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Slide 1 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 1 of 57
PHILIP DUTTON
UNIVERSITY OF WINDSOR
DEPARTMENT OF CHEMISTRY AND
BIOCHEMISTRY
TENTH EDITION
GENERAL CHEMISTRY
Principles and Modern Applications
PETRUCCI HERRING MADURA BISSONNETTE
Thermochemistry 7
Slide 2 of 57
Thermochemistry
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 2 of 57
Slide 3 of 57
6-1 Getting Started: Some Terminology
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-1
Systems and their surroundings
Slide 4 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Energy, U
The capacity to do work.
Work, w
Force acting through a distance.
Kinetic Energy, eK
The energy of motion.
Potential Energy, V
The stored energy has potential to do work.
Slide 5 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Kinetic Energy
ek =
1
2
mv2
[ek ] =
m
s
= Jkg
2
w = F x d
= m x a x d [w ] =
m
s2 = Jm
Work
kg
Slide 6 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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.
Slide 7 of 57
Potential energy (P.E.) and kinetic energy (K.E.)
FIGURE 7-2
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 8 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
7-2 Heat
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.
Slide 9 of 57
Internal energy is 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).
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 10 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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
Slide 11 of 57
Determining the specific heat of lead – Example 7-2 illustrated
FIGURE 7-3
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 12 of 57
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.
System is one gram of substance
Heat capacity
(Mass of system) x specific heat.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
q = mc∆T
q = C∆T
Slide 13 of 57
Law of conservation of energy
In interactions between a system and its
surroundings the total energy remains constant—
energy is neither created nor destroyed.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
qsystem + qsurroundings = 0
qsystem = -qsurroundings
Slide 14 of 57
TABLE 7.1 Some Specific Heat Values, J g-1
°C-1
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 14 of 57
Slide 15 of 57
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.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 16 of 57
Exothermic and endothermic reactions
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 17 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
KEEP IN MIND
that the temperature of a
reaction mixture usually
changes during a reaction, so
the mixture must be returned
to the initial temperature
(actually or hypothetically)
before we assess how much
heat is exchanged with the
surroundings.
Slide 18 of 57
Conceptualizing a heat of reaction at constant temperature
FIGURE 7-4
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 19 of 57
Bomb Calorimetry
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-5
A bomb calorimeter assembly
qrxn = -qcal
qcal = q bomb + q water + q wires +…
Define the heat capacity of the
calorimeter:
qcal = Σmici∆T = Ccal∆Tall i
heat
Slide 20 of 57
The “Coffee-Cup” Calorimeter
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-6
A Styrofoam “coffee-cup” calorimeter
A simple calorimeter.
Well insulated and therefore isolated.
Measure temperature change.
qrxn = -qcal
Slide 21 of 57
7-4 Work
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-7 Illustrating work (expansion) during the chemical reaction
2 KClO3(s) 2 KCl(s) + 3 O2(g)
In addition to heat effects
chemical reactions may
also do work.
Gas formed pushes
against the atmosphere.
The volume changes.
Pressure-volume work.
Slide 22 of 57
Pressure-volume work
FIGURE 7-8
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
(m x g)
= P∆V
w = -Pext∆V
A
× ∆h= x A
x ∆h
w = F x d
= (m x g)
Slide 23 of 57
7-5 The First Law of Thermodynamics
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-9
Some contributions to the internal energy of a system
Internal Energy, U.
Total energy (potential and
kinetic) in a system.
•Translational kinetic energy.
•Molecular rotation.
•Bond vibration.
•Intermolecular attractions.
•Chemical bonds.
•Electrons.
Slide 24 of 57
The 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.
Law of Conservation of Energy
The energy of an isolated system is constant
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
∆U = q + w
Slide 25 of 57
The First Law of Thermodynamics
An isolated system is unable to exchange either
heat or work with its surroundings, so that
∆Uisolated system = 0, and we can say:
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
The energy of an isolated system is constant.
