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Chapter Six:
THERMOCHEMISTRY
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Contentsp228
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6-1 Nature of Energyp229
The Law of Conservation of Energy
Kinetic Energy Heat Work
Pathway State Function or State Property
Chemical Energy
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Initial Position
In the initial position, ball A has a higherpotential energy than ball B.
Figure 6.1
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After A has rolled down the hill, the potential
energy lost by A has been converted to
random heading) and to the increase motions
of the components of the hill (frictional in the
potential energy of B.
p230
Final Position
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Final Position
After A has rolled down the hill, thepotential energy lost by A has beenconverted to random motions of thecomponents of the hill (frictional heading)and to the increase in the potential energyof B.
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System
Surrounding
Exothermic: exo- is meaning “out off”; energy flows out ofthe system.
Endothermic: the heat flow is into a system is endothermic.Reactions that absorb energy fromsurroundings are said to be endothermic.
p231
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Chemical Energy p231
Figure 6.2
The combustion of methane releases the quantity ofenergy Δ (PE) to surrounding via heat flow, This anexothermic process.
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The study of energy and its interconversions is called thermodynamics.The conservation of energy is often called the firstlaw of thermodynamics. The energy of the universeis constant.
The internal energy of E of a system can be definedmost precisely as the sum of the kinetic and potentialenergies of all the “particles”in the system.
Surroundings Surroundings
p232
ΔE = q + w
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Ex 6.1 Internal EnergyCalculate ΔE for a system undergoing an
endothermic process in which 15.6 kJ of heat flows
and where 1.4 kJ of work is done on the system.Solution:
P233
ΔE = q + w
Where q = + 15.6 kJ, since the process isendothermic, and w = + 1.4 kJ, since work isdone on the system. Thus
ΔE = 1.56 kJ + 1.4 kJ = 17.0 kJ
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p233
Work = force x distance = F x △h
Since P = F/A or F = P x A
Work = F x △h = P x A x △h
△V = final volume –initial volume
= A x △h
Work = P x A x △h = P △V
W = - P △V
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Calculate the work associated with the expansion ofa gas from 46 L to 64 L at a constant externalpressure of 15 atm.Solution:
P234Ex 6.2 PV Work
For a gas at constant pressure,
w = - PΔV
In this case P = 15 atm , andΔV = 64 - 46 = 18 L. Hence
w = -15 atm x 18 L = - 270 atm ∙L
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Ex 6.3 Internal Energy, Heat, and WorkA balloon is being inflated to its full extent by heating theair inside it. In the final stages of this process, the volumeof the balloon changes from 4.00 × 106 L to 4.50 × 106 L bythe addition of 1.0 atm, calculate ΔE for the process. (To
convert between L˙atm = 101.3 J.)Solution:
P234
V = 4.50 x 106 L - 4.00 x 106 L = 0.50 x 106 L
Thus w = - (1.0 atm) x (5.0 x 105 L) x (101.3 J/(L· atm)
= - 5.1 x 107 J
Then E = q + w = (+1.3 x 108 J) + ( -5.1 x 107 J)
= 8 x 107 J
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Ex 6.4 EnthalpyWhen 1 mole of methane (CH4) is burned atconstant pressure, 890 kJ of energy is released asheat. Calculate ΔH for a process in which a 5.8-gsample of methane is burned at constant pressure.Solution:
P236
5.8 g ÷(16 g/mole) = 0.36 mol CH4, and
0.36 mol x ( - 890 kJ/mol) = - 320 kJ
Thus, when 5.8-g Sample of CH4 is burned atconstant pressure,
ΔH = heat flow = -320 kJ
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Coffee Creamer Flammability
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Sugar and Potassium Chlorate
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Thermite
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Is the freezing of water an endothermic or
exothermic process? Explain.
React 1
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Classify each process as exothermic or endothermic.
Explain.
a) Your hand gets cold when you touch ice.
b) The ice gets warmer when you touch it.
c) Water boils in a kettle being heated on a
stove.
d) Water vapor condenses on a cold pipe.
e) Ice cream melts.
