ch17 - l 17.100 (a) The SN for each carbon atom in cash...

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Unformatted text preview: l 17.100 (a) The SN for each carbon atom in cash molecule is 4 indicating all C-C-C bond angles should be 1093. (b) chgmg)+%02(g) -+ 3C02®+3H20(g), Ann—e -1957.7k1= [3Dc_c +61)“. #213030]- [6Dc=° +6DO_H] -1957: u = [(3 leDc—c )+ (6 W1)[414%]+(%mol)(498£]] mo] 1:} H - 11101)[799;;a-)+ (6 (3 ml)(Dc.c) -= 889 U D“: = 296 mm: ForCJIs®+602® —+ 4002(g)+4H:O(s). AH... = 4567.6 k1 =' [mmC + sac.“ + 6130.0 ] — [scho + 3130.3] _ .11 15.. 4567.6 k] — [(4 uni)(Dc_c ) + (s mol)[4l4 m J + (6 mol)(498 ml - [(3 mol)[799%)+(8 mol)[463—§’I]] (4mI)(Dc.c)= 12281:] Dcc= 307lemol For (mug) + 13-02(3) —> 5C02(g) + SH20(g). AH“. = 6097.6 k] = [5%.c + 1013c.H +‘ 12105.0 ] - [1013M + 1090.3] 6097.6 1:] = [(5 mol)(D¢_c ) + (10 mol)[414;%) + (11.51 m1)(493_‘9_)] l - [(10 mol)(799%)+(10 [1100(463731-J] m (5 molXDcc) = 1647 Id Dec = 329 iii/mo! For Cal-112(3) + 902(3) —> 6C02(g) + 6H20(g). AH“. = -3685.5kJ= [6Dc_c +1211»;fl +9DM1—[12Dw +12D0_.H] 3685.5 k3== [(6 molXDc_c)+(12 mol)[414%)+(9 mol)(498%)] - [(12 mol)[799£] + (12 mol)(463—E-)] mo] ml (6 molXDcc) = 2008 1:] Doc = 335 kJ/mol (c) Asthe C-C-Cbondanglesapproachthepredictedangleof109.5°thereislessringstrainandbetteroverlapof thesp’ orbitalsusedforbondingbytbecalbonam. Thisladstostronggtbondingandthcaverage carbon- carbonbondencrgyinmeasesmtdingly. IE. Answers To Chapter 17: Thermodynamics Dr SJ Bi-ois systemtocontractundertheinflueneeeofpressmeexenedbytheamundin ° ' " ' _ gs. Themgnofw lspontxve,81nce thesurroundmgs transfers energytothe systemas it compresses it 17.30 AemdingmmthstLawofThemodymmieenE=q+w Inthiscase ~ 27 . , —- . 1 W = -PAV = -(l.02 alm)(3.5 L) = -3.6 L-atm. q k] and ConvertingL-auntoklgivesw=-3.6L-atmxlOl-3L J x 1? =—036k1 -atm 10 J andAE=-2.7lk1+(—0.36kJ)=-3.07 kJ. J 11:] X L-atm 103 J SinceAE=0foranisothermal compression, andAE=q+w, {fin-7.29 kJ. ConvertingL-atm to 1:] gives w = 72.0 L-atmx 101.3 = 7.29 In! (b) Following the plan outlined in the solution for (a) gives: (1) V; at the end ofcompression against 3.00 atm = 3.00 L, w; = -PAV = -(3.00 atm)(3.00 L «9.00 L) = 18.0 Latin or 1.82 [d and q; = 4.82 It]. (2) vgat the end ofoompression from (4.50 L, 3.00 aim) to (2 L, 9.00 atm) = 1.00 L, w; = -(9.00 alm)(1.00 L - 3.00 L) = 18.0 L-alm or 1.82 k! and q; = 4.82 16. Hence,w==w3 +w;-= 1.82 1d+ 1.82 kJ=3.64 Id and q = q, + q, = 4.82 In] +£4.82} kJ = -3.64 1d. _ (c) Following the plan outlined in the solution for (3) gives: _ 1) V; at the end ofoompression against 2.00 atm = 4.50 L, w; = -PAV = -(2.00 ann)(4.50 L - 9.00 L) = 9.00 L-atm or 0.912 k] and q1 = -0.912 is]. 2) Véat the end ofcornpression from (0.50 L, 2.00 31111) to (2 L, 3.00 em) = 3-00 L, w2_= -(3.00 atm)(3.00 L - 4.50 L) = 4.50 L-atm or 0.456 k] and q; = 0456 k]. 