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Chapter 6
Thermochemistry
Chapter 6
Chapter 6
Thermochemistry
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
Chemical Hand Warmers
The Nature of Energy: Key Definitions
The First Law of Thermodynamics: There is no Free Lunch
Quantifying Heat and Work
Measuring DE for Chemical Reactions: Constant – Volume
Calorimetry
Enthalpy: The Heat Evolved in a Chemical Reaction at
Constant Pressure
Constant – Pressure Calorimetry: Measuring DHrxn
Relationships Involving DHrxn
Determining Enthalpies of Reaction from Standard Enthalpies
of Formation
Energy Use and the Environment
2
Chapter 6
Thermochemistry
Matter vs Energy
•
Chapters 1 – 5 have mainly been
concerned with matter
–
–
•
How we describe it
How we measure it
This chapter describes energy
–
–
How we describe it
How we measure it
3
Section 6.1
Chemical Hand Warmers
Thermochemistry
• The study of the relationships between matter
and energy is a field called thermochemistry
• 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s) + heat
– Oxidation of iron
• This reaction takes place inside chemical heat
packs
• Notice heat is written as a product
4
Section 6.1
Chemical Hand Warmers
Thermochemistry
• Another familiar reaction we have seen
• 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) + heat
– Oxidation of octane
• This reaction also generates heat
• The heat is used to do work – we will talk about
this in this chapter
5
Section 6.2
The Nature of Energy: Key Definitions
The Transfer of Energy: Heat and Work
• Energy is something an object has
• Heat and work are ways energy is transferred
• When you hold a warm object in you hand,
energy is transferred from the object to your
hand in the form of heat
• When a ball on a pool table hits another ball
energy is transferred from the first ball to the
second ball in the form of work
6
Section 6.2
The Nature of Energy: Key Definitions
Potential vs Kinetic Energy
• Potential energy is stored energy. Energy of
position.
– The water in a reservoir behind a dam, an
automobile poised to coast downhill, and a coiled
spring have potential energy waiting to be released.
• Kinetic energy is the energy of motion.
– When the water falls over the dam and turns a
turbine, when the car rolls downhill, or when the
spring uncoils and makes the hands on a clock
move, the potential energy in each is converted to
kinetic energy.
7
Section 6.2
The Nature of Energy: Key Definitions
Potential vs Kinetic Energy in Chemistry
• Chemical Potential energy is also stored
energy – the energy stored in chemical bonds
–
–
–
–
CO2 + H2O 
Low Ep
More stable
Lower energy
sugars
hi Ep
less stable
higher energy
• Chemical Kinetic energy is the energy of
motion of the particles
– what we measure when we measure temperature
8
Section 6.2
The Nature of Energy: Key Definitions
The Law of Conservation of Energy
• Law of conservation of energy – energy can
be neither created nor destroyed
– Can assume different forms
• Energy from iron and oxygen in our heat pack example
becomes heat
• Energy from octane becomes heat that is converted to work
• Energy from the sun is converted to complex sugars in plants
9
Section 6.2
The Nature of Energy: Key Definitions
The Law of Conservation of Energy
• Law of conservation of energy – energy can
be neither created nor destroyed
– Can be transferred from one object to another
• In this chapter we are going to learn how to
quantitatively measure the transfer of energy
as either heat or work
10
Section 6.2
The Nature of Energy: Key Definitions
System vs Surroundings
• Energy is exchanged between the system under
investigation and the surroundings
• System – part of the universe on which we wish
to focus attention
– Kind of a weird definition can be just the chemicals
reacting or can include the beaker and liquids in it
• Surroundings – everything with which the
system exchanges energy
• The definition isn’t really the most important thing
– Understanding how they exchange Energy is important
11
Section 6.2
The Nature of Energy: Key Definitions
Exothermic vs Endothermic Reaction
•
Exothermic Reaction
–
–
•
Endothermic Reaction
–
–
•
Heat flows out of the system to the
surroundings
Amount lost by system exactly the same as
amount gained by surroundings *
Heat flows is into a system from the
surroundings.
Amount gained by system is exactly the same
as amount lost by surroundings*
* What law makes this true?*
12
Section 6.3 (preview)
The First Law of thermodynamics: There is no Free Lunch
Heat and Work: Pathways to Energy Change
• A system can exchange energy with its
surrounding through heat, work, or both.
• The change in Energy is the sum of the heat
exchanged and work done DE = q + w
13
Section 6.2
The Nature of Energy: Key Definitions
Units of Energy
• In Chapter 5 we saw that Kinetic Energy was
defined as ½ mv2 where m = mass and v =
velocity
2
The SI unit of Energy is mv
2
m
= kg
 1J
 s 
• 1 J = the joule

