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Chapter 6Chapter 6

EnergyEnergyThermodynamicsThermodynamics pppp

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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 of Made of heatheat and and workwork.. A state function (unlike work and A state function (unlike work and heat).heat).

I.e.I.e., it’s independent of the path, , it’s independent of the path, or how you get from point A to B.or how you get from point A to B.

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

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Work and HeatWork and Heat WorkWork is a force is a force acting overacting over a a distance.distance.

HeatHeat is energy is energy transferredtransferred between objects because of between objects because of temperature difference.temperature difference.

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The universeThe universe Is divided into two halves.Is divided into two halves. The The systemsystem and the and the surroundingssurroundings..

The system is the part you are The system is the part you are concerned with.concerned with.

The surroundings are the rest.The surroundings are the rest. ExothermicExothermic reactions reactions releaserelease energy energy toto the surroundings. the surroundings.

EndothermicEndothermic reactions reactions absorbabsorb energy energy fromfrom the surroundings. the surroundings.

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CH + 2O CO + 2H O + Heat4 2 2 2

CH + 2O 4 2

CO + 2 H O 2 2

Potential energy

Heat

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N + O2 2

Potential energy

Heat

2NO

N + O 2NO2 2 + heat

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DirectionDirection Every energy measurement has Every energy measurement has

three parts.three 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 Positive Positive - - endoendothermicthermic

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System

Surroundings

Energy

E < 0

1010

System

Surroundings

Energy

E >0

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Same rules for heat and workSame rules for heat and work pppp

Heat Heat given offgiven off is is negativenegative.. Heat Heat absorbedabsorbed is is positivepositive.. WorkWork done done byby system system onon surroundingssurroundings is is negativenegative..

Work done Work done onon system system byby surroundingssurroundings is is positivepositive..

ThermodynamicsThermodynamics - The study of - The study of energy energy andand the changes it the changes it undergoes.undergoes.

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First Law of ThermodynamicsFirst Law of Thermodynamics The The energyenergy of the universe is of the universe is constantconstant..

Law of conservation of Law of conservation of energyenergy..

q = heatq = heat w = workw = work E = q + wE = q + w Take the Take the system’ssystem’s point of point of view to decide signs.view to decide signs.

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What is work?What is work? Work is a force acting over a Work is a force acting over a distance.distance.

w= F x w= 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= PVV Work can be calculated by Work can be calculated by multiplying pressure by the change multiplying pressure by the change in volume at constant pressure.in volume at constant pressure.

units of work: liter-atm or L-units of work: liter-atm or L-atmatm

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Work needs a signWork needs a sign If the If the volumevolume of a gas of a gas increasesincreases, , the the systemsystem has done work on the has done work on the surroundingssurroundings (memorize this!). (memorize this!).

And . . . work is And . . . work is negativenegative: w = : w = (-)(-)

w = - Pw = - PV (don’t forget negative V (don’t forget negative sign)sign)

ExpandingExpanding a gas: a gas: Work is negativeWork is negative.. ContractingContracting: : Surroundings do workSurroundings do work on the on the systemsystem and w = (+). and w = (+).

1 L-atm = 101.325 J (need for 1 L-atm = 101.325 J (need for conversions)conversions)

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Fig 6.4 p. 246, Volume Fig 6.4 p. 246, Volume of a Cylinderof a Cylinder

a.a. The piston, moving The piston, moving a distance delta h a distance delta h against pressure P, against pressure P, does work on the does work on the surroundings.surroundings.

b. b. Since V of cylinder = Since V of cylinder = area of base x area of base x height, the delta V height, the delta V of gas = delta h x Aof gas = delta h x A

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ExamplesExamples pppp

What amount of What amount of workwork is done when 15.00 is done when 15.00 L of gas is L of gas is expandedexpanded to 25.00 L at 2.40 to 25.00 L at 2.40 atm pressure? Steps . . .atm pressure? Steps . . .

ww = -P∆V = -2.40 atm • 10.00L = = -P∆V = -2.40 atm • 10.00L = -24 L-atm-24 L-atm If 2.36 If 2.36 kJkJ of of heatheat are absorbed what are absorbed what

is the change in energy? Steps . . .is the 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 gas How much heat to change the gas withoutwithout

changingchanging the internal energy of the the internal energy of the gas?gas?

