chapter 14 chemical equilibrium chemistry ii. speed of a chemical reaction is determined by kinetics...

Chapter 14Chemical EquilibriumChemistry II

Tro, Chemistry: A Molecular Approach

Speed of a chemical reaction is determined by kineticsHow fast a reaction goes

Extent of a chemical reaction is determined by thermodynamicsHow far a reaction goes


Hemoglobinprotein (Hb) found in red blood cells that reacts with O2 enhances the amount of O2 that can be carried through the blood stream

Hb + O2 HbO2

the Hb represents the entire protein it is not a chemical formulathe represents that the reaction is in dynamic equilibrium


Hemoglobin Equilibrium SystemHb + O2 HbO2[Hb], [O2], and [HbO2] are all interdependent

[Hb], [O2], and [HbO2] at equilibrium are controlled by the equilibrium constant, Keqthe larger the value of Keq, the more product is found at equilibrium

changing the concentration of any of these necessitates changing the others to reestablish equilibrium


O2 TransportHb+O2lungs, with high [O2], the equilibrium shifts to combine Hb and O2 to make more HbO2in the cells, with low [O2] (muscles and organs use O2), the equilibrium shifts to break down the HbO2 and increase the [O2]

HbO2O2 in lungsHbHbO2O2 in cells


Tro, Chemistry: A Molecular Approach*HbFHbFetal Hemoglobin, HbFHbF + O2 HbFO2fetal hemoglobins Keq is larger than adult hemoglobinbecause HbF is more efficient at binding O2, O2 is transferred to the HbF from the mothers hemoglobin in the placentaHb+O2HbO2O2HbO2O2Hb+O2HbFO2HbFO2O2


Tro, Chemistry: A Molecular Approach*Oxygen Exchange betweenMother and Fetus


Reaction Dynamicsforward reaction: reactants products

therefore the [reactant] decreases and the [product] increasesas [reactant] decreases, the forward reaction rate decreases

reverse reaction: products reactants assuming the products are not allowed to escapeas [product] increases, the reverse reaction rate increases

processes that proceed in both the forward and reverse direction are said to be reversiblereactants products


Hypothetical Reaction2 Red BlueThe reaction slows over time,But the Red molecules never run out!At some time between 100 and 110 sec,the concentrations of both the Red andthe Blue molecules no longer change equilibrium has been established.Notice that equilibrium does not meanthat the concentrations are equal!Once equilibrium is established, the rateof Red molecules turning into Blue is thesame as the rate of Blue molecules turning into Red

Time[Red][Blue]00.4000.000100.2080.096200.1900.105300.1800.110400.1740.113500.1700.115600.1680.116700.1670.117800.1660.117900.1650.1181000.1650.1181100.1640.1181200.1640.1181300.1640.1181400.1640.1181500.1640.118

Hypothetical Reaction2 Red Blue

Sheet1

Choose the reaction you wish to investigate

Choose the Initial Concentration of Chemicals[A]init =[Y]init =

00

Choose the Celsius temperature you wish to useChoose the time interval between readings

Choose the value of DG, in Joules

Choose the order for each reactant in the forward rate equationOrder of A =0

Choose the order for each product in the reverse rate equationOrder of Y =0

Choose the activation energy, in joules, for the forward reaction

Given the above information, the equilibrium constant, K, for the reaction will equal3.2

Given the above information, the value of the forward reaction rate constant, kf, will equal0.15

Given the above information, the value of the reverse reaction rate constant, kr, will equal0.0475

Data Sheet

Reaction ListCoeff ACoeff BCoeff YCoeff ZCoeffA =2Init Conc[A]init =0.4OrdersA order =2

2 A ----> YA ----> Y1010CoeffB =00.1[B]init =0.10B order =1

2 A ----> Y2010CoeffY =10.2[Y]init =01Y order =1

A ----> 2 Y1020CoeffZ =00.3[Z]init =0.42Z order =0

3 A ----> Y30100.4

A ----> 3 Y10300.5

2 A ----> 3 Y20300.6

3 A ----> 2 Y30200.7

A + B ----> Y11100.8

2 A + B ----> Y21100.9

A + B ----> 2 Y11201

A + B ----> Y + Z1111

2 A + B ----> Y + Z2111

A + B ----> 2 Y + Z1121

Time IntervalTime Interval List, secondsTotal Time

20.550

1100

2200

5500

101000

Temp, CTemp List, CTemp, K

250-100523

0

50R, J/mol-K

1008.314

500

DGDG List, JKx

-5000-250003.2E+000

-20000Kcalc

-150003.1578622343

-10000

-5000

0

5000

10000

15000

20000

25000

Ea forwardEa List, jouleskfkf_calcForward Collision Freq, Afkf/kr

700010001.50E-010.14993768880.753.1578947368

3000

Ea reverse5000krkr_calcReverse Collision Freq, Ar

1200070004.75E-020.04748075680.75

10000

Calc Sheet

Mole MultiplierAvog. Number

1.66E-196.02E+23

Time[Red]0[Blue]0Molecules AMolecules BMolecules YMolecules ZRate ForwardRate ReverseMolecules A ReactMolecules B ReactMolecules Y ReactMolecules Z ReactMolecules A MadeMolecules B MadeMolecules Y MadeMolecules Z Made

