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Resources
Chapter Presentation
Bellringer
Transparencies
Sample Problems
Visual Concepts
Standardized Test Prep
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Chapter 10
Causes of Change
Table of Contents
Section 1 Energy Transfer
Section 2 Using Enthalpy
Section 3 Changes in Enthalpy During Chemical
Reactions
Section 4 Order and Spontaneity
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Chapter 10
Section 1 Energy Transfer
Bellringer
• Look at the words below.
heat
temperature
• Write a definition for each of these words and
describe how they are related.
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Chapter 10
Section 1 Energy Transfer
Objectives
• Define enthalpy.
• Distinguish between heat and temperature.
• Perform calculations using molar heat capacity.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat
• A sample can transfer energy to another sample.
• One of the simplest ways energy is transferred is as
heat.
• Heat is the energy transferred between objects that are at
different temperatures
• Though energy has many different forms, all energy is
measured in units called joules (J).
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
• Energy is never created or destroyed.
• The amount of energy transferred from one sample
must be equal to the amount of energy received by a
second sample.
• The total energy of the two samples remains exactly the
same.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
Temperature
• Temperature is a measure of how hot (or cold)
something is; specifically, a measure of the average
kinetic energy of the particles in an object.
• When samples of different temperatures are in
contact, energy is transferred from the sample that
has the higher temperature to the sample that has the
lower temperature.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
Temperature, continued
• If no other process occurs, the temperature of a
sample increases as the sample absorbs energy.
• The temperature of a sample depends on the
average kinetic energy of the sample’s particles.
• The higher the temperature of a sample is, the faster the
sample’s particles move.
• The temperature increase of a sample also depends
on the mass of the sample.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
Heat and Temperature are Different
• Temperature is an intensive property, which means
that the temperature of a sample does not depend on
the amount of the sample.
• Heat is an extensive property, which means that the
amount of energy transferred as heat by a sample
depends on the amount of the sample.
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Chapter 10
Visual Concepts
Comparing Extensive and Intensive Properties
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Chapter 10
Visual Concepts
Temperature and the Temperature Scale
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
A Substance’s Energy Can Be Measure by Enthalpy
• Measuring the total amount of energy present in a
sample of matter is impossible, but changes in energy
content can be determined.
• These changes are determined by measuring the energy
that enters or leaves the sample of matter.
• If 73 J of energy enter a piece of silver and no change in
pressure occurs, we know that the enthalpy of the silver
has increased by 73 J.
• Enthalpy, which is represented by the symbol H, is
the total energy content of a sample.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
A Substance’s Energy Can Be Measure by Enthalpy,
continued
• Enthalpy is the sum of the internal energy of a
system plus the product of the system’s volume
multiplied by the pressure that the system exerts on
its surroundings
• If pressure remains constant, the enthalpy increase of
a sample of matter equals the energy as heat that is
received.
• This relationship remains true even when a chemical
reaction or a change of state occurs.
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Chapter 10
Section 1 Energy Transfer
Energy as Heat, continued
A Sample’s Enthalpy Includes the Kinetic Energy of
Its Particles
• The particles in a sample are in constant motion.
• These particles have kinetic energy.
• The enthalpy of a sample includes the total kinetic
energy of its particles.
• Both the total and average kinetic energies of a
substance’s particles are important to chemistry,
because these quantities account for every particle’s
kinetic energy.
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity
• Molar heat capacity can be used to determine the
enthalpy change of a sample.
• The molar heat capacity of a pure substance is the
energy as heat needed to increase the temperature of
1 mol of the substance by 1 K.
• Molar heat capacity has the symbol C and the unit
J/Kmol.
• Molar heat capacity is accurately measured only if no
other process, such as a chemical reaction, occurs.
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity, continued
• The following equation shows the relationship between
heat and molar heat capacity, where q is the heat
needed to increase the temperature of n moles of a
substance by T.
q = nCT
heat = (amount in moles)  (molar heat capacity) 
(change in temperature)
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Chapter 10
Section 1 Energy Transfer
Calculating Molar Heat Capacity
Sample Problem A
Determine the energy as heat needed to increase the
temperature of 10.0 mol of mercury by 7.5 K. The
value of C for mercury is 27.8 J/K•mol.
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Chapter 10
Section 1 Energy Transfer
Calculating Molar Heat Capacity, continued
Sample Problem A Solution
q = nCT
q = (10.0 mol )(27.8 J/K•mol )(7.5 K)
q = 2085 J
The answer should only have two significant figures,
so it is reported as 2100 J or
2.1  103 J.
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Sample problem A, practice pg.342
1. The molar heat capacity of tungsten is 24.2
J/K•mol. Calculate the energy as heat needed to
increase the temperature of 0.40 mol of tungsten by
10.0 K.
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Sample problem A, practice pg.342
2. Suppose a sample of NaCl increased in temperature
by 2.5 K when the sample absorbed 1.7 × 102 J of
energy as heat. Calculate the number of moles of NaCl
if the molar heat capacity is 50.5 J/K•mol.
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HomeWork
Practice pg.342
3. Calculate the energy as heat needed to increase the
temperature of 0.80 mol of nitrogen, N2, by 9.5 K.The
molar heat capacity of nitrogen is 29.1 J/K•mol.
