Download Chapter 1: Matter and Measurement

General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 7: Thermochemistry
Philip Dutton
University of Windsor, Canada
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General Chemistry: Chapter 7
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Contents
7-1
7-2
7-3
7-4
7-5
7-6
7-7
Getting Started: Some Terminology
Heat
Heats of Reaction and Calorimetry
Work
The First Law of Thermodynamics
Heats of Reaction: U and H
The Indirect Determination of H: Hess’s Law
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General Chemistry: Chapter 7
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Contents
7-8
7-9
Standard Enthalpies of Formation
Fuels as Sources of Energy
Focus on Fats, Carbohydrates, and Energy
Storage
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6-1 Getting Started: Some Terminology
• System
• Surroundings
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General Chemistry: Chapter 7
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Terminology
• Energy, U
– The capacity to do work.
• Work
– Force acting through a distance.
• Kinetic Energy
– The energy of motion.
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Energy
• Kinetic Energy
ek =
1
2
m· v 2
kg m2
= J
[ek ] =
2
s
• Work
w = F·d
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kg m m
= J
[w ] =
2
s
General Chemistry: Chapter 7
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Energy
• Potential Energy
– Energy due to condition, position, or
composition.
– Associated with forces of attraction or
repulsion between objects.
• Energy can change from potential to kinetic.
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Energy and Temperature
• Thermal Energy
– Kinetic energy associated with random
molecular motion.
– In general proportional to temperature.
– An intensive property.
• Heat and Work
– q and w.
– Energy changes.
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Heat
Energy transferred between a system and its
surroundings as a result of a temperature
difference.
• Heat flows from hotter to colder.
– Temperature may change.
– Phase may change (an isothermal process).
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Heat transfer
Themochemistry
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Units of Heat
• Calorie (cal)
– The quantity of heat required to change the
temperature of one gram of water by one
degree Celsius.
• Joule (J)
– SI unit for heat
1 cal = 4.184 J
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Heat Capacity
• The quantity of heat required to change the
temperature of a system by one degree.
– Molar heat capacity.
• System is one mole of substance.
– Specific heat capacity, c.
q = mcT
• System is one gram of substance
– Heat capacity
• Mass · specific heat.
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General Chemistry: Chapter 7
q = CT
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Conservation of Energy
• In interactions between a system and its
surroundings the total energy remains constant—
energy is neither created nor destroyed.
qsystem + qsurroundings = 0
qsystem = -qsurroundings
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Determination of Specific Heat
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General Chemistry: Chapter 7
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Example 7-2
Determining Specific Heat from Experimental Data.
Use the data presented on the last slide to calculate the specific
heat of lead.
qlead = -qwater
qwater = mcT = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C
qwater = 1423 J
qlead = -1423 J = mcT = (150.0 g)(c)(28.8 - 100.0)°C
clead = 0.1332 Jg-1°C-1
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7-3 Heats of Reaction and Calorimetry
• Chemical energy.
– Contributes to the internal energy of a system.
• Heat of reaction, qrxn.
– The quantity of heat exchanged between a
system and its surroundings when a chemical
reaction occurs within the system, at constant
temperature.
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Heats of Reaction
• Exothermic reactions.
– Produces heat, qrxn < 0.
• Endothermic reactions.
– Consumes heat, qrxn > 0.
• Calorimeter
– A device for measuring quantities of heat.
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Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires +…
Define the heat capacity of the
calorimeter:
qcal = S mici T = C T
heat
all i
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Example 7-3
Using Bomb Calorimetry Data to Determine a Heat of Reaction.
The combustion of 1.010 g sucrose, in a bomb calorimeter,
causes the temperature to rise from 24.92 to 28.33°C. The
heat capacity of the calorimeter assembly is 4.90 kJ/°C.
(a) What is the heat of combustion of sucrose, expressed in
kJ/mol C12H22O11
(b) Verify the claim of sugar producers that one teaspoon of
sugar (about 4.8 g) contains only 19 calories.