00.4000400000000.024038400000019200

20.361725880600.01920668390361600192000.01962684190.000912317528390408014200

40.333406025300.03337161330333290333600.01667393670.00158515162223011022011120

60.311388363200.04438544610311280443700.01454440690.0021083087181101903809060

80.29365219100.0532585340293550532400.01293474140.0025297804151902705407600

100.27899709100.06059108570278900605700.01167590650.0028780766130303507006520

120.266662798700.06676323360266570667400.01066635720.0031712536113704208405690

140.256129132900.07203506820256040720100.00984031990.0034216657100804909805040

160.24702596500.07658665220246940765600.00915327410.003637866904056011204520

180.239103207900.08054803070239020805200.00857555160.0038260315820062012404100

200.23214078500.08402924220232060840000.00808340160.003991389750067013403750

220.225978640600.08711031440225900870800.00765995190.0041377399692072014403460

240.220496732900.08985126820220420898200.00729282140.0042679352643077015403220

260.215605030600.09230212110215530922700.00697282940.0043843508601081016203010

280.211213502300.0945028870211140944700.00669167150.0044888871565085017002830

300.207262127200.09648357630207190964500.00644363840.0045829699534088017602670

320.203680880900.09827419940203610982400.00622288520.0046680245507092018402540

340.200449756500.09989476340200380998600.00602701570.0047450013483095019002420

360.197518736500.101365275101974501013300.00585204770.0048148506462098019602310

380.194857810500.102695738101947901026600.00569543490.00487804764440100020002220

400.192416961100.103916162801923501038800.0055536430.00493601774270103020602140

420.190206191800.105026549201901401049900.00542675930.00498876114130105021002070

440.188175485100.106046904301881101060100.0053115020.0050372284000107021402000

460.186314837600.106977228101862501069400.00520698280.00508141833880109021801940

480.184614245800.10782752401845501077900.0051123630.00512180743770110022001890

500.183043699200.10861779901829801085800.00502574940.00515934553680112022401840

520.181603197900.109338049601815401093000.00494695820.00519355743590114022801800

540.180292741900.109998279401802301099600.00487582090.00522491833520115023001760

560.179072317200.110608491801790101105700.00481003420.00525390343440116023201720

580.177951927300.111168686701778901111300.00475003330.00528051263380117023401690

600.176911565200.111688867701768501116500.00469465530.00530522123320118023601660

620.17595123100.112169034801758901121300.00464382540.00532802923270119023801640

640.175060921200.112619191501750001125800.00459694890.00534941163220120024001610

660.174240635700.113029334201741801129900.00455396990.00536889343170121024201590

680.173490374700.113409466501734301133700.00451483650.00538694973130122024401570

700.172800134400.113759588301727401137200.0044789830.00540358043090123024601550

720.172169915100.114079699701721101140400.0044463720.00541878573060124024801530

740.171589713200.114369800701715301143300.00441645450.00543256553030124024801520

760.171039521800.114649898201709801146100.00438817770.00544587023000125025001500

780.170539347700.114899985201704801148600.00436255040.00545774932970125025001490

800.170069184100.115140068701700101151000.00433852910.00546915332950126025201480

820.169639034400.115360145301695801153200.00431661030.00547960692930126025201470

840.169228891700.115570218401691701155300.00429576270.00548958542910127025401460

860.168858762800.115760284601688001157200.00427699230.00549861352890127025401450

880.16850864100.115940347201684501159000.00425927430.00550716652870128025601440

900.168198533100.116100402901681401160600.0042436120.00551476912850128025601430

920.167908432100.116250455201678501162100.00422898620.00552189662840128025601420

940.167628334700.116390503901675701163500.00421488880.00552854892830129025801420

960.167378247600.116520549201673201164800.00420232170.00553472612810129025801410

980.167148167600.116640590901670901166000.00419077650.00554042812800129025801400

1000.16692809100.116750629201668701167100.00417974810.00554565492790129025801400

1020.166718017900.116860667501666601168200.00416923460.00555088172780130026001390

1040.166537955200.116950698801664801169100.00416023360.00555515822770130026001390

1060.16636789600.117040730201663101170000.00415174150.00555943472760130026001380

1080.166207840300.11712075801661501170800.00414375690.0055632362750130026001380

1100.166057788100.117200785901660001171600.00413627830.00556703732750130026001380