Want ∆E = 0, so q + w = 0Want ∆E = 0, so q + w = 0 since w = -2430 J; q added must = +2430 since w = -2430 J; q added must = +2430

JJ

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6.2 Enthalpy & Calorimetry6.2 Enthalpy & Calorimetry EnthalpyEnthalpy is abbreviated as H is abbreviated as H H = E + PV (that’s the definition H = E + PV (that’s the definition of heat)of heat)

At At constantconstant pressure pressure H = H = E + PE + PVV

The heat at constant pressure qThe heat at constant pressure qpp

can be calculated from . . . can be calculated from . . .

E = qE = qpp + w = q + w = qpp - P - PV (w = - PV (w = - PV)V)

qqpp = = E + P E + P V = V = HH

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CalorimetryCalorimetry Measuring heat -- use a Measuring heat -- use a calorimeter.calorimeter.

Two kinds: Constant P or constant VTwo kinds: Constant P or constant V Constant pressure calorimeter Constant pressure calorimeter (called a coffee cup calorimeter, (called a coffee cup calorimeter, open to outside)open to outside)

Heat capacity for a material, CHeat capacity for a material, Cpp is is calculated. calculated.

CCpp= heat absorbed/= heat absorbed/T = T = H/H/TT Specific heat capacity = Cp/mass Specific heat capacity = Cp/mass

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CalorimetryCalorimetry Specific heat capacity = Cp/mass Specific heat capacity = Cp/mass Heat = specific heat x m x Heat = specific heat x m x TT q = m•∆T•Cq = m•∆T•Cpp

Sometimes q is shown as ∆H Sometimes q is shown as ∆H ∆H = m•∆T•C∆H = m•∆T•Cp p (online HW)(online HW)

MolarMolar heat capacity = C heat capacity = Cpp//molesmoles Heat = molar heat x moles x Heat = molar heat x moles x TT Make the units work and you’ve done the Make 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 uses uses gramsgrams or or molesmoles..

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CalorimetryCalorimetry A coffee cup calorimeter A coffee cup calorimeter measures measures H.H.

It’s an insulated cup, full It’s an insulated cup, full of water. of water.

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

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CalorimetryCalorimetry Specific heat of Specific heat of HH22OO = 1 cal/gºC = 1 cal/gºC

= = 4.184 J/g4.184 J/gooCC

Heat of reaction= Heat of reaction= H = sh x H = sh x mass x mass x TT (see this form on (see this form on online HW)online HW)

Also seen as: q = mAlso seen as: q = mTTCCpp

See, Table 6.1 p. 237 (next page) for See, Table 6.1 p. 237 (next page) for CCpp of some common substances. of some common substances. Note the units! (J/ºC•g)Note the units! (J/ºC•g)

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Z5e 237 Table 6.1Z5e 237 Table 6.1

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ExamplesExamples CCpp of graphite is 0.710 of graphite is 0.710

J/gºC. Calculate q needed J/gºC. Calculate q needed to raise the temperature of to raise the temperature of 75.0 75.0 kg kg of it from 294 K to of it from 294 K to 348 K. . .348 K. . .

Convert Kg to g; q = mConvert Kg to g; q = mTCTCpp . . Ans . . . Ans . . .

= 2 880 000 J = 2880 kJ= 2 880 000 J = 2880 kJ

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ExamplesExamples pp pp

46.2 g of copper heated to 95.4ºC & 46.2 g of copper heated to 95.4ºC & placed in calorimeter with 75.0 g water placed in calorimeter with 75.0 g water at 19.6ºC. Final temp. of both is at 19.6ºC. Final temp. of both is 21.8ºC. What is C21.8ºC. What is Cpp of Cu? . . . of Cu? . . .

q gained by Cu = q lost by Hq gained by Cu = q lost by H22O; set both O; 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 4.18 J/ºC•g (memorize). Use absolute value for ∆T; plug-and-chug to get answer of . . for ∆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 Don’t need to convert ºC to K since both have the same degree increments & we are have the same degree increments & we are only interested in the only interested in the changechange..

