Ch6z5ethermo 110115225500-phpapp02

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Chapter 6Chapter 6
EnergyEnergy
ThermodynamicsThermodynamicspppp

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6.1 The Nature of Energy6.1 The Nature of Energy
Energy is . . .Energy is . . .
The ability to do work.The ability to do work.
Conserved (It’s the Law).Conserved (It’s the Law).
Made ofMade of heatheat andand workwork..
A state function (unlike work and heat).A state function (unlike work and heat).
I.e.I.e., it’s independent of the path, or how, it’s independent of the path, or how
you get from point A to B.you get from point A to B.

Enthalpy is a STATE FUNCTION. State
functions are PATH-INDEPENDENT

44
Work and HeatWork and Heat
WorkWork is a forceis a force acting overacting over a distance.a distance.
HeatHeat is energyis energy transferredtransferred betweenbetween
objects because of temperature difference.objects because of temperature difference.

55
The universeThe universe
Is divided into two halves.Is divided into two halves.
TheThe systemsystem and theand the surroundingssurroundings..
The system is the part you areThe system is the part you are
concerned with.concerned with.
The surroundings are the rest.The surroundings are the rest.
ExothermicExothermic reactionsreactions releaserelease energyenergy toto
the surroundings.the surroundings.
EndothermicEndothermic reactionsreactions absorbabsorb energyenergy
fromfrom the surroundings.the surroundings.

66
CH + 2O CO + 2H O + Heat4 2 2 2→
CH + 2O4 2
CO + 2 H O2 2
Potentialenergy
Heat

77
N + O2 2
Potentialenergy
Heat
2NO
N + O 2NO2 2 + heat →

88
DirectionDirection
Every energy measurement has threeEvery energy measurement has three
parts.parts.
1.1. UnitUnit ( Joules or calories).( Joules or calories).
2.2. NumberNumber - how many.- how many.
3.3. SignSign - to tell direction.- to tell direction.
NegativeNegative -- exoexothermicthermic
PositivePositive -- endoendothermicthermic

99
System
Surroundings
Energy
∆E < 0

1010
System
Surroundings
Energy
∆E >0

1111
Same rules for heat and workSame rules for heat and workpppp
HeatHeat given offgiven off isis negativenegative..
HeatHeat absorbedabsorbed isis positivepositive..
WorkWork donedone byby systemsystem onon surroundingssurroundings
isis negativenegative..
Work doneWork done onon systemsystem byby surroundingssurroundings
isis positivepositive..
ThermodynamicsThermodynamics - The study of energy- The study of energy
andand the changes it undergoes.the changes it undergoes.

1212
First Law of ThermodynamicsFirst Law of Thermodynamics
TheThe energyenergy of the universe isof the universe is constantconstant..
Law of conservation ofLaw of conservation of energyenergy..
q = heatq = heat
w = workw = work
∆∆E = q + wE = q + w
Take theTake the system’ssystem’s point of view topoint of view to
decide signs.decide signs.

1313
What is work?What is work?
Work is a force acting over a distance.Work is a force acting over a distance.
w= F xw= F x ∆∆dd
P = F/areaP = F/area
d = V/aread = V/area
ww = (P x area) x= (P x area) x ∆∆ (V/area)(V/area) = P= P∆∆VV
Work can be calculated by multiplyingWork can be calculated by multiplying
pressure by the change in volume atpressure by the change in volume at
constant pressure.constant pressure.
units of work: liter-atm or L-atmunits of work: liter-atm or L-atm

1414
Work needs a signWork needs a sign
If theIf the volumevolume of a gasof a gas increasesincreases, the, the
systemsystem has done work on thehas done work on the
surroundingssurroundings (memorize this!).(memorize this!).
And . . . work isAnd . . . work is negativenegative: w = (-): w = (-)
w = - Pw = - P∆∆V (don’t forget negative sign)V (don’t forget negative sign)
ExpandingExpanding a gas:a gas: Work is negativeWork is negative..
ContractingContracting:: Surroundings do workSurroundings do work onon
thethe systemsystem and w = (+).and w = (+).
1 L-atm = 101.325 J (need for1 L-atm = 101.325 J (need for
conversions)conversions)