1120.165907735900.117280813701658501172400.00412880650.00557083872740131026201370

1140.165787694100.117340834601657301173000.00412283390.00557368962730131026201370

1160.165677655800.117400855501656201173600.00411736280.00557654062730131026201370

1180.165567617500.117460876401655101174200.00411189540.00557939162720131026201360

1200.165467582700.117510893801654101174700.00410692810.00558176752720131026201360

1220.165367547900.117560911201653101175200.00410196390.00558414332710131026201360

1240.165277516600.117610928601652201175700.00409749860.00558651912710131026201360

1260.165187485200.11766094601651301176200.00409303580.00558889492700131026201350

1280.165107457400.117700959901650501176600.00408907090.00559079562700132026401350

1300.165047436500.117730970401649901176900.00408609840.00559222112700132026401350

1320.164987415600.117760980801649301177200.00408312710.00559364662690132026401350

1340.164937398200.117790991301648801177500.00408065180.00559507212690132026401350

1360.164887380800.117821001701648301177800.00407817730.00559649762690132026401350

1380.164837363400.117851012201647801178100.00407570350.00559792312690132026401350

1400.16478734600.117881022601647301178400.00407323040.00559934862680132026401340

1420.164747332100.117901029601646901178600.00407125250.00560029892680132026401340

1440.164707318100.117921036501646501178800.00406927510.00560124922680132026401340

1460.164667304200.117941043501646101179000.00406729820.00560219962680132026401340

1480.164627290300.117961050401645701179200.00406532170.00560314992680132026401340

1500.164587276400.117981057401645301179400.00406334570.00560410022670132026401340

1520.164557265900.118001064401645001179600.00406186410.00560505062670132026401340

1540.164527255500.118021071301644701179800.00406038270.00560600092670132026401340

1560.16449724500.118041078301644401180000.00405890150.00560695122670132026401340

1580.164467234600.118061085301644101180200.00405742070.00560790152670132026401340

1600.164437224200.118081092201643801180400.00405594010.00560885192670132026401340

1620.164407213700.118101099201643501180600.00405445980.00560980222670132026401340

1640.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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1800.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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1900.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

1920.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

1940.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

1960.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

1980.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

2000.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

2020.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2100.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

2120.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2300.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2500.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3000.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3100.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3360.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3400.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3460.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3480.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3500.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3520.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3540.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3560.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3580.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3600.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3620.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3640.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3660.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3680.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3700.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3720.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3740.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3760.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3780.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3800.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3820.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3840.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3860.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3880.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3900.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3920.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3940.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3960.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3980.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

4000.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

Conc vs Time

0.40

0.36172588060.0192066839

0.33340602530.0333716133

0.31138836320.0443854461

0.2936521910.053258534

0.2789970910.0605910857

0.26666279870.0667632336

0.25612913290.0720350682

0.2470259650.0765866522

0.23910320790.0805480307

0.2321407850.0840292422

0.22597864060.0871103144

0.22049673290.0898512682

0.21560503060.0923021211

0.21121350230.094502887

0.20726212720.0964835763

0.20368088090.0982741994

0.20044975650.0998947634

0.19751873650.1013652751

0.19485781050.1026957381

0.19241696110.1039161628

0.19020619180.1050265492

0.18817548510.1060469043

0.18631483760.1069772281

0.18461424580.107827524

0.18304369920.108617799

0.18160319790.1093380496

0.18029274190.1099982794

0.17907231720.1106084918

0.17795192730.1111686867

0.17691156520.1116888677

0.1759512310.1121690348

0.17506092120.1126191915

0.17424063570.1130293342

0.17349037470.1134094665

0.17280013440.1137595883

0.17216991510.1140796997

0.17158971320.1143698007

0.17103952180.1146498982

0.17053934770.1148999852

0.17006918410.1151400687

0.16963903440.1153601453

0.16922889170.1155702184

0.16885876280.1157602846

0.1685086410.1159403472

0.16819853310.1161004029

0.16790843210.1162504552

0.16762833470.1163905039

0.16737824760.1165205492

0.16714816760.1166405909

0.1669280910.1167506292

0.16671801790.1168606675

0.16653795520.1169506988

0.1663678960.1170407302

0.16620784030.117120758

0.16605778810.1172007859

0.16590773590.1172808137

0.16578769410.1173408346

0.16567765580.1174008555

0.16556761750.1174608764

0.16546758270.1175108938

0.16536754790.1175609112

0.16527751660.1176109286

0.16518748520.117660946

0.16510745740.1177009599

0.16504743650.1177309704

0.16498741560.1177609808

0.16493739820.1177909913

0.16488738080.1178210017

0.16483736340.1178510122

0.1647873460.1178810226

0.16474733210.1179010296

0.16470731810.1179210365

0.16466730420.1179410435

0.16462729030.1179610504

0.16458727640.1179810574

0.16455726590.1180010644

0.16452725550.1180210713

0.1644972450.1180410783

0.16446723460.1180610853

0.16443722420.1180810922

0.16440721370.1181010992

0.16437720330.1181211061

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[Red]