React 2
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Work vs. Energy Flow
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6-2 Enthalpy and Calorimetryp235
Enthalpy H = E + PV
Change in H = (Change in E) + (Change in PV)
△E = qp + W; △E = qp- P△V; qp = △E + P△V
△H = △E + △ (PV); △(PV) = P △V
△H = △E + P△V
qp =△E + P△V
△H = qp
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p236
ΔH = ΔE + Δ (PV), Δ(PV) = P ΔV
ΔH = ΔE + PΔV
qP = ΔE + PΔV
ΔH = qP
ΔH = Hproducts - Hreactants
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Calorimetry p237
Specific heat capacity:its unit L/K ‧ g
Molar heat capacity:it has the units J/K‧ mol
Constant-pressure calorimetryA Coffee Cup CalorimeterMade of Two StyrofoamCups
Figure 6-5
H+(aq) + OH-(aq) → H2O(l)
Heat capacity
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p238Energy released by the neutralization reaction
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p242
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p242Ex 6.6 Constant-Volume CalorimetryIt has been suggested that hydrogen gas obtainedby the decomposition of water might be a substitutefor natural gas (principally methane). To comparethe energies of combustion of these fuels, thefollowing experiment was carried out using a bombcalorimeter with a heat capacity of 11.3 kJ/℃.When a 1.50-g sample of methane gas burned withexcess oxygen in the calorimeter, the temperatureincreased by 7.3℃. When a 1.15-g sample ofhydrogen gas was burned with excess oxygen, thetemperature increase was 14.3℃. Calculate theenergy of combustion (per gram) for hydrogen andoxygen.
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Solution
Energy released in the combustion of 1.5 g CH4
= 11.3 kJ/℃)(7.3℃) = 8.3 kJ
Energy released in the combustion of 1 g CH4
= 83 kJ/(1.5 g) = 55 kJ/g
Similarly, for hydrogen
Energy released in the combustion of 1.15 g H2
= (11.3 kJ/℃)(14.3℃) = 162 kJ
Energy released in the combustion of 1 g H2
= 162 kJ/(1.15 g) = 141 kJ/g
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You have a Styrofoam cup with 50.0 g of water at
10C. You add a 50.0 g iron ball at 90C to the
water. The final temperature of the water is:
a) Between 50°C and 90°C.
b) 50°C
c) Between 10°C and 50°C.
Calculate the final temperature of the water.
React 3
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p242
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6-3 Hess’s Law p242
Figure 6-7 The principle of Hess’slaw. The same change in
enthalpy occurs when nitrogen and oxygen react to form
nitrogen dioxide, regardless of whether the reaction occurs
in one (red) or two (blue) steps.
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p243
N2(g) +2O2(g) → 2NO2 (g), △H1 = 68 kJ
N2(g) + O2(g) → 2NO(g), ΔH2 = 180 kJ
2NO(g) + O2(g) → 2NO2(g), ΔH3 = -112 kJ
Net reaction: N2(g) + 2O2(g) → 2NO2(g), ΔH2 + ΔH3 = 68 kJ
ΔH1 = ΔH2 + ΔH3 = 68 kJ
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Characteristic of Enthalpy Changes p243
1. If a reaction is reversed, the sign of ΔH is also reversed.
2.The magnitude of ΔH is directly proportional to thequantities of reactants and products in a reaction. If thecoefficients in a balanced reaction are multiplied by integer,the value of ΔH is multiplied by the same integer.
Xe(g) + 2F2(g) → XeF4(s), ΔH = -251 kJ
XeF4(s) → Xe(g) + 2F2(g), ΔH = ?
Crystals of xenon tetrafluoride, the first reportedbinary compound containing a noble gas element.
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Hess’s Law
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Ex 6.7 Hess’s Law
Two forms of carbon are graphite, the soft, black,
slippery material used in “lead”pencils and as a lubricant
for locks, and diamond, the brilliant, hard gemstone.
Using the enthalpies of combustion for graphite (-394
kJ/mol) and diamond (-396 kJ/mol), calculate ΔH for the
conversion of graphite to diamond:
P244
)()( sCsC diamondgraphite Solution:
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Hints for Using Hess’s Law p246
1. Work backward from the required reaction, using the
reactants and products to know to manipulate the
o t h e r g i v e n r e a c t a n t s a t y o u r d i s p o s a l .
2. Reverse any reactions as needed to give the required
reactants and products.
3. Multiply reactions to give the correct numbers of
reactants and products.
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6-4 Standard Enthalpies of Formationp246
Cgraphite(s) → Cdiamond(s)
Standard enthalpy of formation is defined as the change
in enthalpy that accompanied the formation of one mole
of a compound from its elements with all substances in
their standard states.
A degree symbol on a thermodynamic function, for
example, ΔH0, indicates that the corresponding process
has been carried out under standard conditions.