Slide 26 of 57
Illustration of sign conventions used in thermodynamics
FIGURE 7-10
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 27 of 57
Functions of State
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Any property that has a unique value for a
specified state of a system is said to be a
function of state or 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.
Slide 28 of 57
Functions of State
U is a function of state.
Not easily measured.
∆U has a unique value
between two states.
Is easily measured.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 29 of 57
Path Dependent Functions
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Changes in heat and work are not functions
of state.
Remember example 7-5,
w = -1.24 x 102
J in a one step expansion of gas:
let us consider a two step process from
2.40 atm to 1.80 atm, and finally to 1.20 atm.
Slide 30 of 57
Path Dependent Functions
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-11
A two-step expansion for the gas shown in Figure 7-8
w = (-1.80 atm)(1.36-1.02)L –
(1.30 atm)(2.04-1.36)L
= -0.61 L atm – 0.82 L atm
= -1.43 L atm
= -1.44 x 102
J
Compared -1.24 × 102
J
for the one stage process
Slide 31 of 57
7-6 Heats of Reaction: ∆U and ∆H
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Reactants → Products
Ui Uf
∆U = Uf - Ui
∆U = qrxn + w
In a system at constant volume (bomb calorimeter):
∆U = qrxn + 0 = qrxn = qv
But we live in a constant pressure world!
How does qp relate to qv?
Slide 32 of 57
Two different paths leading to the same internal energy change in a system
FIGURE 7-13
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 33 of 57
Heats of Reaction
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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 enthalpy be H = U + PV
Then ∆H = Hf – Hi = ∆U + ∆PV
If we work at constant pressure and temperature:
∆H = ∆U + P∆V = qP
Slide 34 of 57
Comparing Heats of Reaction
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-14
Comparing heats of reaction at constant volume and constant pressure for
the reaction 2 CO(g) + 2 O2(g) 2 CO2(g)
qV = ∆U = ∆H - P∆V
= -563.5 kJ/mol
w = P∆V = P(Vf – Vi)
= RT(nf – ni)
= -2.5 kJ
qP = ∆H
= -566 kJ/mol
Slide 35 of 57
Enthalpy Change (∆Η) Accompanying a
Change in State of Matter
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
H2O (l) → H2O(g) ∆H = 44.0 kJ at 298 K
Molar enthalpy of vaporization:
Molar enthalpy of fusion:
H2O (s) → H2O(l) ∆H = 6.01 kJ at 273.15 K
Slide 36 of 57
Standard States and Standard Enthalpy Changes
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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 at a pressure of
1 bar and at the temperature of interest.
Slide 37 of 57
Enthalpy Diagrams
FIGURE 7-15
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 38 of 57
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.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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) ∆H° = -90.25 kJ
Slide 39 of 57
Hess’s Law Schematically
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 40 of 57
• Hess’s law of constant heat summation
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
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) + O2(g) → NO2(g) ∆H° = +33.18 kJ
½N2(g) + O2(g) → NO(g) + ½ O2 (g) ∆H° = +90.25 kJ
NO(g) + ½O2(g) → NO2(g) ∆H° = -57.07 kJ
Slide 41 of 57
7-8 Standard Enthalpies of Formation
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.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
∆Hf
°
The standard enthalpy of formation of a pure
element in its reference state is 0.
Slide 42 of 57
Liquid bromine vaporizing
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Br2(l) Br2(g) ∆Hf° = 30.91 kJ
Slide 43 of 57
Diamond and graphite
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 44 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 45 of 57
Standard Enthalpy of Formation
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 46 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 47 of 57
Some standard enthalpies of formation at 298.15 K
FIGURE 7-18
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
number of carbons
Slide 48 of 57
Standard Enthalpies of Reaction
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-20
Computing heats of reaction from standard enthalpies of formation
∆Hoverall = -2∆Hf
°
NaHCO3
+ ∆Hf
°
Na2CO3
+∆Hf
°
CO2
+ ∆Hf
°
H2O
Slide 49 of 57
Diagramatic representation of equation (7.21)
FIGURE 7-21
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
∆H° = ∑νp∆Hf°(products) - ∑νr∆Hf°(reactants) (7.21)
Slide 50 of 57
Ionic Reactions in Solutions
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 51 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 52 of 57
7-9 Fuels as Sources of
Energy
Fossil fuels.
