Transcript Vann - Chemistry ch. 6.3

PHALL not integrated into presentation
Chapter 6.3-6.4
Thermochemistry
THERMODYNAMICS
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Chapter 6
Table of Contents
•
•
•
•
•
•
6.1
6.2
6.3
6.4
6.5
6.6
The Nature of Energy
Enthalpy and Calorimetry
Hess’s Law
Standard Enthalpies of Formation
Present Sources of Energy
New Energy Sources
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3
Section 6.3
Hess’s Law
Tuesday - November 12, 2013
• CW: Notes 6.3 together Hess’s Law - with new handout
• CW/HW: Practice with Hess’s law w/sheet
• CW: CW: Notes 6.1-6.2 should be finished in class or
homework.
• Calorimetry Lab - Discussion Questions t/g
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4
Section 6.3
Hess’s Law
Calorimetry Lab - Discussion Questions
• 1. Calorimetry is the science of measuring heat. A
calorimeter is an instrument used to measure the
enthalpy of a chemical or physical change.
• 2.a. Calculate the heat of solution for 10.0 g of sodium
carbonate. qsol = (m x c x ΔT)water + (C x ΔT)calorimeter
• q = -(75.0 g x 4.18 J/(°C*g) x 7.4 C + (10 J/°C x 7.4 °C) = -2.4 kJ
• 2b. Calculate the number of moles of sodium carbonate.
• 10.0 g / 106 g/mol = 0.0943 moles
• c. Calculate the enthalpy of solution per mole of sodium carbonate
• ΔH = -2.4 kJ/0.0943 mol = -25 kJ/mol
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5
Section 6.3
Hess’s Law
Calorimetry Lab Discussion Questions #3
• Heat of neutralization
• qneutralization = (m x c x ΔT)water + (C x ΔT)calorimeter
• q = - (110.0 g x 4.18 J/(C*g) x 6.1 C + 10 J/C x 6.1 C)
• q = -2.9 kJ evolved from 60.0 mL of 0.5 M sulfuric acid
and 50.0 mL of sodium hydroxide are mixed.
• 3b. Calculate the number of moles of water
• Sodium hydroxide is limiting reactant, so the number of
moles of NaOH = # moles of water formed from eqn..
• 0.0500 L x 1.0 M NaOH = 0.050 moles of water formed.
• 3c. ethalpy of neutralization per mole of water
• ΔH = -2.9 kJ/0.0500 mol = -58 kJ/mol
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Section 6.3
Hess’s Law
Hess’s Law
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7
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0
Hproducts > Hreactants
H > 0
6.3
Table 6.2 Standard Enthalpies of Formation
for Several Compounds at 25°C
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6–
6.5
Hess’s Law
“In going from a particular
set of reactants to a
particular set of products,
the change in enthalpy is
the same whether the
reaction takes place in one
step or a series of steps.”
np & nr - is the moles of product and reactant respectively.
The other symbol (sigma) means to take the sum of the terms as done in math class.
Hess’s Law Problem Example 1
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2 CH4
Ho
-74.80 kJ
C + O2  CO2
-393.50 kJ
H2 + ½ O2  H2O
-285.83 kJ
CH4 C + 2H2
+74.80 kJ
Step #1: CH4 must appear on the reactant side,
so we reverse reaction #1 and change the sign
on H.