4 . A 0.07 mol sample of octane, C8H18, absorbed
3.5 × 103 J of energy. Calculate the temperature
increase of octane if the molar heat capacity of octane
is 254.0 J/K•mol
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity, continued
Molar Heat Capacity Depends on the Number of Atoms
• One mole of tungsten has a mass of 184 g, while one
mole of aluminum has a mass of only about 27 g.
• You might expect that much more heat is needed to change
the temperature of 1 mol W than is needed to change the
temperature of 1 mol Al.
• The molar heat capacities of all of the metals are nearly the
same.
• The temperature of 1 mol of any solid metal is raised 1 K when
the metal absorbs about 25 J of heat.
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity, continued
Molar Heat Capacity Depends on the Number of Atoms
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity, continued
Molar Heat Capacity Is Related to Specific Heat
• The specific heat of a substance is represented by cp
and is the energy as heat needed to raise the
temperature of one gram of substance by one Kelvin.
• The molar heat capacity of a substance, C, is related to moles
of a substance not to the mass of a substance.
• Because the molar mass is the mass of 1 mol of a
substance, the following equation is true.
M (g/mol)  cp (J/Kg) = C (J/Kmol)
(molar mass)(specific heat) = (molar heat capacity)
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Chapter 10
Visual Concepts
Specific Heat
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Chapter 10
Section 1 Energy Transfer
Molar Heat Capacity, continued
Heat Results in Disorderly Particle Motion
• When a substance receives energy in the form of heat,
its enthalpy increases and the kinetic energy of the
particles that make up the substance increases.
• The direction in which any particle moves is not related
to the direction in which its neighboring particles move.
The motions of these particles are random.
• Other types of energy can produce orderly motion or orderly
positioning of particles.
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Section10.1 Review
8. Calculate the molar heat capacity of diamond,
given that 63 J were needed to heat a 1.2 g of
diamond by 1.0 × 102 K. (6.3j/k mol)
9. Use the molar heat capacity for aluminum from
Table 1 to calculate the amount of energy needed
to raise the temperature of 260.5 g of aluminum
from 0°C to 125°C. (q=29.2 KJ)
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10. Use the molar heat capacity for iron from Table 1
to calculate the amount of energy needed to raise the
temperature of 260.5 g of iron from 0°C to 125°C.
( q=14.6KJ)
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11. A sample of aluminum chloride increased in
temperature by 3.5 K when the sample absorbed
1.67 × 102 J of energy. Calculate the number of
moles of aluminum chloride in this sample. Use
Table 1. (0.52 mole)
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12. Use Table 1 to determine the final temperature
when 2.5 × 102 J of energy as heat is transferred to
0.20 mol of helium at 298 K. ( 258K)
13. Predict the final temperature when 1.2 kJ of
energy as heat is transferred from 1.0 × 102 mL of
water at 298 K. ( 295K)
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14. Use Table 1 to determine the specific heat
of silver. ( 0.234 J/K.g)
15. Use Table 1 to determine the specific heat
of sodium chloride ( 0.864 J/K.g)
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16. Why is a temperature difference the same in Celsius
and Kelvin?
The temperature interval is the same on both scales.
17. Predict the molar heat capacities of PbS(s) and
Ag2S(s).
Using the “ about 25 J/K.mol for each atom in a solid” rule, C values of 50
and 75 J/K.mol, respectively would be predicted.
18. Use Table 1 to predict the molar heat capacity of
FeCl3(s).
C=100J/K.mol
19. Use your answer from item 18 to predict the specific
heat of FeCl3(s).
C = cp .M
cp = C/M
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Chapter 10
Section 2 Using Enthalpy
Bellringer
• Make a list of objects that have absorbed energy and
evidence that the objects have gained energy.
• For example, a burner on an electric stove gains energy
when it is turned on. Evidence of this is that the burner
glows red and it is hot.
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Chapter 10
Section 2 Using Enthalpy
Objectives
• Define thermodynamics.
• Calculate the enthalpy change for a given amount of
substance for a given change in temperature.
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Chapter 10
Section 2 Using Enthalpy
Molar Enthalpy Change
• Because enthalpy is the total energy of a system, it is
an important quantity.
• The only way to measure energy is through a change.
• There’s no way to determine the true value of H.
• H can be measured as a change occurs.
• The enthalpy change for one mole of a pure substance
is called molar enthalpy change.
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Chapter 10
Section 2 Using Enthalpy
Molar Enthalpy Change, continued
Molar Heat Capacity Governs the Changes
• When a pure substance is only heated or cooled, the
amount of heat involved is the same as the enthalpy
change.
• H = q for the heating or cooling of substances
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Chapter 10
Section 2 Using Enthalpy
Molar Enthalpy Change, continued
Molar Heat Capacity Governs the Changes, continued
• The molar enthalpy change is related to the molar heat
capacity by the following equation.
molar enthalpy change = CT
molar enthalpy change = (molar heat capacity)(temperature change)
• This equation does not apply to chemical reactions or
changes of state.
