TEST 4 - {i} What is the s_ign offlS” for the reaetion...

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Unformatted text preview: {i} What is the s_ign offlS” for the reaetion below: D; (g) + 2 F; {g} —+ 2 DE {3} < a} negative 3, h]positire {2} Arrange the following four substances in order of increasing S“ {the one with the lowest standard entropy will he left—most. ete. eto.) l; {g} at 293 HE} at 293 K [1 {s} at 298 K 12 {g} at 498 K a] [2 {g} at 298 K s I2 {g} at 493 K *1 [(g} at 293 K. s“. I2 {5) at 293 K h} lg {s} at 293 K «i 1 {g} at 293 K a“ l3 {3} at 493 K *1 I; {g} at 298 K e} iglfs}at298l( «: Itgimsssn I3 {1;} at toss a lgiglatr-IEFSK to Irgimzssn < igtglatwttli a. IgisjntEUSK c lg[g}at493I-C {3] The Third Law of Themiodyiiamies says that a} a} the entropies of all pure, perfeetly erystnlline substanees are identical at zero K b} the only ways that the energy of the system may change is by enehange of either heat. q. or world or. with the surroundings _ e} for a spontaneous process. {£3535 -:- 51351;”) t? t} d] ifsystem A is in thermal equilibrium with system H. anti system B is in thermal equilibrium 1with system (I. then systems .«"'|. and t.‘ are in thermal equilibrium with eaeh other e} in a eheinioal reaeti on. the sum of the kinetic and potential energies on the produet side must equal the sum ofthe kinetie and potential energies on the- reaetant side [4} USE tabulated thenncrdynamic data to Bfitimuta the normal bttjling paint 0f PCI3. a}-5"C h}4"C h‘nf " . . -3- MM 51?? 'I‘ 1M we. in a“: “AMI-1*“ _._-- “Pr-W. $C‘i 3 f 3 n D * a __ . fifi 7‘ Tbg J a}? gprL1\r-\gr:9hm ‘1 "‘ G 3’ {- N’" z 3 *4 ‘1 K Tr.” F ' ['sul—ztflfiiwq “‘3- w 173 ____________ “2'4 “*1 {5) Use tabulated themflclynamit; data to find the Km ut'stmntium sulfate. SI'SCh, at 293 K. [R=3.314x10' th‘fmoI-KJ] a}fi.?8x m "' h}—1334 c} 3.32:;143'233 d} 118:: IU'“ u} 3.23x1tt" 11th 5 '1‘ —-3 D be??? (_55—j_3 _7q1.‘i‘1 + t$3H\ 3215‘ + ELL—J". Sta? r mama he a- At; ,- $13 E ‘7 {6} For the reaction 211(3) + 59 '32 {E} ‘3' 21309}. at 293 K, find the £15“ with its correct units. 11) — ill? HK - c) +lflfl.2 JIK d} +231? UK e} +ltltl.2 .l-K {T} Combining the results of question # 5 above with whatever other inf have acCeSs to, choose the correct statement below: The formation of zinc oxide at 293 K from its constituent elements in their __ states is 3] uonspontaneous at all temperatures 51-"? b) spontaneous at all temperatures c) spontaneous above some Kelvin temperature, T ', and nonspontaneous beiowT " d] nonspontaneous above some Kelvin temperature T ‘, and spontaneous below T ' e) no information available allows an'j,r conclusion to be made about the spontaneity of this reaction at an}r temperature AR” "'5- "' ‘3' JET-ole. fibula II la. {5, hi. (IL 9- ‘hlebeni‘ut' is f. 1-1.“ '1’: Sean Ask hen-‘1 *1”?53 hues-pun air “1%- 3”“??? The following m questi " -=-.2 to the chemical reaction Fifth {a} r I Stanfield states, the formation of n t-i -.;;':_' 3 0: {g} {3} With dioxygen and s ntaneous stall: " ) non-spontaneous; :I - _ _I G Spfl “I” H 3. W5:- '* .v a: non-spontaneous at low temperatures d} spontaneous at lowlanperatmes, non-spontaneous at high teinperamres q _ e} neither spontaneous nor aomspontaneous, as the reaction above is at - :-'" at all temperatures AH“ -; {1)IIHB): 4: 2% hi 5%“ =- (11 (ma-arm - Lemma-m ._ I37. 