• Another
common unit is the calorie (cal)
– 1 cal = 4.184 Joules
• The Calorie we see on nutritional labels = 1000
cal = kilocal
14
Section 6.2
The Nature of Energy: Key Definitions
Energy Conversion Factors
• Don’t memorize these!
• Be able to use them.
15
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Internal Energy
• The Internal Energy of a system is the sum of
all the kinetic and potential energies of the
particles that compose the system.
• Potential energy is the energy stored in the
chemical bonds
• Kinetic energy is the energy of the motion of the
particles
16
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Internal Energy is a State Function
• A state function is a quantity whose value
depends only on the state of the system not on
the pathway the system arrived in that state.
• What does this mean?
17
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
What is a state function
• The change in altitude is a state function.
• The pathway is not
18
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Change in Internal Energy
• Just like a change in altitude a change in internal
energy is a state function.
 DE = Efinal – Einitial
• In a chemical system the reactants are the initial
state (ground level) and the products are the
final state (peak)
 DE = Eproducts – Ereactants
19
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Change in Internal Energy
• For the reaction C (s) + O2 (g)  CO2 (g)
• We can represent the DE like this
• Since DE = Eproducts – Ereactants DE is negative
20
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Change in Internal Energy
• If the system (the reaction) loses energy
 DE is negative
• Where does the energy go?
• Energy has to be conserved (1st Law of TD)
• Energy is released to the surroundings
21
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Change in Internal Energy
• For the reverse reaction CO2 (g)  C (s) +
O2(g)
• The reaction diagram would look like this
• Since DE = Eproducts – Ereactants DE is positive
22
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Change in Internal Energy
• If the system (the reaction) gains energy
 DE is positive
• Energy is absorbed from the surroundings
23
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Concept Check
Hydrogen gas and oxygen gas react
explosively to form water.
2H 2 (g) + O2 (g)  2H 2O(g) + energy (heat)

Which is lower in energy: a mixture
of hydrogen and oxygen gases, or
water?
24
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Solution
Hydrogen gas and oxygen gas react
explosively to form water.
2H 2 (g) + O2 (g)  2H 2O(g) + energy (heat)

