General Chemistry II
General Chemistry II CHEM 1220
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Date Created: 10/26/15
CHEMISTRY 1220 NOTES CHAPTER 20 ORGANIC CHEMISTRY WHAT39S 80 SPECIAL ABOUT C gt ABUNDANT 2 STRONG BONDS Compare the following C C 347 N N 159 0 0 138 kJmole Si Si 220 Ge Ge 180 9 0 0 chains are stable SnSn 120 gt FAVORABLE 71 BONDING CC CEC CO CN and CEN All are important in organic chemistry KEY PRINCIPLE C forms bonds leading to octet gt May bond to 2 3 or 4 atoms eg iiD Also note CN39 HCH CH4 CH3NC formaldehyde methane 2030 Chapter 20 1 Let39s look at the most important classes of organic compounds 1 Alkanes Saturated Hydrocarbons General formula CnH2n2 gt Only 0 H and C C bonds Examples W W W39 39i 39 39 HrH HsiCeH H Q P CFH H H H H H mam eth e 9ij CH4 c2H6 03H8 Geometries local tetrahedral environment about C H H H HR H C H HCVH C C H st H H H H H H H CH4 C2H6 C3H8 Note Staggering rather than eclipsing of H atoms on adjacent carbons Cyclic species also possible eg CH2n for n 2 3 cyclopropane cyclobutane etc A O ltgt CsHs CsH12 Note H is default substituent Chapter 20 2 Higher alkanes isomers possible Take C4H10 quotbutanequot 39l i39 l W 39i i39 39i Hi EH H ic H H H H H H H Z H H Mane H butane 2 methylpropane normal quotlinearquot iso butane C10H22 C20H42 Others named based on longest C chain C5 pentane 3 l3 th t WI ou C6 hexane C C C C H atoms C7 heptane C CB octane 224trimethylpentane etc standard for octane ratings 100 vs 0 for 39 nheptane cetane rating nC16H34 100 2244688 heptamethylnonane O gt Isomers display differences in chemistry and physical properties Note three CsH12 isomers H H H CH3 H CH3 H3C C 3 Ck3 CIZ CH3 H3CC 3Cll CH3 H3C I3 CH3 l l l i l gm n pentane 2 methylbutane 22dimethylpropane bp 360 280 IO0 Intermolecular interactions proportional to exposed surface area Chapter 20 3 2 Unsaturated Hydrocarbons Alkenes Alkvnes Olefins Acetylenes Alkenes CC bonds Cann Alkynes CEC bonds CnH2n2 Nomenclature alkane alkgne alkyne CC C C CEC C C C C C C C CC C C CEC C n butane 1butene amp 1butyne amp 2 butene 2butyne But CC usually not ethene but ethylene H CEC H CC C either propene or propylene acetylene ethyne Reactivities note bond energies 0 0 347 gt 265 gt Added in bonds not as strong CC 612 as lt5 bond more reactive d 208 gt Origin of the term unsaturated CEC 820 kJmole as in fa HgCCH2 gt polyethylene CH2n Example Decolorize Brg redbrown liquid H H ii ii cc Br2 H c c H Addition Reaction H H Br Br Chapter 20 4 3 Special Case of Unsaturated Hydrocarbons Aromatics Arenes Simplest Benzene CBHG Q lt gt O E recall graphite 1 35cyclohexatriene All C C bonds equivalent resonance forms formal bond order 15 H assumed as default substituents gt Conjugated alternating pi bonds experience special stability aromaticity tend to undergo substitution reactions not addition reactions Typical examples 1 O COH ow ow ow ow 1 OH OCCH3 toluene phenol aniline benzoic acid H quotmethylbenzenequot acidic basic aspirin Others are usually named more systematically eg 121 woe some mothballs o m or pdichlorobenzene ortho meta para 0 not 1 2 m not 1 3 pnot14 Chapter 20 5 Cl OH Cl lt3 but 124trichlorobenzene N02 0 or 2nitrophenol CI Also note polycyclic aromatics naphthalene some mothballs ClOHB and much higher often potent carcinogens Structure is an average of three localized resonance forms biphenyl nonplanar pcb s Naming Organic Substituents Alkane Substituent CH4 met ane CH3 methyl C2H6 emane 02H5 ethyl C3H8 pLopane C3H7 propyl etc CGHG benzene CBH5 phenyl Chapter 20 6 So CH3CHZCH3CIDHCH2CH3 CHZCH3 3 ethylhexane Cl CHZCHg m or 3Chloroethylbenzene Note for propyl and higher differing possibilities CH3CHZCH2 propyl normal CH323 H l propyl iso benzyl actually refers to CgH5CH2 Chapter 20 7 4 OxygenContaining Functional Groups we ve already seen some above General R used as alkyl designation CH3 C2H5 etc R OH R o R R CH R H gt gt alcohol ether aldehyde formaldehyde MTBE e g H3C O CCH33 MTBE RCHZR39 RCICPOH RCHJOR39 H3C39CC2H5 ketone carboxylic ester MEK acid RCH3 acetic HBC39CECHB acetone Some preparations co 2 H2 M CHgOH gasoline syn gas wood alcohol methanol methyl alcohol from C H20 Ni 2 CH4 other metals alkanes CZH5OH ethanol grain alcohol ethyl alcohol 9H3 Other HCISOH isopropanol rubbing alcohol alcohols CH3 Isopropyl alcohol HO CHg CH2 OH ethylene glycol antifreeze Chapter 20 8 9H3 9H CH3CHCH3CH2CHCHZCH3 6 methyI 3 heptanol Esters H20 ll RCOR39 RC OH H OR39 carboxylic acoho acid gt Named similar to salts So if the carboxylic acid is M acetic acid CH3COH and the alcohol is ethyl alcohol ethanol one has CH3 COC2H5 eth3Ll acetate l J Esters are used in perfumes characteristic odors ll ltCH3COCHCH2 vinyl acetate PVAgt ll ltHCOH formic acid gt Chapter 20 9 5 NitrogenContaining Functional Groups Amines Amides RR39RquotN Amine any R can II E RCNR39Rquot Amide any R can ll 33 Important Example H H H H l I H H H2NCH26NH2 gt nylon a polymer Both from NCCH24CN a nitrile ISOMERISM We39ve already seen several examples of structural isomerism as for two C4H10 forms three 05 etc Another type of isomerism is exhibited by alkenes Geometric lsomerism This relates to orientation on opposite sides of a CC bond CH3 CH3 CH3 H C C C C H H H CH3 cis 2 butene trans2 butene gt Physical and chemical properties differ Chapter 20 10 H3C CH 3 c CH3 lI3C CZC H CH2CH3 224trimethyltrans3 hexene Note Addition reactions often follow the Markovnikov rule with H adding to C with more H atoms CH3ZCCH2 HCI gt CH33CCI Additionally there is another type of isomerism Optical lsomerism Requirement usually need 4 different groups attached to carbon eg CHIFOH f 391 These are mirror 0 C0 images F kHOH F quotenantiomersquot Y J of each other try nonsuperimposable t0 COHfirm this 39 nonequivalent isomers However physical properties are identical bp density etc except f effect on planepolarized light One isomer will rotate light39s electric plane clockwise the other counter clockwise 25 Extremely important in biochemistry amp medicinal chemistry Chapter 20 11 Examples LDOPA Parkinson39s naproxen antiinflammatory etc phthalidomide HO VH3 l iO CHZ39HHH39CAEHWHCOZ L DOPA H H3CO S CH3 S Naproxen COZH H H O N O O N phthalidomide 0 Note With 2 or more chiral centers some isomers become non mirror images and therefore have different physical properties become quotdiastereomersquot Chapter 20 12 CHEMISTRY 1220 NOTES CHAPTER 15 