QUIZ - Practice Problems—Chapter 18 I. Which of the...

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Unformatted text preview: Practice Problems—Chapter 18 I. Which of the following species has the highest entropy [3"] at 25°C? a. onioom game} e. Mgflflgts] d. H2013) e. Hits] 2. Arrange the thIowing compounds in order of increasing standard molar entropy-r at 25°C: le-Igtg]. C1H4{g}, 2.115(5). and H100}. 2115(5) ‘~’ 1410111} ‘1' CiHsItl-I} ‘1' C2514th CeHdIg} *1 Ham” “:- CJHSIEI '1 NHCIIS} 3115(5) ‘- CiI'lstg} ’4 C1H4IE}{ Hifltl} CgHgtg} <1 CgI-I4tfg] *1 HECHZI} *i 2115(5) 2118(5} *1 HgOfl} if C2H4tg} “1 CiHstg} 3. Which of the following processes are aoeonipanied by an increase in entropy? tr;- I. zsoitg} + Gite} —) sons} II. Hgflfl} —:r H2015] .L t. III. Emil} —> Bug} LW-L' . 1v. Hgogflj —> HgD(1}+(Ir'EIUsIEI t, it. III. IV 1, H n. m, w III, IV I, iv in. int. ' 4. Determine as“ thr the reaction Sflgtg} + H303] —> HgSO-t 'I 503 H10 H3504 169.2 .Ii’K-niol I343.2 .I.-"K-1‘no1 4159.2 I-"K-ntol —29_4 IIK'moI 29.4 Jr'K'ntol Sutlx'K-moll 2561.2 69.9 156.9 night 138 -'J is.“ : ‘féfimtl PJIS‘ in} 31ml mung trot! HISMGEM mantel-k] «r I[imutt-gflzsiili’mitI 4' that! III-whit ism j/K # [feel TEE”: tel-ti JILL] tt‘: elki’tlffli of huitI' 5. Calculate £8“ at 25°C for 2PhC|{s} + C(s} —> 2Fh[s) + Cflfig] given the following absolute entropies: S” JFK-incl PbC+[s) 69.45 C(s} 5.? Pb{s} 454.39 T CCIfl} 213.6 1-. _ L g _._ ' - } w- L 1 g 5.3 — [final Mlle LE {milky [Imiliiilhlab 11“le (2.") +1983 JfK-mol T— .._ . +4331) JfK-mol " lllmti I was ——— + - L J c. +3516 memo: 919“) [ mm {X173 WW) :1. —293.3 JIK'niol J: ._ e +293.3L’Krmol at; L ME- ‘6- W l”- s. A negative sign for no indicates that, at constant T and P, a. the reaction is exothermic. £5 2 _, '. 31331.11“ngwa _t FEW h. the reaction is endothermic. ' . c. a the reaction is fast. {at Pl/hfilllii "{‘31.L‘fil"fll J (if; the reaction is spontaneous. e. o8 must he 3' CI. 2. Calculate so” for the reaction Ethgtg] + H20“) u) 2HNUJU} + NOtg}. soar llernolI H20“) —23?.2 a _n h _ I 1 .. W Hwogm 49.9 M; :- Zimbl Pliaé; r) — 2 (ma k—llMi is) NC{g} 86.”? N‘Uzig) 51.3 CE} 3.?ka1nol -“_- ’ c H LI.- + imam) lite-"l I — b. 192 lernol M LBW“ HMZlL 7 l'l mol l l L-Wl c. —414 kJi'rnol .. ' - - - c . L . . . a. 492 kL’rnol llama 19511“? W l) + [EH-til +|:.llill"3?5l~2 E 1W1 Win c —155 kJr‘mol G _F_ at r a .“1 Li 8. For the atmospheric reaction: Dfig} + Nfl{g} —> NCrgtg] + {Mg}. Calculate the no“ tier this reaction at 25°C. Ilyl'iHD = —l99 kL’niol, $3“ = —4.1 Jr'K-inol} a _ . I: a. 1929 klr’tnol 566 3 {EH -' l 55. b. 4.22 X lo3 kJi'inol _ _ W T _ e. 2.oax1o3k.wmoi M.