AK7 - 46. (x << 0149, thus the approximation is...

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Unformatted text preview: 46. (x << 0149, thus the approximation is valid) ‘ x = 9.1x 10-6 M 2 [1130*] pH =-—1og(9.1x10-6) 2: 5.04 (M) (a) (b) (c) (d) This part simply involves calculating the pH of a 0.275 MNH3 solution. Equation: NH3 (aq) + H200) : NH4+ (aq) + OH‘ (aq) Initial: 0.275M — 0M z 0M Changes: me — +xM +xM Equil: (0.275—x)M — xM xM NH “‘ OH’ 2 2 Kb:LiJ=l.8XIO'5= x ———z x x=2.2X10_3M=[0H [NH3] - 0.2754—x 0.275 ()6 << 0.275, thus the approximation is valid) pOH : _Iog(2.2 X104) = 2.66 pH = 11.34 This is the volume of titrant needed to reach the equivalence point. The relevant titration reaction is NH 3,(aq) + HI(aq) ——> NH 41(aq) 0.275 mmol NH3 Immol I-H lmL HI soln VHI=20.00mLNH3(aq)x ——X——-—X‘—-—‘ ‘ lInL NH3 soln Immol NH3 0.325m1nolHI VHl =16.9mL HI soln The pOH at the half-equivalence point of the titration of a weak base with a sh ‘ acid is equal to the pr of the weak base. pOH = pr = 4.74; pH =14.00—4.74 = 9.26 NH 4+ is formed during the titration, and its hydrolysis determines the pH ofthe'h solution. Total volume of solution = 20.00mL +16.9 mL : 36.9mL 0.2751nm01NH3 X Immol NH4+ = 5.50 mmolw lmL NH3 soln lmmol NH3 mmol NH4+ = 20.00 mL NH3 (aq)x _ 5.50mmol NH: [NH4+l" 36.9mL soln = O'HQM Equation: NH4+ (aq) + H200) ‘ZLA NH3(aq) + H3O+ (aq) Initial: '0.149M — 0M zOM Changes: — xM — +xM +xM Equil: (0.149 — x) M — x M xM J‘.‘ 21.0x1044- : [NH3 ][H30+] _ x2 x2 1.8 ><10‘S [mg] 0.149 _ x i" 0.149 i _.-.n. V. . I. n I 1 '41 1 _ '1 ..__l,.._1__4 n of a weak acid with a strong base, the middle range of the titration;- 48. (D) In the titratio Ka (= 4.74 for acetic acid), is known as the buffer with the pH within one unit of 13 region. The acid and anion concentration . calculations to determine the amount of base t (a) _ [021130;] pH : pKa +log = 3.85 = 4.74+10g [HCZHSOZ] 2 3 2] EH95} I r [C1H3OZF] :10” = 0.13; .=3£5—4J4 [chngoz] [HCZHJOZ] 10g 0.100 1 HC H O nHC H O = 25.00mLx mmo 33—3—1 : 2.50mmol 1 3 2 lmL a01d on, we can use their The amount of acetate acetic acid are in the same soluti place of their concentrations. is the amount created by the addition of strong base, one for each millimole of strong based added. The amount of acetic acid is r (1 110.11302 +OH' a 0.1130; +1120 Since acetate ion and amounts in millimoles in the same number of millimoles. in 0.200 mmolOl-l‘ l rnmol C2H302’ '1 ' 36 mL base x x _ _ l ranase 1 mmol OH 0.20017 ‘ 0.13 : ffl = 0.200 mmolOH‘] 2.50 . 