Need to know this for online HW05 Q8Need to know this for online HW05 Q8

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CalorimetryCalorimetry Constant Constant volumevolume calorimeter is calorimeter is called a bomb calorimeter.called a bomb calorimeter.

Material is put in a container Material is put in a container with pure oxygen. Wires are used with pure oxygen. Wires are used to start the combustion. The to start the combustion. The container is put into a container is put into a container of water.container of water.

The heat capacity of the The heat capacity of the calorimeter is known and tested.calorimeter is known and tested.

Since Since V = 0, PV = 0, PV = 0, V = 0, E = q E = q

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Bomb CalorimeterBomb Calorimeter thermometerthermometer

stirrerstirrer

full of full of

waterwater

ignition ignition

wirewire

Steel bombSteel bomb

samplesample

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Bomb Calorimetry ExampleBomb Calorimetry Example pp pp

T increases by 3.2ºC when 0.1964 g of quinone T increases by 3.2ºC when 0.1964 g of quinone (C(C66HH44OO22) is burned in a bomb calorimeter with ) is burned in a bomb calorimeter with heat capacity of 1.56 kJ/ºC. What is the heat capacity of 1.56 kJ/ºC. What is the combustion energy of quinone per combustion energy of quinone per gramgram and alsoand also per per molemole??

Heat gained by calorimeter = Heat gained by calorimeter = 1.56kJ/ºC 1.56kJ/ºC x 3.2 ºC = 5.0 kJ = heat loss by x 3.2 ºC = 5.0 kJ = heat loss by quinonequinone

∆∆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 the amount of related to the amount of substance.substance.

E.g.E.g., density, specific heat, , density, specific heat, temperature.temperature.

ExtensiveExtensive property - property - doesdoes depend on the amount of stuff.depend on the amount of stuff.

E.g., E.g., heat capacity, mass, heat capacity, mass, heat from a reaction.heat from a reaction.

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Note on HydratesNote on Hydrates pp pp

A A hydratehydrate means a means a saltsalt (formula unit) (formula unit) that has some that has some water hooked onto itwater hooked onto it..

Copper(II) chloride Copper(II) chloride didihydrate hydrate is . . .is . . .

CuClCuCl22•2H•2H22OO The M of the The M of the hydratehydrate is is 170.49 g/mol170.49 g/mol The molar mass of the The molar mass of the anhydrideanhydride is is 134.45 g/mol134.45 g/mol (170.49 less 2 waters) (170.49 less 2 waters)

You need to know this for the online You need to know this for the online HWHW

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More online HW notesMore online HW notes pp pp

HW05 Q4 - answer in kJ/mol of HW05 Q4 - answer in kJ/mol of Z (not J/g). Find the q, Z (not J/g). Find the q, then ÷ by # of moles.then ÷ by # of moles.

HW05 Q2 & Q8 - use the HW05 Q2 & Q8 - use the examples just done today. examples just done today. Use internet for CUse internet for Cpp of Pb. of Pb.

HW05 Q11 - look at the HW05 Q11 - look at the example problems in your example problems in your assigned reading.assigned reading.

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Note on online HWNote on online HW pp pp

When a question asks, “how much heat When a question asks, “how much heat is is liberatedliberated,” your answer will be ,” your answer will be positive because there is no positive because there is no “negative” heat. “negative” heat.

When a question asks, “what is the When a question asks, “what is the changechange in heat” then you have to in heat” then you have to indicate the change by a (+) or (-) indicate the change by a (+) or (-) sign.sign.

When you use energy or heat in a When you use energy or heat in a mathematical equation (e.g., q = m∆TCmathematical equation (e.g., q = m∆TCpp then you also have to show the sign.then you also have to show the sign.

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6.3 Hess’s Law6.3 Hess’s Law Enthalpy is a state function.Enthalpy is a state function. So, it is So, it is independentindependent of the path. of the path. We can We can addadd the pathway equations the pathway equations to come up with the desired final to come up with the desired final product, and add the pathway product, and add the pathway H’s.H’s.

