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122ch15c_001 - Z 21:59.— 3;5 ’4sz figéggas...

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Unformatted text preview: Z . 21:59.— 3 ;5 ’4sz figéggas 9¢flwgé§ giégafif ' g; ’1'; gg: 2 M3 ;; g 5an —(l”._'9£#a,_7 _ .411 angina-4&2 /C = . .40 {Em W [5: 57m ,7! a _ ,-" ---.-/ r m-~/= iflflL : II, ,1 a, : Plv-laBEl3 is a solid and its concentration is taken as a constant_. C. _ ________ hm'mw— PHaBCla ‘——r PHa + 30:3 '— —‘—““ initial - c o M 0.0256 M * __ change +x M I +x M m———- equil. C x M 0.0256+x M . m-— g \ _Me __ _H‘M _ r4457" 2303 a 25:02 + 4::2 initial 0.01033 0 0 change 2:: +2x _ +x 4 equil. 0.01038-2x . 2x it [equiL] 0.00234 M 0.00804 44 0.00402 M nt = 0.0144 = 0.01038—2x + 2x + x; x = 0.00402 = 0.0040 mol 02 WNW Since the 1volume Is 1 L, the equilibrium molar concentrations are equal to the moles of each component. "0 .=04746 0.047 l|||llllll|||. [$03]2 . _ (0 00234)2 __ Kp‘= K.(RT)"‘n = 4 746 x 10-2 _.(0 08206 x 1100)1 = 4 3 lllll _4)flMi_ce—m~¥m ' i -- ___ l M4254; __ :4; + I4. '_ -.._ ____Lu*_J_ ___..Iflgégaémi'L; . “(£34423 + a___'2: +2: 4/ 344 ‘ ' " ' _” _ _ ____I 饧f {- 7C "MIPII __ “___-.. 7:- —— 4. 34 2/4444. ' ' €29. _, 4 244244.";344—4244244ng 2M— . 7i?" Mfg—Wm £49234 4292749444441? 44974 .# :. 3 At the temperature of the exhaust pipe. the Ni(CO)4 product is a gas and is carried _ into the atmosphere with other exhaust gases. Thus; equilibrium is never__ established (we do not have a closed system) and the reaction proceeds to the right ____ as Ni(CO)4 product is removed. ' u—w b I..." Avg .‘1 4 d ’ 1 I ’ 1'.’ ’41.]. «A ' , mi. 4-)}. 0-». :4, w. j I ' . . , i I / . __ . I 1. It; _' /'#/../4 .11"../_" I- .- 4’ J 5 {44 A J a” ‘4’ -"X 1 (4‘- ; [-1, 4 I ’1 I I 0614(9) ‘—-; 0(5) + 2CI2(g) [(5) {a 2" ,9, [its initial 200 atm 0 atm W. mama—Wear change 4 atm ' ' +2x atm M A” #55 2m ______._ W ; g. ,e _‘ equil. . (ZOO-x) atm 2x atm - P; (2)02 Kp = 0.76 = 2 = PCCI4 (2.00‘” '—w """" ‘ 1.52 - 0.76): = 4x2; 4):2 + 0.76x — 1.52 = 0 Usingthe quadratic formula' 3 = 4, b = 0.76. c = -1.52 _ 2 _ I x- 0.76 d: [(0 76) —4(4)(— —1 .52)_ 0.76 + 4.99 = 0.528? = 053 atm 2(4) 8 — Fraction CC!4 reacted = 4—— : —-—— = 0.264 (26%) ' a 5" PCI: = 2x = 200.5237) = 1.05 atm Pom. = 2.00 - x = 2.00 - 0.5287 = 1.47 atm 'First calculate Kc for the equilibrium —I-l.‘,+I2 1; 2H1 (0.155)? __ [H2][I2] (2.24 x10-2)(2.24 x 10-2) = 47.88 = 47.9 _ The added HI represents a concentration of EM = 0.0200 M. 5.00 L "— _ H2 + 12 : 2HI-- ‘ #1" _ initial 2.24 x 10-2 M - 2.24 x 10-2 M 0.155 + 0.0200 M rm .r/{Wz‘ ._. . - - 7% EH _ change +xM . +xM -2xM <1 ......_ equil. . (2.24 x 10'2 M+ x M 2.24 x102 M+ M (0.175.29M wifzf -_ 0175 — 1’ '5”! . 2.. ( ' 2x) _'=_ 47.88. Take the square root of both sides: #35, _ [0. WI)" (2.24 x10-2+x)2 J 5?" (TH/0.29 _ 0.175 " 2X I 1.12 ' : é/ . —_ = 47.88 = 6.920 = 6.92 . __ 2.24x10-2+x _( ) . ape/(J (41.5%: 0.175 - 2x_= 0.155 + 6.92)(; x = 2.242 X 10'3 = 2.24 X 10'3_ ‘— [12] = [H2] {224 X 10‘2 + 2.24 X 10—3 = 2.464 X 10'2 = 2.46 X 10'2 M __ [Hl]= 0.175-2x= 0.1705= 0.171 M 1919?- f L 1' 2L 1‘ I! a’! 7L 1 La ’2 '- '6/411./ LII/WM” LL LL. -.- / I i '4 4.1 A 4' l" ’1.- A, 4’ JJ-Ltt/A L _ 1 I 14-14;, 14 LILL- ,z- Vé’ l " , 2' 1’ I I I " 6 Using thermodynamic data from Appendix C calculate AH for the reaction _ in part (a). _...——.—-—.—_._._._____ AH’ = AH; Agiaq) + AH; cr(aq) - AH; AgCl(s) ' AH' =_ +1os.eo kJ — 167.2 kJ — (— 127.0 kJ) ,= 4-65.? kJ The reaction is W]; (h_e_a_t is a Ieactgt), so the solubility of AgC_l(s) in H200) will increase with increasing temperature. F- 4W.W, 1.x ' r 1'2"“ ./ {(4.41 .v " - I /""-4‘4 A ’154‘1444-44 .11 ...
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