“ ‘4 16151 M” 5_ _ [Eggpo‘i .1: —i .42 x m3 kJi‘rno] mi: N jam it a. ~193k1i‘nioi my“: “lawman-l flatten/ma 9. Sodium carbonate can be made by heating sodium bicarbonate: ZNaHCDgts} —> N33C03($] + COfig} + Hgfllg) Given that are = 123.9 klflnol and M3" = 33.] kli‘mol at 25°C, above what minimum temperature will the reaction become spontaneous under standard state conditions?I . L. _ - c r c EJ— a. can Mr I M _:M LT LT . izfifl Wat E. if]: so or - M:- :lztfl m 42.: fifl'fi' 1" M“ Mail 5:," an] K T l'l-‘ltl~L Cei/ 525K 2‘1? l: as‘ —- Deal Winn-t 1': salt a] are L It}. The element oxygen was prepared by Joseph Priestley in IT’M by heating mereuryfll] oxide: Hgots] —> Hgfl] + {lf2}01(g}, aH° = acs4 kJe’moL Estimate the temperature at which this reaction will become spontaneous under standard state conditions. I 1).:1 grfl’ll Hit: (33‘: '. fig” ; 37L S°[Hg} = Tom HK-mcl ‘( sarog = scan JFK-mo] w an t Song0}= T1129 JtK'mol Tl’lfl’l - T r ts“ a. 133K T : shall i"l/Ifttt-l a. oar. ——-—r _ F c. scar. _:e3.2.'%1_ tit—1: a. TTSK Lmd iL'tJUT Ce; 84UK i T: EL”) l I. For the reaction Hgtg} + 5(5) —> HgS[g}, d ° = —2f}.2 kamol and = 4-43.]. JEK'niol. Wl-iich of the following statements is true? PVT-ll if“ Firm 'i'TLL' ' a. The reaction is only spontaneous at low temperatures. , The reaction is spontaneous at all temperatures. F Br r.‘ {Lugs 15H = — 41 {fl-fl 1'58 r l- c. d6“ becomes less favorable as temperature increases. d. The reaction is spontaneous only at high temperatures. e. The reaction is at equilibrium at 25°C under standard conditions. 12. The normal freezing point of ammonia is 43°C. Predict the signs of an, as. and so for ammonia when it freezes at —EU°C and l atin: NHgfl) —ar NH3{s} AH g g git : a act-Lin Milghestjtit at dill-QC, a — — a as ' straits: sisal e34.- l'wm .t A 3' b. — + — I c- + - r M: i all ' T1335 usiia=1saiceluim El. + + fl {:3 _ _ _ M lach Tag: is small deal ital-rt- Dwrrt-i-mhlto 9544 45+ 5H [at as nit-Lilli is: b at tat-t- q . f, 1173 't'. 13. At lSUU°C the equilibrium constant for the reaction CD[g} + 2H1{g} 2! CH30H[g} has the value K1, = 1.4 x 1077. lCalculate £6” for this reaction at 150fl°C. a. ma kJe’mo] M} : - ETJbL-i L Egakiilfifll oh"- — [3.971s rims-k)[naaaltmhaxlfl a. —1osamnul _ _€;: assisting] My; +233 tar/H.131 l4. H303 decomposes according to the equation H1030} —> Hgflfl] + [IfflOfigL Calculate KP for this reaction at 25°C. {aHs = —93.2 klfmol, a8” = TUJ L’K-mol] t that? MUM -. . . 1 fl. 3 $63; m—al 33b 5H T-‘lg I W! # tiquMTUIJ /L1HIHJ c. 3.4m to” M“: * H91 Ulnar-J - X 20 - a - 3L - "1 _F "- 'u a? iii; an ' = ~ "Wm" Jim” —. Lana Inflow” at" ' .[ggHT/Wltl Waist] 15. Calculate the equilibrium constant for the decomposition ofwatcr, lefltl} : 2H1[g} + ozrg) at 25°C, given that aomngom} = 4312 iclfrnol. _ . - Ir a m MU: [mt-trains) + [hauling -{2mdHasl'r’flifipl assist“ M”; LHLH LT c. 2.5x 1b"42 - .- . ._'.r' a. 1mm“ ML- 1’23 :ffflw UWMfiT ~ltll.‘§ e. 4.T>< m5 P-T lasts Visa—mom) E: m was 16. The equilibrium constant tor the reaction AgBulIs} 1; Aghfiaq} + Br‘ {aq} is the solubility.r product constant. Ksp = 1? xfilfi H at EST. Calculate at] for the reaction when [Agl'] = Lt] x 10“ M and [Br'] = 1.12} x lfl'J M. is the reaction spontaneous or nonspontaneous at these concentrations? a; _ grim E 5 _ “Far-i 'j/IWI'k-)(ZLI& Him {:Ejiwa'fl) a. at} = 69.1 ka'mol, nonspontaneous {3 LE}; {:6}! [j 3' :17 m 1,: l b. so = —69.i kL’mol. spontaneous c. of}: 915 thnol, spontaneous bi: : ‘7 4r [2 q! Q : llal‘illa l EEG = 443.6 kL’ L ' ‘ _ ' - -' .I-li; J fig 2 £915 : 4.