ACIDS AND BASES DEFINITIONS To begin with what are acids and bases Old definitions from Arrhenius Acid releases H H30 Base releases OHquot What is the origin of this behaviorphenomenon H20 H OH gt Autoionization self ionization of water 2 H20 H30aq OH aq even HgO4 etc KW H30OH39 10 14 at 25 ignore H20 T dependence Note endothermic why More useful definition from Bronsted Lowry Acid proton H donor still Base proton H acceptor Later chapter Lewis definition Chapter 15 1 Examples NH4 H20 NH3 H30 HF H20 7 F39 H30 Acids any charge HSOs39 H20 303239 H30 Bases NH3 H20 NH4 OH Neutral or Anionic 303239 H20 HCOg39 OHquot Note that either definition works for the above examples But BL does not require H20 to be the source of H However what about HF NH3 NH4 Fquot Note All appear above T as acids or bases Proton Proton acceptor here but does donor not release OH39 in this reaction gt BL AcidBase reaction but not Arrhenius Note Related reactantproduct species NH3 NH4 Bases F HF ACIds NH3 and NH4 are referred to as a coniuqate acidbase pair Same for F39HF Thus NH3 is said to be the conjugate base of NH4 etc Chapter 15 2 Also note these relationships differ by H CACB CACB CACB CACB NH4 NH3 NH239 NHZ39 N3 H30 H20 OHquot 02quot NH3 and H20 are said to be amphiprotic as they can serve as either a conjugate acid or as a conjugate base Thus NH3 is the conjugate acid of NH239 but it is also the conjugate base of NH4 AciditiesBasicities are relative what is an acid in one situation can be a base in another What is ob of OH39 MEASURING ACIDITIES AND BASICITIES pH pOH Recall KW 103914 for 2 H20 2 H30 OH collision dependent KW H30OH39 551014 at 50 Question If a solution is neutral ie has equal concentrations of H30 and OH39 what are H30 and OH39 2 H20 H30Jr OH 55 M 2X X x so X2 KW 103914 x 10 7M H30 OH39 Chapter 15 3 What if H30 1 M Hgotnom KW 103914 1OH39 103914OH39 103914M Likewise if OH 1 M H3OOH39 KW 103914 moi1 KW 103914 met 10 14 M IIA Simplified Notation pH oqHi pOH oqOHi p power So for our neutral solution H30 10397M OH39 10397M DH 409107 Z DOH 40910 7 Z Observe H3O 1 10393 10397 10quot11 10 14 M OH39 10quot14 103911 10397 10 3 1 M pOH 14 11 7 3 0 Measure with a pH 0 3 7 11 14 pH meter or use indicators pH gt 7 for approximate values pHlt74 acidic basic Note pH pOH 1 pH x H301 10 Chapter 15 4 IIB Example Generally the pH of seawater 83 gt basic What are H30 OH39 and pOH 83 pH ogH30 logH30 83 Hgo 1083 100710399 510399M Since pH pOH 14 pOH1l4pH1483 OH39 1057 100310 6 210396M sigfigs Question Is it possible to have a negative pH Strong AcidBase Solutions Recall from Chapter 4 Table 42 amp pp 670 687 Strong acids HCl HBr HI HNOg HCIO3 HClO4 H2804 gt One H from each is completely dissociated in aqueous solution Strong bases MOH M Li Na K Rb Cs MOH2 M2 Ca2 Sr2 Ba2 gt M OH39 ions completely dissociated in aqueous solution Chapter 15 5 How does one calculate pH for strong species Examiple What is the pH of 100 L of an aqueous solution which contains 0240 g NaOH Since NaOH is a strong base OH39 is directly related to the number of moles of NaOH MW NaOH 400 gmole 0240 g l N OH mo es 3 400 gmole 60010 3 moles 600 mmoles m 10393egmL Therefore we have 60010393 moles in 100 L gt OH39 60010393M 103914 12 SO H30 16710 M 60010393 pOH log6OO1O393 3 og600 222 M9 Significant 12 figures pH og1673910 12 log167 1178 each Question What if it had been 60010393 moles of BaOH2 Chapter 15 6 WEAK ACIDS Note Substantial difference in dealing with Strong vs Weak Acids Strong acid HXH20 H3oX39 Kgtgt1oo Weak acid HXH20 H30X39 Klt1 Examples of weak acids HN02 H20 H30 N02 Neutral HSO439 H20 H30 804239 Anionic rare NH4 H20 H3O NH3 Cationic AlH2063 H20 Hgo AIH205OH2 gt For these we see partial dissociation as an equilibrium process Note that H20 serves as a base proton acceptor lllA Equilibrium Constants HX H20 H30 X H30X K8 Hx again omit H20 Chapter 15 7 IIIB Some examples pKa logKa H2803 Ka 1740 2 177 also hasa K32 D x g HF Ka 69104 316 HCN Ka 581010 924 Determination of K8 Example Suppose we find that the pH for a 0010 M solution of HCN is 570 What is the pKa and Ka Note for a strong monoprotio acid pH would be 200 HCN H20 2 H30 CN39 Starting point 0010 0 0 Adjust for equil 0010x x x If pH 570 H3O 10570 100301O 600 20106M CM Also HCN 0010 X 0010 2010396 0010 Note differs from value above and piltg 10 log40 940 Could pKa determinations be simplified Chapter 15 8 lllC Determination of pH from ng gt Sometimes straightforward sometimes not Example What would the pH be for a solution produced from 0010 moles of acetic acid HOAc dissolved to a final volume of 10 L Ka 1810395 OAc CHgCOg OAC be H30OAc39 2 1810395 HOAq HOAC H2O t H30 OAC Ka HOAc H30 OAc initial 0010 0 equil 0010 x x 39 2 H3O OAc X 18105 HOAC 0010 X Let39s first assume 2 hope thatx lt lt 0010 If this is valid we get 0010 x 0010 and therefore x2 18103950010 1810397 x 4210394 Was our assumption valid We will assume so if the value of x is s 5 of the value it is being subtracted from Here we get 4210394 0010 0042 42 so valid unless quot 33 Chapter 15 9 H301 OAC39 42104 M pH 4log42 337 Another example This time dissolve 0010 moles of HF in water until V 10 L What is the pH HF H20 H30 F39 Ka 69104 HF H301 F initial 0010 0 equil 0010x x 2 H30F x2 4 39 6910 HF 0010 x Let39s assume again that x lt lt 0010 Ifthis is valid x2 69106 x 00026 00026 H t 026 260 ere we ge 0010 o i Therefore our assumption is n Ot Valid39 How do we deal with this Either exact quadratic or approximate method Chapter 15 10 Exact Method Go back to the original equation X 5 910 4 0010 X I OR x2 6910396 6910394x then rearranging x2 6910394x 6910 6 0 ax2 bxc 0 2 Using the quadratic formula x 123 4aC Here a 1 b 6910394 andc 6910 6 4 42 6 so 5910 i1i6910 416910 21 691o4 i 1476107 27610 5 47610 7 2 00476105 6910394 i 28110 5 2 691O394 i 53403 2 Only 69 answer makes sense 5310393 69104 2 2310 3M Hgot F pH log2310393 3log23 264 Chapter 15 11 Other method Successive Approximation Again let39s go back to the original equation see p 1510 X2 1394 If ltlt gt 0010 69 0 x 0010 x 00026 We can use our first value of X 00026 p 1510 as a first approximation To get a better approximation substitute this value rather than 0 into 0010 x 3 X2 X2 4 2 6 6910 x 5110 0010 x 00074 x 2310393 Since this has changed a little from the previous value of 26103 let39s try one more x2 x2 0910394 x2 5310396 0010x 00077 x 2310393 Since there is no change