Which is lower in energy: a mixture
of hydrogen and oxygen gases, or
water?
Water is lower in energy because a lot of energy was released
when hydrogen and oxygen gases reacted.
25
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Summarizing Energy Flow
• In both cases the DEsys is the opposite of the DEsurr
 DEsys = – DEsurr
• If reactants have higher internal energy than
products
 DEsys is negative energy flows out of the system
• If reactants have lower internal energy than
products
 DEsys is positive and energy flows into the system
26
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Heat and Work: Pathways to Energy Change
• A system can exchange energy with its
surrounding through heat, work, or both.
• The change in Internal Energy is the sum of the
heat exchanged and work done DE = q + w
27
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Heat and Work: Pathways to Energy Change
• When we looked at the change in altitude on the
mountain we saw that the two pathways were
very different.
– The change in altitude was a state function
– The pathways were not state functions
• Same situation for heat and work
 DE = heat + work = q + w
 DE is a state function
– q and w are not
28
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Heat and Work: Pathways to Energy Change
 DE is a state function
– q and w are not
• Think about a gallon of gas
– Burning in an open container (lots of heat, not much
work)
– Moving a car 40 miles (lots of work, less heat)
• Same change in DE but q and w are completely
different depending on the path
29
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Sign conventions for DE, q and w
 DE with no subscript always refers to DE of the
system
– Negative sign means the system loses energy
• What about heat and work?
• Signs are defined from the point of view of the
system.
–
–
–
–
If the system does work w is negative
If the system loses heat q is negative
If work on done on the system w is positive
If the system gains heat q is positive
30
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Concept Check
Determine the value of DE for the
combustion of 1 gallon of gasoline in a
car engine
q = – 8.89 x 104 kJ
w = – 3.81 x 103 kJ
31
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Solution
Determine the value of DE for the combustion
of 1 gallon of gasoline in a car engine
DE = q + w = – 8.89 x 104 kJ + – 3.81 x 103 kJ
DE = – 9.27 x 104 kJ
32
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Concept Check
In this problem for the determination of
the value of DE for the combustion of 1
gallon of gasoline in a car engine, why
are the values of q and w both negative.
q = – 8.89 x 104 kJ
w = – 3.81 x 103 kJ
33
Section 6.3
The First Law of thermodynamics: There is no Free Lunch
Solution
In this problem for the determination of
the value of DE for the combustion of 1
gallon of gasoline in a car engine, why
are the values of q and w both negative.
q = – 8.89 x 104 kJ combustion
produces heat which leaves the system
w = – 3.81 x 103 kJ work is done on the
surroundings
34
Section 6.4
Quantifying Heat and Work
How do we measure Heat and Work
• How do we measure the amount of heat gained
or released or the amount of work done on a
system, or performed by a system?
• We measure changes in temperature and
volume.
• Then use a bunch of equations.
• Lets do heat first.
35
Section 6.4
Quantifying Heat and Work
Heat
• Heat is the exchange of thermal energy between
a system and its surroundings caused by a
temperature difference.
• Heat and temperature are not the same thing.
• Although sometimes we talk about them as if
they were.
• Temperature is a measure of thermal Energy.
• Heat is a transfer of thermal Energy
– Heat is not a substance contained by an object,
although we often talk of heat as if this were true.
36
Section 6.4
Quantifying Heat and Work
Temperature Change and Heat Capacity
•
Substances respond differently to being
heated
 Think of a pool vs a car door on a hot
day
 Metal absorbs heat fast
 Water absorbs heat slowly
•
The Heat capacity (C) is a measure of this
property
C
heat absorbed
increase in tempera ture
37
Section 6.4
Quantifying Heat and Work
Temperature Change and Heat Capacity
heat absorbed
C
increase in tempera ture
• Metal absorbs heat fast
•
Has a low heat capacity
• Water absorbs heat slowly
•
Has a high heat capacity
38
Section 6.4
Quantifying Heat and Work
Heat Capacity
•
Specific heat capacity (Cs):
 The energy required to raise the
temperature of one gram of a
substance by one degree Celsius.
• Molar heat capacity (Cm):
 The energy required to raise the
temperature of one mole of substance
by one degree Celsius.
39
Section 6.4
Quantifying Heat and Work
Specific Heat Capacities
• Specific Heat Capacities
(per g) and Molar heat
capacities (per mole) are
called intensive properties
• Intensive properties of
matter depend on the
identity of the matter – not
the amount
40
Section 6.4
Quantifying Heat and Work
Heat Calculations
• Specific Heat Capacity is used to determine the
relationship between the amount of heat added
to a system and the corresponding change in
temperature.
 DT is always Tfinal - Tinitial
41
Section 6.4
Quantifying Heat and Work
Learning Check
A piece of iron with a mass of 75.0 g at 125.0 °C is allowed to
cool to room temperature of 25.0 °C . Determine the magnitude
and sign of q. The specific heat capacity of iron is 0.449 J/°C g.
42
Section 6.4
Quantifying Heat and Work
Solution
A piece of iron with a mass of 75.0 g at 125.0 °C is allowed to
cool to room temperature of 25.0 °C . Determine the magnitude
and sign of q. The specific heat capacity of iron is 0.449 J/°C g.
q = m × Cs × ΔT
m = 75.0 g
Cs = 0.449 J/°C g.
DT = 25.0 – 125.0 = -100.0°C
0.449 J
q = 75.0 g x 
x - 100.0  C = - 3.37x10 3 J
Cg
43
Section 6.4
Quantifying Heat and Work
Learning Check
The specific heat capacity of silver is 0.24 J/°C g.
a.Calculate the energy required to raise the temperature of 150.0g
Ag from 273 K to 298 K.
b.Calculate the energy required to raise the temperature of 1.0
mol Ag by 1.0°C (called the molar heat capacity of silver).
44
Section 6.4
Quantifying Heat and Work
Solution
The specific heat capacity of silver is 0.24 J/°C °g.
a.Calculate the energy required to raise the temperature of 150.0g
Ag from 273 K to 298 K.
q = m × Cs × ΔT
m = 150.0 g
Cs = 0.24 J/°C g.
ΔT = 298 – 273 = 25°C
0.24 J
q = 150.0 g x o
x 25 oC = 9.0x10 2 J
Cg
45
Section 6.4
Quantifying Heat and Work
Solution
The specific heat capacity of silver is 0.24 J/°C °g.
b. Calculate the energy required to raise the temperature of 1.0
mol Ag by 1.0°C (called the molar heat capacity of silver).
J
107.9 g
J
0.24
x
= 26
gC
mol
mol  C
46
Section 6.4
Quantifying Heat and Work
Thermal Energy Transfer
• If two substances at different temperatures are
combined the heat lost by one substance is
absorbed by the other
• qsys = – qsurr
• For example if we place a piece of hot metal in a
beaker of water
• qmetal = – qwater
• The negative sign here just signifies that the heat is moving in the
opposite direction
47
Section 6.4
Quantifying Heat and Work
Thermal Energy Transfer
• How do we use this in calculations
• qmetal = – qwater
• mmetal × Csmetal × ΔTmetal = – mwater × Cswater ×
ΔTwater
• There are two types of these problems
• 6.3 in the book – unnecessarily complicated – you
are not responsible for any calculations that ask
you for final temperature of both substances
• Other type gives you all variables except one
48
Section 6.4
Quantifying Heat and Work
Learning Check
• An iron bar at 65.5°C is placed in a beaker of
water at 25.0 °C. If the mass of the water is 65.0
g and the final temperature at thermal equilibrium is
32.5 °C. What is the mass of the iron bar? The
specific heat capacity of water is 4.184 J/ °C g.
49
Section 6.4
Quantifying Heat and Work
Solution
• An iron bar at 65.5°C is placed in a beaker of
water at 25.0 °C. If the mass of the water is 65.0
g and the final temperature at thermal equilibrium is
32.5 °C. What is the mass of the iron bar?
• qmetal = – qwater
• mmetal × Csmetal × ΔTmetal = – mwater × Cswater ×
ΔTwater