3) V§atthe end ofcompression from (3.00 L, 3.00 atm) to ('21., 9.00 31111) = 1.00 L, w; = -(9.00 atm)(1.00 I. - 3.00 L) = 18.0 L-atm 011.82 1:3 and q; = -l.82 lrJ. Hence, w = w, + w; + w; = 0.912 k1+ 0.456 1:] + 1.82 k] = 3.19 1:1, and q = q; + q; + q, = 0912 k} + (-0.456) k} + -l.82 k} = «3.19 It}. Theinitalandfinalsmtesformegasarethesamemabmdc. However, oompxessionocmusinaone-step processina, atwo-steppmcessinb,andinathree-stepprocessinc: r a (9.00 L, 1.00 atm) —> (1.00 L, 9.00 atm) b. (9.00 L, 1.00 aim) -~> (3.00 L. 3.00 am) —> (1.00 L, 9 atm) c. (9.00 L, 1.00 atm) —-> (4.50 L, 2.00 am) —) (3.00 L, 3.00 atm) a (1.00 L, 9.00 aim) Sineethe inimlandfinalstatesarefltesanmmaflthreepmcessesandaflmvolveismhermal compressions, AE= 0 for all three. However, w. = +7.29 k1 q. ~4- -7.29 1:] w.=+3.64kJ q..=-3.54k1 w. =+3.19 H q. = .119 k] These calculations illustrate that heat and work (q and w) are path dependent and therefore not state functions whereas AB is independent of the pathway and therefore internal energy (E) is a state function. (d) The calculations for a. b, and 0 indicate that the amount of work done on the system decreases as the number of increments of compression increases. Hence, the minimum amount of work would be done on the system when the gas contracts against infinestimally-small increases in pressure. l 7.34 q... = CA: q... = 5.694.051 x more = 34.23 k] c qo. = «q... q.,. = -34.23 in —34.20 H 60.10 3 03117031 " 1.022 gQH70H x lmol (231-1qu l 17.38 C;I-l-,0H(£)+4§-02(g) -+ 300,(g)+4H,0(2)im1vesadocreaseofg-mo1gas, suAnr-wzimol. AH=AE+PAV=AB+AnRT AH= -2.016 :4 103m» (—%m1)[s.314 >010"3 I 17.46 (a) For 0(5) + 1120(3) —-)- CO(g) + Big), there is an increase of one mole of gas and a decrease of one mole of solid and therefore an accompanying increase in disorder. The sign of the entropy change should be positive, and the magnitude of the entropy change should be rather large. AS”... = [5" 00(3) + 5" His)! - [Sq C(s) + 5" H20(g)I . ' 3°... = [(1 me] x 197.56 J-mol”‘-K") + (1 mo] x 130.57 J-moi"-K“)] - [(1 mol x 5.74 J-mofi-K') + (1 mo! x 188.72 J-mol"-K“)] = 133.67 J-K“ =-2.016 x 103 Id 461 it] moi-K )(298 K) = 4.020 x 10’ kJ in disorder. The sign of the entropy change should be negative, and the magnitude of the entropy change should be rather small. AS“..... = [25° 1465(3)] - [28° N0(g) + 8" 02(3)] = [(2 mo] 3; 239.95 I-mol"-K")] .