Two rules:Two rules: (1) If the reaction is (1) If the reaction is reversedreversed the the signsign of of H is H is changedchanged..

(2) If the reaction is multiplied, (2) If the reaction is multiplied, so isso is HH

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N2 2O2

O2 2NO

68 kJ

2NO2180 kJ

-112 kJ

H (kJ)

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Hints for using Hess’ LawHints for using Hess’ Law pp pp

When using Hess’s Law, work When using Hess’s Law, work backwardsbackwards from the required from the required reaction to decide how to reaction to decide how to manipulate each reaction step so manipulate each reaction step so you can add them up to make it you can add them up to make it look like the answer.look like the answer.

ReverseReverse any reactions any reactions as neededas needed (change (change H sign) & multiply the H sign) & multiply the coefficients so the extra parts coefficients so the extra parts cancelcancel..

Note: HNote: H22O(O(ll) will ) will notnot cancel cancel HH22O(O(gg)!)!

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Example of Hess’ LawExample of Hess’ Law pp pp

Calculate ∆HºCalculate ∆Hºff for for 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) + + 2O2O2(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 as a product, so reverse the 3rd equation & reverse the 3rd equation & change its ∆H sign.change its ∆H sign.

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Example of Hess’ LawExample of Hess’ Law pp pp

Calculate ∆HºCalculate ∆Hºff for for 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 + 1/2 OO2(g)2(g) HH22OO(l)(l) ∆Hº ∆Hºcc = -285.8 kJ/mol = -285.8 kJ/mol COCO2(g)2(g) + + 2H2H22OO(l) (l) CH CH4(g)4(g) + 2O + 2O2(g)2(g) ∆Hº ∆Hºff = = ++890.8 kJ/mol890.8 kJ/mol

HH22O is not in the original equation O is not in the original equation so need to cancel it out.so need to cancel it out.

But, Eq. But, Eq. 22 has only has only 11 water as a water as a productproduct while Eq. while Eq. 33 has has 22 waters as waters as a a reactantreactant..

MultiplyMultiply Eq. 2 by 2 ( Eq. 2 by 2 (includingincluding ∆Hº ∆Hºcc) )

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Example of Hess’ LawExample of Hess’ Law pp pp

Calculate ∆HºCalculate ∆Hºff for for 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 = -393.5 kJ/mol 2H2H22(g)(g) + O + O2(g)2(g) 2 2HH22OO(l)(l)

∆Hº ∆Hºcc = ( = (22)(-285.8 kJ/mol))(-285.8 kJ/mol) COCO2(g)2(g) + 2H + 2H22OO(l) (l) CH CH4(g)4(g) + 2O + 2O2(g)2(g) ∆Hº ∆Hºff = +890.8 = +890.8 kJ/molkJ/mol

AddAdd the 3 equations and cancel the 3 equations and cancel like terms on both sides of the like terms on both sides of the reaction.reaction.

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Example of Hess’ LawExample of Hess’ Law pp pp

Calculate ∆HºCalculate ∆Hºff for for 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 -393.5 kJ/mol 2H2H22(g)(g) + O + O2(g)2(g) 2 2HH22OO(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 kJ74.3 kJ

Since ∆H is (-) the reaction is Since ∆H is (-) the reaction is exoexothermic.thermic.

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

Symbol Symbol HºHº

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H2(g) + 1

2O2(g) H2O(l)

C(s) + O2(g) CO2(g) Hº= -394 kJ

Hº= -286 kJ

C2H2(g) + 5

2O2(g) 2CO2(g) + H2O(l)

ExampleExample

GivenGivenHº= -1300. kJ

2C(s) + H (g) C H (g) 2 2 2Calculate Hº for this reaction Ans. . .

Reverse equation 1; multiply eq. 2 by 2 (include delta H)

Ans. 226 kJ226 kJ, so endothermic

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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= +77.9kJHº= +495 kJHº= +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)

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Standard Enthalpies of FormationStandard Enthalpies of Formation Hess’s Law is much more useful if you Hess’s Law is much more useful if you know lots of reactions.know lots of reactions.