J 

mmetal

 65.0g4.18 o  7.5oC 

 mwater Cs water DTwater 
g C

 = 140 g
=



J 
Csiron DTiron 
o
0.449 o  –33.0 C 


g C




50
Section 6.4
Quantifying Heat and Work
Work
• We know that energy transfer can occur via heat
or work
• We have seen how to determine the amount of
heat transfer by measuring the change in
temperature
• Now we will look at how to determine the
amount of work by measuring the change in
volume
51
Section 6.4
Quantifying Heat and Work
Work: Pressure-Volume Work
• Chemical reactions can do several different
types of work
• We are only going to consider pressure-volume
(or PV) work.
• Work is a force (F) acting through a distance.
• PV work occurs when the force is caused by a
volume change against an external pressure.
– Like in the engine of a car – when the pistons are
pushed outward against the external atmospheric
pressure.
52
Section 6.4
Quantifying Heat and Work
Work: Pressure-Volume Work
• So we define work as Force x Distance or F x D
• W=FxD
• In chapter 5 we defined pressure P = F/A which
we can rearrange to F = P x A
• Substitute (P x A) for F into the definition for
work and we get
• w = P x A x D (pressure x area x distance)
53
Section 6.4
Quantifying Heat and Work
Work: Pressure-Volume Work
• w = P x A x D (pressure x area x distance)
• w = P x A x Dh (distance between initial and final state
• w = PDV (where A x Dh = DV)
54
Section 6.4
Quantifying Heat and Work
Work: Pressure-Volume Work
•
•
•
•
So Pressure volume work is defined
w = PDV
The only thing missing is the sign
As the volume of the cylinder expands, work is
done on the surroundings, so the sign of the work
must be negative according to the way we defined
work and heat, the only way to accomplish this is to
include the negative sign in the formula
• w = – PDV
55
Section 6.4
Quantifying Heat and Work
Concept Check
Which of the following performs more
work?
a) A gas expanding against a pressure
of 2 atm from 1.0 L to 4.0 L.
b) A gas expanding against a pressure
of 3 atm from 1.0 L to 3.0 L.
56
Section 6.4
Quantifying Heat and Work
Solution
Which of the following performs more work?
a)A gas expanding against a pressure of 2 atm
from 1.0 L to 4.0 L.
w = – (2 atm)(4.0 – 1.0) = – 6 L·atm
b) A gas expanding against a pressure of 3 atm
from 1.0 L to 3.0 L
w = – (3 atm)(3.0 – 1.0) = – 6 L·atm
They both perform the same amount of work.
57
Section 6.4
Quantifying Heat and Work
Concept Check
A balloon is being inflated to its full extent by
heating the air inside it. The volume of the
balloon changes from 4.00 x106 L to 4.50
x106 L. Assuming that the balloon expands
against a constant pressure of 1.0 atm,
calculate the work done in this process in
Joules. (To convert between L atm and J,
use 1 L atm =101.3 J.)
58
Section 6.4
Quantifying Heat and Work
Solution
A balloon is being inflated to its full extent by heating the
air inside it. The volume of the balloon changes from 4.00
x106 L to 4.50 x106 L. Assuming that the balloon
expands against a constant pressure of 1.0 atm, calculate
the work done in this process in Joules. (To convert
between L atm and J, use 1 L atm =101.3 J.)
w =  PDV =  1.0 atm x 4.50 x 10 6 L - 4.00 x 10 6 L = - 5.0 x 10 5 L  atm
101.3 J
w =  5.0x10 L  atmx
=  5.1x10 7 J
L  atm
5
59
Section 6.4
Quantifying Heat and Work
Concept Check
The balloon in the previous example was
expanded by heating the air inside it by
the addition of 1.3 x108J of energy as
heat. Calculate DE for the process.
60
Section 6.4
Quantifying Heat and Work
Solution
DE = q+w
Heat = 1.3 x108J (given in problem)
Work = – 5.1 x 107 J (previous problem)
DE = q + w = 1.3x10 J  + 5.1x10 J  8x10 J
8
7
7
61
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Measuring DE - Calorimetry
• We measure heat evolved in a chemical reaction
using calorimetry
• In calorimetry we measure the thermal energy
(heat) exchanged between the reaction (the
system) and the surroundings by measuring the
change in temperature
• Two types of Calorimetry
– Constant Volume – measures DE
– Constant Pressure – measures DH (see slide 86)
62
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Measuring DE - Calorimetry
• We are going to look at Constant Volume
Calorimetry first
• Measure of DE
• Change in Internal Energy of a Reaction
– Remember Internal Energy is the sum of all the
kinetic and potential energy.
63
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Measuring DE
• So we know that systems exchange energy with
their surroundings via heat and work.
64
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Measuring DE
• Change in internal energy during a reaction is a sum
of both the heat and work. DE = q + w
• We could determine q and w separately and then
add them to determine DE.
• It would be easier though if we could force all the
change in internal energy to manifest as heat flow
(q) with no work done. Only 1 measurement.
65
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Constant Volume Calorimetry
How do we force all of the DE to be heat flow (q)
Well DE = q + w
And w = – PDV
So DE = q – PDV
So if we carry out the reaction at constant
volume then DV = 0 and w = 0 so
 DE = q
• This is written as
 DErxn = qv
•
•
•
•
•
66
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Constant Volume Calorimetry
• How do we measure
 DErxn = qv
• Device called a constant
volume (or
bomb)calorimeter.
– Sample is burned in oxygen
gas = system
– Surroundings are the
calorimeter.
• Measure heat flow by
change in temperature
67
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Constant Volume Calorimetry
• DErxn = qv
• Temperature change measured in the bomb
calorimeter is converted to DE using
• qcal = Ccal x ΔT
C = heat capacity of the calorimeter (J/°C)
DT = change in temperature (°C)
The calorimeter absorbs heat so the sign of the
qrxn is reversed to reflect that the system (the
sample) released heat
• qrxn = – qcal
68
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Learning Check
When a 1.50 g sample of methane or hydrogen
gas was burned with excess oxygen in the
calorimeter, the temperature increased by 7.3°C
for methane and 14.3°C for hydrogen.
Calculate qrxn for methane and hydrogen. The
bomb calorimeter has heat capacity of 11.3
kJ/°C.
69
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Solution
When a 1.50 g sample of methane or hydrogen gas was burned with
excess oxygen in the calorimeter, the temperature increased by 7.3°C for
methane and 14.3°C for hydrogen.
Calculate qrxn for methane and hydrogen. The bomb calorimeter has heat
capacity of 11.3 kJ/°C.
11.3 kJ
For Methane q cal = 7.3 C x o
= 82 kJ
C
q rxn =  q cal so q rxn =  82 kJ
11.3 kJ
For Hydrogen q cal = 14.3 oC x o
= 162 kJ
C
q rxn =  q cal so q rxn =  162 kJ
o
70
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
DErxn per mole
• We just calculated the heat of reaction (in
Joules) based on temperature change and heat
capacity of the calorimeter
 DErxn can also be expressed per mole of