- [(2 mol x 210.65 J-mol"-K") + (1 mo! x 205.03 3-mo1'1-K“‘)] = -146.43 J-K" (c) For NaCl(s) —-> Na+(aq) + Cl‘(aq), there is an increase in disorder and therefore an accompanying increaSe in ginopy. The sign of the entropy change should lie-positive, and the maydtnde of the entropy change should small. - . AS“... = [5° Na+(aq) + s° Cl'(aq)] - [5° NaCl(s)] AS”... = [(1 mol 20159.0 smote-Ki) + (1 me] x 56.5 I-mol"-K")] - [(1 me] x 72.13 J-mol"-K")] = 43.4 J-K‘ ' les of gas and an increase of six d ForC5H (g)+ 802(g) —> 5C02(g)+61~120(£), therersadecreaseoffcnrmo iniles of liqliizd andjherefore an accompanying decrease in disorder. The sign of the entropy change should be negative, and the magnitude of the entropy change should be large. Asa. =[ss°c:02 (g) + 68°H20(£)]—[S°05H12(2)— 88°02 (3)] 158°”... = [(511101 x 213.63 J incl“1 ' K"1)+(6mol x 69.91 J . may! .K—1 - [(lmol x 343.40 1 - moi-l -K“ ) + (Sniol x 205.03 I - moi" -K"‘ = -501.03 .LK‘ l 17.54 (a) For 2320(3) + 2012(3) '4 4110(3) + 02%). an?“ =[4AH‘f'Hc1(g)+ 1111302 (g)] - [2011?H20(3) +ZAH¥C12 (3)] 1111;;u ~—— [(411101 x —92.31kJ-m61'1)+(1m61x 0.1141161" - [(21110] x —241.32 111-16014 ) +(2m61 x 0. 1:1 quot" = 114.40 to As?“ a [43°Hc1(g) + 3°02 (9] - [25°H20(g) + 2:1"(212 (g)] 713:“ = {(411101}! 136.30 1-1110]?! -1(")+(1mo1 x 205.03 J-Inolfl ~11“ - [(21110] x 133.72 J- mol'1 .11“! )+(1mot x 222.96 J-mot" -K" = 123.77 1-K“ At 25°C, A6393 = 114.40 1d - (293 10023.37 11:10“3 kJ-K“) = 76.00 1:] ThevalueofAG‘z’98 ispositive,andthereactionisnonspontaneousintheforwarddirecfion. ’I'hisiseonsistent with its unfavorable (MP = +) enthalpy change, but not consistent with its favorable (AS' = +) entropy change (b) For 2C02(g) + 4H20(£ ) *9 2CHaOHU ) + 305(3). 11fo,, =[2111301130132+3AH1'02(1)]—[2AH‘3002(9+4AH?H20(£)] 1111:,m = [(21110] x 433.6611 - mot" ) + (31110] x 0.1:} - mol'l - [(21110] x —393.51 111-11161“ ) +(4m61 x 435.33 11.11161" = 1453.02 111 Ang = [23°c113011u) +33°02 (3)] - [28°002t3) + 4932009)] . Asgm =[(2m61x 126.8 J-mol" -K“‘)+(3mot x 205.03 J-mol" -K“ 633m = [(2m61x 126.3 J-mol'l 31" ) +(3m01 x 205.03 1- mot“ 11“) - [(21661 x 213.63 1 - mot" ~11”I )+(4m61 x 69.91 J - mot" -K"l = 161.79 3-K“ At 25°C. 3039, = 311°- (293 1:) 33° 6639, = 1453.02 1:1 - (2911 10061.79 1: 10“3 kJ-K“) = 1404.31 1:: Thevalue ofAGiQ,8 is positive, andthe reaction is nonspontmeons intheforward direction. This inconsistent with its unfavorable-(AF = +) enthalpy change, but inconsistent with its favorable (AS‘ = +) entropy change. I 17.62 (a) For 290cm) —» 2PCI3(g)+02(g), 3112,... = [2111179101301 + A1170; (g)]— [2AH:P0013 (2)] 1111:“ =[(2m61x -237.0 11.nw1'1)+(1mo1x o. 111 - lawn-[(23661 x —592.7 11 -mol‘1 = 611.4 kJ Ass... = [28°PCIa (11+ 90:01)] — [2870613 (1)] AS?“ = [(23301 x 311.67 1 - 111014 41" ]+(1mo1 x 205.03 1 - m1" at“ -[(2m61 x 324.6 J - mol'l 41" = 179.2 J-K" At 25°C 2163,, = AH°—(298 K)AS°. 