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Example 7-3
Calculate qcalorimeter:
qcal = CT = (4.90 kJ/°C)(28.33-24.92)°C = (4.90)(3.41) kJ
= 16.7 kJ
Calculate qrxn:
qrxn = -qcal = -16.7 kJ per 1.010 g
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Example 7-3
Calculate qrxn in the required units:
-16.7 kJ
qrxn = -qcal =
= -16.5 kJ/g
1.010 g
qrxn
343.3 g
= -16.5 kJ/g
1.00 mol
= -5.65 · 103 kJ/mol
(a)
Calculate qrxn for one teaspoon:
qrxn
4.8 g 1.00 cal
)= -19 cal/tsp
= (-16.5 kJ/g)(
)(
4.184 J
1 tsp
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General Chemistry: Chapter 7
(b)
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Coffee Cup Calorimeter
• A simple calorimeter.
– Well insulated and therefore isolated.
– Measure temperature change.
qrxn = -qcal
See example 7-4 for a sample calculation.
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7-4 Work
• In addition to heat effects chemical reactions
may also do work.
• Gas formed pushes against
the atmosphere.
• Volume changes.
• Pressure-volume work.
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General Chemistry: Chapter 7
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Pressure Volume Work
w=F·d
= (P · A) · h
= PV
w = -PextV
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Example 7-5
7-3
Calculating Pressure-Volume Work.
Suppose the gas in the previous figure is 0.100 mol He at 298 K.
How much work, in Joules, is associated with its expansion at
constant pressure. (Pi = 243.18 kPa, Pf = 131.72 kPa)
Assume an ideal gas and calculate the volume change:
Vi = nRT/Pi = 1.02 L
R = 8.3145 J/mol K
Vf = nRT/Pf = 1.88 L
V = 1.88-1.02 L = 0.86 L
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Example 7-5
7-3
Calculate the work done by the system:
w = -Pf·V
= -(131.72 kPa)(0.86 L)
= -113.6 J
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Hint: If you use
pressure in kPa and
volume in liter (L)
you get Joules
directly.
General Chemistry: Chapter 7
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7-5 The First Law of Thermodynamics
• Internal Energy, U.
– Total energy (potential and kinetic) in a system.
•Translational kinetic energy.
•Molecular rotation.
•Bond vibration.
•Intermolecular attractions.
•Chemical bonds.
•Electrons.
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General Chemistry: Chapter 7
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Maxwell distributions of a gas with M=50
g/mol
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First Law of Thermodynamics
• A system contains only internal energy.
– A system does not contain heat or work.
– These only occur during a change in the system.
U = q + w
• Law of Conservation of Energy
– The energy of an isolated system is constant
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General Chemistry: Chapter 7
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First Law of Thermodynamics
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General Chemistry: Chapter 7
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State Functions
• Any property that has a unique value for a
specified state of a system is said to be a
State Function.
•
•
•
•
Water at 293.15 K and 1.00 atm is in a specified state.
d = 0.99820 g/mL
This density is a unique function of the state.
It does not matter how the state was established.
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Functions of State
• U is a function of state.
– Not easily measured.
• U has a unique value between two states.
– Is easily measured.
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Path Dependent Functions
• Changes in heat and work are not functions of
state.
– Remember example 7-5, w = -113.6 J in a one step
expansion of gas:
– Consider 2.40 atm to 1.80 atm and finally to 1.30 atm.
– 243.18 kPa, 182.39 kPa, and 131.72 kPa
w = –182.39 ·(1.36-1.02) – 131.72 ·(1.88-1.36) [kPa·L]
= -130.8 J
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p*V work of 0.1 mol of He gas: 2 steps
p.V Work, He, 298 K
250
p (kPa)
200
-17.2 J
150
100
-113.6 J
50
0
1.0
1.2
1.4
1.6
1.8
2.0
V (L)
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General Chemistry: Chapter 7
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p*V work of 0.1 mol of He gas: 3 steps
p(kPa)
243.18
182.39
151.99
131.72
V(l)
1.019
1.359
1.630
1.881
Steps W (J)
1
-113.6
2
-130.8
3
-136.3
p.V Work, He, 298 K
250
243.18
222.92
202.65
182.39
200
p (kPa)
Point
1
2
3
4
151.99
150
131.72
100
50
Vfinal
w-
P
(
V
)
dV
ex

0
1.0
1.5 1.630
1.881 2.0
V (L)
Vinitial
Thermochemistry
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7-6 Heats of Reaction: U and H
Reactants → Products
Ui
Uf
U = Uf - Ui
U = qrxn + w
In a system at constant volume:
U = qrxn + 0 = qrxn = qv
But we live in a constant pressure world!