reactant
• Divide qrxn/moles of reactant
71
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Learning Check
Calculate DErxn/mole for methane and hydrogen
from the last problem.
72
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Solution
Calculate DErxn/mole for methane and
hydrogen from the last problem.
For Methane
82 kJ 16.04 g
x
= - 880 kJ/mol
1.50 g
mol
162 kJ 2.016 g
=
x
= - 218 kJ/mol
1.50 g
mol
q rxn =  82 kJ DE rxn =
For Hydrogen q rxn =  162 kJ DE rxn
73
Section 6.5
Measuring DE for Chemical Reactions: Constant-Volume Calorimetry
Two Types of Calorimetry
• On a previous slide we saw
• Two types of Calorimetry
– Constant Volume – measures DE
– Constant Pressure – measures DH
• What is DH
DH is a change in Enthalpy
• What is Enthalpy?
74
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
What is Enthalpy
•
•
Enthalpy (H) is the total energy of a system
Enthalpy is the sum of internal energy and a
quantity PV
–
–
•
•
Internal energy (E) - energy required to create a system
PV - energy required to make room for the system it by
displacing its environment and establishing its volume
and pressure.
So H = E + PV
Enthalpy is a State function just like Internal
Energy
75
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
What is Enthalpy
•
•
•
•
So H = E + PV
Enthalpy represents both internal energy and a
quantity PV
So a change in Enthalpy represents a change
in internal energy and a change in PV
DH = DE +D(PV)
How do we measure Enthalpy?
76
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
How do we measure Enthalpy?
•
•
•
•
•
•
When a reaction occurs in a sealed bomb calorimeter
all the energy exchanged is in the form of heat
Most chemical reactions in the lab don’t occur this way
– they occur in open containers on the bench.
Energy is exchanged as both heat and work
In reality though, most of the time we are not really
interested in the small amount of work the reaction
does expanding against the atmosphere
What we are really interested in is the heat exchanged
How do we determine this value?
77
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Change in Enthalpy is the Heat Flow at Constant
Pressure
• If H = E + PV
• And DH = DE +D(PV)
–
•
So DH = DE +PDV
–
•
•
•
•
At constant P – the only change is volume
And we have already defined ΔE = q + w
So DH = q + w +PDV and –PDV = w so
So DH = q + w – w
And DH = qp
So Enthalpy is a measure of heat exchange only!
78
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Enthalpy vs Internal Energy
DH and DE can seem quite similar
• They both represent state functions
DE is a measure of all the energy (heat and
work) exchanged with the surroundings
DH is a measure of only the heat exchanged
under conditions of constant pressure
79
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Summarizing Enthalpy
•The value of DH is the amount of heat absorbed or
released under conditions of constant pressure
–
Endothermic Reaction
 DH will be positive – heat absorbed
• Reaction absorbs heat from the surroundings
• Feels cool
– Exothermic Reaction
 DH will be negative – heat released
• Reaction gives off heat to the surroundings
• Feels warm
80
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Learning Check
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
DH = –2044 kJ
For this reaction do the products or the reactants
have the higher enthalpy?
Is this reaction exothermic or endothermic?
81
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Solution
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) DH = –2044 kJ
For this reaction do the products or the reactants have the higher
enthalpy?
The ΔH is negative therefore the products have less enthalpy than
the reactants.
Is this reaction exothermic or endothermic?
Exothermic
82
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Endothermic and Exothermic Processes: A Molecular View
• An exothermic reaction releases energy as heat
• Where does that energy come from?
• Two forms of energy kinetic and potential.
– Kinetic is energy of motion – reflected in temperature.
– Exothermic reaction can’t be drawing from the pool of
kinetic energy – otherwise the temperature would go
down.
• In an exothermic reaction the reaction feels
warm
• So the energy must be coming from the stored
potential energy in the bonds of the reactants
83
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
How do we measure Enthalpy
•
•
•
•
•
Total enthalpy (H) cannot be measured directly
We can only measure a change in Enthalpy (DH)
And we do this by measuring heat flow (DT)
We measure a change in enthalpy by measuring
DT at constant pressure
Then we convert the change in temperature to an
amount of heat
84
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
How do we measure Enthalpy vs Internal Energy
• The value of DE for a reaction is the amount of
heat absorbed or released under constant
volume.
– Bomb Calorimeter DE = qv
• The value of DH for a reaction is the amount of
heat absorbed or released under constant
pressure.
– Coffee Cup Calorimeter DH = qp
85
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Coffee-Cup Calorimeter
• Constant Pressure
Calorimeter
• The reaction takes place
in solution inside the
inner cup.
• Change in temperature
of the solution is
measured.
86
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Coffee-Cup Calorimeter
• Temperature change measured in the coffeecup calorimeter is converted to DH using
• qsoln = msoln x Cs,soln x ΔT
msoln = mass of solution
Cs,soln = heat capacity of the solution inside the
calorimeter (J/g°C)
DT = change in temperature (°C)
• qrxn = – qsoln
87
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Learning Check
• Magnesium metal reacts with hydrochloric acid
according to the following equation
• Mg(s) + 2HCl(g)  MgCl2 (aq) + H2 (g)
• 0.316 g of magnesium metal is combined with
HCl to a final volume of 200.0 mL and
completely reacts. The temperature of the
solution changes from 25.6°C to 32.8°C. The
density of the solution is 1.00 g/mL and the
Cs,soln = 4.18 J/g°C. Determine qrxn.
88
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Solution
• Mg(s) + 2HCl(g)  MgCl2 (aq) + H2 (g)
• 0.316 g of magnesium metal is combined with HCl to a final
volume of 200.0 mL and completely reacts. The temperature
of the solution changes from 25.6°C to 32.8°C. The
density of the solution is 1.00 g/mL and the Cs,soln = 4.18
J/g°C. Determine qrxn.
q so ln = msoln x C s,soln x DT =
200.0 g x 4.18
J
o
3
x
7.2
C
=
6.0x
10
J
o
gC
q rxn = – 6.0x 10 3 J
89
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
DHrxn per mole
 DHrxn can also be expressed per mole of
reactant
• Divide qrxn/moles of reactant
90
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Learning Check
• Find DHrxn/mole for the previous problem.
91
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Solution
• Find DHrxn/mole for the previous problem.
qrxn
 6.0x 10 3 J 24.31 g
J
DHrxn =