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Chapter 10
Visual Concepts
Enthalpy Change
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Chapter 10
Section 2 Using Enthalpy
Calculating Molar Enthalpy Change for Heating
Sample Problem B
How much does the molar enthalpy change when ice
warms from 5.4C to 0.2C?
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Chapter 10
Section 2 Using Enthalpy
Calculating Molar Enthalpy Change for Heating
Sample Problem B Solution
Tinitial = -5.4C = 267.8 K and Tfinal = -0.2C = 273.0 K
For H2O(s), C = 37.4 J/Kmol
T = Tfinal  Tinitial = 5.2 K
H = CT
J 

  37.4
 5.2 K 

K  mol 

J
2
 1.9  10
K  mol
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Chapter 10
Section 2 Using Enthalpy
Calculating Molar Enthalpy Change for Cooling
Sample Problem C
Calculate the molar enthalpy change when an
aluminum can that has a temperature of 19.2C is
cooled to a temperature of 4.00C.
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Chapter 10
Section 2 Using Enthalpy
Calculating Molar Enthalpy Change for Cooling
Sample Problem C Solution
Tinitial = -19.2C = 292 K and Tfinal = 4.00C = 277 K
For Al(s), C = 24.2 J/Kmol.
T = Tfinal  Tinitial = 277K  292 K =  15 K
H = CT
 H = (24.2 J/Kmol)( 15 K ) =  360 J/mol
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Section 2 Using Enthalpy
Sample problem B, practice pg. 346
1.Calculate the molar enthalpy change of H2O(l) when
liquid water is heated from 41.7°C to 76.2°C.
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Section 2 Using Enthalpy
Sample problem C, practice pg. 347
2. Lead has a molar heat capacity of 26.4J/K.mol.
What molar enthalpy change occurs when lead is
cooled from 302°C to 275 °C?
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Chapter 10
Section 2 Using Enthalpy
Molar Enthalpy Change, continued
Enthalpy Changes of Endothermic or Exothermic
Processes
• Enthalpy changes can be used to determine if a
process is endothermic or exothermic.
• Processes that have positive enthalpy changes are
endothermic.
• Processes that have negative enthalpy changes are
exothermic.
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Chapter 10
Section 2 Using Enthalpy
Enthalpy of a System of Several Substances
• Thermodynamics is the branch of science concerned
with the energy changes that accompany chemical and
physical changes.
• Enthalpy is one of three thermodynamic properties.
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Chapter 10
Section 2 Using Enthalpy
Enthalpy of a System of Several Substances,
continued
Writing Equations for Enthalpy Changes
• An equation can be written for the enthalpy change that
occurs during a change of state or a chemical reaction.
• The following equation is for the hydrogen and bromine
reaction.
H2(g, 298K) + Br2(l, 298K)  2HBr(g, 298K) H = -72.8 kJ
• The enthalpy change for this reaction and other chemical
reactions are written using the symbol H.
• The negative enthalpy change indicates the reaction is
exothermic.
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HOMEWORK
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Chapter 10
Section 2 Using Enthalpy
SECTION REVIEW, pg. 349
Homework
1. Name and define the quantity represented by H.
2. During a heating or cooling process, how are
changes in enthalpy and temperature related?
3. What is thermodynamics?
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4. A block of ice is cooled from −0.5°C to −10.1°C.
Calculate the temperature change, ΔT, in degrees
Celsius and in kelvins.
5. Calculate the molar enthalpy change when a block
of ice is heated from −8.4°C to −5.2°C.
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6. Calculate the molar enthalpy change when
H2O(l) is cooled from 48.3°C to 25.2°C.
7. The molar heat capacity of benzene,C6H6(l), is 136
J/K•mol. Calculate the molar enthalpy change when
the temperature of C6H6(l) changes from 19.7°C to
46.8°C.
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Homework
8. The molar heat capacity of diethyl ether,
(C2H5)2O(l), is 172 J/K•mol.What is the temperature
change if the molar enthalpy change equals −186.9
J/mol?
9. If the enthalpy of 1 mol of a compound decreases
by 428 J when the temperature decreases by 10.0 K,
what is the compound’s molar heat capacity?
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Bellringer
• Write a short paragrpah describing what you know
about food calories and how they relate to nutrition,
metabolism, weight gain, and weight loss.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Objectives
• Explain the principles of calorimetry.
• Use Hess’s law and standard enthalpies of formation
to calculate H.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Changes in Enthalpy Accompany Reactions
• A change in enthalpy during a reaction depends on
many variables.
• Temperature is one of the most important variables.
• To standardize the enthalpies of reactions, data are
often presented for reactions in which both reactants
and products have the standard thermodynamic
temperature of 25.00C or 298.15 K.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Changes in Enthalpy Accompany Reactions,
continued
• Chemists usually present a thermodynamic value for a
chemical reaction by using the chemical equation.
1
1
H2  g  + Br2  l   HBr  g  H = -36.4 kJ
2
2
• This equation shows that when 0.5 mol of H2 reacts
with 0.5 mol of Br2 to produce 1 mol HBr and all have
a temperature of 298.15 K, the enthalpy decreases
by 36.4 kJ.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry
• For the H2 and Br2 reaction, in which H is negative,
the total energy of the reaction decreases.