1515
Fig 6.4 p. 246, VolumeFig 6.4 p. 246, Volume
of a Cylinderof a Cylinder
a.a. The piston, movingThe piston, moving
a distance delta ha distance delta h
against pressure P,against pressure P,
does work on thedoes work on the
surroundings.surroundings.
b.b. Since V of cylinder =Since V of cylinder =
area of base xarea of base x
height, the delta Vheight, the delta V
of gas = delta h x Aof gas = delta h x A

1616
ExamplesExamplespppp
What amount ofWhat amount of workwork is done when 15.00 Lis done when 15.00 L
of gas isof gas is expandedexpanded to 25.00 L at 2.40 atmto 25.00 L at 2.40 atm
pressure? Steps . . .pressure? Steps . . .
ww = -P∆V = -2.40 atm • 10.00L == -P∆V = -2.40 atm • 10.00L = -24 L-atm-24 L-atm
If 2.36If 2.36 kJkJ ofof heatheat are absorbed what is theare absorbed what is the
change in energy? Steps . . .change in energy? Steps . . .
∆∆EE == qq ++ ww = 2360= 2360 JJ + (-24 L-atm)(101.325+ (-24 L-atm)(101.325 JJ/L-atm)/L-atm)
== -71.8 J-71.8 J
How much heat to change the gasHow much heat to change the gas withoutwithout
changingchanging the internal energy of the gas?the internal energy of the gas?
Want ∆E = 0, so q + w = 0Want ∆E = 0, so q + w = 0
since w = -2430 J; q added must = +2430 Jsince w = -2430 J; q added must = +2430 J

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6.2 Enthalpy & Calorimetry6.2 Enthalpy & Calorimetry
EnthalpyEnthalpy is abbreviated as His abbreviated as H
H = E + PV (that’s the definition ofH = E + PV (that’s the definition of
heat)heat)
AtAt constantconstant pressurepressure ∆∆H =H = ∆∆E + PE + P∆∆VV
The heat at constant pressure qThe heat at constant pressure qpp can becan be
calculated from . . .calculated from . . .
∆∆E = qE = qpp + w = q+ w = qpp - P- P∆∆V (w = - PV (w = - P∆∆V)V)
qqpp == ∆∆E + PE + P ∆∆V =V = ∆∆HH

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CalorimetryCalorimetry
Measuring heat -- use a calorimeter.Measuring heat -- use a calorimeter.
Two kinds: Constant P or constant VTwo kinds: Constant P or constant V
Constant pressure calorimeter (called aConstant pressure calorimeter (called a
coffee cup calorimeter, open to outside)coffee cup calorimeter, open to outside)
Heat capacity for a material, CHeat capacity for a material, Cpp isis
calculated.calculated.
CCpp= heat absorbed/= heat absorbed/∆∆T =T = ∆∆H/H/∆∆TT
Specific heat capacity = Cp/massSpecific heat capacity = Cp/mass

1919
Specific heat capacity = Cp/massSpecific heat capacity = Cp/mass
Heat = specific heat x m xHeat = specific heat x m x ∆∆TT
q = m•∆T•Cq = m•∆T•Cpp
Sometimes q is shown as ∆HSometimes q is shown as ∆H
∆H = m•∆T•C∆H = m•∆T•Cpp (online HW)(online HW)
MolarMolar heat capacity = Cheat capacity = Cpp//molesmoles
Heat = molar heat x moles xHeat = molar heat x moles x ∆∆TT
Make the units work and you’ve done theMake the units work and you’ve done the
problem right (get dates in college).problem right (get dates in college).
Be careful to note if CBe careful to note if Cpp usesuses gramsgrams oror molesmoles..

2020
A coffee cup calorimeter measuresA coffee cup calorimeter measures ∆∆H.H.
It’s an insulated cup, full of water.It’s an insulated cup, full of water.