Hess’s Law Problem Example 1
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
Ho
C + 2H2 CH4
-74.80 kJ
C + O2  CO2
-393.50 kJ
H2 + ½ O2  H2O
-285.83 kJ
CH4 C + 2H2
C + O2  CO2
+74.80 kJ
-393.50 kJ
Step #2: Keep reaction #2 unchanged,
because CO2 belongs on the product side
Hess’s Law Problem Example 1
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2 CH4
Ho
-74.80 kJ
C + O2  CO2
-393.50 kJ
H2 + ½ O2  H2O
-285.83 kJ
CH4 C + 2H2
C + O2  CO2
2H2 + O2  2 H2O
+74.80 kJ
-393.50 kJ
-571.66 kJ
Step #3: Multiply reaction #3 by 2
Hess’s Law Problem Example 1
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2 CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4
C + O2
2H2 + O2
CH4 + 2O2
C + 2H2
 CO2
 2 H2O
 CO2 + 2H2O
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
-571.66 kJ
-890.36 kJ
Step #4: Sum up reaction and H
Hess’s Law Problem Example 2
EXAMPLE - Calculate ΔH for the reaction from 2 steps
N2(g) + 2O2(g) → 2NO2(g)
ΔH1 = 68 kJ
N2(g) + O2(g) → 2NO(g)
ΔH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g)
ΔH3 = – 112 kJ
Section 6.3
Hess’s Law
N2(g) + 2O2(g) → 2NO2(g)
ΔH1 = 68 kJ
•
This reaction also can be carried out in
two distinct steps, with enthalpy changes
designated by ΔH2 and ΔH3.
•
N2(g) + O2(g) → 2NO(g)
kJ
2NO(g) + O2(g) → 2NO2(g)
N2(g) + 2O2(g) → 2NO2(g)
•
•
•
ΔH2 = 180
ΔH3 = – 112 kJ
ΔH2 + ΔH3 = 68 kJ
ΔH1 = ΔH2 + ΔH3 = 68 kJ
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17
Hess’s Law Problem Example 2
EXAMPLE - Calculate ΔH for the reaction from 2 steps
N2(g) + 2O2(g) → 2NO2(g)
ΔH1 = 68 kJ
N2(g) + O2(g) → 2NO(g)
ΔH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g)
ΔH3 = – 112 kJ
N2(g) + 2O2(g) → 2NO2(g)
ΔH2 = 180 kJ + ΔH3 = – 112 kJ
180 kJ + (-112) kJ = 68 kJ showing that either path has the same overall
net change in enthalpy.
Calculation of Heat of Reaction EX. 3
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Substance
Hf
CH4
-74.80 kJ
O2
0 kJ
CO2
-393.50 kJ
H2O
-285.83 kJ
Calculation of Heat of Reaction EX. 3
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Substance
Hf
CH4
-74.80 kJ
O2
0 kJ
CO2
-393.50 kJ
H2O
-285.83 kJ
Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]
Hrxn = -890.36 kJ
Section 6.3
Hess’s Law
The Principle of Hess’s Law
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21
Hess’s Law
Click here to watch visualization.
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6–
Section 6.3
Hess’s Law
Characteristics of Enthalpy Changes
•
•
If a reaction is reversed, the sign of ΔH is
also reversed.
The magnitude of ΔH is directly
proportional to the quantities of reactants
and products in a reaction. If the
coefficients in a balanced reaction are
multiplied by an integer, the value of ΔH is
multiplied by the same integer.
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23
Section 6.3
Hess’s Law
Example 4
•
Consider the following data:
•
Calculate ΔH for the reaction
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24
Section 6.3
Hess’s Law
Problem-Solving Strategy
•
•
•
Work backward from the required
reaction, using the reactants and
products to decide how to manipulate the
other given reactions at your disposal.
Reverse any reactions as needed to give
the required reactants and products.
Multiply reactions to give the correct
numbers of reactants and products.
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25
Section 6.3
Hess’s Law
Example 4
•
Reverse the two reactions:
•
Desired reaction:
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Section 6.3
Hess’s Law
Example 4
•
Multiply reactions to give the correct numbers
of reactants and products:
4(
)
• 3(
)
•
•
) 4(
) 3(
Desired reaction:
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27
Section 6.3
Hess’s Law
Example 4
•
Final reactions:
•
Desired reaction:
•
ΔH = +1268 kJ
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28
Section 6.3
Hess’s Law
Stop here for Hess’s Law practice
• Calorimetry Lab report due on Thursday.