66393 = 611.4 11 ~ (293100792 11 10'3 11-11") = 553.0 k] 1:936 = 61”” =3412K 113° 179.2x10’3 k] K" The reaction is nonspontaneous in the direction written at 293 MAG;98 = +) and experiences no not driving force in either direction at 3412 K0163“2 = 0). Above 3412 K, the reaction bwomes spontaneous I (b) For Pb0(s) + coztg) -+ PbCOsts), A113mm = [Angmo3 (5)] — [Amman + Alrg'co2 (3)] A Airgun = [(lmol x —699.1 kJ~ mot")]—[(1mot x —218.99 1d - mot")+(1m61 x 393.51 kJ- mol")] = -86.6 As};m = [(S°PbC03(s))] u [S°Pb0(s) + s°coz (3)] ' As‘;m1 = [(11661 x 130.96 1:1 mot" - [(16161 x 66.5 [d o mor")+(1mot x 213.63 11. mar1 = -1492 MC" At 25°C A639, =AH°—(29s K)As°. 11039,3 = 416.6 k} - (29s K)(-l49.2 x 10" lei-K“ = -131.1 it] At the temperature at which there is no not driving force in either direction, AG° = 0. .At this temperature, 0 = AH°- TAS°. Hence. . AH' ' 46.6 11.1 T.—.__.___A._._ As” ’ 149.2 31610"3 kLK" » Thereaetion is spontaneous in the direction written at 298 IMAGE;a = -) and experiences no net driving force in either direction at 580 HAG?“ = 0). Above 580 K the reaction becomes nonspontaneous. Hence, it is spontaneous up to 580 K. I 17.74 ' (a) For.2N20(g) + 301;) 2 2142043), 7 Q = =580K (Pup. )2 (0.10)2 (warm? = (1.0.15 3 mo“: 10‘3) AG = AG° + RT 1n Q AG = -6.36 k: + {8.314 x 10‘3 kJ-mol“-K")(298)(ln 2.5 x 10’) = 18.73 11.1 = 2.5 x 10* {b} ThesignofAGispositive,sothereactionisnotsponthneousintheforwarddirection. Rafitcritisepontaneous mthereversedirection. I 17.76 (a) For 2an3) 2 Zn’Yaq} + 2F”(aq), Q = [Zn‘fim’ = (3.5 x 10")(2.3 x 10")2 = 1.9 x 10‘1 AG = AG° + RT 111 12‘ AG = 8.68 kl + (3.314 1: 10° J-moi"-K“)(298 K) in (1.9 x 10") = ~29.“ 11.1. (b) ‘I'hesignothGisncgative.sothereacfionissgontancmrsintheforwnrddirecfion. LEE At 10.0K, AGfo = AH°-(10.0 K)As° A630 = 10.0 x 103 J —{10.0 K)(-100 14(4): 1.10 x 10" I or 11.0 Id "‘3' _AGo _ -(1.10x10‘ J-mol“) K =55?— and — = —1.32 x102. "' RT [8.314 J-niol'1 -K" i000 K) Therefore, Keq = e'mm' = 4.11 x 10'“. At max, Ach = AH°—(100 K)As° AGIoo = 10.0 x 103 I— (100 K)(—100 1.x"): 2.00 x 10" Jor 20.0 k! 3931 _AGa ~—(2.00x10"J-mol'l) _ K =9 RT and = "24.1. 'l'herefiore, K =62“ =3.57x10'“. °" RT is.314 J-morl -K'l i(100 K) “1 At 1,000 K, AGE”, = AH° —(1,000 K)As° Ammo = 10.0 16103 I -(1,000 K)(-100 J-K") = 1.10 x 105 J 01"110 lrJ *M' 5 -l _ _ o —1.10x10 J-mol Ken :9 RT and A5 = ( ) = -13.2. Therefore, Kml = 8—132 = 1.85 x 10‘ RT 3.314 1.111614 -K'l (1,000 K) Note: 'I'hereactionisendothetmic, miner-casingme 10.0Kt0 100Kt01,000Kincreasesthe value «K... from 4.71 1110'“ to 3.51 x 10'" to 1.35 1: 10°. ‘ w... M .‘I H7023 For 1020(3) + H2(g) 2-2 Nng) + 1120“ ), 'fiHin. = [AH‘EN 2 (3) + AH¥H20(£)] - [AH?N20(g) + AHgHZ (9] AH?