2121
Figure 6.5Figure 6.5
A Coffee-CupA Coffee-Cup
Calorimeter Made ofCalorimeter Made of
Two Styrofoam CupsTwo Styrofoam Cups

2222
Specific heat ofSpecific heat of HH22OO = 1 cal/gºC= 1 cal/gºC
= 4.184 J/g= 4.184 J/goo
CC
Heat of reaction=Heat of reaction= ∆∆H = sh x mass xH = sh x mass x ∆∆TT
(see this form on online HW)(see this form on online HW)
Also seen as: q = mAlso seen as: q = m∆∆TTCCpp
See, Table 6.1 p. 237 (next page) forSee, Table 6.1 p. 237 (next page) for CCpp
of some common substances. Note theof some common substances. Note the
units! (J/ºC•g)units! (J/ºC•g)

2323
Z5e 237 Table 6.1Z5e 237 Table 6.1

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ExamplesExamples
CCpp of graphite is 0.710 J/gºC.of graphite is 0.710 J/gºC.
Calculate q needed to raise theCalculate q needed to raise the
temperature of 75.0temperature of 75.0 kgkg of it from 294 Kof it from 294 K
to 348 K. . .to 348 K. . .
Convert Kg to g; q = mConvert Kg to g; q = m∆∆TCTCpp . Ans . . .. Ans . . .
= 2 880 000 J = 2880 kJ= 2 880 000 J = 2880 kJ

2525
ExamplesExamplespppp
46.2 g of copper heated to 95.4ºC & placed in46.2 g of copper heated to 95.4ºC & placed in
calorimeter with 75.0 g water at 19.6ºC. Finalcalorimeter with 75.0 g water at 19.6ºC. Final
temp. of both is 21.8ºC. What is Ctemp. of both is 21.8ºC. What is Cpp of Cu? . . .of Cu? . . .
q gained by Cu = q lost by Hq gained by Cu = q lost by H22O; set bothO; set both
equations equal to each other, solve . . .equations equal to each other, solve . . .
qqCuCu = m= mcucu∆T∆TCuCuCpCpCuCu == mmwatwat∆T∆TwatwatCpCpwatwat = q= qwatwat CpCpwatwat ==
4.18 J/ºC•g (memorize). Use absolute value for4.18 J/ºC•g (memorize). Use absolute value for
∆T; plug-and-chug to get answer of . . .∆T; plug-and-chug to get answer of . . .
0.203 J/gºC for Cu0.203 J/gºC for Cu
Don’t need to convert ºC to K since both haveDon’t need to convert ºC to K since both have
the same degree increments & we are onlythe same degree increments & we are only
interested in theinterested in the changechange..
Need to know this for online HW05 Q8Need to know this for online HW05 Q8

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ConstantConstant volumevolume calorimeter is called acalorimeter is called a
bomb calorimeter.bomb calorimeter.
Material is put in a container with pureMaterial is put in a container with pure
oxygen. Wires are used to start theoxygen. Wires are used to start the
combustion. The container is put into acombustion. The container is put into a
container of water.container of water.
The heat capacity of the calorimeter isThe heat capacity of the calorimeter is
known and tested.known and tested.
SinceSince ∆∆V = 0, PV = 0, P∆∆V = 0,V = 0, ∆∆E = qE = q

2727
Bomb CalorimeterBomb Calorimeter
thermometerthermometer
stirrerstirrer
full of waterfull of water
ignition wireignition wire
Steel bombSteel bomb
samplesample

