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Course: CHEM 122, Fall 2010
School: Simon Fraser
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The 25. reduction potentials for Au3* and Ni2* are as follows: : +1.50 V E :4.23 Y Au nN + U- i2* 2e- -+ Ni R Aur* + 3e- -+ + t = t.\-L'o't)) E" t.Jz",t. Calculate AGo (at 25'C) for t}re reaction: 3Ni-+ 3Ni2* + 2Au 2Au3* + b = -nFE"rd), D0" a)-2l40kJ b) +5.00 x 102 kJ @t.oo x lo3 kJ d) +1.00 x l0'kJ e) -5.00 x 102 kJ Qlc)-v : *br lbut5x ,-tl P'K-Tfr"c...

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The 25. reduction potentials for Au3* and Ni2* are as follows: : +1.50 V E :4.23 Y Au nN + U- i2* 2e- -+ Ni R Aur* + 3e- -+ + t = t.\-L'o't)) E" t.Jz",t. Calculate AGo (at 25'C) for t}re reaction: 3Ni-+ 3Ni2* + 2Au 2Au3* + b = -nFE"rd), D0" a)-2l40kJ b) +5.00 x 102 kJ @t.oo x lo3 kJ d) +1.00 x l0'kJ e) -5.00 x 102 kJ Qlc)-v : *br lbut5x ,-tl P'K-Tfr"c = Jr{ ,9 ul /N. 26. Determine the equilibrium constant at2ioC for the reaction t-- Zn + Ni?+ -+ Zn2+ + Ni Zn?+ + 2e- r@+2e-+ a)4.1 x 103 b) x c) 9.3 6.4 x Y: -0.76v -+@ 108 l0{ E:4.23Y Ni e,t^= -o-rt -Gt'1b) = o'J] v k n Eo \., = olofl lnq I r^o'-' @-o " ol these e) none 27. A concentration cell is constructed using two$! electrodes with Niz* concentrations of 1.0 M and 1.00 x iq the two half-cells. The reduction potential ofNi2* is -{.23 V. Calculate the potential of the cell at25"C. trSo.r ra v Yl.ifl} v d) +0.132 e) +0.0592 V ,' 'i-l Ni't t :e *> ui Ed= -e'>3 f = " - V o'o}? l0{ M Q-o-" 'q?E = =zz1 o. 6 -+
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Simon Fraser - CHEM - 122
25. The reduction potentials for Au3* and Ni2* are as follows:: +1.50 VE :4.23 YAunN +U- i2* 2e- -+ NiRAur* + 3e- -++t = t.\-L'o't)E"t.Jz",t.Calculate AGo (at 25'C) for tre reaction:3Ni-+ 3Ni2* + 2Au2Au3* +b= -nFE"rd),D0"a)-2l40kJb) +5.
Simon Fraser - CHEM - 122
Answers:1.a2.a3.d4.a5.e6.c7-c8.d9.410. d11. d12.a13. c14. a15. a16. d17.b18. d19. c20. b2t.b22.b23.24.25.26.27.bacda
Simon Fraser - CHEM - 122
.$-eaker bondK.=HFHCIFIBri0rHIl0e10.108stronger acid=1:c'BondSrei:e:h LengthBondDissociationEnersr| |t 1tl i Ilvv L,Bond Polarity.Acid strength also correlates to bond polarityIncreasing acid strengthcH, NH3 HrO ItrIncreasing bo
Simon Fraser - CHEM - 122
i-*r.i\'a l,\Y l'i--1-59oc)sdo=Ed.$,O1r-fIttoor. IoES?',6\J.: t!r6l.L/ rrHilOHqoa'Pt:,')otou!oU$.609JH)p"l:EH.gcfw_"390.iC'F ?I-:-lo-qt\.-. o\trcfw_-1\'-sAJt).(|J/r rNooll5('(_\,\I't'll
Simon Fraser - CHEM - 122
$ltr-l'l\Iti\lnI l-\l^f\ |/\ I.o.to+-.ol\16l-.alrXsesol'E,2vIllgE,,aIE.'I,^t^F'1u.9oHrg<(-;xxvY\o_i-.:$aI'<l-I\)-lN:d'11"-s!a'a./','l/q<t!g'V=x'lxJurJlr3r;1/'tESJs.'=f9odN97tsi-+r6l
Simon Fraser - CHEM - 122
Chem L22 Fall-0g Trtorial 4Question 1-: For the gas phase reaction wO(S)+Or(g) - NOz(S)+Or(S),the following experimental values for the temperature dependance wereobtained:p= O ,-Ea/RITemperature (K) ll 195 na 260 298 396k(x10eL.mol-ls-l)ll1.08
Simon Fraser - CHEM - 122
c) What is k at?:50K?B -+ 3 D' the activationn 2z For the uncatal yzed reaction A + catalyst was added, theQuestiokJ. when aenergy was determined to bL 25what is the ratio of the rate constantactivation energy became 20 kJ.uncatalyzed' reaction?