from the previous value we have converged on the real value of x see quadratic result Summary Input x New x Change 0 00026 00026 00026 00023 00003 00023 00023 00000 Chapter 15 12 lllD Polvprotic Weak Acids or Weak Polyprotic Acids Example H3PO4 H3PO4 H20 2 Hgo H2PO439 Km 7110393 H2PO439 H20 H30 HPO42 Ka2 62108 HPO4239 H20 H30 PO43quot K 45103913 Observe that K39s are successively smaller as one is pulling H from ever greater negative charges So let39s take 1000 mole of H3PO4 and add water until V 1000 L Let39s calculate pH and all possible Since we have lots of H3PO4 we start with it It dissociates as indicated above ngo4 H20 s H30 H2PO439 K 7110 3 H30H2PO439 1 X X X H3PO4 x 1 x 7110393 Let39s assume hopex ltlt 1 If so x2 7110 3 x 0084 Mvalid 84 Using the successive approximation method we replace 1 x by 1 0084 0916 Then X2 7110393 gt x2 6510393 x 0081 0916 Chapter 15 13 Since there was little change from the previous x value we can stop Therefore mgm 11 x 1 0081 0919M Hgo HZPo i 0081 M pH logH30 og8110392 109 So far Oif strong We now turn to the second step H2P0439 H20 Hgo HPO4239 K 6210 8 initial 0081 0081 equil 0081 X 0081 x x HPo42Hso x0081 x 52108 H2p0439 0081 X SO Ifx ltlt 0081we getx 62108 assumption valid So for all practical purposes there is no change in H2PO439 or H30 but we see that HPof i 62108 M Finally the third step HPO4239 H20 s H3o PO43 K 451013 Initial 0210 8 0081 equil 0210398 x 0081 x x Chapter 15 14 008 x 451013 HPO42 6210 X Let39s hope x lt lt 0081 and x lt lt 6210398 usually assumptions get better with successive steps x0081 6210 8 If valid 451013 13 8 X 454003261210 34103919 assumption valid Summarizing H3PO4 0919 M H301 H2PO439 0081 M HPO4239 6210quot8 M PO43 34103919 M ZOOmL Chapter 15 15 IV WEAK BASES 9quot React with H20 to generate OH39 This means Hquot is transferred from H20 to the base eg NH4OH39 NH HO NHOH K 3aQ 2 4 b Kb 1810 5 pr 474 0N H20 3 HCN OH39 Kb HCNHOH CN39l Kb 1710 5 pr 477 Note that these reactions generate the conjugate acids of the particular bases and that the water functions as an acid proton donor We can ask if there should be a relationship between the Ka and Kb values for a given conjugate pair For a general HX Xquot H3OX39 HXHO HOX Ka 2 3 HX K H20 OH Kb HXH X adding 2 H20 H30 OH39 KaKb H30OH39 KW Kw Kw Ka w d K 30 Kb an b Ka We see that for a stronger acid greater Ka Kb must be smaller so as one conjugate species gets stronger the other gets weaker The more HX wants to go to Xquot the less X39 wants to go to HX Chapter 15 16 Compare Appendix II C weaker Conj Acid Ka Conj Base Kb H2303 1710 2 Hsog39 59103913 H3P04 71 10 3 H2PO439 1 4103912 HF 6910394 F39 1410 11 HN02 13010394 Noz 1710 11 HOAc 1810395 OAC39 5610391O Hsog39 6010398 8032 1 710397 HCN 58103910 CN 1710395 NHA 56103910 NH3 1810395 stronger Ka39Kb X s X E X E E E XXEXXX E Now an example initial adjust to equil Suppose we have a 10 L solution which was obtained by dissolving 010 moles of KCN in the necessary volume of water What is the pH CN H20 HCN CH ON 010 010X X2 010 x HCN OH39 Chapter 15 17 b ENHOH39l CN39 1710395 x2 17106 ifx ltlt 010 1710quot5 gtx 1310393 assumption valid HCN OH39 so if OH39 1310 3M pOH 3 log13 m pH 14288 m Solution is basic as expected Compare 01 M NaOH 9 pH 13 In other cases one may need to use either the quadratic formula or the method of successive approximations SALTS PREDICTING ACIDBASE BEHAVIOR Suppose we have a salt of general formula MX How do we decide whether this would be acidic or basic or neutral 7 gt We must look at the properties of cations and anions Cations either acidic or neutral quotspectatorquot Anions acidic basic or neutral quotspectatorquot Let39s take cations first Which are neutral 9 LL Na K 9212 Sr2 Ba2 also Rb and Cs Do these ring any bells Recall that their hydroxides are strong bases Connection Only neutral cations would have strongly basic hydroxides Hydroxides of other cations have reduced basicities due to the cations39 acidities All other cations acidic eg NH4 Mg Al3 transition metal cations chargeradius ratio important Chapter 15 18 What about anions Which are neutral 9 CIO439 Clogquot NOg39 Cl39 Br39 139 no H can t be acidic Do these ring any bells Recall that their conjugate acids are all strong Connection Only neutral anions would not reduce the acidities of their conjugate acids the H does not bond as strongly to them as to H20 Note 8042quot not neutral since HSO439 is not a strong acid Which are acidic HSO439 HSOg39 H2PO439 Not many since one is removing H from a 2 charge AH other common anions are basic Let39s confirm this for HPO4239 given Ka39s for H3PO4 H3PO4 H20 s H2PO4 H3O Ka1 7110393 H2P04 H20 2 HPO4239 H30 Ka2 62108 HPo4239 H20 lt PO43quot Hgot K33 4503913 K33 is clearly Ka for HPO4239 How do we get its Kb Kb would involve HPO4239 H20 H2Po4 OH39 Observe that this conjugate pair is found in K82 We should realize that Kb for HPO4239 E 1 Combine KW and K32 to confirm Kb 2 1014 6210398 Chapter 15 19 Since Kb gt Ka HPO4239 is a basic ion as expected try 1 M pH calculation Now that we can assess acidbase properties of individual ions lets look at their combinations ie salts Some combinations are obvious eg Example neutralneutral gt neutral NaCl neutral basic gt basic NaCN neutral acidic gt acidic NaHSO4 acidic neutral gt acidic AlClg NH4CI acidic acidic gt acidic NH4H2PO4 However when one is acidic and the other basic one must compare Ka and Kb 80 Ka Kb NH4CN 5610391O 1710395 basic NH4OAC 5610391O 5610 1O neutral coincidence AiOAc3 1210395 5610391O acidic for AlH20is3 AI3ltaqgt Calculating actual pH values for mixed acidbase salts is complicated we will limit ourselves to simply judging whether the combinations are acidic basic or neutral Chapter 15 20 VI ACIDBASE LEVELING Now that we know something about the acidbase properties of various ions we can take one last look at strongly acidicbasic species Compare Ka Conj acid Conj base Kb very large HCIO4 CIO439 very small very large HCI Cl39 very small very large HN03 NO339 very small H30 H20 6910394 HF F39 14103911 1810395 HOAc OAc39 5610 10 5610 10 NH4 NH3 1810395 H20 OHquot Hra X KOH M OH weakly lt NaOH weakly interact L39OH interact H readily OH readily transferred transferred to H20 t0 H20 hydrogen hydrogen bonding bonding Chapter 15 21 Even more acidicbasic species exist eg SbFsNaH NaNHg But in aqueous solution it is not possible to have significant amounts of