• The energy is released as heat by the system.
• If the reaction was endothermic, energy in the form of
heat would be absorbed by the system and the
enthalpy would increase.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry, continued
• The experimental measurement of an enthalpy change
for a reaction is called calorimetry.
• Calorimetry is the measurement of heat-related constants,
such as specific heat or latent heat
• Combustion reactions are always exothermic.
• The enthalpy changes of combustion reactions are
determined using a bomb calorimeter.
• A calorimeter Is a device used to measure the heat absorbed
or released in a chemical or physical change
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry, continued
• A calorimeter is a sturdy, steel vessel in which the
sample is ignited electrically in the presence of highpressure oxygen.
• The energy from combustion is absorbed by a
surrounding water bath and by the calorimeter.
• The water and the other parts of the calorimeter have
known specific heats.
• A measured temperature increase can be used to calculate the
energy released in the combustion reaction and then the
enthalpy change.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Bomb Calorimeter
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Chapter 10
Visual Concepts
Calorimeter
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry, continued
Nutritionists Use Bomb Calorimetry
• Inside the pressurized oxygen atmosphere of a bomb
calorimeter, most organic matter, including food, fabrics,
and plastics, will ignite easily and burn rapidly.
• Sample sizes are chosen so that there is excess oxygen
during the combustion reactions.
• Under these conditions, the reactions go to completion
and produce carbon dioxide, water, and possibly other
compounds.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry, continued
Adiabatic Calorimetry Is Another Strategy
• Instead of using a water bath to absorb the energy
generated by a chemical reaction, adiabatic calorimetry
uses an insulating vessel.
• The word adiabatic means “not allowing energy to pass
through.”
• No energy can enter or escape this type of vessel.
• The reaction mixture increases in temperature if the
reaction is exothermic or decreases in temperature if
the reaction is endothermic.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Chemical Calorimetry, continued
Adiabatic Calorimetry Is Another Strategy, continued
• If the system’s specific heat is known, the reaction
enthalpy can be calculated.
• Adiabatic calorimetry is used for reactions that are not
ignited, such as for reactions in aqueous solution.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law
• Any two processes that both start with the same
reactants in the same state and finish with the same
products in the same state will have the same enthalpy
change.
• Hess’s law states that the overall enthalpy change in a
reaction is equal to the sum of the enthalpy changes for
the individual steps in the process.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
• When phosphorus is burned in excess chlorine 4 mol of
phosphorus pentachloride, PCl5, is synthesized.
P4(s) + 10Cl2(g)  4PCl5(g)
H = -1596 kJ
• Phosphorus pentachloride may also be prepared in a
two-step process.
Step 1: P4(s) + 6Cl2(g)  4PCl3(g) H = -1224 kJ
Step 2: PCl3(g) + Cl2(g)  PCl5(g) H = -93 kJ
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
• The second reaction must take place four times for
each occurrence of the first reaction in the two-step
process.
• This two-step process is more accurately described by
the following equations.
P4(s) + 6Cl2(g)  4PCl3(g)
H = -1224 kJ
4PCl3(g) + 4Cl2(g)  4PCl5(g) H = 4(-93 kJ) = -372 kJ
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
• So, the total change in enthalpy by the two-step
process is as follows:
(-1224 kJ) + (-372 kJ) = -1596 kJ
• This enthalpy change, H, for the two-step process is
the same as the enthalpy change for the direct route of
the formation of PCl5.
• This example is in agreement with Hess’s law.
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Chapter 10
Visual Concepts
Hess’s Law
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Using Hess’s Law and Algebra
• Chemical equations can be manipulated using rules of
algebra to get a desired equation.
• When equations are added or subtracted, enthalpy
changes must be added or subtracted.
• When equations are multiplied by a constant, the
enthalpy changes must also be multiplied by that
constant.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Using Hess’s Law and Algebra
• The enthalpy of the formation of CO, when CO2 and
solid carbon are reactants, is found using the
equations below.
2C(s) + O2(g)  2CO(g) H = -221 kJ
C(s) + O2(g)  CO2(g) H = -393 kJ
• You cannot simply add these equations because CO2
would not be a reactant.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Using Hess’s Law and Algebra, continued
• You subtract or reverse the second equation, carbon
dioxide will be on the correct side of the equation.
–C(s) – O2(g)  – CO2(g) H = –(– 393 kJ)
CO2(g)  C(s) + O2(g) H = 393 kJ
• Reversing an equation causes the enthalpy of the new
reaction to be the negative of the enthalpy of the
original reaction.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Using Hess’s Law and Algebra, continued
• Adding the two equations gives the equation for the
formation of CO by using CO2 and C.
2C(s) + O2(g)  2CO(g) H = – 221 kJ
CO2(g)  C(s) + O2(g) H = 393 kJ
2C(s) + O2(g) + CO2(g)  2CO(g) + C(s) + O2(g)
H = 172 kJ
• Oxygen and carbon that appear on both sides of the
equation can be canceled.
C(s) + CO2(g)  2CO(g) H = 172 kJ
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Standard Enthalpies of Formation
• The enthalpy change in forming 1 mol of a substance
from elements in their standard states is called the
standard enthalpy of formation of the substance,  H f0
• The values of the standard enthalpies of formation for
elements are 0.