“ = [(1 mol x 0. k1 '-m°1'1)+(1 “101 x 48533 H‘ml-l “[(1 11301 x 32.05 k1-mol")+(1molx 0. kJ-mor‘ = 467.33 1d 35;, = [swam-1 s°H20(0]—[S°N20(g) +S°Hz (1)] ’ Asgm = [(1111on 191.50 1-mo1" ~34 )+(1molx 69.91 1-11.61" -K"1 -[(1me 219.74 J-mol'1,-K"‘)+(1molx 130.57 I - mar? -K'1 = -33.9 1-K" For 25°C, A6398 = AH - (293 1033 = 367.33 x 10’ 1 - (293 K)(-33.9 1-K“) = 3.414 x 105 1 or 4141.4 3.1 —-— 367.33 x 103 1 - (310 K)(-33.9 1-K") = .3403 x 10’ 1 or -340.3 H For N20 (1 atm) + 112 (0.4 atm) 3-2 N; (1 mm) + H200 ) at 310 K, P AG 3 56310 '"' RT 1“ Q AG = 963m + (3.314 1-mol"-K"x310 K) "‘ [PH ] - 2 AG =4 -340.3 kJ+ (3.314 x 10*3 k1.mol-I.K-I)(310 K) In (an; 4)] .__, 43.7.94 U I 17.90 (a) For CH30H(£ 1-3 CH30H(g), K“! = moms) and AHA. = [0111011301101]—_[AH:CH30H(0] 311;.“ =[(1 111013: —200.66 1:1 - moi" -[(1 11101 x -233.66 k1 - mol'l = 33.00 1:1 33;“ = [3“CH300H(g)]—[s°CH3011(1)] 33:1,.n = [(1 111on 239.70 Lama“ -K")}-[(1 molx126.8 I-mol'l -K‘1 = 112.9 1-K" At53°c, 36331 = AH" —(331 10.33“ 3653, = 33.00 x 103 1 -(331 K)(112.9 1-K") = 630 1 -53“ --AG°_ ' 6301.166!" que RT and __ ' =' 41.229. RT 13.314 1- marl -K'l i031 K) Thfireforc, ch = Benson‘s) 8—0229 = 0. atm. (b) For (11301120110 ) —> cngcflzong), 1:...I = Fannie“, and 2111‘;m = [AH§CIi30H20H(g)] — [AH}CH3CH20H(:)] 2111;“ = [(1 1110! x —235.10 16-11101" )]-[(1 mo! x —277.69 111-11101" = 42.59 1:1 113;“, = [S'CH3CH20H(g)]—[S°CH30H20H(£)] 218;”, = [(1 mo} x 232.59 J-moli'l -K")]-[(1 11101 x 160.7 J - moi'l -K'1= 121.89 1-1:" At 29°C, 11630, = 1111” «(302 K)AS° A6502 = 42.59 x 103 J —(302 K)(121.89 1-K"): 5.78 11:103 I —AG° _ -5.78 1:103 I K =3 RT and — ='—2 30 «1 RT 3.314 J-mol" oK" (302 K) Therefore, K“! = chscgjoflm) = 8.230 = 0.100 Still . (c) Fong(£ )—>I-Ig(g), K“! = Pm) and MIL... = [AHEHgts)]— [AHEHsm] 21mm = [(1 mo! x 51.32 111-11101" -—[(1 11101 x 0. k1 - mol'1 2 61.32 1:: = [smug] -[s°H(t)] AS as [(1 mo! x 174.35 I - 11101" .K")]— [(1 11101 x 76.021 - 111.91“1 -K" = 98.83 J¥K'l E“ 3' At 45°C, 116;", = AH" —(313 K)AS° A6513 = 61.32 x 103 J “(313 K)(98.83 1-1:") = 2.939 x 10‘ J .50. o ‘ —AG _ -2.989x10 I KmI =9 RT and — = —11.31. _ RT is.314 J-mol"-K'—1i(318 K) Therefore, If;m1 = PH“) = 9—11-31 = 1.2 x 106 aim . 17.98 ____ _ 0 3 u -l 1 AG “'1‘” 1 mm =3.39x102. 111613111112,192“I mm” K :6 RT 311‘! = “I RT 13.3141-11191" -K" §(298K)- . 3 'l'hevalne:omelisgmatcrthmtl1095'anclcaluuntlaacalculatedusingK“l “33"” onmostcalculators. .543- However, K“. = 101m“ derived fi’om AG" = -2.303 RT log K.q canbe used to obtain the value of Keq: K“! = 10”"-23 =1o°33 x 1o“w = 1.7 x 10147. ’ The equilibrium constant tells us Al3+ - 02' interactions are stonger than Fe.3+ — 02' interactions. This Is to be expected, since A]3+ is smaller than Fe“. ...
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This note was uploaded on 11/19/2010 for the course R 102 taught by Professor Morrison during the Spring '10 term at Rutgers.

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ch17 - l 17.100 (a) The SN for each carbon atom in cash...

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