

Making HMaking H22 from H from H22O involves two steps.O involves two steps.

HH22O(liq) + 44 kJ ---> HO(liq) + 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(liq) + 286 kJ --> HO(liq) + 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 If a rxn. is the sum of 2 or more others, the net others, the net H is the sum of H is the sum of the the H’s of the other rxns.H’s of the other rxns.

Calc. Calc. H for: SH for: S(s)(s) + + 3/23/2 O O2(g)2(g) --> SO --> SO3(g)3(g)

Given:Given:

SS(s)(s) + O + O2(g)2(g) --> SO --> SO2(g)2(g) HH11 = -320.5 kJ = -320.5 kJ

SOSO2(g)2(g) + + 1/21/2 O O2(g)2(g) --> SO --> SO3(g)3(g) HH22 = -75.2 kJ = -75.2 kJ

The two equations add up to give the The two equations add up to give the

desired equation, so -desired equation, so -

HHnetnet = = HH11 + + HH22 = -395.7 kJ = -395.7 kJ

S solid

SO3 gas

SO2 gas

direct path

+ 3/2 O2

H = -395.7 kJ

ENERGY

H 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

H along one path =H along one path =

H along another pathH along another path

HH along one path = along one path = HH along another path along another pathHH along one path = along one path = HH along another path along another path

• This equation is valid This equation is valid because because HH is a is a STATE STATE FUNCTIONFUNCTION

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

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

• Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot measure absolute H. We can measure absolute H. We can only measure only measure H.H.

Standard Enthalpy ValuesStandard Enthalpy Values

Most Most H values are labeled H values are labeled ΔHHoo

Measured under Measured under standard conditionsstandard conditions

P = 1 atmosphere or 1 bar, P = 1 atmosphere or 1 bar, (10(1055Pascals)Pascals)

Concentration = 1 mol/LConcentration = 1 mol/L

T = 25 T = 25 ooCC

with all species in standard stateswith all species in standard statese.g., C = graphite and Oe.g., C = graphite and O22 = gas, etc. = gas, etc.

NIST (National Institute for Standards and NIST (National Institute for Standards and Technology) gives values ofTechnology) gives values of

HHooff = standard molar enthalpy of = standard molar enthalpy of

formationformation

This is the enthalpy change when 1 mole of This is the enthalpy change when 1 mole of compound is formed from elements under compound is formed from elements under standard conditions.standard conditions.

See Table 6.2 and Appendix LSee Table 6.2 and Appendix L

Standard Enthalpy ValuesStandard Enthalpy Values

Hof, standard molar

enthalpy of formationHo

f, standard molar enthalpy of formation

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 = 0

for elements in their for elements in their

standard states.standard states.

Using Standard Enthalpy Values

Use Use HHf’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”)”)

From reference books we find:From 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 O vaporvapor = - 241.8 kJ/mol = - 241.8 kJ/mol

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

HH ff of CO = - 110.5 kJ/mol of CO = - 110.5 kJ/mol

HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) (g) HHoo = = ++241.8kJ241.8kJ

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g)(g) --> CO(g) HHoo = -110.5 kJ = -110.5 kJ

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

HHoonetnet = +131.3 kJ = +131.3 kJ

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

HH22 and CO and CO requiresrequires 131.3 kJ of energy. 131.3 kJ of energy.

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

Using Standard Enthalpy ValuesUsing Standard Enthalpy Values

In general, when In general, when ALLALL enthalpies enthalpies

of formation are known, of formation are known,

HHoorxnrxn = =

HHoof f (products) - (products) - HHoo

f f (reactants)(reactants)

Calculate Calculate H of reaction?H of reaction?

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

methanol, i.e., 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)

HHoorxnrxn = = HHoo

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

Use Appendix LUse Appendix L

49Using Standard Enthalpy Values

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 + (-200.7 kJ) }- { 0 + (-200.7 kJ) }

HHoorxnrxn = - 676.4 kJ per mol of methanol = - 676.4 kJ per mol of methanol

OF2(g) + H2O(g) --> O2(g) + 2 HF(g) Horxn =- 318.0kJ

Hof (H2O) = - 241.8 kJ and Ho

f (HF) = - 271.1 kJ

Calculate Hof (OF2)

Horxn = {Ho

f (O2) + 2 Hof (HF) }

- { Hof (OF2) + Ho

f (H2O) }

-318.0 kJ = {( 0 ) + 2 (-271.1 kJ) }

- {Hof (OF2) + (-241.8 kJ) }

Hof (OF2) = 17.6 kJ /mole

51Measuring Heats of Reaction with aBOMB CALORIMETRY

Coffee CupCalorimeter

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 1300. g waterCalorimeter contains 1300. g water

