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Chapter 15 questions and solutions...

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Chemistry104–Chapter15SummaryandReview–Answers 1. Thereactionofhydrogenandnitrogentoproduceammonia(theHaberprocess)takes placeinthegasphase. a. Whatisthebalancedequationforthisreaction? b. WhatistheexpressionforKPforthisreaction? c. WhatistheexpressionforKcforthisreaction? d. Ataparticulartemperature,iftheequilibriumconcentrationsare:  Gas Concentration/M N2 0.0240 H2 0.0720 NH3 0.0520  WhatisthevalueofKc? e. WhatisthevalueofKPatthissametemperature?   Answer: Parta:3H2(g)+N2(g)  2NH3(g)   2 PNH 3 Partb: KP   PN PH3 2



2

 NH  Partc: Kc   3 

2

N2  H2 

3





 Partd:Let’suse500K:







 Parte:Continuingtouse500K:

NH3  0.0520  3  3  302  N2  H2  0.0240 0.0720  2

Kc 

2

K c  302  KP  RT  



KP  Kc  RT 

n gas

 0.179         

  ngas

1 1  302 0.08206L  atm mol  K  500K 

2



2. Forthesamereaction(asnumber1),ifthepartialpressuresofallthegasesareequalat 0.550atm,isthereactionatequilibrium? a. Ifnot,whichdirectionwillthereactionshifttoachievethisequilibrium? b. SetuptheICEtableforthiscalculationbasedonyourpreviousanswer. c. Ifthetemperatureisdecreased,theconcentrationschangeto:  Concentration/M Concentration/M Gas (500K) (298K) N2 0.0240 0.00179 H2 0.0720 0.00537 NH3 0.0520 0.100  Isthereactionexothermicorendothermic?Usetheenthalpyofformationof ammoniatosupportthisclaim. d. Willincreasingthevolumeofthereactionvesselaltertheamountofammonia produced? e. Howwouldaddingmorehydrogenorremovingammoniashiftthereaction?  2 PNH 0.550  3   3.31 3 PN2 PH2  0.550 0.550 3 2

Answer: Parta:QP 

QP  KP 

No,thereactionisnotatequilibrium.Thereactionwillshifttotheleft (tomakemorereactant)toestablishequilibrium.  



 

Partb:

N2  g  

3H2  g 

initial

0.550atm

0.550atm

0.550atm

change

x

 3x

2 x

equilibrium

0.550  x

0.550  3 x

0.550 2 x



2NH 3 g 



 Partc: Evaluatingtheconcentrations,theproductincreasesinpartialpressure andthereactantsbothdecrease.Therefore,thatsuggeststhat Kc,298K>Kc,500Kwhichalsomeansthatasthetemperatureincreases,the equilibriumconstantdecreases.Thatsuggeststhatthereactionis exothermic.  Let’scheckthatwithacalculationofKc’s:

0.0520 2  3 302  0.0240 0.0720 2 0.100   3.61  10 7 K c,298K  3 0.00179 0.00537    K c,500K 



 K c,298K  K c,500K

Andnowcheckthiswithcalculatingchangeinenthalpy(ΔHo)using appendix2: H

o f

N2  g  

3H2  g  

2NH3  g 

0.0

0.0

 46.3kJ mol1

o H reaction 

nH products  mH reactants   2molNH o f

o f

3

 46.3kJ     92.6kJ  molNH3 

 Partd:Yes,ifthevolumeisincreased,thisfavorsthegreaternumberofmolesof gas.Becausethereare4molesofgasreactingtoform2molesofgas,a greatervolumefavorsthereactantsortheamountofammoniaproduced woulddecrease.  Parte:Addingmorehydrogenwouldshiftthereactiontoproducemore product;removingammoniawoulddothesame.   3. ForthereactionofN2O4toNO2,thereactionisbegunwith20.0gN2O4withnoNO2and thefinalpressureinthe5.0Lcontainerat298Kis1.90atm. a. Whatisthevalueoftheequilibriumconstant,KP? b. IsthevalueofKcthesameasKP?Ifnot,whatisthevalueofKc? c. UsingAppendix2,howwouldincreasingthetemperaturechangeKP?  Answer: Parta:WehavetodeterminewhattheinitialpartialpressureofN2O4whichwas aregoingtodousingwhatweknowaboutformulacalculations(Chapter 3)andgases(Chapter5): 

 molN2O 4  92.02gN 2O 4

 20.0gN2O4   PN2 O4 

nN2 O4 RT



V

   0.217molN 2O 4 initial  

 0.217mol 0.08206L atm mol 1 K 1 298K  5.0L

 1.06atm

 Now,wearegoingtosetupourICEtable: N2O 4  g 

initial change



1.06atm x 1.06  x

equilibrium

2NO 2  g 

0atm 2 x 2x



 Thefinalpressureis1.90atm(PT).FromChapter5,wealsoknow: PT  PN O  PNO  so: 1.90atm  1.06atm  x  2 x x  0.84atm  2 4

2

 Therefore, PN O  0.22atmandPNO  1.68atm  2 4

2

Finally, KP 

 1.68  0.22

2

2 PNO 2

PN 2O 4



 13 

 Partb:KcandKPareequalwhenthenumberofmolesofgasonboththeproduct andreactantsideareequal(orΔngas=0).Because1moleofgas(N2O4) reactstoform2molesofgas(NO2),Δngas≠0andratherequals1.  Therefore,Kc≠KP: 

K c  K P  RT 

n gas

 13  0.08206L atm mol 1 K 1 298K  0.53  1

 Partc: Toanswerthis,thereferencetoAppendix2wasthehint.Weneedto firstcalculatetheenthalpyofreaction.Ifitisendothermic,anincreasein temperaturewillincreasethevalueofK;ifitisexothermic,anincreasein temperaturewilldecreasethevalueofK:  N2 O4  g 



H

o f

1

9.66kJ mol

2NO2  g 33.85kJ mol 1



 33.85kJ   9.66kJ  H oreaction  2molNO2    1molN2 O4    58.04kJ  molNO 2   molN 2O 4 



 Thereactionisendothermic(ΔHo>0),therefore,anincreasein temperaturewouldincreasetheequilibriumconstant....