Combustion is exothermic.
Non-renewable resource.
Environmental impact.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 53 of 57
Global Warming – An Environmental Issue Involving Carbon Dioxide
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
FIGURE 7-22
World primary energy consumption by energy source
petroleum
coal
natural gas
wind
nuclear
Slide 54 of 57
The “greenhouse” effect
FIGURE 7-23
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Slide 55 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
Global Average
Global average atmospheric
carbon dioxide level
Actual and predicted CO2
emissions
Slide 56 of 57
Coal and Other Energy Sources
Gasification of Coal
steam and heated coal
Liquifaction of Coal
convert coal gas to long
chain hydrocarbons using
catalysis
Methanol
prepared from coal or
thermal decomposition of
various organic materials
Ethanol
mostly from ethylene
but also by fermentation
Biofuels
biodiesel is composed
of methyl esters of fatty acids
Hydrogen
excellent fuel but
requires a good production
method
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 56 of 57
Slide 57 of 57
End of Chapter Questions
Seek help with problems
But only after struggling with them.
The initial effort pays off
Understanding occurs when solutions are
explained by peers, tutors or professors.
Do not go straight to the solution manual
You only think that you understand the solution if
you have not worked hard on it first. The
struggle to solve the problem is a key part of
the memory and understanding process.
Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 57 of 57

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Ch07lecture 150104200718-conversion-gate02

  • 1. Slide 1 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 1 of 57 PHILIP DUTTON UNIVERSITY OF WINDSOR DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY TENTH EDITION GENERAL CHEMISTRY Principles and Modern Applications PETRUCCI HERRING MADURA BISSONNETTE Thermochemistry 7
  • 2. Slide 2 of 57 Thermochemistry Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 2 of 57
  • 3. Slide 3 of 57 6-1 Getting Started: Some Terminology Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-1 Systems and their surroundings
  • 4. Slide 4 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Energy, U The capacity to do work. Work, w Force acting through a distance. Kinetic Energy, eK The energy of motion. Potential Energy, V The stored energy has potential to do work.
  • 5. Slide 5 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Kinetic Energy ek = 1 2 mv2 [ek ] = m s = Jkg 2 w = F x d = m x a x d [w ] = m s2 = Jm Work kg
  • 6. Slide 6 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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.
  • 7. Slide 7 of 57 Potential energy (P.E.) and kinetic energy (K.E.) FIGURE 7-2 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 8. Slide 8 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 7-2 Heat 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.
  • 9. Slide 9 of 57 Internal energy is 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). Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 10. Slide 10 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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
  • 11. Slide 11 of 57 Determining the specific heat of lead – Example 7-2 illustrated FIGURE 7-3 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 12. Slide 12 of 57 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. System is one gram of substance Heat capacity (Mass of system) x specific heat. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 q = mc∆T q = C∆T
  • 13. Slide 13 of 57 Law of conservation of energy In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 qsystem + qsurroundings = 0 qsystem = -qsurroundings
  • 14. Slide 14 of 57 TABLE 7.1 Some Specific Heat Values, J g-1 °C-1 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 14 of 57
  • 15. Slide 15 of 57 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. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 16. Slide 16 of 57 Exothermic and endothermic reactions Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 17. Slide 17 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 KEEP IN MIND that the temperature of a reaction mixture usually changes during a reaction, so the mixture must be returned to the initial temperature (actually or hypothetically) before we assess how much heat is exchanged with the surroundings.