[Blue]

Time, sec

Concentration

Concentration vs. Time for 2 Red --> Blue

MBD00000100.unknown

MBD00000240.unknown

MBD000002E0.unknown

MBD00000380.unknown

MBD000003D0.unknown

MBD0057AC9A.unknown

MBD005704B5.unknown

MBD000003CC.unknown

MBD00000330.unknown

MBD0000037C.unknown

MBD0000032C.unknown

MBD00000290.unknown

MBD000002DC.unknown

MBD0000028C.unknown

MBD000001A0.unknown

MBD000001F0.unknown

MBD0000023C.unknown

MBD000001EC.unknown

MBD00000150.unknown

MBD0000019C.unknown

MBD0000014C.unknown

MBD000000A4.unknown

MBD000000FC.unknown

MBD00000050.unknown

Reaction DynamicsTimeRateInitially, only the forwardreaction takes place.As the forward reaction proceedsit makes products and uses reactants.Because the reactant concentration decreases, the forward reaction slows.As the products accumulate, thereverse reaction speeds up.Eventually, the reaction proceedsin the reverse direction as fast asit proceeds in the forward direction.At this time equilibrium is established.Once equilibrium is established,the forward and reverse reactions proceed at the same rate, so theconcentrations of all materialsstay constant.

H2(g) + I2(g) 2 HI(g)Since the [HI] at equilibrium is larger than the [H2] or [I2], we say the position of equilibrium favors productsAs the reaction proceeds, the [H2] and [I2] decrease and the [HI] increasesSince the reactant concentrations are decreasing, the forward reaction rate slows downAnd since the product concentration is increasing, the reverse reaction rate speeds upOnce equilibrium is established, the concentrations no longer changeAt equilibrium, the forward reaction rate is the same as the reverse reaction rate

Dynamic Equilibrium

some reactions reach equilibrium only after almost all the reactant molecules are consumed

equilibrium favors the products

other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed

equilibrium favors the reactants


An Analogy: Pwad(left) Pwad(right)Rules:Each student wads up two paper wads.You must start and stop as the timekeeper says.Throw only one paper wad at a time.If a paper wad lands next to you, you must throw it back.


Equal Number of Students on Each Side of the ClassroomPwad (left) Pwad (right)

Time (s)LeftRight0All05101520

Chart1

640

2939

3232

3034

3133

Left Side (Reactants)

Right Side (Products)

Time (s)

Number of Paper Wads

Sheet1

TimeLeft Side (Reactants)Right Side (Products)

0640

52939

103232

153034

203133

Sheet1

00

00

00

00

00



Time (s)


Sheet2

Sheet3

Most Students on the Left Side 2 Students on the Right SidePwad (left) Pwad (right)


Chart2

064

460

757

955

856



Time (s)


Sheet1


0640

52939

103232

153034

203133


30 students2 students

0064

5460

10757

15955

20856

Sheet1

00

00

00

00

00



Time (s)


Sheet2

00

00

00

00

00



Time (s)


Sheet3

Most Students on the Left Side 2 Students on the Right SidePwad (left) Pwad (right)


Chart3

640

2935

1549

955

1153



Time (s)


Sheet1


0640

52939

103232

153034

203133



0064

5460

10757

15955

20856



0640

52935

101549

15955

201153

Sheet1

00

00

00

00

00



Time (s)


Sheet2

00

00

00

00

00



Time (s)


Sheet3

00

00

00

00

00



Time (s)


Common MisconceptionsEquilibrium means equal amounts of reactant and product. No - A reaction can be at equilibrium and have more paper wads on the product side of the room

A reaction at equilibrium has stopped. No - The paper wads keep flying in both directions even after equilibrium is achieved

Equilibrium can only be achieved by starting with reactants. No - Equilibrium can also be achieved when starting with all of the paper wads on the product side of the room.