x
=  4.6 x 10 5
mole Mg
0.316 g Mg
mole
mole
92
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Conceptual Connection
• Lighters are usually fueled by butane (C4H10).
When 1 mole of butane burns at constant
pressure it produces 2658 kJ of heat and does
3kJ of work. What are the values of DH and DE
for the combustion of one mole of butane.
93
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Solution
• Lighters are usually fueled by butane (C4H10).
When 1 mole of butane burns at constant
pressure it produces 2658 kJ of heat and does
3kJ of work. What are the values of DH and DE
for the combustion of one mole of butane.
• w and q are both negative – work is done on the
surroundings and heat is lost to surroundings
 DE = q + w = – 2661 kJ
 DH = q = – 2658 kJ
94
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Conceptual Connection
• The same reaction with exactly the same
amount of reactants is conducted in a bomb
calorimeter and a coffee cup calorimeter. In one
measurement qrxn = – 12.5 kJ and in the other
qrxn = – 11.8 kJ. Which value was determined in
the bomb calorimeter? (Assume the reaction
has a positive DV in the coffee-cup calorimeter)
95
Section 6.7
Constant-Pressure Calorimetry: Measuring DHrxn
Solution
• The same reaction with exactly the same amount of
reactants is conducted in a bomb calorimeter and a
coffee cup calorimeter. In one measurement qrxn = –
12.5 kJ and in the other qrxn = – 11.8 kJ. Which value
was determined in the bomb calorimeter? (Assume the
reaction has a positive DV in the coffee-cup calorimeter)
• The larger value 12.5 was measured in the bomb
calorimeter which measures DE (heat and work). The
coffee cup calorimeter measures DH (heat only).
96
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Stoichiometry Involving DH: Thermochemical
Equations
• Enthalpy change is also called enthalpy of
reaction or heat of reaction.
• Extensive property – depends on amount of
material undergoing the reaction
– The amount of heat generated depends on the
amount of reactant or product
• Consider the combustion of propane
97
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Example
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
DHrxn = –2044 kJ
This means that 2044 kJ of energy is released for every
mol of propane or every 5 moles of oxygen reacted
98
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Example
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
DHrxn = –2044 kJ
Calculate DH in which 5.00 g of propane is burned in
excess oxygen at constant pressure.
1 mol C 3H 8
 2044 kJ
5.00 g C 3H 8 x
x
= – 232 kJ
44.09 g C 3H 8 mol C 3H 8
99
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Learning Check
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
DHrxn = –2044 kJ
Calculate DH in which 25.0 g of water is released when
propane is burned in excess oxygen at constant
pressure.
100
Section 6.6
Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure
Solution
Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
DHrxn = –2044 kJ
Calculate DH in which 25.0 g of water is released when
propane is burned in excess oxygen at constant
pressure.
1 mol H 2O
1 mol C 3H 8  2044 kJ
25.0 g H 2O x
18.02 g H 2O
x
4 mol H 2O
x
mol C 3H 8
= – 709 kJ
101
Section 6.8
Relationships Involving DHrxn
Characteristics of Enthalpy Changes DHrxn
• The change in enthalpy for a reaction (DHrxn) is
associated with a particular reaction
• If we change the characteristics of the the
reaction then DHrxn changes too
• Lets look at a couple of examples
102
Section 6.8
Relationships Involving DHrxn
Characteristics of Enthalpy Changes
1. If a chemical reaction is multiplied by
some factor then DHrxn is multiplied by
the dame factor.
A + 2B  C DH1
Multiply by 2
2A + 4B  2C DH2 = DH1 x 2
103
Section 6.8
Relationships Involving DHrxn
Characteristics of Enthalpy Changes
2. If a reaction is reversed, then DHrxn
changes sign.
A + 2B  C DH1
Reverse reaction
C  A + 2B DH2 = – DH1
104
Section 6.8
Relationships Involving DHrxn
3. If a chemical reaction can expressed as
the sum of a series of steps, then DHrxn
for the overall equation is the sum of the
heats of reactions for each step.
A + 2B  C
DH1
C  2D DH2
–––––––––––––––––––
A + 2B  2D
DH3 = DH1 + DH2
105
Section 6.8
Relationships Involving DHrxn
Hess’s Law
• This last
relationship
is called
Hess’s Law
106
Section 6.8
Relationships Involving DHrxn
Example of Hess’s Law
• N2(g) + 2O2(g) → 2NO2(g)
ΔH1 = 68 kJ
This reaction also can be carried out in two distinct
steps, with enthalpy changes designated by DH2
and DH3.
N2(g) + O2(g) → 2NO(g)
DH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g)
DH3 = – 112 kJ
N2(g) + 2O2(g) → 2NO2(g)
DH2 + DH3 = 68 kJ
107
Section 6.8
Relationships Involving DHrxn
Example
•
Consider the following data:
1
3
NH3 (g)  
 N2 (g)  H2 (g)
2
2
2 H2 (g)  O2 (g)  
 2 H2O(g)
•
DH = 46 kJ
DH =  484 kJ
Calculate ΔH for the reaction
2 N2 (g)  6 H2O(g)  
 3 O2 (g)  4 NH3 (g)
108
Section 6.8
Relationships Involving DHrxn
Example
1
3
NH3 (g)  
 N2 (g)  H2 (g)
2
2
2 H2 (g)  O2 (g)  
 2 H2O(g)
DH = 46 kJ
DH =  484 kJ
Desired reaction:
2 N2 (g)  6 H2O(g)  
 3 O2 (g)  4 NH3 (g)
Reverse the two reactions:
1
3
N2 (g)  H2 (g)  
 NH3 (g)
DH =  46 kJ
2 H2O(g)  
 2 H2 (g)  O2 (g)
DH = +484 kJ
2
2
109
Section 6.8
Relationships Involving DHrxn
Example
•
Multiply reactions to give the correct numbers
of reactants and products:
1