  • 18. Slide 18 of 57 Conceptualizing a heat of reaction at constant temperature FIGURE 7-4 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 19. Slide 19 of 57 Bomb Calorimetry Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-5 A bomb calorimeter assembly qrxn = -qcal qcal = q bomb + q water + q wires +… Define the heat capacity of the calorimeter: qcal = Σmici∆T = Ccal∆Tall i heat
  • 20. Slide 20 of 57 The “Coffee-Cup” Calorimeter Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-6 A Styrofoam “coffee-cup” calorimeter A simple calorimeter. Well insulated and therefore isolated. Measure temperature change. qrxn = -qcal
  • 21. Slide 21 of 57 7-4 Work Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-7 Illustrating work (expansion) during the chemical reaction 2 KClO3(s) 2 KCl(s) + 3 O2(g) In addition to heat effects chemical reactions may also do work. Gas formed pushes against the atmosphere. The volume changes. Pressure-volume work.
  • 22. Slide 22 of 57 Pressure-volume work FIGURE 7-8 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 (m x g) = P∆V w = -Pext∆V A × ∆h= x A x ∆h w = F x d = (m x g)
  • 23. Slide 23 of 57 7-5 The First Law of Thermodynamics Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-9 Some contributions to the internal energy of a system Internal Energy, U. Total energy (potential and kinetic) in a system. •Translational kinetic energy. •Molecular rotation. •Bond vibration. •Intermolecular attractions. •Chemical bonds. •Electrons.
  • 24. Slide 24 of 57 The 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. Law of Conservation of Energy The energy of an isolated system is constant Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 ∆U = q + w
  • 25. Slide 25 of 57 The First Law of Thermodynamics An isolated system is unable to exchange either heat or work with its surroundings, so that ∆Uisolated system = 0, and we can say: Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 The energy of an isolated system is constant.
  • 26. Slide 26 of 57 Illustration of sign conventions used in thermodynamics FIGURE 7-10 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 27. Slide 27 of 57 Functions of State Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Any property that has a unique value for a specified state of a system is said to be a function of state or 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.
  • 28. Slide 28 of 57 Functions of State U is a function of state. Not easily measured. ∆U has a unique value between two states. Is easily measured. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 29. Slide 29 of 57 Path Dependent Functions Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Changes in heat and work are not functions of state. Remember example 7-5, w = -1.24 x 102 J in a one step expansion of gas: let us consider a two step process from 2.40 atm to 1.80 atm, and finally to 1.20 atm.
  • 30. Slide 30 of 57 Path Dependent Functions Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-11 A two-step expansion for the gas shown in Figure 7-8 w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-1.36)L = -0.61 L atm – 0.82 L atm = -1.43 L atm = -1.44 x 102 J Compared -1.24 × 102 J for the one stage process
  • 31. Slide 31 of 57 7-6 Heats of Reaction: ∆U and ∆H Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Reactants → Products Ui Uf ∆U = Uf - Ui ∆U = qrxn + w In a system at constant volume (bomb calorimeter): ∆U = qrxn + 0 = qrxn = qv But we live in a constant pressure world! How does qp relate to qv?
  • 32. Slide 32 of 57 Two different paths leading to the same internal energy change in a system FIGURE 7-13 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 33. Slide 33 of 57 Heats of Reaction Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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 enthalpy be H = U + PV Then ∆H = Hf – Hi = ∆U + ∆PV If we work at constant pressure and temperature: ∆H = ∆U + P∆V = qP
  • 34. Slide 34 of 57 Comparing Heats of Reaction Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-14 Comparing heats of reaction at constant volume and constant pressure for the reaction 2 CO(g) + 2 O2(g) 2 CO2(g) qV = ∆U = ∆H - P∆V = -563.5 kJ/mol w = P∆V = P(Vf – Vi) = RT(nf – ni) = -2.5 kJ qP = ∆H = -566 kJ/mol
  • 35. Slide 35 of 57 Enthalpy Change (∆Η) Accompanying a Change in State of Matter Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 H2O (l) → H2O(g) ∆H = 44.0 kJ at 298 K Molar enthalpy of vaporization: Molar enthalpy of fusion: H2O (s) → H2O(l) ∆H = 6.01 kJ at 273.15 K
  • 36. Slide 36 of 57 Standard States and Standard Enthalpy Changes Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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 at a pressure of 1 bar and at the temperature of interest.