• Hess’s Law problems due Thursday.
• Finish Notes 6.1-6.3 by Thursday.
• Hess’s Law Lab on Thursday (?)
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29
Need to add 6.4 to student notes
outline next time for heat of
formation
Monday - Nov. 18, 2013
• CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4
intro.)
• QUIZ - 2 questions - Hess’s Law
• CW: Finish Notes 6.1-6.2 from handout sheet
• CW: Hess’s Law problems finish and turn in
• HW: Quia.com Quiz (9 questions)
• HW: Hess’s Law Informal Lab report due Wed.
• (Be sure Calorimetry Lab has been turned in today to not
lose points for being late.)
• TEST ch. 6 - Tuesday - Nov. 26, 2013 before
Thanksgiving Break (will not include ch. 16 for test now)
Section 6.4
KEY - VOCABULARY DEFINITION
Standard Enthalpies of Formation
Standard Enthalpy of Formation (ΔHf°) is the
•
Change in enthalpy that accompanies the
formation of one mole of a compound
from its elements with all substances in
their standard states.
•
(SEE APPENDIX A19-A22 for Heat of
formation, entropy, and Gibbs Free
Energy values for common compounds
and elements.)
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32
Section 6.4
Standard Enthalpies of Formation
Conventional Definitions of Standard States
•
For a Compound
 If compound is a gas, pressure is exactly 1 atm.
 For a solution, concentration is exactly 1 M.
 Pure substance (liquid or solid)
•
For an Element
 The form [N2(g), K(s)] in which it exists at 1 atm
and 25°C.
 Heat of formation is zero at standard state.
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33
Section 6.4
Standard Enthalpies of Formation
A Schematic Diagram of the Energy Changes for the Reaction
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• ΔH°reaction = –(–75 kJ) + 0 + (–394 kJ) + (–572 kJ) = –891 kJ
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34
Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
1. When a reaction is reversed, the
magnitude of ΔH remains the same, but
its sign changes.
2. When the balanced equation for a
reaction is multiplied by an integer, the
value of ΔH for that reaction must be
multiplied by the same integer.
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35
Section 6.4
Standard Enthalpies of Formation
Problem-Solving Strategy: Enthalpy Calculations
3. The change in enthalpy for a given
reaction can be calculated from the
enthalpies of formation of the reactants
and products:
•
H°rxn = npHf(products) nrHf(reactants)
•
4. Elements in their standard states are
not included in the ΔHreaction calculations
because ΔHf° for an element in its
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36
Section 6.4
Standard Enthalpies of Formation
Exercise
•
Calculate H° for the following reaction:
•
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
•Given the following information:
•
Hf° (kJ/mol)
•
Na(s)
0
•
H2O(l)
–286
•
NaOH(aq)
–470
•
H2(g)
0
•H° = (-470 kJ/mol)(2 mol) + 0 - [(-286 kJ/mol)(2mol) + 0]
•–368 kJ
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Return to TOC
Monday - Nov. 18, 2013
• CW: Notes 6.4 (brief) (6.3 done last Tuesday & 6.4
intro.)
• QUIZ - 2 questions - Hess’s Law
• CW: Finish Notes 6.1-6.2 from handout sheet
• CW: Hess’s Law problems finish and turn in
• HW: Quia.com Quiz (9 questions)
• HW: Hess’s Law Informal Lab report due Wed.
• (Be sure Calorimetry Lab has been turned in today to not
lose points for being late.)
• TEST ch. 6 - Tuesday - Nov. 26, 2013 before
Thanksgiving Break (will not include ch. 16 for test now)
Wednesday - Nov. 20, 2013
• TURN IN informal Hess’s Law Lab Report in box.
• TURN IN Calorimetry Lab (formal lab report) in box if
not already done
• TURN IN Hess’s Law problems worksheet with work shown
or attached.