3 {a '3"; £4..- .5 pin: Era“: 1°50 5 I 1L: [9} Calculate M3 at 298 K for this reaction if Pm = flflflll l atm and P01 = 1.5m l _- _ and indicate whether under the conditions the formation of ozone is or is not ' spontaneous. [R= 8.314 x [U '3 k] Hmo] - K}] _ .“L H”; em. :10 a) + ..I . not spontaneous l! c) —326 k], Spontaneous d) —239 kJ. spontaneous an" e} +314 k]. not spontaneous i a a‘afllnl‘a 7. + {2.1- L 61 - L’Q AG be megs +(amaiE-33L1‘l‘5" 3“ 05°) _. 2%.[0 .- (1Q%3(— D" Fi- .. splat) k3 .— [ it'll] A chemical reaction is occurring and all we know about it is that no" is: I]. Ween infer that a} the reaction equilibrium constant K is a negative number; the reaction qllntiun Q is a positive number b} the reaction equilibrium constant K is a large positive number; the reaction quotient Q is a small positive number _ e} the reaction equilibrium constant K is a large positive number; the reaction quotient Q is a large positive number Had) the reaction equilibrium constant K is a large positive number e} the reaction equilibrium constant K is a small positive number (less than I} Hue "ira‘b flo‘é’u' ‘0 I”? 11;. at: it» “- D II'/ '55ng “1'5 Ir J AG ‘: ‘ KT 3“! 1 “at a; lure}?— O (in Am 2 L a AC1" 40' E, 1: K Consider the tblltnving balanced retlox equation to answer the following, tw_o questions (questions # ti and a ll}: *3 +ll. 2 £3th {so} 4- 2 Hgfl {l} + fiCIO'laq} —- H: 1 o Eiffel—)4" {an} .- 3 Cl; {g} 't 4 lLl'l-i {no} [11} 1Which species is being midi-act]? _ ht Ha} e} mo {12} Which species is the reducing agent's1 -"'_—'_"'--._ 11} ago c} [.‘ltJ' t] 3} The redex resetidn helew eeeurs in basic squeeus selutien. 4 "'1 a + {a 30331an + (312th 3042' £an + —i C1 ' {sq} When the equation is halaneed, there will he a} 2 H20 en the Iet‘t-hsnd side, find 2 0H ' en the right—hand side h} 3 H10 en the left-hand side, and 2 DH " en the right-hand side e] I H20 011 the left—hand side, and 2 DH _ en the right-hand side d 2 DH ' en the left-hand side, and | H30 en the right-hand— side e} 2 H + en the left-hand side, and 1 H30 en the right—hand side a as l, f 601:. —*:> so“ + “#10 J" 33:; + sefli F____.. Cr'lll —7}' 1 C3 + ge‘ ”' H ’L' '- ' t ' Cull a. 80.5 -# 10.9. —} i“ 60% art 1 qt. r4 WI '. '. . .1; . _ t .) 3J1!“ { 1| - 11th 1s iht. strongest Jet LEng agent. Hg {1] m fan is] . we :1}Hg{]} 4* - n- - e35" 214.3033 + My rm + as E * Ea: Jr {fa-Hr" + _ gigs“. e (kitten +1e {15] For the: folluwing 113'»de martian {Jamming in basic sfllutifln, calculata E” w“ : MnOz {s} + Ag + [an —1 Mnflg "(11121) + Agfi} [BASICI a} +1311: 2 03014 In} 11.3w _ mam.“ __r_rfl______ c} +11.211.r + 0‘1“; d] 11.10233 a} 41.2111 {16} When :1 clean 111111 nail is placed 111 an 311111211115 snlutiun ufmpper {II} sulfate: the nail beenmes cnnted with a hrnwnifih black material. The material mating the iron nail is awe" 111111“ c} 33 1113042' Hm frfio {HEEL} Fm” l‘t’i: 364. $1. r£§u~L¢5 *TRL C1... - +.: waa‘) {H} |3}-‘1I1:fi11i!ir.1n. 1111c vull '15 llJlL'gI cjll-{J'II'J at] 1J.