Twoissues:SpontaneityofreactionsatnonstandardstateChemicalequlibriumChapter3ChemicalEquilibrium一、非標態下，化學反應方向的判據——化學反應等溫式△rGm=△rGmo+RT㏑QQ：反應商△rGm<0正向自發進行△rGm=0達平衡△rGm>0逆向自發進行aA+bB=gG+hH

例1：C2H5OH(l.)→C2H5OH(g.)(298K)，△rGmo=6.2(KJ/mol)，當P1=1/2Po，P2=1/10Po，P3=1/20Po時，反應能否自發進行？

解：△rGm(1)=△rGmo+RT㏑()=6.2+8.314×298㏑(1/2)=6.2－1.7=4.5(KJ/mol)>0△rGm(2)=6.2+8.314×298×10－3㏑(1/10)=6.2－5.7=0.495(KJ/mol)>0△rGm(3)=6.2+8.314×298×10－3㏑(1/20)=6.2－7.4=－1.2(KJ/mol)<0蒸發可自發進行例2：求298K時，Fe生銹的逆反應發生的條件。解：2Fe(s.)+3/2O2(g.)→Fe2O3(s.)△fGmo(KJ/mol)00－741.0∴△rGmo=－741.0KJ/mol若要逆向進行，需△rGm

<0

即△rGm

=△rGmo+RT㏑Q=-741.0+8.314×298×10－3㏑(PO2/Po)–3/2>O=-741.0—3.72㏑(PO2/Po)>0㏑(PO2/Po)<－199.4PO2/Po<2.52×10－87PO2<2.55×10－82Pa

不可能發生例3：求下述反應逆反應發生的條件。

Zn(s)+Cu2+(aq)=Zn2+(aq)+Cu(s)

解：

Zn(s)+Cu2+(aq)=Zn2+(aq)+Cu(s)

t=01mol/dm30tx1-x△fGmo(KJ/mol)064.77－153.890△rGmo=△fGmo(Zn2+)－△fGmo(Cu2+)=－153.89－64.77=－218.66(KJ/mol)△rGm=△rGmo+RT㏑=－218.66+8.314×298×10－3㏑([Zn2+]/[Cu2+])>0㏑([Zn2+]/[Cu2+])>(218.66/2.48)>0([Zn2+]/[Cu2+])>2.13×1038

即若想使反應逆向進行，[Zn2+]>2.13×1038[Cu2+]

三點說明①當△rGmo數值較大時，可用△rGmo代替△rGm估計/判斷反應方向。

|△rGmo|>41.84KJ/mol②當△rGmo數值較小時，|△rGmo|<40KJ/mol，需用化學反應等溫式計算出△rGm，用△rGm的符號判斷方向。③當△rGm=0時，反應達平衡，由化學反應等溫式△rGmo=－RT㏑Q=－RT㏑Ko∵一定溫度下，一確定反應，其△rGmo是個定數∴此時Q為一常數，稱之為標準平衡常數，用Ko表示。顯然，同樣類型的反應，相同溫度時，Ko越大，表明反應正向進行的程度越大。2.Chemicalequlibrium2.1

Characteristicsofanequilibrium2.2

Equlibriumconstantsandequlibriumconstantexpressions：Ko,Kc,Kp2.3Transferrate平衡轉化率α=(C初－C平)/C初

=△C/C初

3.Shiftsinchemicalequilibium3.1Effectofchangesinconcentrations3.2Effectofchangesinpressure3.3Effectofchangesintemperature——3.4Effectofcatalyst

u

Eachcomponentconcentrationwillnotchangewithtime.Althoughdifferentinitialconcentrationsgivedifferentequilibiumconcentrations,thequantity,[HI]2eq/[H2]eq[I2]eq,isconstant.Kc=[HI]2eq/[H2]eq[I2]eqTheconstantiscalledequilibriumconstantfortheequilibriumandisreprensentedbythesymbolKc.Dynamicequilibrium.Theforwardreactionrateequalstothereversereactionrate.