• CW: Thermodynamics Team Activity Game with Partners
(selected by drawing cards)
• HW: Be sure all ch. 6 notes are caught up 6.1-6.4
• TEST is TUESDAY Nov. 26, 2013 over chapter 6,
calorimetry labs & thermodynamics problems & notes.
• HW/CW: Quia.com Quiz (9 questions)
• HW/CW: Sciencegeek.net Quiz ch. 6 - 15 questions
• Over Thanksgiving Break: Read chapter 16 :)
Friday - Nov. 22 2013
•
•
•
•
•
•
•
•
•
Follow up late items to turn in:
Hess’s Law Informal Lab Report in
Calorimetry Lab (formal lab report)
Hess’s Law problems worksheet with work shown
Hess’s Law - 2 question quiz
To be turned in with TEST on TUESDAY:
1) Thermodynamics Team Activity Game with work shown
2) Interactive ch. 6 notes
3) By Monday Quia.com Quiz (9 questions) at latest - not
accepted late for points.
• TODAY: Show me HW/CW: Sciencegeek.net Quiz AP ch.
6 - 15 question results and work shown if applicable.
• TODAY: Thermochemistry Link Reviews
• TEST is TUESDAY Nov. 26, 2013 over chapter 6,
calorimetry labs & thermodynamics problems & notes.
• Over Thanksgiving Break: Read chapter 16 :)
Section 6.4
Standard Enthalpies of Formation
QUIZ with Hess’s Law
• 1. Find the ΔH for the reaction below, given the following
reactions and subsequent ΔH values:
N2(g) + 2O2(g) → 2NO 2(g)
• N2(g) + 3H2(g) → 2NH3(g)
ΔH = -115 kJ
2NH3(g) + 4H2O(l) → 2NO2(g) + 7H2(g) ΔH = -142.5 kJ
H2O(l) → H2(g) + 1/2O 2(g)
ΔH = -43.7 kJ
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41
Section 6.4
Standard Enthalpies of Formation
QUIZ with Hess’s Law
• 2. Find the ΔH for the reaction below, given the following
reactions and subsequent ΔH values:
CO2(g) → C(s) + O2(g)
• H2O(l) → H2(g) + 1/2O2(g)
ΔH = 643 kJ
C2H6(g) → 2C(s) + 3H 2(g)
ΔH = 190.6 kJ
2CO2(g) + 3H2O(l) → C 2H6(g) + 7/2O2(g) ΔH = 3511.1 kJ
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42
Chapter 6
Questions
Thermochemistry
QUESTION #1
The combustion of a fuel is an exothermic process. This means…
1. the surroundings have lost exactly the amount of energy gained
by the system.
2. the potential energy of the chemical bonds in the products
should be less than the potential energy of the chemical bonds in
the reactants.
3. q must be positive; w must be negative
4. E would have a + overall value because the surroundings have
gained energy.
Copyright © Houghton Mifflin Company. All rights reserved.
CRS Question, 6–
ANSWER
Choice 2 correctly summarizes the connection between reactants and
products for an exothermic process. The exothermic nature of the
process means that the products must be less energetic than the
reactants. The energy descriptions and comparisons are from the view
point of the system.
Section 6.1: The Nature of Energy
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CRS Question, 6–
QUESTION #2
The change in enthalpy, or heat of reaction, for chemical changes is
equal to the heat flow when…
1.
2.
3.
4.
the pressure is constant and only PV work is allowed.
H = –q (constant pressure)
H = E – PV
the pressure is constant and enthalpy = internal energy (E)
Copyright © Houghton Mifflin Company. All rights reserved.
CRS Question, 6–
ANSWER
Choice 1 properly relates enthalpy to q. At constant pressure, any
work is change in volume work. So, H = E + PV and since q (at
constant pressure) = E + PV it follows that H = q (at constant P
where most chemical reactions take place).