-K HIKE-{J {18} A. voltaic cell based on between aqueous Br: and vanadium _ has E“ “if = 1.39 v: Erfiaq} + 2V3+{aqj + salon) —+ 2vo“(aq} + “11an + 1What is E” “mm, the standard electrode half—cell potential for the reduction c51- to v 3* s - a] +2.4filtl'r E cud :— b)+o.32v ‘ 31 '- to? ._= 0,31U c] -2.4ov d -o.64v . ~ J Ea Gar 0*;Aulo‘h "b” .. (3.31“ [19] Which statement is FALSE about galvanic cells? a} the anode is electrically negative relative to the cathode h} the cathode is the electrode at which reduction occurs c] as the cell operates, cations from the salt bridge flow into the reduction half-celi- d] if the cathode is one ofthe products of the cell reaction {that is, an active ester-W they are devices in which non—spontanetms redox reactions are driven forward by an external power supply 6] The next twn guestinns {questions # 2t} and # 21) apply tn the t'nllnwing ait .- -' A vnltaie cell with Mimi 3* and Como 1* half-cells has the fellnwing initial mmmtrations: [Ni 2+] =n.an M; [ [301*] = nan M. [20] Theoretically, what is the reading an the 1irnltn'ieter‘? [Note that Same of n are close in value, an make your calculation with care. R = 3.314 J i“ [met- nnmv e! _ £1— e]fi.53"v' Egan = Ecflt “fl: 3* 0‘ dflfllv ‘1” '— ) Cate—3'3" Ca “WW HE e} [1.4 V Nzfifl‘th‘k‘ + it- '9 EH“: guarantee) .7, 5.05“ {21} What will be the concentrations at" the inns“? . .I (yin a}[Ni2']=t}.09M,[Cn2‘]=[IHIM I'" 53ft” h}[Ni3+1=t}.5UM,[Cn1']= {15th t' i , c] [Ni"]=[l.tl2 M.[Cn"}= naam a} [Ni“]=n.9aM,[CD“]= [1.02M e} [Ni“]=n1aM,[Co“]= 0.31 M The next tw_n tuStiflnS (questiflns # 22 and # 23} pertain m the ihllnwing titmtion: A chemist titrates 25H ml afa [1.155 M aqueous: salution ofthc weak acid NHJUH ' wifll H.103 M NaGH. Ka fm'NI-hCIH + is 1.52 :1 IE! "'5. {22) What is the pH nfthc NHgflH 7(an baforfi 1116 titmtinn hagins‘? a} H.033 b} 2.412 1:} 3.315 a} 1.445 a) 5m N${w'a- if: rJEEJT‘ 5... VJLLE. hLiA ‘arpk‘c‘lgm H + A 0H + 2.0 NH30|4+ + H20 w.— NH’L (“,1 m} “In {2} 1L >‘~ cf) é : Irv/fl fiat-1.51%?" a155,. [23} What i3 the. pH at [11¢ muivalcncc Quint? :1} 4.59”) b} 5.313 c} :13” d} {1353 e) I I '44” '— J- .3 £5— WA 0:5: ‘1‘». am“ far “Mi Ner Manx L3“; L0 0150L3(fl JS'E‘N W: .5 pogfii meh F .12 omul‘fi‘e F‘fl-JM 01-1 M 50 ale-BM ' FL NfiaH m’uel j. T- 5" 955$! u + NH DH ‘2» at 3w whaukefix“m NHsOH h +» ’1- 3., JPN- w‘kruk in #urn AJSW-wtb fl EH? - 1 1:1“; 1515:15— T. “FL—"I 5%‘4Efi “hi j'filE'E W” {a .# GIG?“ [(1-54 35"“175'3 L" {’9' R f R Fetthe hypothetical aeitaieetzfi‘ Reta} | Rb+[aq} || Znhfaq} l' [21) Write the ehemieai eqeaetae fetthe hatatteeti eeii teaetiee, being aate te i .5 phase labels and charges {New} for all species. .— + fiktefi "‘9‘ E + R‘b NJ“ Ehaituw‘ + REL- .9 Ehrbh) (“tel-l “WU- ” 1'": 1 + 9‘5"!“ {AAA + + (b) Using information attaiiabie ta you fi'flm APPENDIX 3, find E“ eeit at 293: sure to include correct unite. Shim your Work. ' AG "* eve (et(—1”-=~‘J '(*'”""‘“ A} 1. wtlfla‘ 5” . q ._ e a .— H31 S E 5:“ I: 1 Problem 2 Find the current {mulmnhs I 3%} nan-dad to deposit 3.1341} g [If gold metal [Au (5)) by electrolyzing an aqueous solution anuC]; fin-2:111} min. 1% a“ 1*“ “J 5'" K - . Au. 3w 6) Fw- * fl; #1 W‘Q .2; 10 GM?" E09 liq—IEALK ...
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This note was uploaded on 05/19/2008 for the course CHEM 1212 taught by Professor Dockery during the Spring '08 term at Kennesaw.

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TEST 4 - {i} What is the s_ign offlS” for the reaetion...

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