an acid stronger than H30 or a base stronger than OHquot leveling Addition of such species to H20 leads to either H30 or OHquot eg NaH H20 gt NaOHaq H2 H20 gt H30 33904 Attachment of H20 to OH39 not usually indicated Chapter 15 22 CHEMISTRY 1220 NOTES CHAPTER 18 ELECTROCHEMISTRY Objectives 1 How can we generate energy electrical from spontaneous reac ons 9 VOLTAIC GALVANIC CELL 2 How can we bring about nonspontaneous reactions through the input of electrical energy 9 ELECTROLYTIC CELL 3 How can we relate voltages to spontaneity Before we address these issues let s be sure we can balance the sorts of reactions we need to deal with Example Feza1 Cr2072 aq Feeraq Cr3aOI A HalfReaction Method Fe2 gt Fe3 Cr2072 gt Cr3 Step 1 Balance non O H atoms Fe2 gt Fe3 Cr2072 gt ZCr3 Chapter 18 1 Step 2 Balance e Fe2 gt Fe3 e Cr2072 6e gt 2 Cr3 Cr6 Cr Step 3 Balance charges by H acidic media or CH basic media Fe Fe e Cr2072 14 H 6e 2 Cr3 each 6 Step 4 Balance H O by H20 Fe Fe e Cr2072 14 H 6e 2 Cr 7 H20 If an overall reaction is desired combine these in the proper ratio for the e to cancel 6 Fe Cr2072 14 Hquot 6 Fe 2 Cr 7 H20 B Single Reaction Step 1 The same balance non O H atoms but for entire reaction Fe or207 Fe 2 Cr Step 2 Again balance electrons different here we adjust stoichiometry so that e do not remain 6 e Fe e Fe Cr2072 2Cr Cr6 Chapter 18 2 6 Fe2 Cr2072 6 Fe3 2Cr3 Step 3 Balance charges Add 14 Hr to left or 14 OH on right although in basic media Cr2072 is converted to CrO42 6 Fe2 Crzo72 14 H 6 Fe3 2 or3 Step 4 Balance H O by H20 6 Fe2 orzo72 14 H 6 F63 2 or3 7 H20 We will see some important differences between the galvanic and electrolytic types of cells however common to both are the following Two electrodes e transferred from one to other QATHODE 9 Reduction occurs added electrons attract gations ANODE 9 Oxidation occurs loss of electrons attracts anions Start with voltaic cells eg batteries anZn2l lCu2lCu ANODE CATHODE 39l l l phase boundary Zn 4 CU H saltbridge Zn2aq L 1 CU2aQ Chapter 18 3 Will this work gt Add salt bridge Note advantage over thermochemistry reaction does not need to go to completion Spontaneous reaction occurs Zn Cu2aq gt Zn2aq Cu What if we just mixed them e go from Zn electrode 9 OXlDlZED ANODE to Cu electrode 9 REDUCTION CATHODE WHAT ARE ELECTRODE CHARGES AS ON BATTERIES gt Observe that e39 wants to go from Zn anode to Cu cathode 9 This would only occur be spontaneous if the Zn anode is negatively polarized vs the Cu cathode see on diagram reversed for electrolysis 3 To balance charge anions flow in other direction cations also move We can define halfreactions for each side Anode Zns gt Zn2aq 2e oxidation 69 Zn aZHSOA Cathode Cu2aq 2e gt Cus reduction CuSO4 gtCu Overall Zns Cu2aq Zn2aq Cus Chapter 18 4 Cell Potentials We have already observed that different spontaneous reactions have different driving forces AG39s K39s etc How is this reflected in voltaic cells voltage potential quotProblemquot We can39t measure a potential for a single halfcell one must measure one vs another Therefore we need a standard halfcell Pt H2 H or H H H2 Pt SHE anode cathode gt For this an inert quotmetalquot electrode is required Pt Ni Cr C 2 H 2e gt H2 E red 0000V by definition 1 M 1 atm gt Tabulate standard reduction voltagespotentials relative to HH2 Result When two halfcells are coupled together the net potential is equal to the difference between the Eoce Eocathode 39 Eoanode Zn Zn2 H H2 Pt 0762 v We can tell from direction of current flow which is Pt H2 H 002 Cu 0339 v anodecathode Example std pot O762 V ZnIZn2Cu2Cu observe 110v 0762 0339 Chapter 18 5 Note 1 These are independent of the number of e transferred as long as no e are left in the final reaction Changing number of e changes current not potential which reflects relative pushpull on e doubling molecules will double heat but not push behind each electron 2 Spontaneity observed for cell potentials unlike AG Greater potential 9 more favorable reaction Other examples 2 Li a 2 Liaq 2e 3040 F2 2e gt 2 F aq 2889 2 Li F2 gt 2 Liaq 2 F aq 5929V Cuaq gt Cu2aq e O161 e Cuaq gt Cus 0518 2Cuaq gt Cus Cu2aq 0357V disproportionation Als A3aq 3e 1680 3 e 3 Cr3aq gt 3 Cr2 0408 Als 3Cr3aq Al3aq sor2aq 1272v Again the number of e is not important for the overall V Chapter 18 6 Now how can we relate E0 to AG and K Do not need to go to completion Result p 834 AG nFE lt cell voltage e Faraday constant exchanged conversion factor between units F 9648104 Cmole Coulomb unit of charge 9648104 Jmole VEnergy Vit t in sec watts coulombs gt 9648n kJmole V For standard concentrations AG nFE Since AG can be related to K AG RTEnK so can E gt EnK AG FEO E RT orlong Eu RT RT 00257 v 0 0592 So for 2cUaq Cus Cu2aq 0357v n 1Why A60 nFE 19648 m 0357 V 344 kJmole mole EnK nE 10357V 1389 00257 V 00257 V K 108106 E G Kgt 1 How about Kfor1 V Chapter 18 7 What about nonstandard conditions Recall AG AGO RTEnQ nFE nFE gt dividing by nF E E0 E gEnQ Nernst Equation Eo std E nonstd Now we can substitute for RF or since most electrochemistry is done at room temp ma gt E E0 n nQ E0 ogQ Q usual reaction quotient Application Suppose we want to generate 1000 V from the reaction below Let39s see if we can do it 2CU Cds Cd2 2CUS E0 0921 v E E0 1000 0921 wig VJIogo Why doesn 2 20079 00592IogQ IogQ 267 L9G2alt1l Q 21410393 Cuaq2 gt Cd2aq 21410 3cUquotaq2 Chapter 18 8 Some possibilities icd2l 1 00021 M 10 M would not last long low capacity 10 M 216 M 9 possible 010 M 68 M possible 0050 M 48 M possible Do these trends make sense Reaction 2CUaq Cds 2CUS Cd2aq E0 0921v By increasing reactant driving force is increased Note that one can also use electrochemical measurements to obtain thermodynamic information on other types of reactions eg AgIs e gt Ags 0152V Agls gt Ag e 0799 AgIs gt Ag I O951 nE 1 0951 1609 39OgK 00592 00592 Ksp 810 17 Also can be used for Kf note that M in water E MH20X Chapter 18 9 lon Selective Electrodes Consider this Pt H2 H H H2 Pt E0 00000 for each half cell and for entire cell How can this do anything H e gt 12H2 cathode Recall E E0 W39IOQQ H2 He39 anode H HH2 gt HH2 E 000592 Vlog l i i i H CP422U c a a c With PH2a PHZC 1 atm W E 00592 V I a Ogiiwic E If H 1ME 00592l HR 00592 H H la ogl l p p 00592 HC 1O005EQZV 3 Every 00592 volts represents a change in H by a factor of 10 1 pH unit Also works for P eg oxygen sensors Electrolytic Cells Applied voltage by a battery