The Equilibrium Constant[reactants] and [products] are not equal at equilibrium, there is a relationship between them Law of Mass Action: aA + bB cC + dD,

Keq (or Kc) is called the equilibrium constantUnits vary from reaction to reaction, unitless in this case


The Equilibrium Constantso for the reaction: 2 N2O5 4 NO2 + O2 Units: mol3/L3


The Equilibrium ConstantSignificance of KeqKeq >> 1, when the reaction reaches equilibrium there will be more product than reactant moleculesthe position of equilibrium favors products

Keq

A Large Equilibrium ConstantRemember: does not tell us about how fast, only how far

A Small Equilibrium Constant

Relationships between K and Chemical Equationsreaction is written backwards, the equilibrium constant is invertedfor the reaction aA + bB cC + dDthe equilibrium constant expression is:for the reaction cC + dD aA + bBthe equilibrium constant expression is:


Relationships between K and Chemical Equationscoefficients of an equation are multiplied by a factor, K is raised to that factorfor the reaction aA + bB cC equilibrium constant expression is:for the reaction 2aA + 2bB 2cC equilibrium constant expression is:


Relationships between K and Chemical Equationswhen you add equations to get a new equation, Keq for the new equation is the product of the equilibrium constants of the old equationsfor the reactions (1) aA bB and (2) bB cC the K expressions are:for the overall reaction aA cC the K expression is:


Kbackward = 1/Kforward, Knew = KoldnEx 14.2 Compute the equilibrium constant at 25C for the reaction NH3(g) 0.5 N2(g) + 1.5 H2(g) for N2(g) + 3 H2(g) 2 NH3(g), Keq = 3.7 x 108 at 25C

Keq for NH3(g) 0.5N2(g) + 1.5H2(g), at 25C Solution:Concept Plan:

Relationships:Given:

Find:N2(g) + 3 H2(g) 2 NH3(g)K1 = 3.7 x 1082 NH3(g) N2(g) + 3 H2(g) NH3(g) 0.5 N2(g) + 1.5 H2(g)


Equilibrium Constant in Terms of PressureReactions Involving Gasesthe concentration of a gas in a mixture is proportional to its partial pressure

aA(g) + bB(g) cC(g) + dD(g)or


Kc and Kpwhen calculating Kp, partial pressures are always in atmthe values of Kp and Kc are not necessarily the samebecause of the difference in units

the relationship between them is:n = c + d (a + b) or no. moles of reactants minus no. Moles productsWhen n = 0, Kp = Kc


Deriving the Relationshipbetween Kp and Kc

Deriving the RelationshipBetween Kp and Kcfor aA(g) + bB(g) cC(g) + dD(g)substituting

Ex 14.3 Find Kc for the reaction 2 NO(g) + O2(g) 2 NO2(g), given Kp = 2.2 x 1012 @ 25CK has units L mol-1since there are more moles of reactant than product, Kc should be larger than Kp, and it isKp = 2.2 x 1012 atm-1Kc Check:Solution:Concept Plan:


Find:2 NO(g) + O2(g) 2 NO2(g)Dn = 2 3 = -1


Heterogeneous Equilibria: Reactions Involving Solids and Liquids

aA(s) + bB(aq) cC(l) + dD(aq)

Concentrations of solids and liquids doesnt changeits amount can change, but the amount of it in solution doesnt because it isnt in solution

solids and liquids are not included in the Keq expression


Heterogeneous EquilibriaThe amount of C is different, amounts of CO and CO2 remain the same. C has no effect on the position of equilibrium.

Calculating Equilibrium Constants from Measured Equilibrium ConcentrationsTo find K measure the amounts of reactants and products in a mixture at equilibrium

may have different amounts of reactants and products, but the value of the equilibrium constant will always be the sameas long as the temperature is kept constant


*Initial and Equilibrium Concentrations forH2(g) + I2(g) 2HI(g) @ 445C

InitialEquilibriumEquilibriumConstant[H2][I2][HI][H2][I2][HI]0.500.500.00.110.110.780.00.00.500.0550.0550.390.500.500.500.1650.1651.171.00.50.00.530.0330.934

Calculating Equilibrium Constants from Measured Equilibrium ConcentrationsStoichiometry can be used to determine the equilibrium [reactants] and [products] if you know initial concentrations and one equilibrium concentratione.g.2A(aq) + B(aq) 4C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M.+0.50-(0.50) -(0.50) 0.750.88

[A][B][C]initial molarity1.001.000change in concentrationequilibrium molarity0.50


Calculating Equilibrium Concentrationse.g.2A(aq) + B(aq) 4C(aq)+0.50-(0.50) -(0.50) 0.750.88Referred to as ICE table (initial, change, equilibrium)

Keq = [C]4/[A]2[B] = 0.13

[A][B][C]initial molarity1.001.000change in concentrationequilibrium molarity0.50


Ex 14.6 Find the value of Kc for the reaction2 CH4(g) C2H2(g) + 3 H2(g) at 1700C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M+0.035

Construct an ICE table for the reactionfor the substance whose equilibrium concentration is known, calculate the change in concentration

[CH4][C2H2][H2]initial0.1150.0000.000changeequilibrium0.035


Ex 14.6 Find the value of Kc for the reaction2 CH4(g) C2H2(g) + 3 H2(g) at 1700C if the initial [CH4] = 0.115 M and the equilibrium [C2H2] = 0.035 M+0.035-2(0.035) +3(0.035) 0.0450.105

use the known change to determine the change in the other materialsadd the change to the initial concentration to get the equilibrium concentration in each columnuse the equilibrium concentrations to calculate Kc

[CH4][C2H2][H2]initial0.1150.0000.000changeequilibrium0.035


The following data were collected for the reaction 2 NO2(g) N2O4(g) at 100C. Complete the table and determine values of Kp and Kc for each experiment.