0.200x .: 1 ‘ 2.50 mmol HC2H3,O2 — [x mL base x mease 0.33 = 0.200;: + 0.026}: = 0;. "If 0.200x : 0.l3(2.50«—0.200x) = 0.33 — 0.026x x:—(—)—'§§—=1.5 mLofbase 0.226 {b} This is the same set-up as part (a), except for a differe C H 0 ' C H O ' [ 2 3 2] 1 [ 2 3 2] 5.25fl4.74=0.51 nt ratio of concentrations. H: 5.25:4.74 +1 M p 0g [11911302] Og[HC2HBQZ] [czHgoz’] : 100.1: 3.2 32 2 0.200x _ [HC1H302] 2.50— 0.200x 3.: 0.200x = 3.2(2.50 — 0.200 x) = 8.0 — 0.64): 8.0 : 0.200x + 0.64): = 0.84x 8.0 . . . . . x = m = 9.5 mL base. Thls lsclose to the equivalence pomt, which is reached by adding 12.5 ml. of base. ' (c) This is after the equivalence point, where the pH is determined by the excess added base. pOH = 14.00 a pH =14.00w—11.10 = 2.90 [011‘] : 10‘!”OH =10”0 = 0.0013M 0.200rmnoion; [OH] = 00013 M : lmease _: 01200x xmL+(12.50mL+ 25.00 mL) 37.50+x 0.049 0.200x=0.0013(37.50+x)=0.049+0.0013x x= 0200—00013 12.5 mL to equivalence point +0.25 mL excess = 12.8 ml. 7... xmLx = 0.25mL excess Total base added = (M) In this problem we are given I and gas pentane at 298.15 K and asked to estimate the normal boiling pomt of pentane, AG?” and furthermore comment on the significance of the sign of AGMP. The general strategy in solving this problem is to first determine AH“;p from the l 5 known enthalpies of formation. Trouton’s rule can then be used to determine the . i standard enthalpies of the formation (AH?) of liquid ; l l l l normal boiling point of pentane. Lastly, AG‘j’ap‘zggx can be calculated using AGZaP : M331} -- TASSap ' Stepwise approach: Calculate MS” from the known values of AH)“r (part a): l l 5 CsHiza) : C5H12(g) . AH; 473.5 kJmol'l 446.9 kJmol'l ' M301, = 446.9 — (—173.5)kJm61" = "6.6kJmol'l Determine normal boiling point using Trouton’s rule (part a): 0 o __ WP _ -l -1 mp —— — 87Jmol K T nbp AHSW, d 26.6Umor‘ r = _ 4————— "5P 4539p 87kIK"mol“ 1000 = 306K Tm = 32.9”C Use AGCap = AH:ap — T A83” to calculate AGZap,298K (part b): AGvap = AHvap —- T AS“; 8'l'kJmol'lK'l AG“ = 26.6kJmo1'1— 298.-15K >< V3p,298K AGjapm = 0.661(Jmol‘1 Comment on the value of A63”,298K (part c): The positive value of 136;? indicates that normal boiling (having a vapor pre of 1.00 atm) for pentane should be non-spontaneous (will not occur) at 298. vapor pressure of pentane at 298 K should be less than 100 atm. ' " ‘ Conversion pathway approach: C5H12(1) :— C5H12(g) AH; 473.5 kJmor‘ 446.9 kJmor' my”; 2 —146.9 - (—173.5)kJmol'1 : 26.6kJmol‘1 AH” 26.6kJmol‘l AH“ AS“ = “a” = 87Jmol"K'1 :> r := W = —-—— = 306K w Tap "hp A334 M 1000 4 4 - AG;P = AHCap - TASS... = 26.6kJmol" — 298.15K x W— : 0.661<Jm91'1 1000 17. (E) (3) AH0 < 0 and A50 < 0 (since Angas < 0) for this reaction. Thus, this reaction?