Simon Fraser - CHEM - 122
Question 4: The following partial pressures were observed for the reac-tion at50OK:2 NH3 (g)*N, (*)'PNH'atm0.01*3H, (*)PNz0.03(3)Puzatm0.01 atma) Calculate the value of K, and the total pressure at 50O K.tr= fry'Pll'IpNHl'-cfw_,+4 f
Simon Fraser - CHEM - 122
Question 5: For the reactionNH4OCONH, (")*-'2 NH3(r)*CO,(*)(5)at 25 oC, the equilibrium constant in terms of activity is 0.03. A certainamount of NH'OCONH, is allowed to come to equilibrium with theabove products. Calculate the total pressure.
Simon Fraser - CHEM - 122
Chem 122 Fa11-09 Week 6, Tutoriall.5Calculate the value of K fbrOz(e)* O(*)<)O:G)given thatNOzG)NO19*O(e)Or(g)* NO(e)K=6.8x10-aeNO21g)* Oz(e)9 +N,t: = lV,Oz tl^lz = r.tC a JgK=5.8x10-34t. cfw_ o x ,"*1-)K = "y, tit' )2. Calculate the va
Simon Fraser - CHEM - 122
3. Consider the decomposition of HOCI at298K2HOCl6y)H2O19 +ClzOlr;K = 0.090If l.200atm of HOCI, 6.000atm of H2O and 5.500atm of CL2O are placedin a vessel and allowed to equilibrate what would be the o/o change in theamount of HOCI?,4"l,z, H0l
Simon Fraser - CHEM - 122
4. Given the reaction<)Fe3'1"4 * A8(.)Fe2.1ud* Ag.(uq)K=1.10x10-2a*'^tCalculate the amount of Ag'("d in moVL in solution if 0.60 M ofand 0.50 M Fe2t1"q; are put into solution with Agt,l.Jc-rXo-JtXc-xE o'6-xoxO.L/(6'r *x) xk=l.lo y /D
Simon Fraser - CHEM - 122
,Ar 1'fFyt?c$'ll* -tucfw_al^r= o'zb1q) =at* -t" -LhPcrcfw_u(1) + U->t!)cfw_ ^i,pe4bwe;Eaccfw_r41n1*l1i*;"^t9W-iX+^'"t2.oo ab^)nx c- 'j-,fdt +CtPalsJr"o()C-XKxtQ-xY=xn-x JYrk- 2.oo?r=-xPv=nKT-)n="ffCr( r.v7q+-)'
Simon Fraser - CHEM - 122
5. consider the followingon'reaction.,-' ta&A e BGl * 2ctt.-b/'fvYkt'"'/.- l^tl^ *t[v^'l#1" * *ihcfw_,/NIIL'"?org)fl;ft,,i14+yrygvo+,l,wn,uFor each of the followiig sets of initiat conditions, a! 25"C, in whichdirection will the reaction p
Simon Fraser - CHEM - 122
+:.EilCiii;E?E=i!=EI9-E. E -e =.t.J ZG o.-6<-_z,. _iti=2 oS.?Y=ao?+il=ts;=*= " F?f3 gfF=+f; f:;?Etafr ziir?:;4=E *li iAn=37;tiEiri?u ? ^:I=*iii+;*'q06=-J'-FII-IqLIFll-l?iI"rT I:;EA I!i ll|c,:lglglI z-"'"r-=1
Simon Fraser - CHEM - 122
\cba\Z,.Nen\ Fos.?fila(E,FtrFo=tf;'=.=C\C\FIE6=E,|lYr?\r-r/-oIrF.()Exdb ._=oq.Sr^-.,OSXo-.EgEEo.=.=88=FEL-'IJJ=I--'FJr(JJ.o-r+.l-Ja.Fcfw_IJJtra'lJ-3=IFzIYJIu=lC)a4>cfw_l-cfw_63r-1odt1>cfw
Simon Fraser - CHEM - 122
&._3$.r-toorl]X-IaJ-AHN)F(D(Dr-tLrlw5F ? 8?8 J6 | ? 5 r.E os.ZFoNg303'P +og -o;a6d.Aii;FI0)rdiatIo3oa,ECLIJg,Ez+iI36'av)\-/-Ecfw_JEIEagsEO)EhEUiJ[B$ IsP.vE'GF;s'8 x s-9lpc)ev) SodX
Simon Fraser - CHEM - 122
(fCoKnrratio.s (ml/L)tJzbJsaJI'JozzzP-OOAld_=.IIo5orh+goF+II0a='!N)ilililooobbbooo3EoaN]oaAAzoNoFA)iltDFtv0qolliivirl.t!doi+cfw_CD(DJr+tiJCDoiJov)CD)(+H.Hi)CDFr.HiJPO\0H.(+
Simon Fraser - CHEM - 122
Ncfw_ILUDecomposition ofTODAY,S LECTURE:2NO2(g) -+ 2NO(g) + Oztg)*REACTION KINETICS'.Instantaneous Rates.NO2Rate LawstDetermining the Rate Law!:o.quMethod of Initial Rateslnstantaneous RatesRate of Product FormationInstantaneous rate of
Simon Fraser - CHEM - 122