The amount of heat needed to form The amount of heat needed to form 11 mole mole of a compound from its elements in their of a compound from its elements in their standardstandard states; e.g., H states; e.g., H22O(l) v. HO(l) v. H22O(g)O(g)

Standard states are 1 atm, 1Standard states are 1 atm, 1MM and 25ºC and 25ºC

(Std T = 0ºC for gas laws like PV (Std T = 0ºC for gas laws like PV = nRT)= nRT)

For an element For an element HHffº = 0 (don’t º = 0 (don’t forget!!)forget!!)

See Table 6.2 p. 247; also Appendix 4 See Table 6.2 p. 247; also Appendix 4 (pg A19)(pg A19)

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6.4 Standard Enthalpies of Formation6.4 Standard Enthalpies of Formation Need to be able to write the Need to be able to write the equations.equations.

What is equation for formation of What is equation for formation of NONO22 ??

½N½N2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) because . . . (g) because . . .

Have to make Have to make one mole one mole to meet the to meet the definition. So, what coeff. for definition. So, what coeff. for NN22??

Use 1/2 for coefficient of NUse 1/2 for coefficient of N22

1/21/2NN2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) (g)

Why not NWhy not N2 2 (g) + (g) + 22OO22 (g) (g) 22NONO22 (g) ? (g) ? Because Because by definitionby definition only looking at only looking at oneone mole of NO mole of NO22

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Standard Enthalpies of Formation cont.Standard Enthalpies of Formation cont. Write equation for formation of Write equation for formation of methanol, methanol, CHCH33OH OH 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 of Note: heat of combustioncombustion is the is the reversereverse of heat of of heat of formationformation (sign (sign also changes). This will be on the also changes). This will be on the test!!test!!

The above reaction showing heat of The above reaction showing heat of combustioncombustion instead of formation instead of formation is . . .is . . .

CHCH33OH OH C(s) + 2HC(s) + 2H22(g) + 1/2 O(g) + 1/2 O22(g) (g)

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Since we can manipulate the Since we can manipulate the equationsequations

We can use heats of formation We can use heats of formation to figure out the heat of to figure out the heat of reaction.reaction.

Let’s look at the schematic Let’s look at the schematic (see fig. 6.9, p. 248, next (see fig. 6.9, p. 248, next slide) . . .slide) . . .

48

Figure 6.9. A Schematic Diagram of the Energy Changes for the Reaction

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

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Since we can manipulate the equationsSince we can manipulate the equations We can use heats of formation We can use heats of formation to figure out the heat of to figure out the heat of reaction.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 • -394 + 3 • -2862 • -394 + 3 • -286) - () - (1• -278 + 3 • 01• -278 + 3 • 0) ) = = -1368 kJ-1368 kJ = exothermic = exothermic

Oxygen = O since element; liquid Oxygen = O since element; liquid water value is not same as gaseous water value is not same as gaseous waterwater

Which leads us to this rule:Which leads us to this rule:

( H products) - ( H reactants) = Hfo

fo o

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Example 1Example 1 pp pp

Using the standard enthalpies Using the standard enthalpies of formation listed in Table of formation listed in Table 6.2, p. 247, calculate the 6.2, p. 247, calculate the enthalpy change for the enthalpy change for the following reaction . . . following reaction . . . StepsSteps

4NH4NH3(g)3(g) + 7O + 7O2(g)2(g) 4NO 4NO2(g)2(g) + + 6H2O6H2O(l)(l)

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Example 1Example 1 pp pp

4NH4NH3(g)3(g) + 7O + 7O2(g)2(g) 4NO 4NO2(g)2(g) + 6H2O + 6H2O(l)(l)

Step 1Step 1: : DecomposeDecompose NHNH3(g)3(g) into its into its elements. 4NHelements. 4NH3(g)3(g) 2N 2N2(g)2(g) + 6H + 6H2(g)2(g)

The The reversereverse of this is the of this is the formationformation reaction for NH reaction for NH3(g)3(g)

1/2N1/2N2(g)2(g) + 3/2H + 3/2H2(g)2(g) NHNH3(g)3(g) HHffºº = = --

46 kJ/mol46 kJ/mol We need We need 4 times4 times the the reversereverse of the of the formationformation reaction for our reaction for our equation above.equation above.