• Heat capacity of bomb = 427 J/KHeat capacity of bomb = 427 J/K

Measuring Heats of ReactionMeasuring Heats of Reaction

CALORIMETRYCALORIMETRY

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

q = (4.18 J/g•K)(1300. g)(8.20 K) = 44,600 Jq = (4.18 J/g•K)(1300. g)(8.20 K) = 44,600 J

Step 2Step 2 Calc. heat transferred from reaction to Calc. heat transferred from reaction to

bomb. bomb.

q = (bomb heat capacity)(q = (bomb heat capacity)(T)T)

= (427 J/K)(8.20 K) = 350= (427 J/K)(8.20 K) = 3500 J J

Step 3Step 3 Total heat evolved = Total heat evolved =

44,600 J + 35044,600 J + 3500 J = 48,100 J J = 48,100 J

Heat of combustion of 1.00 g of octane = Heat of combustion of 1.00 g of octane = -- 48.1 kJ 48.1 kJ

CALORIMETRYCALORIMETRYCALORIMETRYCALORIMETRY

Calculate heat of combustion of ethanol.Calculate heat of combustion of ethanol.

CC22HH55OH + 3 OOH + 3 O22 --> 2 CO --> 2 CO22 + 3 H + 3 H22OO

•• Burn 0.888 g of ethanolBurn 0.888 g of ethanol

• Temp rises from 25.00 Temp rises from 25.00 ooC to 35.00 C to 35.00 ooCC

• Calorimeter contains 210. g waterCalorimeter contains 210. g water

• Heat capacity of bomb = 0.950 kJ/KHeat capacity of bomb = 0.950 kJ/K

q = (4.18 J/g•K)(210. g)(10.00 K) = 8.80 kJq = (4.18 J/g•K)(210. g)(10.00 K) = 8.80 kJ

bomb. bomb.

= (0.950 kJ/K)(10.00 K) = 9.50 kJ= (0.950 kJ/K)(10.00 K) = 9.50 kJ

8.80 kJ + 9.50 kJ = 18.30 kJ8.80 kJ + 9.50 kJ = 18.30 kJ

(-18.30 kJ/0.888g)(46.0 g/mole) = (-18.30 kJ/0.888g)(46.0 g/mole) = - - 948 kJ/mole948 kJ/mole

A 0.105 g sample of C2H4 is burned in a

calorimeter with a heat capacity of 0.47 kJ/oC and containing 2,000. g of water. The temperature increased from 25.00oC to 27.14oC. Calculate the heat of combustion of C2H4 in kJ/mole.

Sample ProblemSample Problem

q = (4.18 J/g•K)(2000. g)(2.14 K) = 17,900 Jq = (4.18 J/g•K)(2000. g)(2.14 K) = 17,900 J

bomb. bomb.

= (0.47 kJ/K)(2.14 K) = 1.0 kJ= (0.47 kJ/K)(2.14 K) = 1.0 kJ

17.9 kJ + 1.0 kJ = 18.9 kJ17.9 kJ + 1.0 kJ = 18.9 kJ

(-18.9 kJ/0.105g)(28.0 g/mole) = (-18.9 kJ/0.105g)(28.0 g/mole) = - - 5040 kJ/mole5040 kJ/mole

Practice Problems

1. How many kcal is 75.0 kJ?

2. Convert 88.6 kcal/mole to kJ/mole.

3. Calculate the kinetic energy of a 2000. lb car traveling at 50. miles per hour.

4. Is the kinetic energy of a 500. g object moving at 50. km/hr greater, less than, or the same as that of a 1.00 kg object moving at 25 km/hr?

5. How much heat is required to raise the temperature of 204 grams of lead from 22.8oC to 64.9oC? (C= 0.038 cal/gK)

Practice Problems

6. If the specific heat of silver is 0.0573 cal/gK, how much heat would it take to raise the temperature of 1 kg of silver from -50.oC to 150.oC?

7. To what temperature will one pound of nickel be raised, beginning at 25.0oC, if 750 calories are absorbed by it? (c = 0.106 cal/gK)

8. A 50. g chunk of unknown metal X is heated to 98.5oC and then dropped into 450 g of water initially at 25.00oC. The water temperature is observed to rise to 26.47oC. Calculate the specific heat of X.

Practice Problems

9. 2 H2O (l) --> 2 H2 (g) + O2 (g) H = 273.2 kcal

How many mLs of water will be decomposed to hydrogen and oxygen by 85 kcal?

10. CaCO3 (s) --> CaO (s) + CO2 (g) H = 42.5 kcal

How much heat is required to react 186 g CaCO3?

11. C6H12O6 (s) + 6 O2 (g) --> 6 H2O (g) + 6 CO2 (g)

H = -673 kcal. How much heat is evolved in the oxidation of 240 g of C6H12O6?