• The following equation is used to determine the
enthalpy change of a chemical reaction from the
standard enthalpies of formation.
Hreaction = Hproducts - Hreactants
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Hess’s Law, continued
Standard Enthalpies of Formation, continued
Standard Enthalpies of Formation
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Calculating a Standard Enthalpies of
Formation
Sample Problem D
Calculate the standard enthalpy of formation of pentane,
C5H12, using the given information.
(1) C(s) + O2(g)  CO2(g)  H f0 = -393.5 kJ/mol
(2) H2(g) + ½O2(g)  H2O(l)  H f0 = -285.8 kJ/mol
(3) C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(l)
H = -3535.6 kJ/mol
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Calculating a Standard Enthalpies of
Formation
Sample Problem D Solution
(1)
(2)
5C(s) + 5O2 (g)  5CO2 (g)
H = 5(-393.5 kJ/mol)
6H2 (g) + 3O2 (g)  6H2O(l)
H = 6(-285.8 kJ/mol)
(3) 5CO2 (g) + 6H2O(l)  C5H12 (g ) + 8O2 (g)
5C(s) + 6H2 (g)  C5H12 (g)
H = 3536.6 kJ/mol
Hf0 = -145.7 kJ/mol
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Calculating a Reaction’s Change in Enthalpy
Sample Problem E
Calculate the change in enthalpy for this reaction.
2H2(g) + 2CO2(g)  2H2O(g) + 2CO(g)
State whether the reaction is exothermic or endothermic.
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Chapter 10
Section 3 Changes In Enthalpy
During Chemical Reactions
Calculating a Reaction’s Change in Enthalpy,
continued
Sample Problem E Solution
Standard enthalpies of formation as follows:
H2O(g),  H f0 = -241.8 kJ/mol
CO(g),  H f0 = -110.5 kJ/mol
H2(g),  H f0 = 0 kJ/mol
CO2(g),  H f0 = -393.5 kJ/mol.
H = (2 mol)(-241.8 kJ/mol) + (2 mol)(-110.5 kJ/mol) (2 mol)(0 kJ/mol) - (2 mol)(-393.5 kJ/mol)
= 82.4 kJ
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Sample Problem D , practice pg. 356
1) Calculate the enthalpy change for the following
reaction. NO(g) + 1/2O2(g) → NO2(g)
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2) Calculate the enthalpy change for the combustion
of methane gas, CH4 to form CO2(g) and H2O(l).
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Sample problem E, practice pg.357
1)Use data from Table 2 to calculate ΔH for the
following reaction.
C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g)
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2)The exothermic reaction known as lime slaking is
CaO(s) + H2O(l) →Ca(OH)2(s). Calculate ΔH from the
data in Table 2.
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HW.
Section Review, pg.357,
Q (1-5)
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Chapter 10
Section 4 Order and Spontaneity
Bellringer
• Make a list of examples that show both increasing
and decreasing order.
• For example, making your bed or baking a cake are
examples of increasing order. Breaking a dish or shredding
paper are examples of decreasing order.
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Chapter 10
Section 4 Order and Spontaneity
Objectives
• Define entropy, and discuss the factors that influence
the sign and magnitude of S for a chemical reaction.
• Describe Gibbs energy, and discuss the factors that
influence the sign and magnitude of G.
• Indicate whether G values describe spontaneous or
nonspontaneous reactions.
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Chapter 10
Section 4 Order and Spontaneity
Entropy
• Some reactions happen easily, but others do not.
• Sodium and chlorine react when they are brought together.
• Nitrogen and oxygen coexist in the air you breathe without
forming poisonous nitrogen monoxide, NO.
• One factor you can use to predict whether reactions
will occur is enthalpy. A reaction is more likely to occur
if it is accompanied by a decrease in enthalpy or if H
is negative.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
• Another factor known as entropy can determine if a
process will occur.
• Entropy, S, is a measure of the disorder in a system
and is a thermodynamic property.
• Entropy is not a form of energy and has the units
joules per kelvin, J/K.
• A process is more likely to occur if it is accompanied
by an increase in entropy.
• S is positive.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Factors That Affect Entropy
• Ions in a solution disperse throughout the solution.
• This process of dispersion is called diffusion and causes the
increase in entropy.
• Entropy also increases as solutions become more
dilute or when the pressure of a gas is reduced.
• In both cases, the molecules fill larger spaces and so
become more disordered.
• Entropies also increase with temperature, but this
effect is not great unless a phase change occurs.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Factors That Affect Entropy, continued
• The entropy can change during a reaction.
• The entropy of a system can increase when the total
number of moles of product is greater than the total
number of moles of reactant.
• Entropy can increase in a system when the total
number of particles in the system increases.
• Entropy also increases when a reaction produces
more gas particles, because gases are more
disordered than liquids or solids.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Factors That Affect Entropy, continued
• Entropy decreases as sodium chloride forms: 2 mol of
sodium combine with 1 mol of chlorine to form 2 mol
of sodium chloride.