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Example 1Example 1 pp pp

4NH4NH3(g)3(g) + 7O + 7O2(g)2(g) 4NO 4NO2(g)2(g) + + 6H2O6H2O(l)(l)

Step 2Step 2: Since O: Since O2(g)2(g) is in is in standard state its standard state its HHffº = 0.º = 0.

We now have the elements NWe now have the elements N2(g)2(g), , HH2(g)2(g), and O, and O2(g)2(g), which can be , which can be combined to form the combined to form the productsproducts of the overall reaction as of the overall reaction as follows: . . .follows: . . .

53

Figure 6.10A Pathway for the Combustion of Ammonia

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Example 1Example 1 pp pp

4NH4NH3(g)3(g) + 7O + 7O2(g)2(g) 4NO4NO2(g)2(g) + 6H + 6H22OO(l)(l)

So far, we’ve figured So far, we’ve figured HHffº for º for the the reactantsreactants..

Now we do the same for the Now we do the same for the productsproducts..

Step 3Step 3: Synthesis of NO: Synthesis of NO22. We . We need 4 times the formation need 4 times the formation reaction (from Table 6.2).reaction (from Table 6.2).

Step 4Step 4: Need 6 times formation : Need 6 times formation reaction for H2Oreaction for H2O(l)(l) ( (notnot H H22OO((gg))).).

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Example 1 cont.Example 1 cont. pp pp

4NH4NH3(g)3(g) + 7O + 7O2(g)2(g) 4NO4NO2(g)2(g) + + 6H2O6H2O(l)(l)

Use:Use:

[[4x(34kJ) + 6x(-286 kJ)4x(34kJ) + 6x(-286 kJ)] - ] - [[4x(-46kJ)4x(-46kJ)] = ] = -1396 kJ-1396 kJ..

( H products) - ( H reactants) = Hfo

fo o

Calculating Horeaction using standard enthalpies

of formation -

Values for Hof are looked up in tables

such as Appendix 4 in your book

Calculate Horxn for

CaCO3(s) CaO(s) + CO2(g)

compound Hof (kJ/mol)

CaCO3(s) -1206.9

CaO(s) -635.1 CO2(g) -393.5

[-635.1 + -393.5] - [-1206.9] = 178.3 = endothermic

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Calculation of Ho

Ho = Hfoproducts - Hf

oreactants

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Example What is the value of Hrx for the reaction:

2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)

C6H6(l) Hfo = + 49.0 kJ/mol

Use above plus Appendix 4, p. A19 to get . . .

O2(g) Hfo = 0 kJ/mol

CO2(g) Hfo = - 393.5 kJ/mol

H2O(g) Hfo = - 241.8 kJ/mol

Hrx Hfoproduct - Hf

oreactants

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ExampleWhat is the value of Hrx for the reaction:

2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)

from Appendix 4 p. A21ffC6H6(l) Hf

o = + 49.0 kJ/mol; O2(g) Hfo = 0

CO2(g) Hfo = - 393.5; H2O(g) Hf

o = - 241.8 Hrx Hf

oproduct - Hforeactants

Hrx - 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 Hof 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.

Horxn 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 mol61

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.

Horxn 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

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6363

Note on online HWNote on online HW pp pp

When a question asks, “how much heat When a question asks, “how much heat is is liberatedliberated,” your answer will be ,” your answer will be positive because there is no positive because there is no “negative” heat. “negative” heat.

When a question asks, “what is the When a question asks, “what is the changechange in heat” then you have to in heat” then you have to indicate the change by a (+) or (-) indicate the change by a (+) or (-) sign.sign.

When you use energy or heat in a When you use energy or heat in a mathematical equation (e.g., q = m∆TCmathematical equation (e.g., q = m∆TCpp then you also have to show the sign.then you also have to show the sign.