- [E (I. l} 1. 1L ) L a. = eras—st TIL-ad = (like lib-1t“) 1?. For the reaction 2C£graphitejr — Hfig} —:> CgHflg], a °= +3192 klhnol at 259C. If 19(ng = IUD. atm. and FTC:ng = {llfl atm, calculate so for this reaction. . r; _ I _ F’ i: ~. I -. . a kJIIranI {5 T.- i- u - L. ; LE— : _ b. +2253 kJe’mol T in 1 LED. fl . .— cur,- +I92J kamo] 5 ., : Err“: L {Em} J AD _ 1_ Ca +113 kJr‘mol L G Wu; l H ‘r l? -%lli)lfi.'15§)uLUt Lucio) e. —1s.s kJr‘mol m:— = "4512.1 l'tT/mcii 13. The reaction rates ofnian}r spontaneous reactions are actuallyr tier}:r slow. Which of the following is the best explanation for this olzrseriration'."r a. lip for the reaction is less than one. (13) The activation energy,r of the reaction is large. c. M3“ for the reaction is positive. d. Such reactions are endothermic. e. The entropy,r change is negative. 19. For the reaction CuS{s} + Hflg} 1; H38[g} ~1- Cu{s} nosfmus} = —53.6 kli’mol nG°j~(HgS) = 4:315 klfniol nH°g-[CuS) = —53.1 klfniol nH‘JrflHgS] = — 213.6 lei-“moi a. Calculate m3" and nil-1" at 298 K and I atrn pressure. _ _ I. . Mi ; [inii1Ht3]{~ sst “(will + “pinball-l - liln’lllllslli')+ llmilwgll_9g'l‘hl?‘] Ms; M m“ max—mt; Handyman-tag“) —ll_rl’ttti Hilts) + [mm {LUSH-Eel fir—L l itH‘J: '32s It]— ‘o. Will this reaction proceed spontaneously at 298 K and 1 atm pressure? ME by: 4.113; mm is Liti spilflia'fi'rtL-ti . c. Calculate the equilibrium constant for this reaction at 298 K. El . idmk: Egali- :'_ xlnglJF Ll er . 1 ttsiti T/mmflfifi'lil - an“! lit: siztiu" d. Calculate 5.3“ at 298 K and l atrn pressure. at“: Mil-T55.“ s'e s3“: estate“ __ 532.9 I-:T-2i'.'l.{}L1_ rebut) T h saga LT MC": are 3/14 2D. A reaction that is spontaneous as 1written a. is very rapid will proceed without outside intervention E: is also spontaneous in the reverse direction d. has an equilibrium position that lies far to the left e. is veryr slow 2t. Which one of the following is always positive when a Spontaneous process occurs? 3L fissystem ll flSsurroundings fisuniverse [Ir LaltHuniverse Er flHsurroundings 22. Of the following, the entropy,-r of gaseous is the largest at 25 “C and l atin. a. H: LED ‘52wa C. CQHE d. CH4 e. CgHa 23. The standard Gibbs free energyr of tomiation of is zero. l. rno{n ll. I‘M— “L H? Le leer-itier a. I orin b. [I only c. III onl‘j.r (ED Ilmfllfl e. LIL and III 24. If M3“ For a reaction is greater than zero= then a. K = [l b. K=1 at}: -- Eli-latit— C. K 3’ l 5 l I h «i i V CE“; gel ass-:0 =>_itattmill—wpt-a l, so ld e l 25. For the reaction P: (g) + 3C1; [g] —» 2PCI3 {g} no” = -fi42.9 kJ.-’n'iol at 298 K. The 1value of at} at 298 K for a reaction mixture that consists of 1.5 atm P2? [.6 atm Cl:= and {165 atrn PC13 is lemol_ 5. I. -. — at r Mr + Platte-1Q ,i'. a. 44.2 3 P )L b. "sass 1o 1 7 Po; ' alas“ c. -?.23><1fl3 [1 E I P? " U: a : fillet? a. area *1 in U}: )LLt l- @ '649'5 M t“ MT '35") -; rlcll- 'l '1'" . . 's. not? at H J i {contrasted ...
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QUIZ - Practice Problems—Chapter 18 I. Which of the...

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