-is 2 of Table 19~l It is spontaneous at low - - - tern eratur _ .- temperatures. p 65 and “011 Spontaneous at h1 I) W 3 ' - o . _ _ _ 5:". ( ) flare unable to predict the Slgn of AS for thls reaction, Slnce Angas = 0_ Thais?- strong prediction as to the temperature behavior of this reaction can be made sifi i .1] _ AH” > 0 we can however 1 ' ' E , , , conc ude that the reaction W111 be non-s g temperatures. pontaneou (c) AH > O and AS” > 0 (since Angas > 0) for this reaction. This is case 3 of Tablet] 26.. (M) (a) AS” = 5" [3:2 (1)] + 25° [HNO2 (aq)] —-2 8° (aqfl — 2 8“ [Br‘ (aq)] 0 = 152.2 J/K +2(135.6 J/K)—2(0 J/K)—2(82.4 J/K)—2(240.1 AG :AH —TAS° m—61.6><103J—(1298K)(—221.6 J/K)=+4.4><103 ' products are in their stand ' ard states, because the value of AG0 IS greater than‘z'er - -:‘ ' r "-r h d - ---' ' ' a ‘, (M) We combine the reactions in the same way as for any Hess's law calculations. (a) COS(g)+2C02(g)-—> so: (g)+ 3c0 (g) AG” = —(~246,4 1d): +2486 kJ 2C0(g)+2H20(g)—-> zco2 (g)+ 2H2 (g) AG“ = 2(——28.6 Id): —57.2 k] COS(g)+ 21820 (g)—) so2 (g)+c0 (g)+ 2in2 (g) AG" = +246.6 _ 57.2 = +1894 kJ This reaction is spontaneous in the reverse direction, because of the large positive value of AG” (b) COS(g)+ 2co2 (g) —> so2 (g)+ 3co(g) AG” = {—2464 1a): +2466 kJ 3C°<g>+3e0£g>+3WM HHS-81d cos(g)+3H20(g)-> (301(g)+ so2 (g)+ 3H2 (g) AG“ = +246.6 — 85.8 =+160.8 kJ direction, because of the large positive value This reaction is spontaneous in the reverse of AG”. (c) cos(g)+ir2 (g) —> CO(g)+HZS(g) AG° ; _(+1.4) k] C0(g)+H20(g) —5 C02 (g)+H2 (g) AG" = —28.6 k] = —28.6 k] #W COS(g)+H20(g)—> co2 (g)+HZS(g) AG° : —1.4 U — 28.6 k] = —30.0 k] for this reaction indicates that it is spontaneous in the The negative value of the AG” forward direction. _: Gibbs free energy for the reaction between carbon monoxide and hydrogen to yi _ methanol. The equilibrium concentrations of each reagent at 483K were provided proceed by first determining the equilibrium constant. Gibbs-free energy can be 1 calculated using AG“ = eRTan. Stepwise approach: First determine the equilibrium constant for the reaction at 483K: C0(g)+2H2(g) «=‘= CH30H(g) [CH3OH(g)] : 0.00392 2 14 5 [CO(g)][H2(g)l 0.0911 x 0.08222 ‘ Now use AG” 2 "RT an to calculate the change in Gibbs free energy at 483 K' AG0 = —RT an AG” = —8.3 14 x 483 x 1n(14.5)Jmor‘ = —1.1><104Jmol'1 AG” = —11k.lmol’1 Conversion pathway approach: _ [CH3OH(g)] _ 0.00892 _14 S — [CO(g)][H2(g)] _ 0.0911 x 0.08222 AG” = —RTan = —8.314><_ 483 ><lr1(l4.5).lmol‘1 = —1.1x104Jmor‘ AG” = —11k.lrnol'1 -. l| g; (M) In this problem we are asked to determine the equilibrium constant and the ch ‘ (M) Gibbs free energy