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Bomb Calorimetry ExampleBomb Calorimetry Examplepppp
T increases by 3.2ºC when 0.1964 g of quinoneT increases by 3.2ºC when 0.1964 g of quinone
(C(C66HH44OO22) is burned in a bomb calorimeter with heat) is burned in a bomb calorimeter with heat
capacity of 1.56 kJ/ºC. What is the combustioncapacity of 1.56 kJ/ºC. What is the combustion
energy of quinone perenergy of quinone per gramgram and alsoand also perper molemole??
Heat gained by calorimeter =Heat gained by calorimeter = 1.56kJ/ºC x1.56kJ/ºC x
3.2 ºC = 5.0 kJ = heat loss by quinone3.2 ºC = 5.0 kJ = heat loss by quinone
∆∆EEcombcomb = -5.0kJ/0.1964 g = -25kJ/= -5.0kJ/0.1964 g = -25kJ/gg
∆∆EEcombcomb = -25kJ/g x 108.09g/mol == -25kJ/g x 108.09g/mol = --
2700kJ/2700kJ/molmol
Text problem #56 is like this (HW)Text problem #56 is like this (HW)
(q = m∆TC(q = m∆TCpp + heat gained by calorimeter)+ heat gained by calorimeter)

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PropertiesProperties
IntensiveIntensive properties:properties: NotNot related to therelated to the
amount of substance.amount of substance.
E.g.E.g., density, specific heat, temperature., density, specific heat, temperature.
ExtensiveExtensive property -property - doesdoes depend ondepend on
the amount of stuff.the amount of stuff.
E.g.,E.g., heat capacity, mass, heat from aheat capacity, mass, heat from a
reaction.reaction.

3030
Note on HydratesNote on Hydratespppp
AA hydratehydrate means ameans a saltsalt (formula unit)(formula unit)
that has somethat has some water hooked onto itwater hooked onto it..
Copper(II) chlorideCopper(II) chloride didihydrate is . . .hydrate is . . .
CuClCuCl22•2H•2H22OO
The M of theThe M of the hydratehydrate isis 170.49 g/mol170.49 g/mol
The molar mass of theThe molar mass of the anhydrideanhydride isis
134.45 g/mol134.45 g/mol (170.49 less 2 waters)(170.49 less 2 waters)
You need to know this for the onlineYou need to know this for the online
HWHW

3131
More online HW notesMore online HW notespppp
HW05 Q4 - answer in kJ/mol of Z (notHW05 Q4 - answer in kJ/mol of Z (not
J/g). Find the q, then ÷ by # of moles.J/g). Find the q, then ÷ by # of moles.
HW05 Q2 & Q8 - use the examples justHW05 Q2 & Q8 - use the examples just
done today. Use internet for Cdone today. Use internet for Cpp of Pb.of Pb.
HW05 Q11 - look at the exampleHW05 Q11 - look at the example
problems in your assigned reading.problems in your assigned reading.

3232
Note on online HWNote on online HWpppp
When a question asks, “how much heat isWhen a question asks, “how much heat is
liberatedliberated,” your answer will be positive,” your answer will be positive
because there is no “negative” heat.because there is no “negative” heat.
When a question asks, “what is theWhen a question asks, “what is the changechange inin
heat” then you have to indicate the change byheat” then you have to indicate the change by
a (+) or (-) sign.a (+) or (-) sign.
When you use energy or heat in aWhen you use energy or heat in a
mathematical equation (e.g., q = m∆TCmathematical equation (e.g., q = m∆TCpp thenthen
you also have to show the sign.you also have to show the sign.

3333
6.3 Hess’s Law6.3 Hess’s Law
Enthalpy is a state function.Enthalpy is a state function.
So, it isSo, it is independentindependent of the path.of the path.
We canWe can addadd the pathway equations tothe pathway equations to
come up with the desired final product,come up with the desired final product,
and add the pathwayand add the pathway ∆∆H’s.H’s.
Two rules:Two rules:
(1) If the reaction is(1) If the reaction is reversedreversed thethe signsign ofof
∆∆H isH is changedchanged..
(2) If the reaction is multiplied,(2) If the reaction is multiplied, so isso is ∆∆HH

3434
N2 2O2
O2
2NO
68 kJ
2NO2
180 kJ
-112 kJ
H(kJ)

3535
Hints for using Hess’ LawHints for using Hess’ Lawpppp
When using Hess’s Law, workWhen using Hess’s Law, work
backwardsbackwards from the required reactionfrom the required reaction
to decide how to manipulate eachto decide how to manipulate each
reaction step so you can add them up toreaction step so you can add them up to
make it look like the answer.make it look like the answer.
ReverseReverse any reactionsany reactions as neededas needed
(change(change ∆∆H sign) & multiply theH sign) & multiply the
coefficients so the extra partscoefficients so the extra parts cancelcancel..
Note: HNote: H22O(O(ll) will) will notnot cancel Hcancel H22O(O(gg)!)!