  • 37. Slide 37 of 57 Enthalpy Diagrams FIGURE 7-15 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 38. Slide 38 of 57 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. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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) ∆H° = -90.25 kJ
  • 39. Slide 39 of 57 Hess’s Law Schematically Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 40. Slide 40 of 57 • Hess’s law of constant heat summation Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 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) + O2(g) → NO2(g) ∆H° = +33.18 kJ ½N2(g) + O2(g) → NO(g) + ½ O2 (g) ∆H° = +90.25 kJ NO(g) + ½O2(g) → NO2(g) ∆H° = -57.07 kJ
  • 41. Slide 41 of 57 7-8 Standard Enthalpies of Formation 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. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 ∆Hf ° The standard enthalpy of formation of a pure element in its reference state is 0.
  • 42. Slide 42 of 57 Liquid bromine vaporizing Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Br2(l) Br2(g) ∆Hf° = 30.91 kJ
  • 43. Slide 43 of 57 Diamond and graphite Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 44. Slide 44 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 45. Slide 45 of 57 Standard Enthalpy of Formation Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 46. Slide 46 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 47. Slide 47 of 57 Some standard enthalpies of formation at 298.15 K FIGURE 7-18 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 number of carbons
  • 48. Slide 48 of 57 Standard Enthalpies of Reaction Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-20 Computing heats of reaction from standard enthalpies of formation ∆Hoverall = -2∆Hf ° NaHCO3 + ∆Hf ° Na2CO3 +∆Hf ° CO2 + ∆Hf ° H2O
  • 49. Slide 49 of 57 Diagramatic representation of equation (7.21) FIGURE 7-21 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 ∆H° = ∑νp∆Hf°(products) - ∑νr∆Hf°(reactants) (7.21)
  • 50. Slide 50 of 57 Ionic Reactions in Solutions Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 51. Slide 51 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 52. Slide 52 of 57 7-9 Fuels as Sources of Energy Fossil fuels. Combustion is exothermic. Non-renewable resource. Environmental impact. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 53. Slide 53 of 57 Global Warming – An Environmental Issue Involving Carbon Dioxide Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 FIGURE 7-22 World primary energy consumption by energy source petroleum coal natural gas wind nuclear
  • 54. Slide 54 of 57 The “greenhouse” effect FIGURE 7-23 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7
  • 55. Slide 55 of 57 Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7 Global Average Global average atmospheric carbon dioxide level Actual and predicted CO2 emissions
  • 56. Slide 56 of 57 Coal and Other Energy Sources Gasification of Coal steam and heated coal Liquifaction of Coal convert coal gas to long chain hydrocarbons using catalysis Methanol prepared from coal or thermal decomposition of various organic materials Ethanol mostly from ethylene but also by fermentation Biofuels biodiesel is composed of methyl esters of fatty acids Hydrogen excellent fuel but requires a good production method Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 56 of 57
  • 57. Slide 57 of 57 End of Chapter Questions Seek help with problems But only after struggling with them. The initial effort pays off Understanding occurs when solutions are explained by peers, tutors or professors. Do not go straight to the solution manual You only think that you understand the solution if you have not worked hard on it first. The struggle to solve the problem is a key part of the memory and understanding process. Copyright © 2011 Pearson Canada Inc.General Chemistry: Chapter 7Slide 57 of 57

Editor's Notes

  1. In this chapter, you will learn how to deduce and write chemical formulas and how to use the information incorporated into chemical formulas. The chapter ends with an overview of the relationship between names and formulas—chemical nomenclature.