12. MgCO3(s)+2 HCl(g)->MgCl2(s)+CO2(g) + H2O(g) H = -5.6 kcal. How many grams of MgCl2 will be produced if 80.0 kilocalories are released?

Practice Problems

13. 3 Cu2S(s) + 16 HNO3(l) --> 6 Cu(NO3)2(s) +

3 S(s) + 4 NO(g) + 8 H2O(l) H = -543 kcal/mole

How much energy will be released by the reaction of 325 grams of Cu2S with HNO3.

14. 2 B5H9 (l) + 12 O2 (g) --> 9 H2O (l) + 5 B2O3 (s)

H = -4188 kcal/mole. How much B5H9 must be burned to produce 4500 kcal of energy?

15. Upon dissolving 8.63 g of ammonium dihydrogen phosphate in 100. mL of water, it was found that T= - 2.7oC. Calculate the value of the molar heat of reaction.

Practice Problems16. A 0.25 mole sample of ammonium nitrate was

dissolved in 500. mL of water at 21.0oC. As the solute dissolved the temperature of the solution dropped to a minimum value of 18.0oC. Calculate the heat of reaction in kcal/mole.

17. Given the following data: H(kcal)

MnO2 + CO --> MnO + CO2 -36.0

Mn3O4 + CO --> 3 MnO + CO2 -13.0

3 Mn2O3 + CO --> 2 Mn3O4 + CO2 -34.0

Find H in kcal for the following reaction:

2 MnO2 + CO --> Mn2O3 + CO2

Practice Problems

18. Given the following two reactions: H(kcal)

2 Na(s) + 2HCl(g) --> 2 NaCl(s) + H2(g) -152.34

H2(g) + Cl2(g) --> 2 HCl(g) -44.12

What is Hfo,in kcal/mole for NaCl(s)?

19. Calculate the Hfo of glycerol,

(CH2OH)2CHOH, knowing that its heat of combustion is -397.0 kJ/mole.

20. Find the H for the reaction:

2 SO2(g) + 2 H2O(g) + O2(g) --> 2 H2SO4 (l)

Practice Problems

21. Determine the enthalpy of formation in kcal/mole for chloroethane, C2H5Cl, from the following data:

H(kcal)

H2 + 1/2 O2 --> H2O -68.3

C + O2 --> CO2 -94.1

C2H5Cl --> C2H4 + HCl 17.2

2 CO2 + 2 H2O --> C2H4 + 3 O2 337.2

H2 + Cl2 --> 2 HCl -44.2

Practice Problems

22. The combustion of 0.450 g of ethene (C2H4) causes a temperature rise of 2.00oC in a bomb calorimeter that has a specific heat capacity of 600. cal/oC and containing 500. g of water. What is the heat of combustion in kJ/mole for ethene?

Practice Problems23. A bomb calorimeter has a heat capacity of

385 J/oC and containing 200. g of water. A temperature rise of 3.577oC is observed when the calorimeter is used in the combustion of a 0.7600 g sample of powdered tantalum in excess oxygen to product Ta2O5. Find Hf

o of Ta2O5.

24. How many calories would be required to change the temperature of 750 g of water from 15.0 oC to 90.0 oC?

25. How many joules of heat would be liberated if the temperature of 300. g of iron were changed from 75 oC to 17 oC?

Practice Problems

26. The heat of formation of water is -68.4 kcal/mole. How many joules would be liberated by burning 12 g of hydrogen?

27. 4 Fe + 3 O2 --> 2 Fe2O3 + 398 kcal

4 Al + 3 O2 --> 2 Al2O3 + 798 kcal

Calculate H for the reaction

Fe2O3 + 2 Al --> Al2O3 + 2 Fe

Practice Problems

28. A 0.757 g sample of C was placed in a calorimeter with a specific heat capacity of 1550 cal/oC and containing 1,000. G of water and burned. The temperature rose from 22.54 oC to 24.87 oC. Calculate the heat of combustion in kcal/mole?

Practice Problems Answers

1. 17.9 kcal 2. 370. kJ/mole

3. 2 x 105 J 4. Greater

5. 330 cal 6. 10,000 cal

7. 41oC 8. .8 J/gK

9. 11 mL 10. 79.0 kcal

11. 9.0 x 102 kcal 12. 1400 g

13. 370. Kcal 14. 140 g

15. 15,000 J/mole 16. 6.0 kcal/mole

17. -52.0 kJ 18. -98.23kcal/mole

Practice Problems Answers

19. -1750.8 kJ 20. -550.682 kJ

21. -26.9 kJ 22. -572 kJ/mole

23. -2080 kJ/mole 24. 56000 cal

25. 7800 J 26. 1.7 x 106 J

27. -200 kcal 28. -94.2 kcal/mole

How much heat is required to raise the temperature of 57 g of water from 25.6oC to 66.5oC?

