42
Ch. 17: Thermochemistry 17.1The Flow of Energy (Heat and Work) exothermic/endothermic calorie/joule heat capacity/specific heat 17.2Measuring and Expressing Enthalpy Changes calorimetry enthalpy thermochemical equation 17.3Heat in Changes of State heat of fusion/solidification heat of vaporization/condensation Honor’s only 17.4Calculating Heats of Reaction

Ch. 17: Thermochemistry 17... · 17.4—Calculating Heats of Reaction. 17.1The Flow of Energy - Heat & Work The temperature of lava from a volcano ranges from 550°C to 1400°C. As

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Ch 17 Thermochemistry

171ndash The Flow of Energy (Heat and Work)

exothermicendothermic

caloriejoule

heat capacityspecific heat

172ndash Measuring and Expressing Enthalpy Changes

calorimetry

enthalpy

thermochemical equation

173ndash Heat in Changes of State

heat of fusionsolidification

heat of vaporizationcondensation

Honorrsquos only

174mdash Calculating Heats of Reaction

171 The Flow of Energy - Heat amp Work

The temperature of lava

from a volcano ranges

from 550degC to 1400degC As

lava flows it loses heat

and begins to cool You

will learn about heat flow

and why some substances

cool down or heat up

more quickly than others

Thermochemistry

Energy Transformations

Energy Transformations

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

171 The Flow of Energy - Heat amp Work

The temperature of lava

from a volcano ranges

from 550degC to 1400degC As

lava flows it loses heat

and begins to cool You

will learn about heat flow

and why some substances

cool down or heat up

more quickly than others

Thermochemistry

Energy Transformations

Energy Transformations

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Thermochemistry

Energy Transformations

Energy Transformations

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Energy Transformations

Energy Transformations

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Energy Transformations

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heat Flow q

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heat Flow

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Specific heat typical unit of measure is Jg˚C

degrees Celsius is used

Specific Heat

q = mcΔT

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Water releases a lot of heat as it cools During freezing weather farmers protect citrus crops by spraying them with water

Because it is mostly

water the filling of a hot

apple pie is much more

likely to burn your

tongue than the crust

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Measuring amp Expressing Enthalpy Changes

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Calorimetry

qsys = -qsurr

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Enthalpy

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Calorimeter

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Sample Problem

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Thermochemical Equations

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Thermochemical Equations

Exothermic Reaction

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Thermochemical Equations

Endothermic Reaction

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

172 Sample Problem

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Thermochemical Equations

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Try It

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

17 3 Heat in Changes of State

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heats of Fusion and Solidification

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heats of Fusion and Solidification

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heats of Vaporization and

Condensation

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heats of Vaporization amp

Condensation

∆Hvap = ndash∆Hcond

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Heats of Vaporization and Condensation

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Phase Change Equations and Constants

Heat Constants for Watercs = 2060 JgdegCHfusi = 334 Jgcl = 4180 JgdegCHvap = 2260 JgCg = 187 JgdegC

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

ΔHfus is given in textbook

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Try It

Calculate the amount of heat absorbed to liquefy 150 g of methanol (CH3OH) at its melting point The molar heat of fusion for methanol is 316 kJmol

a) 148 kJ

b) 474 kJ

c) 152 103 kJ

d) 475 kJ

Convert 15 g to moles

molg

molg468750

32

15

Use Qfusion = fusionHm

(046875)(316)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Honorrsquos only

174 Calculating Heats of Reaction

Formation of gems like diamonds and emeralds

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Hessrsquos Law

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Hessrsquos Law in Action

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Hessrsquos Law conti

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Standard Heats of Formation ΔHf˚

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)

Standard Heats of Reaction

ΔH˚ = ΔHf˚(products ) - ΔHf˚ (reactants)