Expt 1

Expt 2

initial [N2O4]

0

0.0200

initial [NO2]

0.0200

0

change [N2O4]

change [NO2]

equilibrium [N2O4]

0.00452

equilibrium [NO2]

0.0172

The following data were collected for the reaction 2 NO2(g) N2O4(g) at 100C. Complete the table and determine values of Kp and Kc for each experiment.

Expt 1

Expt 2

initial [N2O4]

0

0.0200

initial [NO2]

0.0200

0

change [N2O4]

+0.0014

-0.0155

change [NO2]

-0.0028

+0.0310

equilibrium [N2O4]

0.0014

0.00452

equilibrium [NO2]

0.0172

0.0310

The Reaction Quotientreaction mixture not at equilibrium; how can we determine which direction it will proceed?

the answer is to compare the current concentration ratios to the equilibrium constant

reaction quotient, Q for the gas phase reactionaA + bB cC + dDthe reaction quotient is:


The Reaction Quotient:Predicting the Direction of Changeif Q > K, the reaction will proceed fastest in the reverse directionQ must decrease, the [products] will decrease and [reactants] will increase

if Q < K, the reaction will proceed fastest in the forward directionQ must increase, the [products] will increase and [reactants] will decrease

if Q = K, the reaction is at equilibriumthe [products] and [reactants] will not change

if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward directionif a reaction mixture contains just products, Q = , and the reaction will proceed in the reverse direction


*Q, K, and the Direction of Reaction

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverseEx 14.7 For the reaction below, which direction will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl = 0.355 atm for I2(g) + Cl2(g) 2 ICl(g), Kp = 81.9

direction reaction will proceed Solution:Concept Plan:


Find:I2(g) + Cl2(g) 2 ICl(g)Kp = 81.9since Q (10.8) < K (81.9), the reaction will proceed to the right


Ex 14.8 If [COF2]eq = 0.255 M and [CF4]eq = 0.118 M, and Kc = 2.00 @ 1000C, find the [CO2]eq for the reaction given.Units & Magnitude OKCheck:Check: Round to 1 sig fig and substitute back inSolution:Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amountsConcept Plan:

Relationships:Strategize: You can calculate the missing concentration by using the equilibrium constant expression2 COF2 CO2 + CF4[COF2]eq = 0.255 M, [CF4]eq = 0.118 M [CO2]eqGiven:

Find:Sort: Youre given the reaction and Kc. Youre also given the [X]eq of all but one of the chemicals


A sample of PCl5(g) is placed in a 0.500 L container and heated to 160C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635

A sample of PCl5(g) is placed in a 0.500 L container and heated to 160C. The PCl5 is decomposed into PCl3(g) and Cl2(g). At equilibrium, 0.203 moles of PCl3 and Cl2 are formed. Determine the equilibrium concentration of PCl5 if Kc = 0.0635PCl5 PCl3 + Cl2

equilibriumconcentration, M?

Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressuresfirst decide which direction the reaction will proceedcompare Q to Kdefine the changes of all materials in terms of xuse the coefficient from the chemical equation for the coefficient of xthe x change is + for materials on the side the reaction is proceeding towardthe x change is for materials on the side the reaction is proceeding away fromsolve for xfor 2nd order equations, take square roots of both sides or use the quadratic formulamay be able to simplify and approximate answer for very large or small equilibrium constants


Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations since Qp(1) < Kp(81.9), the reaction is proceeding forward

[I2][Cl2][ICl]initial0.1000.1000.100changeequilibrium

Construct an ICE table for the reactiondetermine the direction the reaction is proceeding


Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations +2xxx0.100x0.100x0.100+2x


represent the change in the partial pressures in terms of xsum the columns to find the equilibrium concentrations in terms of xsubstitute into the equilibrium constant expression and solve for x


Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations +2xxx0.100x0.100x0.100+2x


substitute into the equilibrium constant expression and solve for x


Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations +2xxx0.100x0.100x0.100+2x0.0270.0270.246-0.0729-0.07292(-0.0729)


substitute x into the equilibrium concentration definition and solve


-0.0729Ex 14.11 For the reaction I2(g) + Cl2(g) 2 ICl(g) @ 25C, Kp = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations 0.0270.0270.246-0.07292(0.0729)Kp(calculated) = Kp(given) within significant figures


check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K


For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?(Hint: you will need to use the quadratic formula to solve for x)

For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?since [I]initial = 0, Q = 0 and the reaction must proceed forward

[I2][I]initial0.5000change-x+2xequilibrium0.500- x2x

*For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?