is used to force a nonspontaneous reaction to proceed e are taken from the anode gt something oxidized there and pushed to cathode gt something reduced there Chapter 18 10 Battery anode cathode oxidation U U reduction Electrolysis Use same half reactions as before E applied voltage E will be negative if we are driving a nonspontaneous reaction Eo E oc nQ Electrical Measurements We already know voltage is a measure of the force behind the electrons A more precise definition of potential is 1 volt 1 JouleCoulomb push behind each e39 But what is a Coulomb Unit of charge 1 ampsec See p 187 96485104 Coulombs 1 mole of e39 Faraday constant 1 Coulomb 6241018equot 80 96480 Coulombs 96480 ampseconds 1 mole e Coulombs currentampstime flow Circuit breakers set for current Chapter 18 11 Finally Power Watts ampsvolts Joules wattsec T recall Energy qV Vit Energy Example How much H2 and CI2 could be produced from 1 kwhr of electricity 10 using a potential of 20 volts DC 2 H 2e H2 0000 2 CI Cl2 2e 4360 2 H 2Cl gt H2 CI2 1360 need extra over voltage 1 kwhr 1O3 watthrs 103 voltamp hrs 36106 voltampsec 36106 voltCoulombs J Since we are using 2 V this means we have 18106 Coulombs We can now use the relationship between Coulombs and moles A Faraday Constant 1mo4e e converS39On 964810 Coulomb factor 18106 Coulomb 1m039e 9 187 moles e 9648104 Coulomb Since we need 2H9 we get 93 moles H2 and Cl2 each 2 Chapter 18 12 lnterconversions of Electrical Units moles e F q charge Coulombs 39t vots i Current Amperes voltssec gt Energy Joules t 361O6 volts sec 361O6 kilowatth rs Power watts Resistance not often an issue so voltage typically comes from E tables Combaining Inalf reactions in general gt note AG n o E0 nEO MnO439 4 H 3equot gt Mn02 2 H20 169 507 Mno2 4H 2e39 Mn2 2 H20 123 246 MnO439 8H 5e Mn2 4H20 151 753 V 5 Chapter 18 13 CHEMISTRY 1220 NOTES CHAPTER 14 CHEMICAL EQUILIBRIUM In many chemical reactions one side or the other is highly favored and one may see no evidence of the species on the disfavored side eg AH AH Ea Exothermic Endothermic product favoredlt generally reactant favored In other cases there can be an observable balance eg 2 N02g N204g Observe 1 Start with pure N02 brown one forms some N204 colorless 0 2 Start with pure N204 one forms some N02 gt Each proceeds until a balance is reached So if one puts 1 g of N02 in a1 L flask or 1 g of N204 in a1 L flask eventually both flasks contain the same mass of N02 and the same mass of N204 if T is constant When this happens one says that a dynamic reversible equilibrium has been established Chapter 14 1 In such a situation 1 The position of equilibrium is independent of the direction of approach Useful check in practice 2 Both the fonNard and reverse reactions continue to occur ie dynamic vs static but we see no change 3 The rates of these reactions may be slow or fast could take 10 6 sec or less to 1010 years or more to reach equilibrium Ea T control these rates Critical Question How do we define the equilibrium state gt Equilibrium Constant For 2 N02g r A N204g PN204 one observes that 2 is constant for a given T N02 Thus if we take 3 flasks and add N02 andor N204 we might find the following 1 2 3 PN204atm 022 007 042 PN02 156 088 216 K1 022 0090 K2 007 0090 K3 042 0090 1562 0882 2162 Note Units ignored assume atm Chapter 14 2 In general for a reaction such as k we bBltggt K11 coo dDg K PCCPDd products PAaPBb reactants Note 1 For solution phases one replaces P by 2RT 2 The equilibrium state is independent of the path by which it is reached and depends only on energy differences between products and reactants OthenNise catalyst would change K Since the equilibrium state does not depend on path we can consider both forward and backward processes as though they were single elementary steps with the following rates From Chapter 14 gt forward k1PAaPBb equal at equilibrium reverse k1PCCPDd At equilibrium these rates must be the same so k1PAaPBb k1PcCPDd c d Note relationship PDb K ltbetween kineticsgt k1 PAaPB and equilibrium In cases in which fonNard and reverse reactions involve multi step processes math still works out the same Chapter 14 3 Examples Hg1 2HIg K PZHI 2 29 PH239PI2 P2002 1 2009 029 2C02g K1 PZCOP02 OR P 2 2 009 1202g C02g K2 PooP402 Note Adding rxn2 rxn2 rxn1 but K1 K239K2 gt When reactions are added together the new 5 is the product 9fihe gq Q Similarly reversing a reaction requires inverting K since reactants and products are changing places So if the equilibrium constant for A B is K what is the equilibrium constant for nA nB Example At 700 1 8029 72029 803w K1 L031 22 PsogP202 P P3922 2 N02g NOltggt 12049 K2 FIN 00 40 12 8029 N02g 1 8039 NOg Chapter 14 4 P303PNo I 4 Does this e ual K K P802 39 PN02 q 1 2 P303 PNoP02 PSOsPNO K1K2 pSO2 szOZ PN02 P802 PN02 gt Yes Note K K1K2 2240 88 Question What if we are dealing with a heterogeneous reaction often the case in industry 20 GDP First consider vapor pressures as an equilibrium 1Lflask watert waterg 100 L t A m W3 er at RT PH209 175 mm Hg This is K What if we add more H2002 1 L flask AZOO mL water H2005 H209 K PHZO gt We still observe PH20g 175 mm Hg Note relative humidity Therefore we see that the amount of a heterogeneous substance does not affect the equilibrium pressures of any other species as long g some of that substance is present Chapter 14 5 The same would be true of a more complicated reaction eg 3029 H2g quot1 CO9 H2003 reverse of water gas shift reaction H20 In such a case we do know that since H20 is volatile there will be some H20 in the gas phase However as long as there is liquid H20 present PH2O9 vapor PH20 a constant at that temperature In essence we can simplify K for the following 0029 H2g 1 CO9 H20g PCO39PHZO intoanew K K POO K39 whose K P002 PHz PH2o P002 PHz If the reaction occurs in the liguid phase the volume of the solution is important not for k or rate but for conversion larger volume more gas dissolved to react What would K then be What about equilibria in other phases Gases still appear as partial pressures usually atmosphere units Henry39s Law p 533 gt Pure heterogeneous solidsliquids do not appear nor do solvents for dilute solutions in K HgO 555 M gt Reactants in solution appear as molar concentrations Chapter 14 6 So for me 2 Haq Zn2ltaqgt H2ltggt PHz Zn2 W2 K Or for Zns 2 H30 Zn2aq H2g 2 H20 PH2 Zn2 H30 L2 K same DETERM39NATLQNQF K Simplest situation one is