  2. Potassium reacts with water, liberating sufficient heat to ignite the hydrogen evolved. The transfer of heat between substances in chemical reactions is an important aspect of thermochemistry. Thermochemistry is the branch of chemistry concerned with the heat effects that accompany chemical reactions. To understand the relationship between heat and chemical and physical changes, we must start with some basic definitions. We will then explore the concept of heat and the methods used to measure the transfer of energy across boundaries. Another form of energy transfer is work, and, in combination with heat, we will define the first law of thermodynamics.
  3. Open system. The beaker of hot coffee transfers energy to the surroundings—it loses heat as it cools. Matter is also transferred in the form of water vapor. Closed system. The flask of hot coffee transfers energy (heat) to the surroundings as it cools. Because the flask is stoppered, no water vapor escapes and no matter is transferred. Isolated system. Hot coffee in an insulated container approximates an isolated system. No water vapor escapes, and, for a time at least, little heat is transferred to the surroundings. (Eventually, though, the coffee in the container cools to room temperature so this system is not completely isolated.)
  4. Click 1 exposes the work formulae Click 2 exposes the units of work, notice they are the same as the units of kinetic energy
  5. Energy changes continuously from potential to kinetic. Energy is lost to the surroundings.
  6. A 150.0 g sample of lead is heated to the temperature of boiling water (100°C). A 50.0 g sample of water is added to a thermally insulated beaker, and its temperature is found to be 22°C. The hot lead is dumped into the cold water, and the temperature of the final lead–water mixture is 28.8 °C.
  7. Can we explain why liquid water has a high specific heat? The answer is most certainly yes, but the explanation relies on concepts we have not yet discussed. The fact that water molecules form hydrogen bonds (which we discuss in Chapter 12) is an important part of the reason why water has a large specific heat value. Energy is stored in hydrogen bonds, particularly strong intermolecular interactions. High Specific Heat of Compounds Because of their greater complexity at the molecular level, compounds generally have more ways of storing internal energy than do the elements; they tend to have higher specific heats. Water, for example, has a specific heat that is more than 30 times as great as that of lead. We need a much larger quantity of heat to change the temperature of a sample of water than of an equal mass of a metal. Lake Effect An environmental consequence of the high specific heat of water is found in the effect of large lakes on local climates. Because a lake takes much longer to heat up in summer and cool down in winter than other types of terrain, lakeside communities tend to be cooler in summer and warmer in winter than communities more distant from the lake.
  8. (a) An exothermic reaction. Slaked lime, Ca(OH)2 is produced by the action of water on quicklime, (CaO). The reactants are mixed at room temperature, but the temperature of the mixture rises to 40.5°C (b) An endothermic reaction. Ba(OH)2 • 8 H2O(s) and NH4Cl are mixed at room temperature, and the temperature falls to 5.8°C in the reaction which produces BaCl2•2 H2O and 2 NH3 and releasing 8 H2O.
  9. see EXAMPLE 7-3 Using Bomb Calorimetry Data to Determine a Heat of Reaction An iron wire is embedded in the sample in the lower half of the bomb. The bomb is assembled and filled with O2(g) at high pressure. The assembled bomb is immersed in water in the calorimeter, and the initial temperature is measured. A short pulse of electric current heats the sample, causing it to ignite. The final temperature of the calorimeter assembly is determined after the combustion. Because the bomb confines the reaction mixture to a fixed volume, the reaction is said to occur at constant volume. The significance of this fact is discussed on page 259.