Section 6.2: Enthalpy and Calorimetry
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CRS Question, 6–
QUESTION #3
Acetylene torches used for welding work so effectively because of
the large release of energy (H) when acetylene, C2H2, reacts with
oxygen. For the combustion of acetylene, H is approximately
–1,300 kJ/mol. What is the value for H when 1.0 gram of
acetylene reacts in this way?
1.
2.
3.
4.
Approximately +50. kJ
Approximately –50. kJ
Approximately +34,000 kJ
Approximately –34,000 kJ
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CRS Question, 6–
ANSWER
Choice 2 indicates the proper sign, value, and label. The reported
–1,300 kJ is for the combustion of one mole. After 1.0 gram is
properly converted to moles, (1.0/26.04) multiplying by –1,300
kJ/mol will yield the answer. The negative sign indicates that heat is
released to the surroundings.
Section 6.2: Enthalpy and Calorimetry
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CRS Question, 6–
QUESTION #4
The instant cold packs associated with
athletic injuries makes use of the heat
change when NH4NO3 dissolves. If 8.0
grams of NH4NO3 (molar mass = 80.1)
were placed in 100.0 mL of water such as
shown in the constant pressure
calorimeter, and the temperature change
was – 6.0°C, what would you calculate as
the H for dissolving one mole of
NH4NO3?
1.
2.
3.
4.
–2,500 kJ/mol
+2,500 kJ/mol
–25 kJ/mol
+25 kJ/mol
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CRS Question, 6–
ANSWER
Choice 4 shows both the proper sign for this endothermic process
and the proper value. The constant 4.184 J/°C g was used to
determine the heat absorbed from the 100.0 mL of water as its
temperature dropped by 6.0°C, during dissolving. Once this was
known for 8.0 grams, a ratio to 80.1 grams was calculated.
Section 6.2: Enthalpy and Calorimetry
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CRS Question, 6–
QUESTION #5
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CRS Question, 6–
QUESTION (continued)
The heat capacity of a calorimeter of the type shown on the previous
slide must be determined before other investigations, using the
instrument, can be performed. When 0.5000 grams of pure carbon
are combusted in the “bomb” the temperature rose by 1.842°C.
(Note: use –393.5 kJ/mol as the heat of combustion for carbon.)
What is the calorimeter’s heat capacity?
1.
2.
3.
4.
30.20 kJ/°C
106.8 kJ/K
8.894 kJ/°C
35.60 kJ/°C
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CRS Question, 6–
ANSWER
Choice 3 is correct. The heat released per mole must be converted to
heat released for 0.5000 gram (0.5000/12.01)  –393.5. This should be
divided by the number of degrees the calorimeter registered and the
sign changed to be positive.
Section 6.2: Enthalpy and Calorimetry
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CRS Question, 6–
QUESTION #6
A certain small piece of candy is made of 1.525 grams of sucrose
(C12H22O11; molar mass = 342.30 g/mol). When placed in a bomb
calorimeter with a specific heat of 5.024 kJ/°C, how much should the
temperature rise when the candy undergoes combustion? (Note: use
–5,640 kJ/mol as the heat released for the combustion of one mole of
sucrose.)
1.
2.
3.
4.
5.00°C
126°C
25.1°C
Couldn’t I just eat the candy, I am not sure what to do with the
calculations.
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CRS Question, 6–
ANSWER
Choice 1 (although choice 4 might be tastier) is correct. Since it is
given that one mole of sucrose releases 5,640 kJ, the determination
of the number of moles combusted would enable a ratio to find the kJ
released from 1.525 g (1.525/342.30)  5,640). Then the heat released
by the combustion would be divided by the kJ/°C for the calorimeter;
the result would be the change in temperature.
Section 6.2: Enthalpy and Calorimetry
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CRS Question, 6–
QUESTION #7
The following reaction shows the conversion of ozone to oxygen
and provides the standard enthalpy change for the process:
2 O3  3 O2; H = –427 kJ
What is the H value for the conversion of one mole of O2 to O3?
1.
2.
3.
4.