2Na(s) + Cl2(g)  2NaCl(s)
S = -181 J/K
• This decrease in entropy is because of the order
present in crystalline sodium chloride.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Factors That Affect Entropy, continued
• Entropy increases when 1 mol of sodium chloride
dissolves in water to form 1 mol of aqueous sodium
ions and 1 mol of aqueous chlorine ions.
NaCl(s)  Na+(aq) + Cl–(aq)
S = 43 J/K
• This increase in entropy is because of the order lost when a
crystalline solid dissociates to form ions.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Factors That Affect Entropy, continued
Standard Entropy Changes for Some Reactions
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Chapter 10
Visual Concepts
Relating Enthalpy, Entropy, and Free Energy
Changes
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Hess’s Law Also Applies to Entropy
• The decomposition of nitrogen triiodide to form
nitrogen and iodine creates 4 mol of gas from 2 mol of
a solid.
2NI3(s)  N2(g) + 3I2(g)
• This reaction has such a large entropy increase that
the reaction proceeds once the reaction is initiated by
a mechanical shock.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Hess’s Law Also Applies to Entropy, continued
• Molar entropy has the same unit, J/K•mol, as molar
heat capacity.
• Molar entropies can be calculated from molar heat
capacity data.
• Entropies can also be calculated by using Hess’s law
and entropy data for other reactions.
• You can manipulate chemical equations using rules of
algebra to get a desired equation.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Hess’s Law Also Applies to Entropy, continued
• When equations are added or subtracted, entropy
changes must be added or subtracted.
• When equations are multiplied by a constant, the
entropy changes must also be multiplied by that
constant.
• Atoms and molecules that appear on both sides of the
equation can be canceled.
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Hess’s Law Also Applies to Entropy, continued
• The standard entropy is represented by the symbol S0.
• The standard entropy of the substance is the entropy of 1 mol of a
substance at a standard temperature, 298.15 K.
• Elements can have standard entropies that have values other than
zero.
• Most standard entropies are positive; this is not true of standard
enthalpies of formation.
• The entropy change of a reaction can be calculated by using the
following equation.
Sreaction = Sproducts - Sreactants
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Chapter 10
Section 4 Order and Spontaneity
Entropy, continued
Hess’s Law Also Applies to Entropy, continued
Standard Entropies of Some Substances
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Chapter 10
Section 4 Order and Spontaneity
Hess’s Law and Entropy
Sample Problem F
Calculate the entropy change that accompanies the
following reaction.
1
1
1
1
H2 (g ) + CO2 (g )  H2O(g ) + CO(g )
2
2
2
2
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Chapter 10
Section 4 Order and Spontaneity
Hess’s Law and Entropy, continued
Sample Problem F Solution
The standard entropies are as follows.
For H2O, S0 = 188.7 J/K•mol.
For CO, S0 = 197.6 J/K•mol.
For H2, S0 = 130.7 J/K•mol.
For CO2, S0 = 213.8 J/K•mol.
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Chapter 10
Section 4 Order and Spontaneity
Hess’s Law and Entropy, continued
Sample Problem F Solution, continued
S = S(products) + S(reactants)
1

1

S =   mol  (188.7 J/K mol) +  mol  (197.6 J/K mol)
2

2

1

1

-  mol  (130.7 J/K mol) -  mol  (213.8 J/K mol)
2

2

S = 94.35 J/K + 98.8 J/K - 65.35 J/K - 106.9 J/K
S = 193.1 J/K - 172.2 J/K
S = 20.9 J/K
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy
• The tendency for a reaction to occur depends on both
H and S.
• If H is negative and S is positive for a reaction, the
reaction will likely occur.
• If H is positive and S is negative for a reaction, the
reaction will not occur.
• How can you predict what will happen if H and S
are both positive or both negative?
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
• Gibbs energy is the energy in a system that is
available for work represented by the symbol G.
G = H – TS
• Another name for Gibbs energy is free energy.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Gibbs Energy Determines Spontaneity
• A spontaneous reaction is one that does occur or is
likely to occur without continuous outside assistance,
such as input of energy.
• A reaction is spontaneous if the Gibbs energy change is
negative.
• A nonspontaneous reaction will never occur without
assistance.
• If a reaction has a G greater than 0, the reaction is
nonspontaneous.
• If a reaction has a G of exactly zero, the reaction
is at equilibrium.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Entropy and Enthalpy Determine Gibbs Energy
• Reactions that have large negative G values often
release energy and increase disorder.
• The vigorous reaction of potassium metal and water is an
example of this type of reaction.
2K(s) + 2H2O(l)  2K+(aq) + 2OH–(aq) + H2(g)
H = -392 kJ
S = 0.047 kJ/K
• The change in Gibbs energy for the reaction is
G = H – T S
= -392 kJ - (298.15 K)(0.047 kJ/K) = -406 kJ
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Entropy and Enthalpy Determine Gibbs Energy,
continued
• You can calculate G in another way because lists of
standard Gibbs energies of formation exist.
• The standard Gibbs energy of formation, G0f, of a
substance is the change in energy that accompanies
the formation of 1 mol of the substance from its
elements at 298.15 K.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Entropy and Enthalpy Determine Gibbs Energy,
continued
• The standard Gibbs energies of formation can be
used to find the G for any reaction.