given all partial pressures eg NOzN204 equilibrium on p 14 2 Then one simply plugs the values into the K expression What about the following 2 HIg H2g I29 Suppose we fill a flask at 520 with HI until PHI 100 atm assuming no equilibration yet to H2 and 12 One then allows the system to reach equilibrium and finds that P12 010 atm What is K PH2 39 P12 PZHI K so we need P12 and PHI Since H2 and 12 are formed in a 11 ratio P12 PH2 010 atm Chapter 14 7 Since it takes two HI to make one H2 and one 12 there must be a loss in PHI of 2010 020 atm PHI 100 020 080 atm Total Pstill 1 atm K 010010 0016 szI 0802 This is fine if we know we are at equilibrium but Applications Direction and Extent of Reaction Considera system aAg bBg cCg dDg with general pressures which might or might not be at equilibrium We can define a reaction guotient Q analogous to K except that observed or hypothetical perhaps nonequilibrium pressures are employed Q PCC39PDd products PAa PBb reactants If a given set of pressures leads to a Q identical to K the system is at equilibrium lfthe calculated Q lt K the system will have to bring about conversion of reactants 9 products ifQ gt K products must go back to reactants Example For C029 H2g 1quot C09 H2Og 900 K K 064 Suppose we start with P002 PH2 Poo PH 0 1 atm 2 Chapter 14 8 What will these pressures become at equilibrium PcoPHzo 11 Q 1 NoteQgtK PCOzPHz Therefore this must shift to the left C02 H2 H20 initial 1 1 1 1 Adjusted 1X 1X 1x 1x K 064 02 1X2 1X O6412 080 1x O801x 1 x 080 080X 020 180x x 011 77PCO2 PH2 111atmPco PHZO 089 atm i 2 Check Q 0802 054 IIX Chapter 14 9 Second Example Consider N204g 2N02g K 110 100 Start with PNzo4 100 atm and allow the system to reach equilibrium Determine PN02 and PN204 PN204 PN02 initial 100 000 2 K M33 adjusted 100x 2x PN204 However since K gt 1 we might expect the equilibrium to favor the products So let39s shift the starting point to the other side Sometimes this simplifies the math This would give us initial 000 200 PA F X 05 10 adjusted X 2002x 10 141 20 20 P2N02 22X2 80 40 K 11390 PN204 x 11quot 500 100 Could use method of successive approximation see text 2 2x2 110x 4x2 8x 4 110x a b c x2x 0 gtquadratic ax2 bx c 0 Chapter 14 10 mi mgr 444 2a 8 192 361 193291 19i1723 8 8 453 or 022 Since X 453 would lead to a negative PNOZ x 022 must be the solution PN02 2 2x 156 atm not 2x 2 PN204 X E22 atm confirmz 1662 078 change from other conditions Altered Equilibria There are a number of ways in which a system may be perturbed from an equilibrium state 1 Adding or removing a reactant or product in the Q expression 2 Compressing or expanding the system P or V change Not including P of nonparticipants 3 Changing the temperature Only this changes K To determine what would happen after such a perturbation one can consider le Chatelier39s Principle When the conditions of a system are altered the system responds in such a way as to oppose the change Let39s look at these three possible perturbations Chapter 14 11 First Adding or Removing a Reactant or Product Consider aA bB cC dD PcquotF Dd K Q or PAaPBb At equilibrium K Q If we now add B too much reactant Q lt K and we are out of equilibrium The system can be restored to equilibrium by then shifting reactants including B to products Note by increasing reactant concentrations reactants collide m with each other so the rate of product formation increases Note Agreement with Le Chatelier39s Principle We added B and the system responded by reducing B Second Pressure or V Effects Very important reaction N2g 3 H2g 2 NH3g at 25 C K 6105 Nature can do Commercially the reaction rate is too slow at 25 due to high Ea Going to higher T increases rate but at 400 K 1510394 What can be done Catalyst helps some Observe PZNHs a2 PN2 P3H2 sz Q such thata PNH3 b PNZ c PH2 gives Q K under some equilibrium condition eg a 3 b 60 and c 10 at 400 Let s now compress the system by 10 fold thereby increasing each P by 10 Chapter 14 12 2 2 Ner 3 10b1Oc3 100 bc3 We now see that Q lt K so some reactants must be converted to products Or according to E Chatelier39s Principle since the pressure of the system was increased the system responds to reduce the pressure which is accomplished by converting 4 reactant molecules to 2 product molecules One can also remove products to drive reaction Finally Temperature Effects Note In these cases K changes unlike first two cases Temperature effects can also be treated using Le Chatelier39s Principle One simply regards heat as a product of an exothermic reaction or a reactant of an endothermic one Note N2g 3 H2g J 2 NH3g AHO 92 kJmole 30 N2 3 H2 TT 2 NH3 heat Therefore at higher T the reaction shifts to the left accounting for smaller K previous page Similarly heat N204g 2 N02g AH 57 kJmole colorless brown heat absorbed So at higher T the right side is favored Higher the T the less important energy differences are more energy around to break bonds Chapter 14 13 The Change in K can be related to T by the van t Hoff equation K2 AHO 1 1 En C or K KJ R T1 T2 compare to p 1320 Example Calculate Kat 200 C T2 for N204 2 N02 given AH above and K100 11 T1 K1 Eng 57kJmole 1 1 373 473 T1 T2 K1 m 68591030002681 0002114 3889 e3889 101689 2 4886 K1 K2 4886K1 488611 540 92303 101 Chapter 14 14 CHEMISTRY 1220 NOTES CHAPTER 13 CHEMICAL KINETICS Suppose we are interested in a chemical reaction eg A B gt C D There are 2 chemical aspects that need to be considered 1 Is the reaction favored to go Are the products better than the reactants gt Thermodynamics if unfavored thermodynamics might tell you how to bend the rules 2 Even if it should go will it go gt Kinetics often provides insight into how a reaction proceeds ie mechanism of reaction We will focus first on the 2nd question So how does kinetics give us such information gt One measures rates of chemical reactions as a function of concentrations time and temperature less often pressure Example for the reaction 2N205 gt 4N02 02 Chapter 13 1 We can measure the rate as AN20s dN20s rate alwa s At dt y Observe Rate of reaction slows with time N205 M 7 Two ways to estimate rate average vs instantaneous rates time We ll see a better way later Note that the rate could also have been expressed as AN02 AO2 T or T Not necessarily equal Forthe above suppose that att 40 min N205 122 M and then att 55 min N205 110 M Average rate AN205 At N205f N205i 110 122M 55 40min 15 min 0008 