  10. see EXAMPLE 7-4 Determining a Heat of Reaction from Calorimetric Data
  11. This figure is demonstrated in EXAMPLE 7-5 Calculating Pressure–Volume Work In this hypothetical apparatus, a gas is confined by a massless piston of area A. A massless wire is attached to the piston and the gas is held back by two weights with a combined mass of 2M resting on the massless pan. The cylinder is immersed in a large water bath in order to keep the gas temperature constant. The initial state of the gas is P= 2Mg/A with a volume Vi at temperature, T. When the external pressure on the confined gas is suddenly lowered by removing one of the weights the gas expands, pushing the piston up by the distance Dh, The increase in volume of the gas (DV) is the product of the cross-sectional area of the cylinder (A) and the distance (Dh). The final state of the gas is Pf = Mg/A, Vf, and T. Negative sign is introduced for Pext because the system does work ON the surroundings. When a gas expands V is positive and w is negative. When a gas is compressed V is negative and w is positive, indicating that energy (as work) enters the system. In many cases the external pressure is the same as the internal pressure of the system, so Pext is often just represented as P
  12. The models represent water molecules, and the arrows represent the types of motion they can undergo. In the intermolecular attractions between water molecules, the symbols ∂+ and ∂- signify a separation of charge, producing centers of positive and negative charge that are smaller than ionic charges. These intermolecular attractions are discussed in Chapter 12.
  13. Arrows represent the direction of heat flow (red arrow) and work (violet arrow). In the left diagram, the minus signs(-) signify energy leaving the system and entering the surroundings. In the right diagram the plus signs (+) refer to energy entering the system from the surroundings. These sign conventions are consistent with the expression U = q + w. EXAMPLE 7-6 Relating deltaU, q, and w Through the First Law of Thermodynamics illustrates these conventions.
  14. Obtain three different samples of water: 1) one purified by extensive distillation of groundwater; 2) one synthesized by burning pure H2(g) in pure O2(g) and 3) one prepared by driving off the water of hydration from CuSO4•5H2O and condensing the gaseous water to a liquid. The densities of the three different samples for the state that we specified will all be the same: 0.99820 g/mL Thus, the value of a function of state depends on the state of the system, and not on how that state was established.
  15. The difference in internal energy between these two states also has a unique value, ΔU = U2-U1 and this difference is something that we can precisely measure. It is the quantity of energy (as heat) that must be transferred from the surroundings to the system during the change from state 1 to state 2.
  16. In the initial state there are four weights of mass M/2 holding the gas back. In the intermediate state one of these weights has been removed and in the final state a second weight of mass M/2 has been removed. The initial and final states in this figure are the same as in Figure 7-8. This two-step expansion helps us to establish that the work of expansion depends on the path taken.
  17. In path (a), the volume of the system remains constant and no internal energy is converted into work—think of burning gasoline in a bomb calorimeter. In path (b), the system does work, so some of the internal energy change is used to do work—think of burning gasoline in an automobile engine to produce heat and work.
  18. No work is performed at constant volume because the piston cannot move because of the stops placed through the cylinder walls; qp = deltaU = -563.5 kJ (b) When the reaction is carried out at constant pressure, the stops are removed. This allows the piston to move and the surroundings do work on the system, causing it to shrink into a smaller volume. More heat is evolved than in the constant-volume reaction; qp = deltaH = -566.0 kJ. Note that w is small. In most cases we use deltaH except for combustion reactions which use qv=deltaU
  19. When a liquid is in contact with the atmosphere, energetic molecules at the surface of the liquid can overcome forces of attraction to their neighbors and pass into the gaseous, or vapor, state. We say that the liquid vaporizes. If the temperature of the liquid is to remain constant, the liquid must absorb heat from its surroundings to replace the energy carried off by the vaporizing molecules.
  20. *The International Union of Pure and Applied Chemistry (IUPAC) recommended that the standard- state pressure be changed from 1 atm to 1 bar about 25 years ago, but some data tables are still based on the 1 atm standard. Fortunately, the differences in values resulting from this change in standard-state pressure are very small—almost always small enough to be ignored.
  21. Horizontal lines represent absolute values of enthalpy. The higher a horizontal line, the greater the value of H that it represents. Vertical lines or arrows represent changes in enthalpy (delta H). Arrows pointing up signify increases in enthalpy—endothermic reactions. Arrows pointing down signify decreases in enthalpy—exothermic reactions.