3636
Example of Hess’ LawExample of Hess’ Lawpppp
Calculate ∆HºCalculate ∆Hºff forfor CC(s)(s)
+ 2H+ 2H2(g)2(g) →→ CHCH4(g)4(g) given:given:
CC(s)(s) + O+ O2(g)2(g) →→
COCO2(g)2(g) ∆Hº∆Hºcc = -393.5 kJ/mol H= -393.5 kJ/mol H2(g)2(g) + 1/2 O+ 1/2 O2(g)2(g)
→→ HH22OO(l)(l) ∆Hº∆Hºcc = -285.8 kJ/mol CH= -285.8 kJ/mol CH4(g)4(g) + 2O+ 2O2(g)2(g) →→ COCO2(g)2(g)
+ 2H+ 2H22OO(l)(l) ∆Hº∆Hºcc == --890.8 kJ/mol890.8 kJ/mol
Need CHNeed CH44 as a product, so reverse the 3rdas a product, so reverse the 3rd
equation & change its ∆H sign.equation & change its ∆H sign.

3737
CC(s)(s) + O+ O2(g)2(g) →→
COCO2(g)2(g) ∆Hº∆Hºcc = -393.5 kJ/mol H= -393.5 kJ/mol H2(g)2(g) + 1/2 O+ 1/2 O2(g)2(g)
→→ HH22OO(l)(l) ∆Hº∆Hºcc = -285.8 kJ/mol= -285.8 kJ/mol COCO2(g)2(g) + 2H+ 2H22OO(l)(l) →→ CHCH4(g)4(g)
+ 2O+ 2O2(g)2(g) ∆Hº∆Hºff == ++890.8 kJ/mol890.8 kJ/mol
HH22O is not in the original equation soO is not in the original equation so
need to cancel it out.need to cancel it out.
But, Eq.But, Eq. 22 has onlyhas only 11 water as awater as a productproduct
while Eq.while Eq. 33 hashas 22 waters as awaters as a reactantreactant..
MultiplyMultiply Eq. 2 by 2 (Eq. 2 by 2 (includingincluding ∆Hº∆Hºcc))

3838
CC(s)(s) + O+ O2(g)2(g) →→
COCO2(g)2(g) ∆Hº∆Hºcc = -393.5 kJ/mol= -393.5 kJ/mol 2H2H22(g)(g) + O+ O2(g)2(g) →→
22HH22OO(l)(l) ∆Hº∆Hºcc = (= (22)(-285.8 kJ/mol))(-285.8 kJ/mol) COCO2(g)2(g) + 2H+ 2H22OO(l)(l)
→→ CHCH4(g)4(g) + 2O+ 2O2(g)2(g) ∆Hº∆Hºff = +890.8 kJ/mol= +890.8 kJ/mol
AddAdd the 3 equations and cancel like termsthe 3 equations and cancel like terms
on both sides of the reaction.on both sides of the reaction.

3939
+ 2H+ 2H2(g)2(g) →→ CHCH4(g)4(g) given :given :
CC(s)(s) + O+ O2(g)2(g) →→
COCO2(g)2(g) ∆Hº∆Hºcc = -393.5 kJ/mol= -393.5 kJ/mol 2H2H22(g)(g) + O+ O2(g)2(g) →→
22HH22OO(l)(l) ∆Hº∆Hºcc = (= (22)(-285.8 kJ/mol))(-285.8 kJ/mol) COCO2(g)2(g) + 2H+ 2H22OO(l)(l)
→→ CHCH4(g)4(g) + 2O+ 2O2(g)2(g) ∆Hº∆Hºff = +890.8 kJ/mol= +890.8 kJ/mol
CC(s)(s) + 2H+ 2H2(g)2(g) →→ CHCH4(g)4(g) ∆Hº∆Hºff = -74.3 kJ= -74.3 kJ
Since ∆H is (-) the reaction isSince ∆H is (-) the reaction is exoexothermic.thermic.