Method of lnitial RatesMethod of lnitial RatesExperimental data: Initial, instantaneous ratesWhat if the reaction rate depends on morethan one reactant?Concentration (M)RunRate (decreaseof[No]olHzln[NO]) mol L-r s-rMethod of initial rates:#t
Simon Fraser - CHEM - 122
Lcfw_\The lntegrated Rate LawT.DAY,S LE,CTURE:..,REACTION KINETICS"Differential Rate LawHow rate changes with concentration.lntegrated Rate LawThe Integrated Rate LawHow concentration changes with timeFirst Order Rate LawsFirst-order Rate Proc
Simon Fraser - CHEM - 122
Example: 1"t Order Rate Law2N2O5Questionl:-++ 024NO2.Is this a First-Order Reaction?Question 2: What is the rate constant?Answer: We can't tell without an experiment!+Half-life of lst Order Reactions.Measure how concentration varies with tim
Simon Fraser - CHEM - 122
2ndOrder lntegrated Rate LawSecond-Order Rate Processes1laA + productsDifferential rate law:Rste:-(Ddt4ldt=tAI-[A]oIf [A]o andkareknown, [A] canbe= klAf'calculated at any later timeRearranging,Second-order Integrated Rate Law: equation
Simon Fraser - CHEM - 122
.!rIZero-Order Rate ProcessesTODAY'S LECTURE,REACTION KINETIGS"I.Differential rate law+aAproductsIIiRate.Zero Order Reactions.Rate Law for >1 Reactant.The Isolation Method.- M:klAfo-kdtReaction MechanismslAl:-frr+[A]o_"v2 [A]
Simon Fraser - CHEM - 122
What lf We Have > 1 Reactant?Summary: Kinetics for Reactionsof the TYPe: aA + ProductsFor more complex reactions such as:5\^RaE:lawfRate =klAlRate =vtlAI2l,t!=-rt+lt\ r[z],=-tr+t'['a] k=".frIntegratedRate LawPlot need togive astraight li
Simon Fraser - CHEM - 122
Reaction MechanismsReaction Mechanism:series of steps that make upthe overall chemical reactionGOAL: To determine the reaction mechanism fromexperimental kinetic data that we have measuredOverall mechanism: composed ofa sequenceof"elementary rea
Simon Fraser - CHEM - 122
Rates and Molar RatiosL5ToDAY,S LECTURE:Reaction rate oC-.REACTION KINETICS'..Number of molecules consumedor produced by each reaction stepExample: For the first order reaction2A-+28+CRates and Molar RatiosdTCl.Rates and equilibrium.dtPr
Simon Fraser - CHEM - 122
Pre-equilibriumIn reactions with a fast initial equilibrium step:FastSlowk'k,A+B -i- lntermediate-a>k_rExample: Pre-Equilibrium.Decomposition of Ozone2ot@) -+ProductsExperimentally determined Rate Law:FastRate:DeterminingltcO *, and *-,
Simon Fraser - CHEM - 122
Steady-State Approxi mati ontL$ T.DAY,S LECTURE:Inmultiple step reaction:k,k,A+B _-i* Intermediar"t.REACTION KINETICS'IaproductsCan't choose a rate determining step?Steady-state approxim ation+Temperature Dependence ofReaction Ratesuse"S
Simon Fraser - CHEM - 122
Testing the Rate LawExperimentallySteady-State Approxi mationStep-by-StepVary reactant concentrations -+GOAL: To write the rate law without including anyintermediatesDoes rate law predict experiment result?O High tHzlk2lH2l=>O Write reaction m
Simon Fraser - CHEM - 122
Ghemical ExampleBiological ExampleThermal decomposition of acetaldehydeCH3CHO -+ CFL + CORare = _dtcH_3_cHolDatak) vs 1000I'\700t6mol-t stl0.01l -y7300.0357ffi0_1057m0338100-7E9840t4s 419x llTo>FgU,Yio]Goolrj201000L2-t7