+427 kJ
+641 kJ
+142 kJ
+285 kJ
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CRS Question, 6–
ANSWER
Choice 3 provides both the proper sign and value for the new
reaction. The original reaction must be reversed in order to show
oxygen forming ozone. This means the given sign for H must also
be reversed. Since the reversed reaction shows the process for 3
moles of oxygen, the H value must be reduced by 1/3 to determine
the value for only one mole.
Section 6.3: Hess’s Law
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CRS Question, 6–
QUESTION #8
The following reaction depicts one way in which nitric acid could be
formed:
N2O5 + H2O  2 HNO3
H = ?
Use the following information, and apply Hess’s law, to determine
the missing value.
H2 + 1/2 O2  H2O; H = –285.8 kJ
½ N2 + 3/2 O2 + ½ H2  HNO3; H = –174.1 kJ
2N2 + 5 O2  2 N2O5; H = + 28.4 kJ
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CRS Question, 6–
QUESTION (continued)
1.
2.
3.
4.
–362.4 kJ
–69.5 kJ
–76.6 kJ
I am having difficulty rearranging the given equations to produce
the “target” equation.
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CRS Question, 6–
ANSWER
Choice 3 is the answer when the three given equations are properly
arranged. The arrangement that, when added, yields the desired
equation is: reverse the first given equation, double the next one, and
finally take ½ and reverse the third equation. Each manipulation
should be done to the respective H values before tallying those to
yield the final H value.
Section 6.3: Hess’s Law
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CRS Question, 6–
QUESTION #9
Using information from Appendix Four, determine the standard
enthalpy change at 25°C for the combustion of one mole of heptane
(C7H16) sometimes used in gasoline fuel mixtures. The balanced
equation is shown here:
C7H16 (g) + 11 O2 (g)  7 CO2 (g) + 8 H2O (l)
(Note: the standard molar enthalpy change of formation of C7H16
(g) is –264 kJ/mol)
1.
2.
3.
4.
–4430 kJ
–4780 kJ
–5310 kJ
This can’t be solved without the heat of formation of oxygen.
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CRS Question, 6–
ANSWER
Choice 2 provides the correct sign and value. The moles of each
reactant must be multiplied by the heat of formation of each
component (noting their physical state), then the sum of the enthalpy
for reactants is subtracted from the sum of the enthalpies of the
products (noting that oxygen has a zero for enthalpy of formation).
Section 6.4: Standard Enthalpies of Formation
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CRS Question, 6–
QUESTION #10
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CRS Question, 6–
QUESTION (continued)
After examining the graph shown here, which energy source
would you like to see increase and which would you like to see
decrease? Be prepared to explain your choice.
1.
2.
3.
4.
Wood increase; nuclear decrease because…
Hydro and nuclear increase, petroleum decrease because…
Coal increase, hydro and nuclear decrease because…
I have another choice besides these, it is…
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CRS Question, 6–
ANSWER
While there are a variety of opinions, the decrease or increase of
energy sources will depend on a complex combination of politics,
life style choices, economics and technology developments.
Section 6.5: Present Sources of Energy
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CRS Question, 6–
QUESTION #11
One source of energy being considered as a replacement for
petroleum fuels is the combustion of hydrogen. There are several
technical considerations to bring this about, but the heat of
combustion reaction is part of the determination. How much heat,
under standard conditions and 25°C, could be obtained from the
combustion of 2.00 kg of H2?
H2 (g) + ½ O2 (g)  H2O (l)
1.
2.
3.
4.
–572,000 kJ
–484,000 kJ
–284,000 kJ
–242,000 kJ
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CRS Question, 6–
ANSWER
Choice 3 has taken into account the kg to g to mole conversion
for 2.00 kg of H2 and the stoichiometry. Since combustion of one
mole of H2 in the equation yields 286 kJ, the number of moles of
H2 multiplied by the H value for the equation (using H2O (l))
will yield the correct value.
Section 6.6: New Energy Sources
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CRS Question, 6–