4040
Standard EnthalpyStandard Enthalpy
The enthalpy change for a reaction atThe enthalpy change for a reaction at
standard conditions (25ºC, 1 atm , 1standard conditions (25ºC, 1 atm , 1 MM
solutions). (Std T = 0 ºC when workingsolutions). (Std T = 0 ºC when working
with the gas laws!)with the gas laws!)
SymbolSymbol ∆∆HºHº

4242
H2(g)+
1
2
O2(g)→H2O(l)
C(s)+O2(g)→CO2(g) ∆Hº= -394 kJ
∆Hº= -286 kJ
C2H2(g) +
5
2
O2(g) →2CO2(g) +H2O(l)
ExampleExample
GivenGiven
∆Hº= -1300. kJ
2C(s) + H (g) C H (g)2 2 2→
Calculate ∆Hº for this reaction Ans. . .
Reverse equation 1; multiply eq. 2 by 2 (include delta H)
Ans. 226 kJ226 kJ, so endothermic

4343
ExampleExample
O (g) + H (g) 2OH(g)2 2 →
O (g) 2O(g)2 →
H (g) 2H(g)2 →
O(g) + H(g) OH(g)→
Given
Calculate ∆Hº for this reaction. Ans. . .
∆Hº= +77.9kJ
∆Hº= +495 kJ
∆Hº= +435.9kJ
Cut all reactions in 1/2
Reverse #s 2 & 3
Add up to = -426 kJ-426 kJ = exothermic (not -426.5 d/t SF)

4444
Standard Enthalpies of FormationStandard Enthalpies of Formation
Hess’s Law is much more useful if you knowHess’s Law is much more useful if you know
lots of reactions.lots of reactions.
The amount of heat needed to formThe amount of heat needed to form 11 mole ofmole of
a compound from its elements in theira compound from its elements in their
standardstandard states; e.g., Hstates; e.g., H22O(l) v. HO(l) v. H22O(g)O(g)
Standard states are 1 atm, 1Standard states are 1 atm, 1MM and 25ºCand 25ºC
(Std T = 0ºC for gas laws like PV = nRT)(Std T = 0ºC for gas laws like PV = nRT)
For an elementFor an element ∆∆HHffº = 0 (don’t forget!!)º = 0 (don’t forget!!)
See Table 6.2 p. 247; also Appendix 4 (pgSee Table 6.2 p. 247; also Appendix 4 (pg
A19)A19)

4545
6.4 Standard Enthalpies of Formation6.4 Standard Enthalpies of Formation
Need to be able to write the equations.Need to be able to write the equations.
What is equation for formation ofWhat is equation for formation of NONO22 ??
½N½N22 (g) + O(g) + O22 (g)(g) →→ NONO22 (g) because . . .(g) because . . .
Have to makeHave to make one moleone mole to meet theto meet the
definition. So, what coeff. fordefinition. So, what coeff. for NN22??
Use 1/2 for coefficient of NUse 1/2 for coefficient of N22
1/21/2NN22 (g) + O(g) + O22 (g)(g) →→ NONO22 (g)(g)
Why not NWhy not N22 (g) +(g) + 22OO22 (g)(g) →→ 22NONO22 (g) ?(g) ?
BecauseBecause by definitionby definition only looking atonly looking at oneone
mole of NOmole of NO22

4646
Standard Enthalpies of Formation cont.Standard Enthalpies of Formation cont.
Write equation for formation ofWrite equation for formation of
methanol,methanol, CHCH33OHOH from its elementsfrom its elements..
Answer . . .Answer . . .
C(s) + 2HC(s) + 2H22(g) + 1/2 O(g) + 1/2 O22(g)(g) →→ CHCH33OHOH
Note: heat ofNote: heat of combustioncombustion is theis the reversereverse ofof
heat ofheat of formationformation (sign also changes).(sign also changes).
This will be on the test!!This will be on the test!!
The above reaction showing heat ofThe above reaction showing heat of
combustioncombustion instead of formation is . . .instead of formation is . . .
CHCH33OHOH →→ C(s) + 2HC(s) + 2H22(g) + 1/2 O(g) + 1/2 O22(g)(g)

