Chapter5Hess’sLaw

http://genchem1csustan.wikidot.com/

Energyunits:

WhichstatementA–Daboutenergyunitsisnot correct?a. TheSIunitofenergyistheJoule(J).b. IntermsofSIbaseunits,1J=1kgm2 s–2.c. Anutritionalcalorie(1Cal)is4.184kJ.d. Anutritionalcalorie(1Cal)isequalto1,000cal.e. StatementsA–Dallarecorrect.

Joule

TheSIunitforEnergy1J=1kgm2 s–2

1J=1kgm2/s2

calories

AcalorieistheenergyrequiredtoraiseonegramofwateronedegreeCelcius.1calorie=4.184J

DietaryCalories:1Calorieofdietpepsi =1kcal=1000calories100CaloriesofOreos=100kcal=1x105 calories1kcal=4.184kJ

WhichstatementA–Daboutenergyunitsiscorrect?a. TheSIunitofenergyisthekilojoule(kJ).b. IntermsofSIbaseunits,1J=1kgm2 s2.c. Anutritionalcalorie(1Cal)is4.184kJ.d. Anutritionalcalorie(1Cal)isequalto100cal.e. StatementsA–Dallarecorrect.

Whathasthehigherenthapy content?A. Aswimmingpool375m3 at27°CB. Acupofcoffee(237mL)at80°CC.They arethesame.

Whathasthehigherenthapy content?A. Aswimmingpool375m3 at27°CB. Acupofcoffee(237mL)at80°CC.They arethesame.

ΔH =massxCsxΔT375m3 x1000L x1000mL x1g x4.184J x(27°C-25°C)=3.13x109 J

m3 LmLg°C

237mLx x1g x4.184 J x(80°C-25°C)=5.45x104 JmLg°C

Hess’sLaw:addupreactions,addupenthalpyΔHo

f is the energy gained or lost when a compound or molecule is formed from the elements in their natural state.Calculate ΔHo

f for SO3 in kJ/mol given the following data:

Need:S(s) + 1.5*O2(g) àSO3(g) *Can have fractions of molecules

Have:S(s) + O2(g) à SO2(g) ΔHo = -296.8 kJ

SO2(g) + ½ O2(g) à SO3(g) ΔHo = -98.9 kJ

Adding these:

S(s) + O2(g) + SO2(g) + ½ O2(g) à SO2(g) + SO3(g) ΔHo= -395.7kJ

Cancel out if on both sides, add up common molecules:

S(s) + 1.5*O2(g) àSO3(g) ΔHo= -395.7kJ

Usethefollowinginformationtodeterminethestandardenthalpychangewhen1mol ofPbO(s)isformedfromleadmetalandoxygengas:Pb(s)+!

"O2(g)à PbO(s)

PbO(s)+C(graphite)à Pb(s)+CO(g)ΔHo =107kJ2C(graphite)+O2(g)à 2CO(g) ΔHo =–222kJ

Need:Pb(s)+!"O2(g)à PbO(s)

Have:PbO(s)+C(graphite)à Pb(s)+CO(g)ΔHo =107kJ2C(graphite)+O2(g)à 2CO(g) ΔHo =–222kJ

Pb(s)+CO(g)àPbO(s)+C(graphite) ΔHo =-107kJC(graphite)+!

"O2(g)à CO(g) ΔHo =–111kJ

ΔHo =-218kJ

Determine thechangeinenthalpyforthefollowingreactionfromtheenthalpiesofformationforthereactantsandproducts.(ΔHf forpureelements iszero.)

Cl2(g)+C2H2Cl2(g)à C2H2Cl4(g)C2H2Cl2,ΔHf =4.27kJ/molC2H2Cl4,ΔHf =–155.6kJ/mol

AddupΔHof for reactants

Add up ΔHof for products

Products – Reactants = ΔHorxn

(rxn = reaction)

Determinethechangeinenthalpyforthefollowingreactionfromtheenthalpiesofformationforthereactantsandproducts.(ΔHf forpureelementsiszero.)

Cl2(g)+C2H2Cl2(g)à C2H2Cl4(g)C2H2Cl2,ΔHf =4.27kJ/molC2H2Cl4,ΔHf=–155.6kJ/molProducts=-155.6=-155.6kJReactants=0+4.27kJProducts-Reactants=-155.6- 4.27=-160kJ

CompoundN2O4(g) H2O(l) HNO3(aq) NO(g)

ΔHof (kJ/mol) +11.1 −285.8 −207.4 +91.3

Given the standard heats of formation in the table below, what is ΔHo for the following reaction: 3 N2O4(g) + 2 H2O(l) à 4 HNO3(aq) + 2 NO(g)

-108.7kJ