3
4  N2 (g)  H2 (g)  
 NH3 (g) 
2

2

3 2 H2O(g)  
 2 H2 (g)  O2 (g)
•
4 DH =  46 kJ
 3 DH = +484 kJ
Desired reaction:
2 N2 (g)  6 H2O(g)  
 3 O2 (g)  4 NH3 (g)
110
Section 6.8
Relationships Involving DHrxn
Example
•
•
Final reactions:
2 N2 (g)  6 H2 (g)  
 4 NH3 (g)
DH =  184 kJ
6 H2O(g)  
 6 H2 (g)  3 O2 (g)
DH = +1452 kJ
Desired reaction:
2 N2 (g)  6 H2O(g)  
 3 O2 (g)  4 NH3 (g)
DH = +1268 kJ
111
Section 6.8
Relationships Involving DHrxn
Problem-Solving Strategy
•
•
•
Work backward from the required
reaction, using the reactants and products
to decide how to manipulate the other
given reactions at your disposal.
Reverse any reactions as needed to give
the required reactants and products.
Multiply reactions to give the correct
numbers of reactants and products.
112
Section 6.8
Relationships Involving DHrxn
Why?
•
Why on earth would we calculate DH this way?
113
Section 6.8
Relationships Involving DHrxn
Learning Check
Two forms of carbon are graphite, the soft, black, slippery material used in
“lead" pencils and as a lubricant for locks, and diamond, the brilliant, hard
gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and
diamond (-396 kJ/mol), calculate ΔH for the conversion of graphite to diamond.
Cgraphite (s)  Cdiamond (s)
Cgraphite (s) + O2 (g) CO 2 (g)
DH =  394 kJ
Cdiamond (s) + O2 (g) CO2 (g)
DH =  396 kJ
114
Section 6.8
Relationships Involving DHrxn
Solution
Two forms of carbon are graphite, the soft, black, slippery material used in
“lead" pencils and as a lubricant for locks, and diamond, the brilliant, hard
gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and
diamond (-396 kJ/mol), calculate ΔH for the conversion of graphite to diamond.
Cgraphite (s)  Cdiamond (s)
Cgraphite (s) + O2 (g) CO 2 (g)
DH =  394 kJ
CO2 (g)  Cdiamond (s) + O2 (g)
Cgraphite (s)  Cdiamond (s)
DH =  396 kJ 
DH = 2 kJ
The ΔH for the conversion of graphite to diamond is 2 kJ/mole graphite
115
Section 6.8
Relationships Involving DHrxn
Learning Check
• Find DHrxn for the reaction
• N2O(g) + NO2(g)  3NO(g)
• Using
• 2 NO(g) + O2(g)  2 NO2(g)
• N2(g) + O2(g)  2 NO(g)
• 2 N2O(g)  2 N2(g) + O2 (g)
DH = – 113.1 kJ
DH = 182.6 kJ
DH = –163.2 kJ
116
Section 6.8
Relationships Involving DHrxn
Solution
• Desired Reaction
N2O(g) + NO2(g)  3NO(g)
• Reverse eq 1, multiply eq 1 and 3 by 1/2
• ½[2NO2(g)  2 NO(g) + O2(g)] ½[DH = + 113.1 kJ]
• N2(g) + O2(g)  2 NO(g)
DH = 182.6 kJ
• ½[2 N2O(g)  2 N2(g) + O2(g)] ½[DH = –163.2 kJ]
117
Section 6.8
Relationships Involving DHrxn
Solution
• Desired Reaction
N2O(g) + NO2(g)  3NO(g)
• Reverse eq 1, multiply eq 1 and 3 by 1/2
• NO2(g)  NO(g) + ½ O2(g)
• N2(g) + O2(g)  2 NO(g)
• N2O(g)  N2(g) + ½ O2(g)
½[DH = + 113.1 kJ]
DH = 182.6 kJ
½[DH = –163.2 kJ]
118
Section 6.8
Relationships Involving DHrxn
Solution
• Desired Reaction
N2O(g) + NO2(g)  3NO(g)
• Reverse eq 1, multiply eq 1 and 3 by 1/2
• NO2(g)  NO(g) + ½ O2(g)
• N2(g) + O2(g)  2 NO(g)
• N2O(g)  N2(g) + ½ O2(g)
•
½ + 113.1 = 56.55 kJ
182.6 kJ
½ –163.2 kJ = –81.6 kJ
157.6 kJ
119
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
How many ways are there to determine DHrxn?
• So far we have looked at two different ways to
determine DHrxn
• Calorimetry – coffee cup calorimeter
– qp = DHrxn
• Hess’s Law – where we infer DHrxn by knowing
the values of DHrxn of our reaction from other
reactions with known values
• Now we will look at a third way
120
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
• The third method to determine DHrxn uses tables
of Enthalpies (or Heats of Formation)
• This method determines the amount of heat
required to make all the reactants and products
in something called the “standard state” and
then compares them to each other.
• So lets talk about standard state.
121
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
 DH is the change in enthalpy for a chemical
reaction
– The difference in enthalpy between the products and
reactants
 DH = Hproducts – Hreactants
• So the difference in enthalpy is an absolute
value (like the difference in altitude)
• But enthalpy itself (like altitude) is a relative
quantity – defined relative to some standard