4747
Since we can manipulate theSince we can manipulate the
equationsequations
We can use heats of formation to figureWe can use heats of formation to figure
out the heat of reaction.out the heat of reaction.
Let’s look at the schematic (see fig. 6.9,Let’s look at the schematic (see fig. 6.9,
p. 248, next slide) . . .p. 248, next slide) . . .

48
Figure 6.9. A Schematic Diagram of the Energy
Changes for the Reaction
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

4949
Since we can manipulate the equationsSince we can manipulate the equations
We can use heats of formation to figureWe can use heats of formation to figure
out the heat of reaction.out the heat of reaction.
Lets do it with this equation.Lets do it with this equation.
CC22HH55OHOH(l)(l) +3O+3O2(g)2(g) →→ 2CO2CO2(g)2(g) + 3H+ 3H22OO(l)(l) ((2 • -3942 • -394
+ 3 • -286+ 3 • -286) - () - (1• -278 + 3 • 01• -278 + 3 • 0) =) = -1368 kJ-1368 kJ = exothermic= exothermic
Oxygen = O since element; liquid water value isOxygen = O since element; liquid water value is
not same as gaseous waternot same as gaseous water
Which leads us to this rule:Which leads us to this rule:
∑ ∑( H products) - ( H reactants) = Hf
o
f
o o
∆ ∆ ∆

5050
Example 1Example 1pppp
Using the standard enthalpies ofUsing the standard enthalpies of
formation listed in Table 6.2, p. 247,formation listed in Table 6.2, p. 247,
calculate the enthalpy change for thecalculate the enthalpy change for the
following reaction . . . Stepsfollowing reaction . . . Steps
4NH4NH3(g)3(g) + 7O+ 7O2(g)2(g) →→ 4NO4NO2(g)2(g) + 6H2O+ 6H2O(l)(l)

5151
Step 1Step 1:: DecomposeDecompose NHNH3(g)3(g) into itsinto its
elements. 4NHelements. 4NH3(g)3(g) →→ 2N2N2(g)2(g) + 6H+ 6H2(g)2(g)
TheThe reversereverse of this is theof this is the formationformation
reaction for NHreaction for NH3(g)3(g) 1/2N1/2N2(g)2(g) ++
3/2H3/2H2(g)2(g) →→ NHNH3(g)3(g) ∆∆HHffºº == --46 kJ/mol46 kJ/mol
We needWe need 4 times4 times thethe reversereverse of theof the
formationformation reaction for our equationreaction for our equation
above.above.

5252
Step 2Step 2: Since O: Since O2(g)2(g) is in standard state itsis in standard state its
∆∆HHffº = 0.º = 0.
We now have the elements NWe now have the elements N2(g)2(g), H, H2(g)2(g), and, and
OO2(g)2(g), which can be combined to form the, which can be combined to form the
productsproducts of the overall reaction asof the overall reaction as
follows: . . .follows: . . .

53
Figure 6.10
A Pathway for the Combustion of Ammonia

5454
4NH4NH3(g)3(g) + 7O+ 7O2(g)2(g) →→ 4NO4NO2(g)2(g) + 6H+ 6H22OO(l)(l)
So far, we’ve figuredSo far, we’ve figured ∆∆HHffº for theº for the
reactantsreactants..
Now we do the same for theNow we do the same for the productsproducts..
Step 3Step 3: Synthesis of NO: Synthesis of NO22. We need 4. We need 4
times the formation reaction (fromtimes the formation reaction (from
Table 6.2).Table 6.2).
Step 4Step 4: Need 6 times formation reaction: Need 6 times formation reaction
for H2Ofor H2O(l)(l) ((notnot HH22OO((gg))).).