Simon Fraser - CHEM - 122
Gollision Model for KineticsActivation Energyi/7 TODAY'S LECTIJRE:IIINO + NrO -+ NO, +ry.REACTION KINETIGS"IIiI-s,ffi. Collision Model for Kinetics4h\Reaction ProfilesIn this reaction N-O bond must be brokenTransition StatesN-O bond d
Simon Fraser - CHEM - 122
Reaction Profiles andTransition StatesMulti-Step Reaction Profiles. Rate of reaction determined by activationRate of reaction determined by activation energy, Euenergy, E o (and frequency factor, A)(and frequency factor, A).Fgr single step reactio
Simon Fraser - CHEM - 122
EnzymesEnzyme example: GatalaseEnzymes = biological catalystsVery efficient catalysis under bioloeical conditionsDecomposition of hydrogen peroxide (toxic)2HrO, -+ 2H2O + 02Example: Conversion of N, to NH,:Humans: Haber processNatureN2+3H2#flb2N
Simon Fraser - CHEM - 122
Question 3: An "exothermic reaction follows the two step mechanism:A+B -+ CC+D-+E(1)(2)The activation energy for the first step is L25 kJ /mol an for the secondstep is 200 kJ/mol.a) Draw a rough sketch of the reaction profile on the provided graph.
Simon Fraser - CHEM - 122
Relationship between K and KpEquilibrium ExpressionsFor Gases.To describe equilibria involving gases can use:To describe equilibria involving gases usually usepartial pressures instead of concentrationsPartial pressure:independent pressure exerte
Simon Fraser - CHEM - 122
Heterogeneous EquilibriaReactants and/or products in difrerent phasesFrom experiment:Position of heterogeneous equilibria do notdepend on the amounts of pure liquids or solidsWhy? Activities of pure solids & liquids =Example:CaCO3(s)=-CaO(s) + CO
Simon Fraser - CHEM - 122
Equilibrium GalculationsExample 1, Continued.Equilibrium CalculationsExample I6Define change to get to equilibrium and writeexpressions for equilibrium conc.st.O Write balanced equation for reaction:Hr(s)+ rz@) :2HI (g)Lt.c.B.O Write equilibr
Simon Fraser - CHEM - 122
Approximation Method forSmall Equilibrium Constants2NzG)+OzG).= Simpliff:2N2O(g)terrrs in denominator:(0.0482-2x) = 0.0482 (0.0933-x) = 0'0933Q'+t)-' "':l:)&>)-. o.cl3)'.ffisubsurute@checkvalue2X/o-rriro-)ru'is correctdLlsA=g-gygv-ZX
Simon Fraser - CHEM - 122
h,$qluhility Equilibria6',a1ie\th3TODAY'S LECTUREIonic salts dissolved in water are inequilibrium between the solid and the ions"Ghemical Equilibria".Zn(OH), (s).Solubility EquilibriaEquilibriumconstant: t\:- lzn2.lloH-1'?lz(orD,l.Solubility
Simon Fraser - CHEM - 122
Effect of Change in Concentration(at Constant Tem perature)Add reactant or product+system shifts away from added componentChange in Equilibrium ConcentrationsExample: Effect of change in concentration2SOr(g)Remove reactant or product+ Oz(g) =:Equ
Simon Fraser - CHEM - 122
Effect of Change inTemperatureChenge in Temperature -+ Change inThe Effect of a Gatalyston Equilibriaff.Hgiveoutheatenergy*)Catalyst reduces.Treat heat energy as product or reactantSpeeds+%(absorbheatenergy)-fheat energt-+Products\k)Equ