• Hess’s law also applies when calculating G.
Greaction = Gproducts – Greactants
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Standard Gibbs Energies of Formation
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Chapter 10
Visual Concepts
Equation for Free Energy Change
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Chapter 10
Section 4 Order and Spontaneity
Calculating a Change in Gibbs Energy from
H and S
Sample Problem G
Given that the changes in enthalpy and entropy are –
139 kJ and 277 J/K respectively for the reaction given
below, calculate the change in Gibbs energy. Then,
state whether the reaction is spontaneous at 25C.
C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)
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Chapter 10
Section 4 Order and Spontaneity
Calculating a Change in Gibbs Energy from
H and S, continued
Sample Problem G Solution
H = –139 kJ
S = 277 J/K
T = 25C = (25 + 273.15) K = 298 K
G = H - T S
G = (-139 kJ) - (298 K)(277 J/K)
G = (-139 kJ) - (83 kJ)
G = -222 kJ
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Chapter 10
Section 4 Order and Spontaneity
Calculating a Gibbs Energy Change using
0
 Gf Values
Sample Problem H
Calculate G for the following water-gas reaction.
C(s) + H2O(g)  CO(g) + H2(g)
Is this reaction spontaneous?
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Sample Problem H Solution
For H2O(g), Gf0 = -228.6 kJ/mol.
For CO(g),  Gf = -137.2 kJ/mol.
0
For H2(g),  Gf0 = 0 kJ/mol.
For C(s) (graphite),  Gf0 = 0 kJ/mol.
G = G(products) - G(reactants)
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Chapter 10
Section 4 Order and Spontaneity
Calculating a Gibbs Energy Change using
0
 Gf Values, continued
Sample Problem H Solution, continued
G = G(products) - G(reactants) =
[(mol CO(g))( Gf0 for CO(g)) + (mol H2(g))(Gf0 for H2(g))] –
[(mol C(s))( Gf0 for C(s)) + (mol H2O(g))( Gf0 for H2O(g))] =
[(1 mol)(-137.2 kJ/mol) + (1 mol)(0 kJ/mol)] –
[(1 mol)(0 kJ/mol) – (1 mol)(-228.6 kJ/mol)] =
(-137.2 + 228.6) kJ = 91.4 kJ
The reaction is nonspontaneous under standard conditions.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Predicting Spontaneity
• Does temperature affect spontaneity?
• Consider the equation for G.
G = H - TS
• The terms H and S change very little as
temperature changes.
• The presence of T in the equation for G indicates
that temperature may greatly affect G.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Predicting Spontaneity, continued
• Suppose a reaction has both a positive H value and
a positive S value.
• If the reaction occurs at a low temperature, the value for TS
will be small and will have little impact on the value of G.
• The value of G will be similar to the value of H and will
have a positive value.
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Chapter 10
Section 4 Order and Spontaneity
Gibbs Energy, continued
Predicting Spontaneity
• When the same reaction proceeds at a high enough
temperature, TS will be larger than H and G will
be negative.
• Increasing the temperature of a reaction can make a
nonspontaneous reaction spontaneous.
• A nonspontaneous reaction cannot occur unless some
form of energy is added to the system.
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Chapter 10
Standardized Test Preparation
Understanding Concepts
1. Which of these thermodynamic values can be
determined using an adiabatic calorimeter?
A. ∆G
B. ∆H
C. ∆S
D. ∆T
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Chapter 10
Standardized Test Preparation
Understanding Concepts
1. Which of these thermodynamic values can be
determined using an adiabatic calorimeter?
A. ∆G
B. ∆H
C. ∆S
D. ∆T
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Chapter 10
Standardized Test Preparation
Understanding Concepts
2. Which of these statements about the temperature of a
substance is true?
F. Temperature is a measure of the entropy of the
substance.
G. Temperature is a measure of the total kinetic
energy of its atoms.
H. Temperature is a measure of the average kinetic
energy of its atoms.
I. Temperature is a measure of the molar heat
capacity of the substance.
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Chapter 10
Standardized Test Preparation
Understanding Concepts
2. Which of these statements about the temperature of a
substance is true?
F. Temperature is a measure of the entropy of the
substance.
G. Temperature is a measure of the total kinetic
energy of its atoms.
H. Temperature is a measure of the average kinetic
energy of its atoms.
I. Temperature is a measure of the molar heat
capacity of the substance.
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Chapter 10
Standardized Test Preparation
Understanding Concepts
3. Which of the following pairs of conditions will favor a
spontaneous reaction?
A. a decrease in entropy and a decrease in enthalpy
B. a decrease in entropy and an increase in enthalpy
C. an increase in entropy and a decrease in enthalpy
D. an increase in entropy and an increase in enthalpy
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Chapter 10
Standardized Test Preparation
Understanding Concepts
3. Which of the following pairs of conditions will favor a
spontaneous reaction?
A. a decrease in entropy and a decrease in enthalpy
B. a decrease in entropy and an increase in enthalpy
C. an increase in entropy and a decrease in enthalpy
D. an increase in entropy and an increase in enthalpy
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Chapter 10
Standardized Test Preparation
Understanding Concepts
4. What is the standard enthalpy of formation of N2?
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Chapter 10
Standardized Test Preparation
Understanding Concepts
4. What is the standard enthalpy of formation of N2?
Answer: Because the nitrogen molecule is the normal
form of the element nitrogen, its standard enthalpy of
formation is defined as zero.