Itisaconditionalequilibrium.Whentheconcentration,pressureandtemperaturechange,theequilibriumpositionwillchange.2.2EqulibriumconstantsandEqulibriumconstantexpressions：Ko，Kc，KpaA+bB=gG+hH2.2.1

Experimentalequilibriumconstants(1)ConcentrationequilibriumconstantsKc

Kc=

C:mol/L

(2)PartialpressureequilibriumconstantsKp

Kp=P：Pa(3)HeterogeneousequilibiumCaCO3(s)=CaO(s)+CO2(g)Kc=[CO2]Kp=PCO2CaCO3(s)+H2O(l)+CO2(g)=Ca2+(aq)+2HCO3-(aq)

2.2.2Standardequlibriumconstants(1)Insolution,Ko=(2)Ingasphase,Ko=(3)Inheterogeneousphases

CO2(g)+H2O+Ca2+=CaCO3(s)+2H+(aq)Ko=Example1:Atthetemperatureof1000K,put1.00molSO2and1.00molO2inaclosedcontainerof5.00dm3，whenequilibiumisreached,0.85molSO3isformed.（1）CalculateKo,KpandKcofthereaction2SO2+O2=2SO3atthetemperatureof1000K;（2）

UsethermodynamicdatatocalculateKoatthetemperatureof298K.Solution：(1)2SO2(g)+O2(g)=2SO3(g)t=0(mol)1.001.000teq1.00－0.851.00－0.85/20.85=0.15=0.575PSO2=nRT/V=0.15×8.314×1000/(5×10－3)=2.49×105(Pa)PO2=0.575×8.314×1000/(5×10－3)=9.56×105(Pa)PSO3=0.85×8.314×1000/(5×10－3)=1.41×106(Pa)∴Kp=PSO32/(PSO22PO2)=(1.41×106)/[(2.49×105)×9.56×105]=3.4×10－5(Pa)Ko==3.4×10－5×1.013×105=3.4Kc=[SO3]2/([SO2]2[O2])=(0.85/5)2/(0.15/5)2(0.575/5)=0.852×5/(0.152×0.575)=279.2(dm3/mol)(2)2SO2(g)+O2(g)=2SO3(g)△fGmo(KJ/mol)－300.70－370.37∴△rGmo=2△fGmoSO3(g)－2△fGmoSO2(g)=－370.37×2+300.37×2=－140(KJ/mol)∵△rGmo=－RT㏑Ko∴lgKo=△rGmo/2.303RT=140×10－3/(2.303×8.314×298)=240Ko(298K)=3.44×1024

Comments:①

Equilibriumconstantsdependonthetemperature,andhavenothingtodowithconcentrationsorpressures.②

Equilibriumconstantsexpresshowfarawaytheequilibiumproceeds.ThebiggerthevalueofKo,themorecompletethereactionproceeds.③

Forthesamereactionwrittenindifferentways,thecorrespondingequilibriumconstantsaredifferent.H2+I2=2HIKo11/2H2+1/2I2=HIKo22HI=H2+I2Ko3Ko1=Ko22=(Ko3)

－1

IfEquation(3)=Equation(1)+Equation(2),Ko3=Ko1·Ko2Equation(3)=Equation(1)-Equation(2),Ko3=Ko1/Ko2

H2S=H++HS－

Ko1=HS－

=H++S2－

Ko2=H2S=2H++S2－

Ko=Iftherearetwoormorethantwoequilibriaexistinginareactionsystem,itiscalledmultipleequilibria.Acomponenthasonlyoneconcentrationwhenitappearsindifferentequations.Example2:(1)NO(g)+1/2Br2(l)=NOBr(g)(25℃)溴化亞硝酰Ko1=3.6×1015，vaporpressureofBr2(l)is28.4KPaatthetemperatureof25℃；（2）Br2(l.)=Br2(g.)Ko2=P/Po=28.4/101.325=0.28PleasecalculatetheKoofthefollowingreaction:(3)NO(g)+1/2Br2(g)=NOBr(g)Solution：(2’)1/2Br2(l)=1/2Br2(g)K2o’=K2o1/2=0.529(1)－(2’)=(3)，Ko3=Ko1/Ko2’=3.6×10－15/0.529=6.8×10－15