q = (57 g)(4.18 J/g K)(40.9oC)

q = 9700 J

Energy and ReactionsEnergy and Reactions

Reaction-->

Endothermic Exothermic

A + B + energy -->C + DA + B + energy -->C + D W + X -->Y + Z + energyW + X -->Y + Z + energy

A+BA+B Y+ZY+Z

C+DC+D W+XW+X

1. The reaction of 14 g of hydrogen will produce how much energy?

3 H2 + N2 --> 2 NH3 + 92 kJ

14 g ? kJ

= 210 kJ14 g H2 mole H2 92 kJ 2.0 g H2 3 mole H2

2. If 126 kJ are produced how many grams of ammonia are produced?

3 H2 + N2 --> 2 NH3 + 92 kJ

? g 126 kJ

126 kJ 2 mole NH3 17.0 g NH3

92 kJ mole NH3

NN22 (g) + O (g) + O22 (g) --> 2 NO (g) (g) --> 2 NO (g) H = 180.8 kJH = 180.8 kJ

NN22 (g) + 2 O (g) + 2 O22 (g) --> 2 NO (g) --> 2 NO22 (g) (g) H = 67.7 kJH = 67.7 kJ

Find Find H for the following equation:H for the following equation:

2 NO (g) + O2 NO (g) + O22 (g) --> 2 NO (g) --> 2 NO22 (g) (g)

2 NO (g) --> N2 NO (g) --> N22 (g) + O (g) + O22 (g) (g) H = -180.8 kJH = -180.8 kJ

NN22 (g) + 2 O (g) + 2 O22 (g) --> 2 NO (g) --> 2 NO22 (g) (g) H = 67.7 kJH = 67.7 kJ

Hess’s LawHess’s Law

2 NO (g) + O2 NO (g) + O22 (g) --> 2 NO (g) --> 2 NO22 (g) (g) H = -113.1 kJH = -113.1 kJ

2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O H = -3394 kJ

2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O H = -3120. kJ

Find H for the following equation:

C2H4 + H2 --> C2H6

2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O H = -3394 kJ

4 CO2 + 6 H2O --> 2 C2H6 + 7 O2 H = +3120. kJ

C2H4 + H2 --> C2H6 H = -137 kJH = -137 kJ

1 1 7/2 2 3 -1697 kJ1 1 7/2 2 3 -1697 kJ

2 3 1 7/2 +1560. kJ2 3 1 7/2 +1560. kJ

1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 H = -101 kJ

2 NH + H2 --> N2H4 H = -567 kJ

2 N2O5 --> 2 N2 + 5 O2 H = 23 kJFind H for the following equation:

N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH

1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 H = -101 kJ

N2H4 --> 2 NH + H2 H = 567 kJ

2 N2O5 --> 2 N2 + 5 O2 H = 23 kJ

N2H4 + 2 N2O5 --> 2 HNO3 + 2 NO2 + 2 NH

H = 388 kJH = 388 kJ

1 2 5 2 2 -202 kJ1 2 5 2 2 -202 kJ

Calculate the H for the following reaction:

Al2O3 (s) + 3 CO (g) --> 3 CO2 (g) + 2 Al (s)

Horxn = { 3 Ho

f (CO(CO22)) + 2 Hof (Al)(Al) }

- { Hof (Al(Al22OO33)) + 3 Ho

f (CO)(CO) }

= { 3 (-393.5 kJ) + 2 ( 0 )}

- { -1675.7 kJ + 3 (-110.5 kJ) }

Horxn = 826.7 kJ

C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) H = -1323.0 kJ

2 C + 2 H2 --> C2H4 H = 52.3 kJ

H2 + 1/2 O2 --> H2O (g) H = -241.8

Calculate Hf for CO2.

C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) H = -1323.0 kJ

2 C + 2 H2 --> C2H4 H = 52 kJ

H2O (g) --> H2 + 1/2 O2 H = 241.8 kJ C + O2 --> CO2

HHff = -393.7 kJ = -393.7 kJ

1/2 3/2 1 1 -661.5 kJ1/2 3/2 1 1 -661.5 kJ

1 1 1/2 26.0 kJ1 1 1/2 26.0 kJ

C2H4 + 3 O2 --> 2 CO2 + 2 H2O (g) H = -1323.0 kJ

2 C + 2 H2 --> C2H4 H = 52.3 kJ

H2 + 1/2 O2 --> H2O (g) H = -241.8 kJ

Calculate Hf for CO2. ff

Horxn = { 2 Ho

f (CO(CO22)) + 2 Hof (H(H22O)O) }

- {Hof (C(C22HH44)) + 3 Ho

f (O(O22)) }

-1323.0 kJ = { 2 Hof (CO2) + 2 ( -241.8 kJ )}

- { 52.3 kJ + 3 ( 0 ) }Ho

f (CO(CO22)) = -393.6 kJ

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