5555
Example 1 cont.Example 1 cont.pppp
Use:Use:
[[4x(34kJ) + 6x(-286 kJ)4x(34kJ) + 6x(-286 kJ)] - [] - [4x(-46kJ)4x(-46kJ)] =] =
-1396 kJ-1396 kJ..
∑ ∑( H products) - ( H reactants) = Hf
o
f
o o
∆ ∆ ∆

Calculating ∆Ho
reaction using standard enthalpies
of formation -
Values for ∆Ho
f are looked up in tables
such as Appendix 4 in your book

Calculate ∆Ho
rxn for
CaCO3(s) CaO(s) + CO2(g)
compound ∆Ho
f (kJ/mol)
CaCO3(s) -1206.9
CaO(s) -635.1
CO2(g) -393.5
[-635.1 + -393.5] - [-1206.9] = 178.3 = endothermic
57

58
Calculation of ∆Ho
∆Ho
= Σ ∆Hf
o
products - Σ ∆Hf
o
reactants

59
Example
What is the value of ∆Hrx for the reaction:
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)
C6H6(l) ∆Hf
o
= + 49.0 kJ/mol
Use above plus Appendix 4, p. A19 to get . . .
O2(g) ∆Hf
o
= 0 kJ/mol
CO2(g) ∆Hf
o
= - 393.5 kJ/mol
H2O(g) ∆Hf
o
= - 241.8 kJ/mol
∆ Hrx = [Σ ∆ Hf
o
]product - [Σ ∆ Hf
o
]reactants

60
Example
What is the value of ∆Hrx for the reaction:
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)
from Appendix 4 p. A21ff
C6H6(l) ∆Hf
o
= + 49.0 kJ/mol; O2(g) ∆Hf
o
= 0
CO2(g) ∆Hf
o
= - 393.5; H2O(g) ∆Hf
o
= - 241.8
∆ Hrx = [Σ ∆ Hf
o
]product - [Σ ∆ Hf
o
]reactants
∆ Hrx = [12(- 393.5) + 6(- 241.8)]product
- [2(+ 49.0 ) + 15(0)]reactants kJ
= - 6.27 x 103
kJ

Nitroglycerine is a powerful explosive, giving four
different gases when detonated
C3H5(NO3)3(l) 3 N2(g) + O2(g) + 6 CO2(g) + 5 H2O(g)
Given that ∆Ho
f for nitroglycerine is -364 kJ/mol,
and consulting a table for the enthalpies of formation
of the other compounds, calculate the enthalpy
change when 10.0 g of nitroglycerine is detonated.
∆Ho
rxn and Stoichiometry
Find enthalpy for 1 mole of NTG (use previous method)
Then convert 10.0 g NTG to moles and use mole ratio
between that and 1 mole. 10g ÷ 227 g/mol = 0.044 mol
61

C3H5(NO3)3(l) 3 N2(g) + O2(g) + 6 CO2(g) + 5 H2O(g)
calculate ∆Hrxn when 10.0 g of NTG is detonated.
∆Ho
rxn and Stoichiometry
[6•-394 + 5•-242] - [-364] = -3.21 x 103
kJ/mol
This is for 1 mol. Have 10.0 g, convert to moles
10g ÷ 227 g/mol = 0.044 mol
Then 0.044 mol • -3.21 x 103
kJ = -141 kJ exothermic
62

6363
Note on online HWNote on online HWpppp
When a question asks, “how much heat isWhen a question asks, “how much heat is
liberatedliberated,” your answer will be positive,” your answer will be positive
because there is no “negative” heat.because there is no “negative” heat.
When a question asks, “what is theWhen a question asks, “what is the changechange inin
heat” then you have to indicate the change byheat” then you have to indicate the change by
a (+) or (-) sign.a (+) or (-) sign.
When you use energy or heat in aWhen you use energy or heat in a
mathematical equation (e.g., q = m∆TCmathematical equation (e.g., q = m∆TCpp thenthen
you also have to show the sign.you also have to show the sign.

Ch6z5ethermo 110115225500-phpapp02

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