Simon Fraser - CHEM - 122
TODAY'S LECTURE"Acids and Bases"The Nature of Acids and Bases.Arrhenius concept:(H') ItBases produce hydroxide ions (oHJfAcids produce hydrogen ions*tto.The Nature of Acids and Bases.Acid Strength.BasesAcids in Aqueous SolutionsWater acts
Simon Fraser - CHEM - 122
Acid StrengthHA(aq)Acid:<lT1 <IE.lct-l arrrConjugate baseRelativeIo)ilconjugatebase strength()4a0JnqEE.H*(aq1+ A-(aq)Relativeacid strengtho!T1Bv!)+)jt\v/CSP'-rO(D=av.< +EH6 ri 5,=v)>-8!-*,Ashong acidhasaweakScD
Simon Fraser - CHEM - 122
roThe pH ScaleI pg:- tog\Ffl and poH: -log[oHlpKo: -logKo and pKr:-logKupKw: -logK*Kw:tH+ltOH1:l0-r4+ pK*:pH Scale : -log Scale14.mt.iffi.ffipH(Vinegar):3[fI*]:tOH-l:10-7,pH:7, pOH:pH(Stomach7acidl:2[H+]sto*"r,"ia: 10 x [F]vio"*_
Simon Fraser - CHEM - 122
roThe pH ScaleI pg:- tog\Ffl and poH: -log[oHlpKo: -logKo and pKr:-logKupKw: -logK*Kw:tH+ltOH1:l0-r4+ pK*:pH Scale : -log Scale14.mt.iffi.ffipH(Vinegar):3[fI*]:tOH-l:10-7,pH:7, pOH:pH(Stomach7acidl:2[H+]sto*"r,"ia: 10 x [F]vio"*_
Simon Fraser - CHEM - 122
LtlropAYSoLECTuRE:.'ACIDS AND BASES'..Water as an Acid and a BaseAutoionaation;ffirttora -+' Kw: [Icfw_3O*][OH-]: 1g-r+ @25 C. True in Neutral and Acid andBasicsolutionspH of Mixtures of AcidsFor:Any%tct=kb=CWdlDHA(aq)+ Icfw_2o(DLHAA-(
Simon Fraser - CHEM - 122
pH of Weak Acid Solution. Problem: Calculate the pH of a 0.100 MpH of Strong BasesExample: NaOH o'oz4Strong Base: Dissociatessolution of acetic acid.Stratery:+NaOH(aq)O Identifr Major Species in SolutionIdentify the major species in solutionO
Simon Fraser - CHEM - 122
?,>Yt -w f,rt^)LtAH+=zxrd'T)te [Hl - tcfw_Tl 1>1= o,otlt + bz:Acids =o.:1/b o'-tirrt z+ flWv-PolyproticPolyprotic Acids (e.g. Icfw_2CO:, Icfw_rPOo)dissociate in a stepwise manner'AGIDS AND BASES'..o@r-r+-4_v@ FIPO+2-$.7*.10'6rup
Simon Fraser - CHEM - 122
Salts That Produce AcidicSolutionsSalts That Produce BasicSolutions1) Anion = conjugate base of weak acid?2) Cation has no effect?r) Cation = conjugate acid of weak base?2) Anion has no effect?Acidic solution=*) e4ut'cfw_'rvttnkP;ample: NHoBr
Simon Fraser - CHEM - 122
o.Summary: Acid-BaseProperties of SaltsAcid-Base Properties of AqueousSolutions of Various Types of SaltsI*cfw_ !.1C-r-lsircfw_-q.4Ci.rri-rql*ahtXHO-i(-.-i-rIocfw_rc*i*-ri:qrl)$(I NrNO6\rCrH.O!(]L iirF\lL(lxll$s!\ldcfi,O5:cfw_cfw_dcfw_-
Simon Fraser - CHEM - 122
Goncept of EquilibriumLgState in which [Reactantsl and [Products]remain constant with time.T'DAY,S LECTURE:*CHEMICAL EQUILIBRIUM'.Equilibrium is dynamic.Law of mass action.BUT Equilibrium is \namic not StaticExpressions involving pressure.Fo
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How does the legislative branch provide checks and balances against the abuse of power?1.Since a government needs a majority of votes in both houses of Parliament to pass its bills and sincegovernment rarely control both houses, a review of bills is li
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