For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?x = 0.002160.500 0.00216 = 0.498[I2] = 0.498 M2(0.00216) = 0.00432[I] = 0.00432 M

[I2][I]initial0.5000change-x+2xequilibrium0.4980.00432

Approximations to Simplify the Mathwhen the equilibrium constant is very small, the position of equilibrium favors the reactantsfor relatively large initial [reactants], the [reactant] will not change significantly when it reaches equilibriumthe [X]equilibrium = ([X]initial ax) [X]initial we are approximating the equilibrium concentration of reactant to be the same as the initial concentrationassuming the reaction is proceeding forward


Checking the Approximation and Refining as Necessarywe can check our approximation afterwards by comparing the approximate value of x to the initial concentrationif the approximate value of x is less than 5% of the initial concentration, the approximation is valid


Tro, Chemistry: A Molecular Approach*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.since no products initially, Qc = 0, and the reaction is proceeding forward

[H2S][H2][S2]initial2.50x10-400changeequilibrium

Construct an ICE table for the reactiondetermine the direction the reaction is proceeding


Tro, Chemistry: A Molecular Approach*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.50x10-42x2xx

[H2S][H2][S2]initial2.50x10-400changeequilibrium

represent the change in the partial pressures in terms of xsum the columns to find the equilibrium concentrations in terms of xsubstitute into the equilibrium constant expression


*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.50E-42x2xx2.50E-4

[H2S][H2][S2]initial2.50E-400changeequilibrium

since Kc is very small, approximate the [H2S]eq = [H2S]init and solve for x


*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2xx2.50E-4the approximation is not valid!!


check if the approximation is valid by seeing if x < 5% of [H2S]init


*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.50E-42x2xxxcurrent = 1.38 x 10-5


if approximation is invalid, substitute xcurrent into Kc where it is subtracted and re-solve for xnew


Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.50E-42x2xxxcurrent = 1.27 x 10-5since xcurrent = xnew, approx. OK


Substitute xcurrent into Kc where it is subtracted and re-solve for xnew. If xnew is the same number, you have arrived at the best approximation.


Tro, Chemistry: A Molecular Approach*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.50E-42x2xxxcurrent = 1.28 x 10-52.24E-42.56E-51.28E-5


substitute xcurrent into the equilibrium concentration definitions and solve


*Ex 14.13 For the reaction 2 H2S(g) 2 H2(g) + S2(g) @ 800C, Kc = 1.67 x 10-7. If a 0.500 L flask initially containing 1.25 x 10-4 mol H2S is heated to 800C, find the equilibrium concentrations.+x+2x2x2.24E-42.56E-51.28E-5Kc(calculated) = Kc(given) within significant figures


check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K


For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?(use the simplifying assumption to solve for x)

For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?since [I]initial = 0, Q = 0 and the reaction must proceed forward


Tro, Chemistry: A Molecular Approach*For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?the approximation is valid!!



For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10-5 M/L at 1000 K. If 1.00 moles of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?x = 0.002170.500 0.00217 = 0.498[I2] = 0.498 M2(0.00217) = 0.00434[I] = 0.00434 M


Disturbing and Re-establishingEquilibrium once a reaction is at equilibrium, the concentrations of all the reactants and products remain the samehowever if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-establishedthe new concentrations will be different, but the equilibrium constant will be the sameunless you change the temperature


Le Chteliers PrincipleLe Chtelier's Principle:if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open

Concentration, temperature, volume, pressure


An Analogy: Population Changes

The Effect of Concentration Changes on EquilibriumAdding reactant :

aA + bB cC + dD

System shifts in a direction to minimize the disturbance

increases the amount of products until a new equilibrium is found that has the same K


The Effect of Concentration Changes on EquilibriumRemoving product:

aA + bB cC + dD

System shifts in a direction to minimize the disturbance

will increase the amounts of products and decrease the amounts of reactantsyou can use this to drive a reaction to completion!


Disturbing Equilibrium:Adding or Removing Reactantsafter equilibrium is established, how will adding a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?


Disturbing Equilibrium:Adding Reactantsadding a reactant initially increases the rate of forward reaction, but has no initial effect on the rate of reverse reaction. the reaction proceeds to the right until equilibrium is re-established. at the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the additionat the new equilibrium position, [reactants] and [products] will be such that the value of Keq is the same


Disturbing Equilibrium:Adding or Removing Reactants

after equilibrium is established, how will removing a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?