Accordingtothedefinition,

lgKo=△rGmo/2.303RTAccordingtheruleofmultipleequilibria,fromtheknownKo

2.3

Equilibiriumtransformrate平衡轉化率

Def:α=(nini－neq)/nini=△n/nini）×100%whenvolumeisconstant，α=（Cini－Ceq）/Cini×100%Comparedwithequilibriumconstant,αchangeswiththeinitialandequilibriumconcentrations,whileequilibriumconstantonlychangeswiththetemperatureandhasnothingtodowithconcentrations.3.Shiftsofchemicalequilibrium△rGm=△rGmo+RT㏑Q=－RT㏑Ko+RT㏑Q=RT㏑(Q/Ko)△rGm=RT㏑(Q/Ko)△rGm<0，Q<Ko，shifttotheright,forwardreaction△rGm=0，Q=Ko，atequilibrium△rGm>0，Q>Ko，shifttotheleft,reversereaction3.1EffectofChangesinConcentrationChangesinconcentrationsdonotchangethevalueoftheequilibriumconstant.Example3：ReactionCO(g)+H2O(g)=H2(g)+CO2(g)Kc=9t=0(mol/L)0.0200.02000t’=0(mol/L)0.0201.0000teq（mol/L）0.020－x0.020－xxxteq（mol/L）0.020－y1.00－yyyKc=[H2][CO2]/([CO][H2O]=x/(0.02－x)2=9x=0.015(mol/L)CO的轉化率α=0.015/0.02×100%=75%Inthesameway，Kc=y2/[(0.02－y)(1.00－y)]=9y=0.01995CO的轉化率α=0.01995/0.020×100%=99.8%Example4：Attemperatureof298K，mixequalvolumeof0.100mol/LAgNO3，0.100mol/LFe(NO3)2and0.0100mol/LFe(NO3)3together.Fe2++Ag+=Fe3++Ag(s)Ko=2.98Calculatetosolve：（1）whichdirectionwillthereactionshift?（2）whenequilibriumisreached,concentrationsofeachcomponent(3)TransformefficiencyofAg+(4)[Ag+]、[Fe3+]remainthesame，[Fe2+]=0.300mol/dm－3，TransformefficiencyofAg+Solution:=0.0100/(0.100×0.100)=1<KoTheequilibriumwillshifttotheright.Forwardreactionwilltakeplace.(2)Fe2++Ag+=Fe3++Agt=0(mol/L)0.1000.1000.100teq0.100－x0.100－x0.100+x K=(0.0100+x)/(0.100－x)=2.98x=0.0127(mol/L)∴[Fe2+]=[Ag+]=0.0837mol/L[Fe3+]=0.0227mol/L

(3)Ag+

的轉化率α=0.0127/0.100=12.7%(4)設此時Ag+的轉化率為α，則

Fe2++Ag+=Fe3++Agt=00.3000.1000.0100teq0.300－0.100α0.100(1－α)0.0100+0.0100αKo=(0.0100+0.100α)/{(0.300－0.100α)[0.100(1－α)]}=2.98α=0.381=38.1%3.2EffectofchangesinpressureChangingthepressuredoesnotchangethevalueoftheequilibriumconstant.Example5：N2(g)+3H2(g)=2NH3(g)Whenthetotalpressureistwotimesoftheoriginalpressure,Equilibriumwillshifttotheright.Example6:CO(g)+H2O(g)=CO2(g)+H2(g)Whenthetotalpressureistwotimesoftheoriginalpressure,例1：合成氨N2(g)+3H2(g)=2NH3(g)原料氣（Air）循環使用，當惰氣和積累過多時，影響氨的生成，此時，需放空，補充新鮮氣例2：乙烷裂解制乙烯C2H6(g)=C2H4(g)+H2(g)恒T、P時，采用加入H2O(g.)的方法提高乙烯的產率。

3.3Effectofchangesintemperature

∵△rGmo（T）

=△rHmo－T△rSmo△rGmo（T）

=－RT㏑Ko∴－RT㏑Ko=△rHmo－T△rSmo㏑Ko=[－△rHmo（298K）

/RT]+[△rSmo（298K）/R]△rHmo<0，T2>T1，Ko2<Ko1△rHmo<0，T2>T1，Ko2<Ko1Toincreasethetemperature,reactionwillshifttotheendothermicreaction.Example7:Calculate:（1）△rGmat427℃（700K）forthereactionN2（g）+3H2(g)→2NH3(g)IfPN2=33.0atm，PH2=99.0atm，andPNH3=2.0atm,(△rHmo=－92.22KJ/mol，△rSmo=－198.53J/mol·K，△rGmo=－32.96KJ/mol)predictthedirectionofspontaneouschange.（2）Ko（298K）,Ko（700K）,Ko（773K）（3）Discussthesuitablereactionconditions.△rGm=△rGmo+RT㏑Q=46.75+8.314×700×10－3㏑=46.75+8.314×700×10－3㏑=46.75+5.82×(－15.896)=46.75－92.51=－45.76(KJ/mol)<0Solution：(1)△rGmo(700K)=△rHmo－T·△rSmo=－92.22－700×(－198.53×10－3)=46.7I5(KJ/mol)>0Theforwardreactionisnonspontaneousunderstandardstate.Theforwardreactionisspontaneous.lgK773o?