for the reaction (A6” = AH” — T AS") can be calculated using AH; ‘ and S" values for CO(g), H2(g) and CH3OH(g) from Appendix D. AH” = AH}(CH30H(E)) - {AH}(C0(g)) + ZAHHHZ (25)] AH" = —200.7ker:.r1 -— (—1 10.51ch01'1 + OkJmol‘l) = ~90.2kJmol"l AS” = 5°(CH30H(E)) r [3°(C0(g)) + 25°(H2(g)] AS” = 239.8JK’1mol" — (197.7}K’lmor‘ + 2 x 1307314411101") = ~219.3JK'1mol'1 483K x (—219.3)1<JI<;"morl AG“ = —90.2kJmol" — = 15.7kJmor‘ 1000 , Equilibrium constant for the reaction can be calculated using AG” = —RT 111K ‘ —AG° 415.7 1 OOJ r1 an= :>an:= Z 0 4m" =—3.9:¢K:e‘3'9 =2.0x10‘? RT 8.314JK ma! x 483K The T\(alues are different because in this case, the calculated K is the thermodynamic equilibrium constant that represents the reactants and products in their standard states. In Exercise 35, the reactants and products were not in their standard states. = 246: [N205 (g)]— 246: [N204 (gm ~46}; [02 (g)] = 2(1 15.1 kJ/mol) — 2(97.89 kJ/mol)—-(0.00 kJ/moi) = 34.4 kJ/mol AG° 34.4x103 J/mol = 43.9 D_ :_________ ______.._—-w——-——"_ 0’) AG *‘RTh‘Kn My RT 8.3145Jm01*‘K*‘x298K _‘(M) (a) AG” (D) (a) We determine the values of AH” and [35° from the data in Appendix D, and then the value of AG” at 25“ C I 298 K. AH“ : AH; [C11,0H(g)] + AH; [H,o(g)] a AH; [co2 (g)] #3 AH; [H, (g)] =—200.7 kJ/mo] +(g241.8 kJ/rnol) —(s393.5 kJ/rnol)—3 (0.00 kJ/I'nol) = —49.0 kJ/rnol AS“ = 8° [CH,OH(g)] + 3° [H2O(g)]-S° [(30, (g)] "3 3° [H2 (g)] = (239.8 +188.8—213.7 m3xl30.7)J mor‘K“ = —177.2 J incl—1K“ AG° : AH" 4M5" : 419.0 kJ/mol—298 K(—0.1772 kJ mol‘1K_l)= +3.81 kJ/mol Because the value of AG" is positive, this reaction does not proceed in the forward direction at 25“ C. (b) Because the value of AH" is negative and that of AS“ is negative, the reaction is non— spontaneous at high temperatures, if reactants and products are in their standard states. The” reaction will proceed slightly in the forward direction, however, to produce an equilibrium mixture with small quantities of CH3OH(g) and H20 Also, because the forward reaction is exothermic, this reaction is favored by lowering the temperature. That is, the value of K increases with decreasing temperature. (c) 130;,“ : AH” — T AS" = 49.0 kJ/m014500.1<( —0.1772 'kJ mol"K") = 39.6 kJ/mol r I =39.6><103 J/mol=~RTanp 1.. _ ° .. 3 .‘ In K = =_ = KP = 6—9.53 :7.3X10*5 p RT ' 8.3145 J mOl K x500. K .f‘ (d) Reaction: CO. (g)+ 3142(8) : CH.OH(8) +H20(g) ii ‘Initial: 1.00 atm 1.00 atm '0 atm 0 atm Changes: ——x atm' —3x atm +x'atm +x atm Equil'. (1.00 —:c) atm (1.00—Bx) atm x atm x atm P CH OH P H 0 . i” Kp:7.3x105:w{ 3 } { 2 }——H~—————u-—x x zxz , l. P{CO,}P{H2}3 _ (1.00—x)(1.00—-3x)3 rji x = V7.3 x10‘5 = 8.5 ><10‘3 atm = P CH OH Our assumption, that 3x <<1.00 atm, is valid. 