1. Chapter 7: Thermochemistry
Chemistry 140 Fall 2002
CHEMISTRY
Ninth
Edition
GENERAL
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
Chapter 7: Thermochemistry
Iron rusts
Natural gas burns
Represent chemical reactions by chemical equations.
Relationship between reactants and products is STOICHIOMETRY
Many reactions occur in solution-concentration uses the mole concept to describe solutions
Philip Dutton
University of Windsor, Canada
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General Chemistry: Chapter 7
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2. Contents 7-1 Getting Started: Some Terminology 7-2 Heat
7-3 Heats of Reaction and Calorimetry
7-4 Work
7-5 The First Law of Thermodynamics
7-6 Heats of Reaction: U and H
7-7 The Indirect Determination of H: Hess’s Law
7-8 Standard Enthalpies of Formation
7-9 Fuels as Sources of Energy
Focus On Fats, Carbohydrates and Energy Storage
General Chemistry: Chapter 7
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3. 6-1 Getting Started: Some Terminology
System
Surroundings
General Chemistry: Chapter 7
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4. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Terminology
Energy, U
The capacity to do work.
Work
Force acting through a distance.
Kinetic Energy
The energy of motion.
General Chemistry: Chapter 7
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5. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Energy
Kinetic Energy 2 1 m
ek =
mv2
[ek ] = kg
= J 2 s
w = Fd
[w ] =
kg m s2
= J m
Work
General Chemistry: Chapter 7
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6. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
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.
Energy changes continuously from potential to kinetic.
Energy is lost to the surroundings.
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7. Potential and Kinetic Energy
General Chemistry: Chapter 7
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8. 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.
General Chemistry: Chapter 7
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9. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
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).
General Chemistry: Chapter 7
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10. General Chemistry: Chapter 7
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 = J
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11. General Chemistry: Chapter 7
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  specific heat.
q =
mcT
q =
CT
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12. 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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13. Determination of Specific Heat
General Chemistry: Chapter 7
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14. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
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)( )°C
qwater = 1.4103 J
qlead = -1.4103 J = mcT = (150.0 g)(clead)( )°C
clead = 0.13 Jg-1°C-1
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15. 7-3 Heats of Reaction and Calorimetry
Chemistry 140 Fall 2002
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.
General Chemistry: Chapter 7
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16. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Heats of Reaction
Exothermic reactions.
Produces heat, qrxn < 0.
Endothermic reactions.
Consumes heat, qrxn > 0.
Calorimeter
A device for measuring quantities of heat.
Add water
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General Chemistry: Chapter 7
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17. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Bomb Calorimeter
qrxn = -qcal
qcal = q bomb + q water + q wires +…
Define the heat capacity of the calorimeter:
qcal = miciT = CT
all i
heat
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18. General Chemistry: Chapter 7
EXAMPLE 7-3
Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of g sucrose, in a bomb calorimeter, causes the temperature to rise from 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.
General Chemistry: Chapter 7
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19. General Chemistry: Chapter 7
EXAMPLE 7-3
Calculate qcalorimeter:
qcal = CT = (4.90 kJ/°C)( )°C = (4.90)(3.41) kJ
= 16.7 kJ
Calculate qrxn:
qrxn = -qcal = kJ
per g
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20. General Chemistry: Chapter 7
EXAMPLE 7-3
Calculate qrxn in the required units:
-16.7 kJ
qrxn = -qcal =
= kJ/g
1.010 g
343.3 g
1.00 mol
= kJ/g
=  103 kJ/mol
qrxn
(a)
Calculate qrxn for one teaspoon:
4.8 g
1 tsp
= (-16.5 kJ/g)(
qrxn
(b) )(
)= -19 kcal/tsp
1.00 cal
4.184 J
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21. 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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22. General Chemistry: Chapter 7
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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23. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Pressure Volume Work
w = F  d
= (m  g)
 h
(m  g)
 A =
 h A
= PV
w = -PextV
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
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24. General Chemistry: Chapter 7
EXAMPLE 7-5
Calculating Pressure-Volume Work. Suppose the gas in the previous figure is mol He at 298 K and the each mass in the figure corresponds to an external pressure of 1.20 atm. How much work, in Joules, is associated with its expansion at constant pressure.
Assume an ideal gas and calculate the volume change:
Vi = nRT/P
= (0.100 mol)( L atm mol-1 K-1)(298K)/(2.40 atm)
= 1.02 L
Vf = 2.04 L
V = 2.04 L-1.02 L = 1.02 L
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25. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
EXAMPLE 7-5
Calculate the work done by the system:
w = -PV
= -(1.20 atm)(1.02 L)(
=  102 J
Hint: If you use pressure in kPa you get Joules directly.