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Chapter 10
Standardized Test Preparation
Understanding Concepts
5. What are the circumstances that cause a
nonspontaneous reaction to occur?
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Chapter 10
Standardized Test Preparation
Understanding Concepts
5. What are the circumstances that cause a
nonspontaneous reaction to occur?
Answer: Nonspontaneous reactions require an input of
energy, such as heat or light.
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Chapter 10
Standardized Test Preparation
Understanding Concepts
6. Based on changes in entropy and enthalpy, predict
whether this reaction is spontaneous or
nonspontaneous:
2AB(s)  A2(g) + B2(g) + 799 kJ
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Chapter 10
Standardized Test Preparation
Understanding Concepts
6. Based on changes in entropy and enthalpy, predict
whether this reaction is spontaneous or
nonspontaneous:
2AB(s)  A2(g) + B2(g) + 799 kJ
Answer: ∆H is negative and ∆S is positive. Therefore ∆G
is negative and the reaction is spontaneous.
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Chapter 10
Standardized Test Preparation
Reading Skills
Read the passage below. Then answer the questions.
Almost all of the organisms on Earth rely on the
process of photosynthesis to provide the energy needed
for the functions of living. During photosynthesis carbon
dioxide and water combine to form glucose and oxygen.
The photosynthesis reaction is represented by the
chemical equation:
6CO2(g) + 6H2O(l) + energy  C6H12O6(s) + 6O2(g). The
source of energy for the reaction is light from the sun.
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7. If the absolute value of T∆S is smaller than the
absolute value of ∆H, is photosynthesis a spontaneous
reaction? Explain your answer.
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7. If the absolute value of T∆S is smaller than the
absolute value of ∆H, is photosynthesis a spontaneous
reaction? Explain your answer.
Answer: The reaction is endothermic, requiring an input
of energy. If the absolute value of T∆S is smaller than
that of ∆H, then the Gibbs free energy is positive,
whether ∆S is positive or negative, so the reaction is
nonspontaneous.
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8. Based on the nature of the reactants and products, what can be
deduced about the entropy change during photosynthesis?
F. Entropy increases because one of the products is an
element.
G. Entropy decreases because there are substantially
fewer molecules of product than of reactants.
H. Entropy increases because one of the products is a solid
while one of the reactants is a liquid.
I. Entropy does not change during the reaction because both
sides of the reaction include the same number of atoms.
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8. Based on the nature of the reactants and products, what can be
deduced about the entropy change during photosynthesis?
F. Entropy increases because one of the products is an
element.
G. Entropy decreases because there are substantially
fewer molecules of product than of reactants.
H. Entropy increases because one of the products is a solid
while one of the reactants is a liquid.
I. Entropy does not change during the reaction because both
sides of the reaction include the same number of atoms.
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9. If a manufacturing process was developed to make
glucose from carbon dioxide and water, using heat as
the energy source, how would the amount of energy
required compare to that of the process that plants
use?
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9. If a manufacturing process was developed to make
glucose from carbon dioxide and water, using heat as
the energy source, how would the amount of energy
required compare to that of the process that plants
use?
Answer: The same amount of energy would be needed.
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The table below shows molar heat capacities (joules per
kelvins  mole) of elements and compounds. Use it to
answer questions 10 through 13.
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10. What is the specific heat of aluminum chloride, which
has a molar mass of 133.4?
A. -1.45 J/K•g
B. -0.69 J/K•g
C. 0.69 J/K•g
D. 1.45 J/K•g
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10. What is the specific heat of aluminum chloride, which
has a molar mass of 133.4?
A. -1.45 J/K•g
B. -0.69 J/K•g
C. 0.69 J/K•g
D. 1.45 J/K•g
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11. What is the relationship between the number of
atoms per unit of an ionic compound and its molar
heat capacity?
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11. What is the relationship between the number of
atoms per unit of an ionic compound and its molar
heat capacity?
Answer: The molar heat capacity is about 25 J/K•mol for
each atom in a mole of the ionic compound.
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12. Which of these statements best describes the
general relationship of molar heat capacity of two
different metals?
F. The molar heat capacity of the two metals is about the
same.
G. The difference in molar heat capacity of two
metals depends on the temperature.
H. The molar heat capacity of the metal with the lower
atomic mass is generally smaller.
I. The molar heat capacity of the metal with the higher
atomic mass is generally smaller.
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12. Which of these statements best describes the
general relationship of molar heat capacity of two
different metals?
F. The molar heat capacity of the two metals is about the
same.
G. The difference in molar heat capacity of two
metals depends on the temperature.
H. The molar heat capacity of the metal with the lower
atomic mass is generally smaller.
I. The molar heat capacity of the metal with the higher
atomic mass is generally smaller.
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13. How many joules of heat are required to raise the
temperature of one mole of liquid water by 2.00°C?
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13. How many joules of heat are required to raise the
temperature of one mole of liquid water by 2.00°C?
Answer: 150.6 J
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