Chemical Equilibriumq

ChemicalEquilibriumq

Reaction Dynamicswhen a reaction starts, the reactants are consumed and products are madeforward reaction = reactants productstherefore the reactant concentrations decrease and the product concentrations increaseas reactant concentration decreases, the forward reaction rate decreaseseventually, the products can react to reform some of the reactantsreverse reaction = products reactantsassuming the products are not allowed to escapeas product concentration increases, the reverse reaction rate increasesprocesses 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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1700.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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1800.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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1880.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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

2140.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2300.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

2320.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2500.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2700.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2800.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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2980.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3000.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3020.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3040.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3060.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3100.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3120.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3200.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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3300.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3320.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3340.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3360.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3380.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3400.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3420.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3440.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3460.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3480.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3500.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

3520.164377203300.118121106101643201180800.00405297970.00561075252660133026601330

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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

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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.

Dynamic Equilibriumas the forward reaction slows and the reverse reaction accelerates, eventually they reach the same ratedynamic equilibrium is the condition where the rates of the forward and reverse reactions are equalonce the reaction reaches equilibrium, the concentrations of all the chemicals remain constantbecause the chemicals are being consumed and made at the same rate

H2(g) + I2(g) 2 HI(g)at time 0, there are only reactants in the mixture, so only the forward reaction can take place[H2] = 8, [I2] = 8, [HI] = 0at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H2] = 6, [I2] = 6, [HI] = 4

H2(g) + I2(g) 2 HI(g)at time 32, there are now more products than reactants in the mixture the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated[H2] = 4, [I2] = 4, [HI] = 8at time 48, the amounts of products and reactants in the mixture havent changed the forward and reverse reactions are proceeding at the same rate it has reached equilibrium

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

Equilibrium Equalthe rates of the forward and reverse reactions are equal at equilibriumbut that does not mean the concentrations of reactants and products are equalsome reactions reach equilibrium only after almost all the reactant molecules are consumed we say the position of equilibrium favors the productsother reactions reach equilibrium when only a small percentage of the reactant molecules are consumed we say the position of equilibrium favors the reactants

An Analogy: Population Changes

Equilibrium Constanteven though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between themthe relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action for the general equation aA + bB cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the balanced chemical equationalways products over reactants K is called the equilibrium constantunitless

Writing Equilibrium Constant Expressionsfor the reaction aA(aq) + bB(aq) cC(aq) + dD(aq) the equilibrium constant expression is so for the reaction 2 N2O5 4 NO2 + O2 the equilibrium constant expression is:

What Does the Value of Keq Imply? when the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant moleculesthe position of equilibrium favors productswhen the value of Keq

A Large Equilibrium Constant

A Small Equilibrium Constant

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

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

Kbackward = 1/Kforward, Knew = KoldnCompute 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), K = 3.7 x 108 at 25C

K 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 Constants for Reactions Involving Gasesthe concentration of a gas in a mixture is proportional to its partial pressuretherefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gasesfor aA(g) + bB(g) cC(g) + dD(g) the equilibrium constant expressions areor

Kc and Kpin calculating Kp, the partial pressures are always in atmthe values of Kp and Kc are not necessarily the samebecause of the difference in unitsKp = Kc when Dn = 0the relationship between them is:Dn is the difference between the number of moles of reactants and moles of products

Deriving the Relationshipbetween Kp and KcKonsentrasi nitrogen 75% = misal terdapat 100 mol gas yang ada dalam udara, 75 mol merupakan gas nitrogen dalam udara, 25 molMerupakan gas h dan o

% volum merupakan Banyaknya volume gas dalam volum gas total

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

Find Kc for the reaction 2 NO(g) + O2(g) 2 NO2(g), given Kp = 2.2 x 1012 @ 25CK is a unitless numbersince there are more moles of reactant than product, Kc should be larger than Kp, and it isKp = 2.2 x 1012Kc Check:Solution:Concept Plan:


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

Heterogeneous Equilibriapure solids and pure liquids are materials whose concentration doesnt change during the course of a reactionits amount can change, but the amount of it in solution doesnt because it isnt in solutionbecause their concentration doesnt change, solids and liquids are not included in the equilibrium constant expressionfor the reaction aA(s) + bB(aq) cC(l) + dD(aq) the equilibrium constant expression is:

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

Calculating Equilibrium Constants from Measured Equilibrium Concentrationsthe most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibriumactually, you only need to measure one amount then use stoichiometry to calculate the other amountsthe equilibrium mixture 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 constantthe value of the equilibrium constant is independent of the initial amounts of reactants and products

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 ConcentrationsStoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentrationsuppose you have a reaction 2 A(aq) + B(aq) 4 C(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

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

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 Quotientif a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine which direction it will proceed?the answer is to compare the current concentration ratios to the equilibrium constantthe concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the 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 directionthe [products] will decrease and [reactants] will increaseif Q < K, the reaction will proceed fastest in the forward directionthe [products] will increase and [reactants] will decreaseif Q = K, the reaction is at equilibriumthe [products] and [reactants] will not changeif 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, reverseFor 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

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

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

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

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

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

-0.0729For 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 concentrations of reactants, the reactant concentration 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

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]initial2.50E-400changeequilibrium

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

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-42x2xx


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

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


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

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

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

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.

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

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


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 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 guides us in predicting the effect various changes in conditions have on the position of equilibrium it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open

An Analogy: Population Changes

The Effect of Concentration Changes on EquilibriumAdding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same KRemoving a product will increase the amounts of the other products and decrease the amounts of the reactants.you can use this to drive a reaction to completion! Equilibrium shifts away from side with added chemicals or toward side with removed chemicalsRemember, adding more of a solid or liquid does not change its concentration and therefore has no effect on the equilibrium

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?as long as the added reactant is included in the equilibrium constant expressioni.e., not a solid or liquidafter 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:Adding Reactantsadding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the 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, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

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

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 EquilibriumWhen NO2 is added, some of it combines to make more N2O4

The Effect of Concentration Changes on EquilibriumWhen N2O4 is added, some of itdecomposes to make more NO2

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

Disturbing Equilibrium:Changing the Volumeafter equilibrium is established, how will decreasing the container volume affect the total pressure of solids, liquid, and gases? How will it affect the concentration of solids, liquid, solutions, and gases? What will this cause? How will it affect the value of K?

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

The Effect of Volume Changes on Equilibrium

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 concentrations of products and increase the concentrations of reactantsadding 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 concentrations of reactants and increase the concentrations of productsadding 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 PrincipleThe reaction 2 SO2(g) + O2(g) 2 SO3(g) with DH = -198 kJ is at equilibrium. 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 PrincipleThe reaction 2 SO2(g) + O2(g) 2 SO3(g) with DH = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established?adding more O2 to the containershift to SO3condensing and removing SO3shift to SO3compressing the gasesshift to SO3cooling the containershift to SO3doubling the volume of the containershift to SO2warming the mixtureshift to SO2adding helium to the containerno effectadding a catalyst to the mixtureno effect

Contoh soal :0,1 mol HI dimasukkan dalam tabung 1 lt dan terurai sesuai reaksi : 2HI H2 + I2. Jika I2 yang terbentuk adalah 0,02 mol, berapa harga K?

2. Tetapan kesetimbangan untuk reaksi : A + 2B AB2 adalah 0,25. Berapa jumlah mol A yang harus dicampurkan pada 4 mol B dalam volume 5 lt agar menghasilkan 1 mol AB2.

Jawaban no 1.

Jawaban no.2Misal mol A mula-mula = x mol A + 2B AB2Mula-mula: x 4Terurai : 1 2Setimbang: x-1 4-2 = 2 1

[AB2] = mol / lt = 1 / 5 lt = 1/5[A] = mol / lt = x-1 / 5 lt = (x-1)/5[2B] = mol / lt = 2 / 5 lt = 2/5K = [AB2] = 1/5 x = 26 [A] [B]2 (x-1)/5 (2/5)2

Soal-soal :Tetapan kesetimbangan untuk reaksi : 2HBr H2 + Br2 adalah . Hitunglah mol H2 yang dihasilkan jika 2 mol HBr dimasukkan dalam tabung 2 liter.

1 mol A dan 1 mol B direaksikan dalam tabung 1 liter, sesuai dengan reaksi A + B C + D ternyata pada saat setimbang diperoleh 0,33 mol A. Berapa harga K?

Soal-soal :Tetapan kesetimbangan CO + H2O CO2 + H2 adalah 0,1. Berapa jumlah mol CO yang harus dicampurkan pada 3 mol H2O dalam volume 1 liter agar menghasilkan 2 mol H2.

*****************************************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*********************************************

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