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Conventional Definitions of Standard states p246
For a compound
1. The standard state of a gaseous substance is a pressure ofexactly 1 atmosphere.
2. For a pure substance in a condensed state (liquid or solid),the standard state is the pure liquid or solid.
3. For a substance present in a solution, the standard is aconcentration of exactly 1 M.
For an Element
The standard state of an element is the which the element existsunder conditions of 1 atmosphere and 25 C.( The standard state foroxygen is O2(g) at a pressure of 1 atmosphere; the standard state forsodium is Ma(s); the standard for mercury is Hg(l); and so on. )
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p247
)(2H)(C)(CH 24 gsg
kJ/mol75ΔΗ)(CH)(2H)(C 0f42 ggs
kJ/mol286)(OH)(O)(H 0f222
2
1 Hlgg
?)(OH)(CO)(O2)(CH oreaction2224 Hlggg
kJ/mol394)(CO)(O)(C of22 Hggs
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A Schematic Diagram of the Energy Changesfor the Reaction
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
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p247
H°reaction = npHf(products) - nrHf(reactants) (6.1)
Change in enthalpy can be calculated from enthalpiesof formation of reactants and products.
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Standard States
Compound
For a gas, pressure is exactly 1 atmosphere.
For a solution, concentration is exactly 1 molar.
Pure substance (liquid or solid)
Element
The form [N2(g), K(s)] in which it exists at 1 atmand 25°C.
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Using the standard enthalpies of formation listed in
Table 6.2, calculate the standard enthalpy change for
the overall reaction that occurs when ammonia is
burned in air to form nitrogen dioxide and water. This
is first step in the manufacture of nitric acid.
P249Ex 6.9 Enthalpies from Standard Enthalpiesof Formation
)(6)(4)(7)(4 2223 lOHgNOgOgNH Solution:
ΔH0reaction = { - 4 mol [-(46 kJ/mol)]} + [ - 7 mol ( 0
kJ/mol )] + [ 4 mol ( + 34 kJ/mol)] +
[ 6 mol ( - 286 kJ)]
= - 1396 kJ
![Page 43: Chapter Six - 國立臺北科技大學Taipei Techchpro/Chem/Chap6.pdfThus, when 5.8-g Sample of CH4 is burned at constant pressure, ΔH = heat flow = -320 kJ 15 Coffee Creamer Flammability](https://reader031.vdocuments.pub/reader031/viewer/2022022016/5b6007297f8b9ac1478b64fb/html5/thumbnails/43.jpg)
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Ex 6.11 Enthalpies from Standard of Formation III
Methanol (CH3OH) is often used as fuel in high-performance
engines in race cars. Using the data in Table 6.2, compare the
standard enthalpy of combustion per gram of methanol with
that per gram of gasoline. Gasoline is actually a mixture of
compounds, but assume for this problem that gasoline is
liquid octane (C8H18).
Solution )(O4H)(CO2)(O3)(OHCH2 2223 lggl
p252
![Page 44: Chapter Six - 國立臺北科技大學Taipei Techchpro/Chem/Chap6.pdfThus, when 5.8-g Sample of CH4 is burned at constant pressure, ΔH = heat flow = -320 kJ 15 Coffee Creamer Flammability](https://reader031.vdocuments.pub/reader031/viewer/2022022016/5b6007297f8b9ac1478b64fb/html5/thumbnails/44.jpg)
p252For Methanol
For Octane
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6-5 Present Sources of Energy p252
Petroleum andNatural Gas
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p257
Figure 6.14
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Hydrogen as a Fuel
HgOHgCH 3)()( 24
kJ286ΔHO(l)H(g)O(g)H 0222
2
1
p257
CO(g)(g)3HO(g)H(g)CH 224
![Page 48: Chapter Six - 國立臺北科技大學Taipei Techchpro/Chem/Chap6.pdfThus, when 5.8-g Sample of CH4 is burned at constant pressure, ΔH = heat flow = -320 kJ 15 Coffee Creamer Flammability](https://reader031.vdocuments.pub/reader031/viewer/2022022016/5b6007297f8b9ac1478b64fb/html5/thumbnails/48.jpg)
Wind turbines to create electricity.
p259
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49
Ex 6.12 Enthalpies of CombustionCompare the energy available from the combustion of given
volume of methane and the same volume of hydrogen at the
same temperature and pressure.
P260
Solution:
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50
Ex 6.13 Comparing Enthalpies of CombustionAssuming that the combustion of hydrogen gas
provides three times as much energy per gram as
gasoline, calculate the volume of liquid H2 (density
= 0.0710 g/mL) required to furnish the energy
contained in 80.0 L (about 20 gal) of gasoline
(density = 0.740 g/mL). Calculate also the volume
that this hydrogen would occupy as a gas at 1.00
atm and 25℃.
P261
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Solution: p261
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52
Energy Sources Used in the United States
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53
The Earth’sAtmosphere