-101 J )
1 L atm
A negative value signifies that work is done ON the surroundings
Where did the conversion factor come from?
Compare two versions of the gas constant and calculate.
Note that this conversion factor is simply kPa/atm.
J/mol K ≡ L atm/mol K
1 ≡ J/L atm
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26. 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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27. 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
U = q + w
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28. First Law of Thermodynamics
Chemistry 140 Fall 2002
First Law of Thermodynamics
Sign convention used in thermodynamics
General Chemistry: Chapter 7
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29. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Functions of State
Any property that has a unique value for a specified state of a system is said to be a State Function.
Water at K and 1.00 atm is in a specified state.
d = g/mL
This density is a unique function of the state.
It does not matter how the state was established.
General Chemistry: Chapter 7
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30. General Chemistry: Chapter 7
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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31. Path Dependent Functions
Changes in heat and work are not functions of state.
Remember example 7-5, w =  102 J in a one step expansion of gas:
Consider 2.40 atm to 1.80 atm and finally to 1.20 atm.
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32. Path Dependent Functions
w = (-1.80 atm)( )L – (1.30 atm)( )L
= L atm – 0.82 L atm = L atm
=  102 J
Compared  102 J for the two stage process
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33. 7-6 Heats of Reaction: U and H
Reactants → Products
Ui Uf
U = Uf - Ui
 = 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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34. General Chemistry: Chapter 7
Heats of Reaction
General Chemistry: Chapter 7
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35. General Chemistry: Chapter 7
Heats of Reaction
qV = qP + w
We know that w = - PV and U = qP, therefore:
U = qP - PV
qP = U + PV
These are all state functions, so define a new function.
Let H = U + PV
Then H = Hf – Hi = U + PV
If we work at constant pressure and temperature:
H = U + PV = qP
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36. Comparing Heats of Reaction
qP = -566 kJ/mol
= H
PV = P(Vf – Vi)
= RT(nf – ni)
= -2.5 kJ
U = H - PV
= kJ/mol
= qV
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37. 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) H = 6.01 kJ at K
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38. General Chemistry: Chapter 7
EXAMPLE 7-8
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)(4.184 J/g °C)( )°C +
50.0 g
18.0 g/mol
44.0 kJ/mol
= 3.14 kJ kJ = 125 kJ
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39. Standard States and Standard Enthalpy Changes
Chemistry 140 Fall 2002
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 at a pressure of 1 bar and at the temperature of interest.
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40. General Chemistry: Chapter 7
Chemistry 140 Fall 2002
Enthalpy Diagrams
General Chemistry: Chapter 7
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41. 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 = kJ
½N2(g) + ½O2(g) → NO(g) H = kJ
H changes sign when a process is reversed
NO(g) → ½N2(g) + ½O2(g) H = kJ
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42. General Chemistry: Chapter 7
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 = kJ
NO(g) + ½O2(g) → NO2(g) H = kJ
½N2(g) + O2(g) → NO2(g) H = kJ
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43. Hess’s Law Schematically
General Chemistry: Chapter 7
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44. 7-8 Standard Enthalpies of Formation
Chemistry 140 Fall 2002
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.
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.
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45. General Chemistry: Chapter 7
Standard States
General Chemistry: Chapter 7
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46. Standard Enthalpy of Formation
General Chemistry: Chapter 7
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47. Standard Enthalpies of Formation
General Chemistry: Chapter 7
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48. Standard Enthalpies of Reaction
H overall = -2Hf°NaHCO3+ Hf°Na2CO3
+ Hf°CO2 + Hf°H2O
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49. General Chemistry: Chapter 7
Enthalpy of Reaction
Hrxn = Hf°products- H f°reactants
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50. Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions
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51. 7-9 Fuels as Sources of Energy
Fossil fuels.
Combustion is exothermic.
Non-renewable resource.
Environmental impact.
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52. World Primary Energy Consumption
petroleum
coal
natural gas
wind
nuclear
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53. General Chemistry: Chapter 7
Greenhouse Effect
General Chemistry: Chapter 7
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54. General Chemistry: Chapter 7
CO2 in the Atmosphere
Global Average
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55. Actual and Predicted Emission of CO2
total
petroleum
coal
natural gas
General Chemistry: Chapter 7
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56. Focus on: Fats Carbohydrates and Energy Storage
General Chemistry: Chapter 7
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57. Energy Values of Fats and Carbohydrates
1 cal = J
General Chemistry: Chapter 7
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58. 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.
General Chemistry: Chapter 7
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