at 475 min Chapter 13 2 Alternatively if one sets rate AN02 AN2052moles N02 At At mole N205 0 008WW I L min mole N205 0016 L min Or if one sets A02 AN205 1mole 02 rate 7 At At 2 moles N205 0 m0e N205 1m0e 02 39 Lmin 2 moles N205 moles 02 L min 0004 usually based on reactant DEPENDENCE g RATES CONDITIONS Reaction rates depend greatly on several factors 1 Concentrations gt Rates of reactions often depend on collision rates the more often collisions occur the faster the reaction rate 39 higher 9 higher rate usuallyl Chapter 13 3 2 Temperature Paper wood gasoline etc stable at room temperature but burn once lit cf food 3 Catalysts Do not change favorability of a reaction but increase the rate at which it occurs forward and reverse Observations for 2 and 3 can be explained by an energy diagram reactants energy of r activation effect of catalyst energy d t pro ucs AH quotreaction coordinatequot Let s look at Concentration Dependence 1St Example 2N205 gt 4N02 02 INgO5 Rate 034 M 00014 molesL min 068 M 00028 molesL min Chapter 13 4 gt Observe that when N205 was doubled so was the rate Rate kN205 termed first order In general for a reaction aAdD catalyst XX Rate kAmD Catp rate law Rate constant We will not consider For reversible andor stepwise reactions X can also be involved can get fractional or negative exponents m n p overall order mth orderin A etc 2 d Example 2 NOg C2g 2 NOCKQ Wig NO Cl2 Rate w Msec 025 M 025 M 14106 x2 X4 050 J 025sz 57106sz 025 050 2910396 050 050 11410396 gt Rate kNO2C2 third order Might be interpreted as indicating that the reaction takes place by way of a collision of2 NO and 1 Gig in a single step Chapter 13 5 inal Example 0 CH3C OCH3 OH gt CH3C O39 CH30H R moles CH30OCH3 OH Rate L sec 005 M 005 lVlj 000034X 2 005 010 000069 gt 2 gtx 010 010 000137 9 gt Rate k CH3C OCH3 OH39 likely single step k Reflects dependence of reaction rate on collisions Q Q CH3COCH3 CH3COCH3 OH39 OH39 1 collision 4 collisions ltBoth concentrations gt doubled rate quadrupled Now that we have a general idea of how rate laws can be derived crudely from comparisons of rates at different concentrations more sophisticated and detailed information can be obtained graphically Chapter 13 6 INTEGRATED RATE LAWS GRAPHICAL TREATMENT E DATA Examgle 1St Order Reactions Rate law rate kR R for reactant gt E kdt dt R Integrate gt nRt t nRo kt nRt kt t nRo k has time391 units nRoquotquot MR 0 t gt Straight ine makes it easierto derive k gt First order reactions including radioactive decay characterized by a halflife 2nd Examgle What is the halflife for A ifk 54010392hr391 cyclopropane propene both CsHe Chapter 13 7 Note After one halflife R is decreased by 50 Le Ro gt 12 Re nRt nRo Eng kt Re2 R01 39 39kt 2n 12 kt 2 2n O693 kt12 0693 0693 128 hr t for99 k 540 10392 hr391 t completion Now let s look at 3 Order Rates rate kR2 k has units like LmoIesec kdt R12 gt i 17 kt and i kt L Slopeintercept 1 intercept R0 t 0 1 R gtAgain slope of line 1 provides k readily time Chapter 13 8 Example Forthe reaction CO N02 gt C02 NO k 190 Lmolehr at 540 K And the reaction is first order in both reactants Suppose we start with N02 CO 100 M How much time must pass before N02 CO 0500 M How much longer until N02 CO 0250 M What are the reaction rates at each time Since the N02 and CO concentrations are equal and will remain so during their 11 reaction we can simplify the rate expression as follows rate kCONOz kCO2 We can use the appropriate integrated rate formula 1 1 7 7 kt 2nd order reaction th Ro Plugging in CO values for the first interval i 190t 0500M 1M mole hr 210 100 39 100 39 190th mole mole mole hr t Lhr 0526 hr 190 Chapter 13 9 Now for the second interval 1 1 7 7 Rh R10 1 1 L L L 250 500mole 239Oomole 13990moIe hrt 200 t 190hr 1052hr Note There is no constant halflife Corresponding reaction rates Initial CO 100 M mole hr rate kCO2 190 100 M2 190mOIe Lhr When CO 0500 M 2 L 2 mole rate kCO 190m0lehr0500 M 0475Lhlr When CO 025M L 2 rate kCO 190 molehr 2 mole 0250 M 0119hr Chapter13 10 2nd Example The dimerization of butadiene C4H5 to vinylcyclohexene has been followed by pressure measurements at 326 C reactants and products are gasses Looking at the apparent halflives of this reaction decide whetherthe reaction is more likely first or second order Then try to verify the conclusion via an appropriate plot and also determine the rate constant time min Pbutadiene mm Hg 1P mm391 0 316 000316 680 1589 000630 1767 893 001120 373 411 002433 See it took 68 min for first halflife Second halflife would be reached when P 79 mm However 1087 min after 1St halflife P 893 mm meaning the second half life is much longerthan the first this is notfirstamper Let s see if we get a linear 1P vs t plot expected for 2nd order 00120 176397 001120 mm391 00060 gtgoodfit 00030 I 000316 0 100 200 time A1P 1 slope k 2 002433 000316 mm At 373 0 min Chapter13 11 56710395 mmquot1 min391 better value via leastsquares Summam of Graphical Approaches Egan kdt As we have just seen graphical plots provide a much more effective method for determining k s Let s summarize the results for 1St and 2nd order reactions 0 order relatively unimportant First Order Second Order Rh 1 1 1 39kt kt Rio th Rio nRt nRo kt 1 kt th Rio nRt kt nRo 39 i vs t slo e k plot nR vs t slope k quot 39039 R 39 p intercept nRo intercept 4 Rio Since we have seen a 2nd order example above let s try a reaction expected to be 1St order C5H5N2CI39 C5H5C N2 Let s try to confirm that it is first order then determine k Time min mLN2 CeH5N2Cl39 nC6H5N2CI 0 0 00700 2659 116 97 00587 2836 355 263 00393 3237 481 337 00307 3483 Chapter13 12 00 600 00000 25 266 EHC5H5N2CI 3 481 3 I 348 39 I 0 100 200 300 400 500 time min nRt nRo slope k At 348 266 082 481 0 min 481 min gt k 000170 min391 approx What is tyz 0693 0693 tyz k 000170 min1 408 mm looks right Could also get t90 etc Chapter13 13 Temperature Dependence Q Chemical Reactions Observe N2 3 H2 2 NH3 Nature at 25 C 1 world energy Man Haber Process Nobel Prize 1918 17109 lbsyr in USA but 2 4501Q needed for us to get a reasonable reaction rate 80 what is the origin of this temperature dependence We can start by examining some implicit assumptions about chemical reactions excluding unimolecular 1 Molecules must collide with each other for reaction to occur 2 Molecules must collide with the proper orientation 3 Collisions must provide enough energy for