3 60' (M) (3) AH” ‘= M1? [C02 (gll‘LAH; [H2 (all—AH; [(3006)] "AH; [H20(g)] ‘7: == —393.5 kJ/rnol — 0.00 kJ/mol -— (—1 10.5 kJ/mol) —(—241.8 kJ/mol) 4'4 85“ : s" [c0,(g)]+S°[H,(g)]—S°[co(g)]—S°[H,0(g)] ' = 213.7 J mor‘ K" + 130.7 J mor‘ K" —197.7 J mor1 K“ — 188.8 J 11101—1le = —42.1 J mol’] K"l AG” : AH” — TAS” = —41.2kJ/rnol — 298.15K x (—42.1 x 10*)kJ/molK AG” = —41.2kJ/rnol+12.6kJ/mol= —28.6kJ/mol (b) AG° = AH“ w r AS” = —41.2 kJ/mol — (875 K)(—4_2.1><10'3 kJ morl K"') _, = —41.2 kJ/mol +36.8 kJ/rnol : —4.4 kJ/mol = —RT In Kp AG" —4.4x103J/rnol ‘1 111K =— =———-—-——-—-———-——~——=+0.60 K = +08“ = P RT 8.3145 J mol"K‘1 x 875K P e 1'8 -- m WW,JIMLQLJDJ AW; u ‘ 146'“, vii/"'91 " C'Ho‘g W/mni “r104 HAmok) : 161393 KT/mek A grin 1 24’: j/mm- (nmoiflmlmiqé 37an [a] 3/111ng ; «‘34.,w/mk u » c) . _ __ v _ LT I _ A qufadt :- leKfl ... TASHJI ? “I65 ‘ “LT/ka ,. Liqbg) (—54b3/m0‘_fi) floocfl) 57 \gfiji_lifllflgi_iiflfiihngn cafldflmtlky,{;v {yns [(acfiun asguwm) ‘1“? gmn!a'! side of bLuHrOAdt‘nYJi \s vb wag”, Fressom)l _________r——-——r AGO ’P'T Ink? ' 52 ’2 MAM " (36“ j/md-H'k‘ “K000: )(zan) (\n H) Ii? - \ 4H muff] (°.( \qxxxo cum 5) ‘ - 4- - - ‘ wen ‘3') “(\{e an efivrech‘bn {1:( {NEE .tcémue CG 03' _____________.__.—_._ fl____.___,_,_._’i/ YOU" answer m 0:.) __,_____.._...——-—-—-—-—-—‘____ ch The hjdréefmqlaMLEg-gfiflx ts usuallj‘ abnducfcc‘ M 223%: thfi {3 +he eg'je‘flflbvw‘m (.vfl‘a'kufi, \cPl Mdhcd ‘ 3 e A“ m ML.» _ _L._.. “1 K4 H, m \n him?) I ( 298?— 41321) ABBESFTImo‘ 1.. _ if. ) “‘K‘W ‘ \“(V‘fl' “061V (WETSMK $ka 4?“: AGqHL = naggifiymm - (4-?-aa)(’g4¢,I/mol‘k_)( uzywm) 7 8' M0 LS/mul “’6 '- " HTML 8% KTAmi (F8314 no": “JV/mo; k) (+13 M Ina-13 [137—], 1-2.0 e) ? r {1 f?- ’: :1oocfifm. when +h€ machovlfitygfijg‘1433wh’bn'um LEW.G'H¢I we. «Awe; H 7 H:.:,CCH=C:H1 v (,0 + H1 E: ‘HaCCHzCLHZ ((1:07 H L '00 :00 1m O C —K --5< w yt ‘r‘fl 6 100 vx ICC «x. 1‘00"!- x: k? ' (loo ActoTx—TGGTW 3 ([00 33 0‘20 allow’ ‘= 131:; O 4452. (wt :0 4 LM'M 04453 =1 f> 57:018-0 PW?“ PM 7 m :OOOIX) [23g mwmu (s v; jiilfljfliwmémbmmlamn __ 149' \dc‘LRTYm an: "‘5 \{F1- \CLCRT) 3 kc, ” KPUZTP O \1Oodcm‘3[(o C'Ezocofi'eg C; L I i 7010 LEI/mo _-.-—...._.._._._. .-_..—- / 3 ...
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This note was uploaded on 06/10/2011 for the course CHEM 2090 taught by Professor Zax,d during the Spring '07 term at Cornell University (Engineering School).

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AK7 - 46. (x << 0149, thus the approximation is...

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