chapter 14 chemical equilibrium chemistry ii. speed of a chemical reaction is determined by kinetics...
TRANSCRIPT
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Chapter 14Chemical EquilibriumChemistry II
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
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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
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Tro, Chemistry: A Molecular Approach*Oxygen Exchange betweenMother and Fetus
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
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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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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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2080.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2100.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2120.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2140.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2180.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2240.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2280.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2300.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2320.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2360.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2380.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2400.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2480.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2500.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
2520.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2700.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2800.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2900.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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2960.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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
3080.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3100.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3120.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3140.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3160.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3180.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3200.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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3260.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3280.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
3300.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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3360.164377203300.118121106101643201180800.00405297970.00561075252660133026601330
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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
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
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
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0.16437720330.1181211061
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0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
0.16437720330.1181211061
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0.16437720330.1181211061
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0.16437720330.1181211061
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0.16437720330.1181211061
0.16437720330.1181211061
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0.16437720330.1181211061
0.16437720330.1181211061
[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
Tro, Chemistry: A Molecular Approach
-
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.
Tro, Chemistry: A Molecular Approach
-
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
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet2
Sheet3
-
Most Students on the Left Side 2 Students on the Right SidePwad (left) Pwad (right)
Time (s)LeftRight00All5101520
Chart2
064
460
757
955
856
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet1
TimeLeft Side (Reactants)Right Side (Products)
0640
52939
103232
153034
203133
TimeLeft Side (Reactants)Right Side (Products)
30 students2 students
0064
5460
10757
15955
20856
Sheet1
00
00
00
00
00
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet2
00
00
00
00
00
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet3
-
Most Students on the Left Side 2 Students on the Right SidePwad (left) Pwad (right)
Time (s)LeftRight0All05101520
Chart3
640
2935
1549
955
1153
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet1
TimeLeft Side (Reactants)Right Side (Products)
0640
52939
103232
153034
203133
TimeLeft Side (Reactants)Right Side (Products)
30 students2 students
0064
5460
10757
15955
20856
TimeLeft Side (Reactants)Right Side (Products)
30 students2 students
0640
52935
101549
15955
201153
Sheet1
00
00
00
00
00
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet2
00
00
00
00
00
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
Sheet3
00
00
00
00
00
Left Side (Reactants)
Right Side (Products)
Time (s)
Number of Paper Wads
-
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.
Tro, Chemistry: A Molecular Approach
-
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
The Equilibrium Constantso for the reaction: 2 N2O5 4 NO2 + O2 Units: mol3/L3
Tro, Chemistry: A Molecular Approach
-
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:
Tro, Chemistry: A Molecular Approach
-
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:
Tro, Chemistry: A Molecular Approach
-
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:
Tro, Chemistry: A Molecular Approach
-
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)
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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:
Relationships:Given:
Find:2 NO(g) + O2(g) 2 NO2(g)Dn = 2 3 = -1
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
*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
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
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
-
Tro, Chemistry: A Molecular Approach
-
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:
Tro, Chemistry: A Molecular Approach
-
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
Tro, Chemistry: A Molecular Approach
-
*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:
Relationships:Given:
Find:I2(g) + Cl2(g) 2 ICl(g)Kp = 81.9since Q (10.8) < K (81.9), the reaction will proceed to the right
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
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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?
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
[I2][Cl2][ICl]initial0.1000.1000.100changeequilibrium
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
Tro, Chemistry: A Molecular Approach
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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
[I2][Cl2][ICl]initial0.1000.1000.100changeequilibrium
substitute into the equilibrium constant expression and solve for x
Tro, Chemistry: A Molecular Approach
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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)
[I2][Cl2][ICl]initial0.1000.1000.100changeequilibrium
substitute x into the equilibrium concentration definition and solve
Tro, Chemistry: A Molecular Approach
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-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
[I2][Cl2][ICl]initial0.1000.1000.100changeequilibrium
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K
Tro, Chemistry: A Molecular Approach
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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)
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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
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*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]?
[I2][I]initial0.5000change-x+2xequilibrium0.500- x2x
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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
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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
Tro, Chemistry: A Molecular Approach
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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
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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
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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
Tro, Chemistry: A Molecular Approach
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*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
Tro, Chemistry: A Molecular Approach
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*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!!
[H2S][H2][S2]initial2.50E-400changeequilibrium
check if the approximation is valid by seeing if x < 5% of [H2S]init
Tro, Chemistry: A Molecular Approach
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*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
[H2S][H2][S2]initial2.50E-400changeequilibrium
if approximation is invalid, substitute xcurrent into Kc where it is subtracted and re-solve for xnew
Tro, Chemistry: A Molecular Approach
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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
[H2S][H2][S2]initial2.50E-400changeequilibrium
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
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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
[H2S][H2][S2]initial2.50E-400changeequilibrium
substitute xcurrent into the equilibrium concentration definitions and solve
Tro, Chemistry: A Molecular Approach
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*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
[H2S][H2][S2]initial2.50E-400changeequilibrium
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K
Tro, Chemistry: A Molecular Approach
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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)
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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
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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!!
[I2][I]initial0.5000change-x+2xequilibrium0.500- x2x
Tro, Chemistry: A Molecular Approach
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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
[I2][I]initial0.5000change-x+2xequilibrium0.500- x2x
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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
Tro, Chemistry: A Molecular Approach
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Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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An Analogy: Population Changes
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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
Tro, Chemistry: A Molecular Approach
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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!
Tro, Chemistry: A Molecular Approach
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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?
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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?
Tro, Chemistry: A Molecular Approach
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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
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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
Tro, Chemistry: A Molecular Approach
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The Effect of Concentration Changes on EquilibriumN2O4(g) NO2(g) Describe the effect of adding more NO2(g) to the following reaction:
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The Effect of Concentration Changes on EquilibriumWhen NO2 is added, some of it combines to make more N2O4Q = KQ > KQ = K
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The Effect of Concentration Changes on EquilibriumN2O4(g) NO2(g) Describe the effect of adding more N2O4(g) to the following reaction:
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The Effect of Concentration Changes on EquilibriumWhen N2O4 is added, reaction shifts to make more NO2N2O4(g) NO2(g)
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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*The Effect of Volume Changes on Equilibrium
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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The Effect of Temperature Changes on Equilibrium
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
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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
Tro, Chemistry: A Molecular Approach
**************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************************