Disturbing Equilibrium:Removing Reactantsremoving a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reversethe reaction proceeds to the left until equilibrium is re-established. at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before the removalat the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same


Tro, Chemistry: A Molecular Approach*The Effect of Adding a Gas to a Gas Phase Reaction at Equilibriumadding a gaseous reactant increases its partial pressure, causing the equilibrium to the rightincreasing its partial pressure increases its concentrationdoes not increase the partial pressure of the other gases in the mixtureadding an inert gas to the mixture has no effect on the position of equilibriumdoes not effect the partial pressures of the gases in the reaction


The Effect of Concentration Changes on EquilibriumN2O4(g) NO2(g) Describe the effect of adding more NO2(g) to the following reaction:

The Effect of Concentration Changes on EquilibriumWhen NO2 is added, some of it combines to make more N2O4Q = KQ > KQ = K

The Effect of Concentration Changes on EquilibriumN2O4(g) NO2(g) Describe the effect of adding more N2O4(g) to the following reaction:

The Effect of Concentration Changes on EquilibriumWhen N2O4 is added, reaction shifts to make more NO2N2O4(g) NO2(g)

SummarizingIf a chemical system is in equilibrium:Inc. [reactant] (Q < K) reaction shifts to rightInc. [product] (Q > K) reaction shifts to leftDec. [reactant] (Q > K) reaction shifts to leftDec. [product] (Q < K) reaction shifts to right


Effect of Volume Changeon Equilibriumdecreasing the size of the container increases the concentration of all the gases in the containerincreases their partial pressuresif their partial pressures increase, then the total pressure in the container will increaseaccording to Le Chteliers Principle, the equilibrium should shift to remove that pressurethe way the system reduces the pressure is to reduce the number of gas molecules in the containerwhen the volume decreases, the equilibrium shifts to the side with fewer gas molecules


*The Effect of Volume Changes on Equilibrium

Disturbing Equilibrium: Reducing the Volumefor solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibriumdecreasing the container volume will increase the total pressureBoyles Lawif the total pressure increases, the partial pressures of all the gases will increaseDaltons Law of Partial Pressuresdecreasing the container volume increases the concentration of all gasessame number of moles, but different number of liters, resulting in a different molaritysince the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas moleculesshift toward the side with fewer gas moleculesat the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same


SummarizingIf a chemical system is in equilibrium:Dec. volume causes reaction to shift in direction that has fewer moles of gas particlesInc. volume causes reaction to shift in direction that has the greater number of moles of gas particlesIf equal moles of gas on both sides, no effectAdding an inert gas has no effect


The Effect of Temperature Changes on Equilibrium Positionexothermic reactions release energy and endothermic reactions absorb energyif we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Chteliers Principle to predict the effect of temperature changeseven though heat is not matter and not written in a proper equation


The Effect of Temperature Changes on Equilibrium for Exothermic Reactionsfor an exothermic reaction, heat is a productincreasing the temperature is like adding heataccording to Le Chteliers Principle, the equilibrium will shift away from the added heatadding heat to an exothermic reaction will decrease the [products] and increase the [reactants]adding heat to an exothermic reaction will decrease the value of Khow will decreasing the temperature affect the system?aA + bB cC + dD + Heat


The Effect of Temperature Changes on Equilibrium for Endothermic Reactionsfor an endothermic reaction, heat is a reactantincreasing the temperature is like adding heataccording to Le Chteliers Principle, the equilibrium will shift away from the added heatadding heat to an endothermic reaction will decrease the [reactants] and increase the [products]adding heat to an endothermic reaction will increase the value of Khow will decreasing the temperature affect the system?Heat + aA + bB cC + dD


The Effect of Temperature Changes on Equilibrium

Not Changing the Position of Equilibrium - Catalysts catalysts provide an alternative, more efficient mechanismworks for both forward and reverse reactionsaffects the rate of the forward and reverse reactions by the same factortherefore catalysts do not affect the position of equilibrium


Practice - Le Chteliers Principle2 SO2(g) + O2(g) 2 SO3(g) DH = -198 kJ How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?

adding more O2 to the containercondensing and removing SO3compressing the gasescooling the containerdoubling the volume of the containerwarming the mixtureadding the inert gas helium to the containeradding a catalyst to the mixture


Practice - Le Chteliers Principle2 SO2(g) + O2(g) 2 SO3(g) adding more O2 to the containercondensing and removing SO3compressing the gasescooling the containerdoubling the volume of the containerwarming the mixtureadding helium to the containeradding a catalyst to the mixtureshift to SO3shift to SO3shift to SO3shift to SO3shift to SO2shift to SO2no effectno effect


**************Try this also with more reactants than products*Try this also with more reactants than products*Try this also with more reactants than products***********************************if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1***************************http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf************************

chapter 14 chemical equilibrium chemistry ii. speed of a chemical reaction is determined by kinetics...

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