(such as sea level in the case of altitude)
122
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
• So what are the standards that we use when talk
about enthalpy
• Well there are three – but they are related
• The standard state
• The standard enthalpy change (DH°)
• The standard enthalpy of formation (DHf°)
123
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
• Standard State
• For a gas – pure gas at 1 atm of pressure
• For a liquid or solid – pure substance in its most
stable form at 1 atm of pressure and 298 K (25
°C)
• For a solution – 1 M solution
124
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
• Standard Enthalpy Change (DH°)
• The change in enthalpy for a process when all
reactants and products are in their standard
states.
• Standard Enthalpy of Formation (DHf°)
• For a pure compound – the change in enthalpy
when 1 mole of the compound forms from its
constituent elements in their standard states.
• For a pure element in its standard state DHf°= 0
125
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard States and Standard Enthalpy Changes
• When we assign a value of 0 for DHf° to a pure
element in its standard state it is the same thing
as assigning sea level an altitude of 0.
• The interesting thing to notice about the table of
enthalpies (or heats) of formation for compounds
is that most of the values are negative
– Below sea level
• This actually means that compounds have less
enthalpy (more stable) that the original elements
that comprised them.
126
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
127
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Standard Enthalpy of Formation (DHf°)
We can use DHf° to calculate DH for a reaction
Add up all the DHf° for the products
– Because the are being formed
• Subtract all the DHf° of the reactants
– Because they are being broken down
• Pay attention to stoichiometry of equation
•
•
 DHrxn° = SnpDHf° (products) – SnpDHf°
(reactants)
128
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Example
Calculate DH° for the following reaction:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Given the following information:
DHf° (kJ/mol)
Na(s)
0
H2O(l)
–286
NaOH(aq)
–470
H2(g)
0
•[2(–470) + 0] – [0 + 2(–286)] = –368 kJ DH = –368 kJ
129
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Learning Check
Calculate DH° for the following reaction:
4 NH3 (g) + 7O2(g) → 4 NO2(g) + 6H2O(l)
Use the Table on the next slide.
130
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
131
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Solution
Calculate DH° for the following reaction:
4 NH3 (g) + 7O2(g) → 4 NO2(g) + 6H2O(l)
[4(34) + 6(–286)] – [4(–46) + 0] = –1396 kJ
DH = –1396 kJ
132
Section 6.9
Determining Enthalpies of Reaction from Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
1. When a reaction is reversed, the magnitude of ΔH
remains the same, but its sign changes.
2. When the balanced equation for a reaction is multiplied
by an integer, the value of ΔH for that reaction must be
multiplied by the same integer.
3. The change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of the
reactants and products:
DH°rxn = SnpDHf° (products) - SnrHf° (reactants)
4. Elements in their standard states are not included in the
DHreaction calculations because DHf° for an element in its
standard state is zero.
133
Section 6.10
Energy Use and the Environment
Energy Consumption
• Combustion of Fossil fuels is Highly Exothermic
• Coal C (s) + O2 (g)  CO2 (g)
 DHrxn° = – 393.5 kJ
• Natural Gas CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
 DHrxn° = – 802.3 kJ
• Petroleum C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (g)
 DHrxn° = – 5074.1 kJ
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Section 6.10
Energy Use and the Environment
Environmental Problems Associated with
Fossil Fuel Use
• Non renewable
• Produce greenhouse gases (CO2, H2O)
• Produce other gases from side reactions due to
impurities
– Sulfur oxides – contribute to acid rain
– Nitrogen oxides – contribute to SMOG
– Ozone – ground level ozone is a dangerous pollutant
135
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