How does qp relate to qv?
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Heats of Reaction
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Heats of Reaction
qV = qP + w
We know that w = - P · V and U = qv, therefore:
U = qP - P · V
qP = U + P · V
These are all state functions, so define a new function.
Let
H=U+P·V
Then
H = Hf – Hi = U + (P · V)
If we work at constant pressure and temperature:
H = U + P · V = qP
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Comparing Heats of Reaction
qP
= -566 kJ/mol
= H
P · V = P ·(Vf – Vi)
= RT· (nf – ni)
= -2.5 kJ
U = H - P · V
= -563.5 kJ/mol
= qV
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Changes of State of Matter
Molar enthalpy of vaporization:
H2O (l) → H2O(g)
H = 44.0 kJ at 298 K
Molar enthalpy of fusion:
H2O (s) → H2O(l)
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H = 6.01 kJ at 273.15 K
General Chemistry: Chapter 7
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Example 7-8
7-3
Enthalpy Changes Accompanying Changes in States of Matter.
Calculate H for the process in which 50.0 g of water is
converted from liquid at 10.0°C to vapor at 25.0°C.
Break the problem into two steps: Raise the temperature of
the liquid first then completely vaporize it. The total enthalpy
change is the sum of the changes in each step.
Set up the equation and calculate:
qP = mcH2OT + n·Hvap
50.0 g
= (50.0 g)(4.184 J/g °C)(25.0-10.0)°C +
44.0 kJ/mol
18.0 g/mol
= 3.14 kJ + 122 kJ = 125 kJ
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Standard States and Standard Enthalpy
Changes
• Define a particular state as a standard state.
• Standard enthalpy of reaction, H°
– The enthalpy change of a reaction in which all
reactants and products are in their standard states.
• Standard State
– The pure element or compound in its reference
state (g, l, s) at standard pressure and temperature
(101.3 kPa, 0°C) (STP).
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Enthalpy Diagrams
2H2 + O2
2H2O
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7-7 Indirect Determination of H:
Hess’s Law
• H is an extensive property.
– Enthalpy change is directly proportional to the amount of
substance in a system.
N2(g) + O2(g) → 2 NO(g)
H = +180.50 kJ
½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
• H changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g)
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H = -90.25 kJ
General Chemistry: Chapter 7
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Hess’s Law
• Hess’s law of constant heat summation
– If a process occurs in stages or steps (even hypothetically),
the enthalpy change for the overall process is the sum of
the enthalpy changes for the individual steps.
½N2(g) + ½O2(g) → NO(g)
H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
H = +33.18 kJ
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Hess’s Law Schematically
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7-8 Standard Enthalpies of Formation
Hf°
• The enthalpy change that occurs in the
formation of one mole of a substance in the
standard state from the reference forms of
the elements in their standard states.
• The standard enthalpy of formation of a
pure element in its reference state is 0.
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Standard Enthalpies of Formation
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Standard Enthalpies of Formation
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2C(grafit) + 3H2(g) + ½O2(g)  C2H5OH(l)
Termokémia
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H2O (g) and (l)
• 298 K
• 1 atm
Termokémia
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C(graphite) + O2 (g) = CO2(g)
• 298 K
• 1 atm
Termokémia
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Standard Enthalpies of Reaction
Hoverall = -2Hf°NaHCO3+ Hf°Na2CO3
+ Hf°CO2 + Hf°H2O
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General Chemistry: Chapter 7
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Enthalpy of Reaction
Hrxn = ∑Hf°products- ∑Hf°reactants
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General Chemistry: Chapter 7
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Table 7.3 Enthalpies of Formation of Ions
in Aqueous Solutions
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7-9 Fuels as Sources
of Energy
• Fossil fuels.
– Combustion is exothermic.
– Non-renewable resource.
– Environmental impact.
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General Chemistry: Chapter 7
Slide 56 of 50
Chapter 7 Questions
1, 2, 3, 11, 14, 16,
22, 24, 29, 37, 49,
52, 63, 67, 73, 81
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General Chemistry: Chapter 7
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