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)