reaction to occur to overcome an energy barrier Why First requirement is rather obvious Second requirement like burning gasoline even a favorable process may need help to get started to achieve the energy of activation Third requirement Consider Chapter13 14 NOO3 gtN02Oz Possible approaches more than one could work 0 O O O O O N O O O O N IO N O O O N O 0 good m worse worst However even for correct approaches there can be a substantial barrier why We have already seen that there can be attractive intermolecular interactions between molecules dipoledipole dispersion forces etc However these occur at rather long distances compared to bond lengths Example 34A 1 A 103 cm 34 A separation van der Waals distance ideal for C C attractive intermolecular interaction But for a chemical bond 154 A 80 in this case to get a normal bond to form one must push the nonbonded atoms 2 A closer than optimal for a nonbonded interaction In the process one could generate significant Chapter13 15 repulsions between the electron clouds of the two atoms ie an energy barrier is encountered We can quantify such phenomena by use of Energy Profiles eg Typical diagram Less common 132 kJmole 4 kJmole Q 03 NO N02 200 CO 226 kJmole kJmole NO co2 02 N02 Reverse reaction unfavorable Reverse reaction again unfavorable J Note Very small barrier for second reaction 9 one would have to be careful to carry out such a reaction Finally H H H3C H CC gt CC H30 CH3 H CH3 cis trans gt A reversible reaction barriers similar in either direction Barriers are large however reaction generally m slow 262 kJmole Cl kJmole 39 dpter 13 16 L1 Of course more complex behavior is possible eg Intermediate r species may be observable or even isolable Now we are in a position to understand the origin of the dependence of reaction rates on temperature low T quotBoltzmann distributionquot high T probability lt energy of activation kinetic energy Suppose this is the amount of energy required to pass over the barrier energy of activation clearly at higher T more molecules can overcome the barrier Arrhenius Svante developed an effective model for the relationship between reaction rate and temperature and proposed global Chapter13 17 warming gt Arrhenius Equation k Ae39EaRT R 831510393 kJmoleK l Preexponential term or frequency factor frequency of collisions having the correct geometric approach if all reactants are present in 1 M concentration not easily modeled Fraction of molecules having at least the minimum kinetic energy content Ea for reaction Compare T effect on e39Ea RT values EakJmole 60 120 240 T C 300 41011 11021 21042 400 1510 8 210 16 51oa32 Application of Arrhenius Equation k AeEaRT Bnk EnA R T T T y b mx gt plot Bnk vs 1T K not C l slope EaR intercept EnA can provide some insight into order etc of reaction Chapter13 18 same asfnkifT 00 Examgle CH3I NaOC2H5 CH3OC2H5 NaI T K 1T k Lmole sec nk 273 0003663 560105 979 279 0003584 11810395 904 285 0003509 245105 831 291 0003436 48810395 763 297 0003367 100105 691 303 0003300 208105 618 0003300 618 6 0000363 0003663 0003300 12nk 979 E 8 618 33961 1 G 39 0003663 979 00033 00034 00035 00036 00037 1T Least squares line y 264 9890 K391x 96 0999 i i znA EaR E8 9890R 9890831510393kJmoIeK lt9945fr0mgt Chapter13 19 tWO POIH ES 822 kJmole Alternatively if data for only 2 temperatures are available E E EkBA a39 kEA a quot2 quot RT2 n1 quot RT1 Subtracting Ea1 1 k k n2 n1 RT2 T1 k2 Ea 1 1 BH k1 R T2 T1 ly With above numbers Now let s ask if there could be a way of increasing reaction rates otherthan by increasing temperature or D In some cases y Through catalysis uncatalyzed Reduction of barrier with a catalyst increases rate of forward and backward reactions Relative favorability of reactants and products unchanged Chapter13 20 Extremely important industrially Heterogeneous N2 3 H2 gt 2 NH3 17109 kgyrin USA Haber Process Fe203 NH3 L NO gtHN03 8109 kgyr in USA Pt FischerTropsch CO H2 gt hydrocarbons CO H20 gt 002 H2 WGSWater GasShift Homogeneou Monsanto i CH30H co gt CH3C OH 21109kgyrin USA 0 I c 1 Via Rh now Ir 0 1 o ZieglerNatta catalysis olefin alkene polymerization Chapter13 21 I I uumI 02H4 Ti 151 09 kgyr 107 l O O O I quotmu I 7 Reaction Mechanisms We have seen how some simple reactions like H2 12 gt 2 HI seem to behave as expected kinetically so that in this case rate kH212 suggesting a onestep process In this case the overall reaction might appear to be a single elementam step In some other reactions things are not so simple For example H2 Brz 2 HBr M No absolute HBr rate 39 1 k39HBrBr2 or Br2 order Explainable via Brz ZBr elementary Br H2 HBr H steps HBr2 gt HBrBr Sometimes simple appearing rate laws can be deceptive Chapter13 22 however Important to realize that while the overall reaction may not be directly reflected by the rate equation the rates of the individual elementary steps are directly and predictably related to the species contained in each Example Elementary Step Rate Reaction Order 1 A gt B kA unimolecular 2 28 gt C k B2 bimolecular 3 A B c D k ABC termolecularuncommon Overall 4 A gt D 31 2 3 Could give inverse dependence since this could prevent step 3 Now let s take a look at a real example Mno439 8 H 5 Fe2 Mn2 5 Fe3 4 H20 gt Very rapid yet obviously could not involve a simultaneous collision of 14 ions In this and many other cases a complex series of elementary steps is actually involved In real life the determination of mechanisms can be difficult and controversial Much easier to disprove than to prove a mechanism In ideal cases one might be able to assign a mechanism rather unambiguously Not always going to be so easy Chapter13 23 Example 0 2NO Br2 gt 2NOBr N Br rate kNO2Br2 Like Cl analogue p 135 Does this demonstrate a termolecular process so that the overall reaction and the single elementary step are equivalent No The actual rates are too high to be consistent with a termolecular process So the rate equation here is deceptive Can be explained by NO Br2 gt NOBr2 2 NO gt NO2 NOBr2 NO gt 2 NOBr NO2 Br2 gt 2 NOBr ALther example 2NO2 F2 gt 2NO2F rate kNO2F2 gt Seems to implicate a slow ratedetermining step in which NO2 and F2 form some sort of intermediate eg elementary NO2 F2 gt NO2F F slow some bond breakage steps NO2 F gt NO2F fast no bond breakage 2 N02 F2 gt 2 NO2F overall reaction Now let s look at NO2 CO gt NO CO2 Chapter13 24
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