  22. Note that in summing the two equations NO(g), a species that would have appeared on both sides of the overall equation was canceled out. Also, because we used an enthalpy diagram, the superfluous term, ½ O2(g), entered in and then canceled out. We have just introduced Hess’s law, which states the following principle:
  23. Absolute enthalpy cannot be determined. H is a state function so changes in enthalpy, H, have unique values. Reference forms of the elements in their standard states are the most stable form of the element at one bar and the given temperature. The superscript degree symbol denotes that the enthalpy change is a standard enthalpy change and The subscript “f” signifies that the reaction is one in which a substance is formed from its elements.
  24. Although we can obtain bromine in either the gaseous or liquid state at 298.15 K, Br2 (l) is the most stable form. Br2(g) if obtained at 298.15 K and 1 bar pressure, immediately condenses to Br2(l).
  25. Standard enthalpies of formation of elements are shown in the central plane, with ΔH°f=0. Substances with positive enthalpies of formation are above the plane, while those with negative enthalpies of formation are below the plane.
  26. Enthalpy is a state function, hence ΔH° for the overall reaction 2 NaHCO3(s)is the sum of the enthalpy changes for the two steps shown.
  27. The symbol Σ (Greek, sigma) means “the sum of.” The terms that are added together are the products of the standard enthalpies of formation (∆H˚f) and their stoichiometric coefficients (ν), One sum is required for the reaction products (subscript p), and another for the initial reactants (subscript r). The enthalpy change of the reaction is the sum of terms for the products minus the sum of terms for the reactants. Equation (7.21) avoids the manipulation of a number of chemical equations. The state function basis for equation (7.21) is shown in Figure 7-21 and is applied in Example 7-11 Calculating ∆H˚ from Tabulated Values of ∆H˚f .
  28. We should also be able to calculate this enthalpy of neutralization by using enthalpy of formation data in expression (7.21), but this requires us to have enthalpy of formation data for individual ions. And there is a slight problem in getting these. We cannot create ions of a single type in a chemical reaction. We always produce cations and anions simultaneously, as in the reaction of sodium and chlorine to produce Na+ and Cl- in NaCl. We must choose a particular ion to which we assign an enthalpy of formation of zero in its aqueous solutions. We then compare the enthalpies of formation of other ions to this reference ion. The ion we arbitrarily choose for our zero is H+(aq) Now let us see how we can use expression (7.21) and data from equation (7.22) to determine the enthalpy of formation of OH-(aq)
  29. These graphs show the history of energy consumption since 1970, with predictions to 2025. Petroleum (dark blue line) is seen to be the major source of energy for the foreseeable future, followed by coal (yellow) and natural gas (pink), which are about the same. Other sources of energy included are wind power (purple) and nuclear power (light blue). The unit BTU is a measure of energy and stands for British thermal unit (see Exercise 95). [Source: www.eia.doe.gov/oiaf/ieo/pdf/ieoreftab_2.pdf]
  30. (a) Some incoming radiation from sunlight is reflected back into space by the atmosphere, and some, such as certain UV light, is absorbed by stratospheric ozone. Much of the radiation from sunlight, however, reaches Earth’s surface. (b) Earth’s surface re-emits some of this energy as infrared radiation. (c) Infrared radiation leaving Earth’s atmosphere is less intense than that emitted by Earth’s surface because some of this radiation is absorbed by CO2 and other greenhouse gases and warms the atmosphere.
  31. (a) The global average atmospheric carbon dioxide level over a 50-year span, expressed in parts per million by volume, as measured by a worldwide cooperative sampling network. (b) The actual and predicted CO2 emissions for a 55-year span due to the combustion of natural gas (pink line), coal (yellow), and petroleum (dark blue), together with the total of all CO2 emissions (light blue). The CO2 content of the atmosphere continues to increase, from approximately 375 ppm in 2003 to 385 ppm in 2008.