Hwk 3 Solution EnvE 321 Sp09

Hwk 3 Solution EnvE 321 Sp09 - SOLUTIONS FOR CHAPTER 2 2 ....

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Unformatted text preview: SOLUTIONS FOR CHAPTER 2 2 . l Combustion of propane: a. C3H.8 + 5 02 ~—> 3 C02 + 4 H20 b. 5 moles of 02 needed per mole of propane. lmol CRH8 SmolOz 32gO2 C. 100 g Cngx x X =363.6gO., " 44 g C3Hh. mol C3H8 molO2 ' l 3 (1. 363.6 go. x m x 22.414x10'3 in- : 0.255 m30. " 32 g mol ‘ ~ m3 air 0.255 m" o, x —-———.——— = 1.213 m3 air - 0.21 m' 02 3 mo] C02 100 g C311,. X22.414x10‘3 m3 x =0.153 m3 co, mol ‘CRHS 44 g/mol C3H8 mol ' 2.2 4 C7H§;N3O6 + 2102 —> 28 C02 +10 H20 + 6 N2 mol wt of TNT = 7x12 + 5x.l + 3x14 + 6x16 2 227 g/mol 100 gTNT 2lmolO2 32gO2 _ x x 227 g/mol 4 mol TNT mol 0 2 74.0 g O2 2.11 a. methanol(l): CH30H +3/2 02 ——> C02 + 2H20 (-238.6) (0) (—393.5) 2(—285.8) (1) gross gross: AHO = 2x(—285.8) + (-393.5) - (-238.6) = -726.5 kJ/mol HHV =726.5-l(-J—- x mOl x IOOOg x 6.7—12— x ' Btu mol 32 g 2.2 lb gal 1.055 U = 65,537 Btu/ gal b. ethanol (1): C2H50H +302 -—) 2C02 + 3 H20 (—277.6) (0) 2(-393.5) 3(—285.8) (1) gross gross: AHO : 3x(—285.8) + 2(-393.5) - (-277.6) 2 4366.8 kJ/mol HHV = l366.8£~ x m—Ol x 1000 g x 6.6—l—b— x Btu mol 46 g 2.2 lb gal 1.055 U = 84,492 Btu/ gal cl propane: C3H8 + 5 02 —9 3 C02 + 4 H20 (-103.8) (0) 3(—393.5) 4(-285.8) (1) gross gross: AHO = 4x(—i285.8) + 3(—393.5) - (—103.8) 2 -2219.9 kJ/mol kJ mol 1000 g lb Btu x x 4.1—— x mol 44g x 2.21b gal 1.0551d HHV = 2219.9 =89,123 Btu/gal 2.12 21. methanol (1): CH30H +3/2 02 ~—> C02 + 2H20 (—238.6) (0) (-393.5) 2(—24l.8) (1) net net: AHO : 2x(—24l.8) + (393.5) — (—238.6) :2 -638.5 kJ/mol LHV =638.5-——kJ x ml x 10mg 673)- Btu mol 32g 2.21b x ' gal x 1.055kJ = 57,598 Btu/gal b. ethanol (1): C2H50H + 3 02 ——> 2 C02 + 3 H20 (~277.6) (0) 2(-393.5) 3(-241.8) (1) net net: AHO : 3x(-241.8) + 2(—393.5) - (-277.6) = -1234.8 kJ/mol k l 1000 0 lb Bt LHV =1234.8——J— x “‘0 x ° x 6.6— x “ mol 46 g 2.2 lb gal 1.055 U = 76,332 Btu/ gal Pg 2.4 + 5 02 ——> 3 C02 (0) 3(—393.5) net: AHO : 4x(—241.8) + 3(»—393.5) - (-103.8) 2 —2043.9 kJ/mol + 4 H20 4(—241.8) (1) net 6. propane: C3H8 (— I 03.8) t LHV’ =2043.9-—k—J—- x ml x 10mg x4.l-IE x B_“ =82,057Btu/gal mo] 44 g 2.2 lb gal 1.055 k] 2.13 03 + hv —> 02 + 0* (142.9) (0) (438) AHC‘ : 438- 1429:2951 kJ/mol (endothermic) ~ .4 r _ A St 1.19x10Aij m/mol (2.12) . ‘4 , _ s. Wit =0.403x10’6m 20.403pm 295.1kJ/mol 2.14 N02 —> NO + O (33.9) (90.4) (247.5) (enthalpies from Table 2.2) AH" 2 904+ 247.5— 33.9: 304 kJ/mol (endothermic) .4 r _ A f; 1.19x10 1:] m/mol (212) AH —4 ; 1‘19—"19mm=0.39x10*m=0.39pm 304kJ/m01 2.15 HQ) —-> H+ + C1- 2559—51 x ~3— x -——“3°—1—— = 6.8587x 10'4mol/L = [W] L 1000 mg 36.45 g HCl pH 4 log [n+1 :— log[6.858’7x 10‘4]=3.16 2. 18 a. CH3COOH + 2 02 ———> 2 CO2 + 2 H2O Acetic acid mol wt :2 2x12 +4x1 + 2x16 2 6O g/mol £9.33; LXMX%X%XM:ZI3mg/L ThOD = 200 x 1000 mg 60 g molAA molO2 g b. C2H50H + 3 02 —> 2 C02 + 3 H2O ethanol mol wt :46 g/mol “OD =3OE§flEx__§_x_291_x3_flx§flxmz 62_6mg,L L 1000 mg 46 geth moleth molO2 g c. C6H1206 + 602 —> 6CO2 + 61-120 sucrose mol wt: 180 g/mol “OD =mwaxmxmxwxmfl3mgm L 1000 mg 180 g suc mol suc molO2 g P0 2.6 2.19 HOCI c: H+ + 00‘ K : 2.9 x 10-8 [H+]10CI'] 8 ——————~ = 2.9x10‘ [HOCI] f_ [HOCI] 1 1 1 "—‘7—_____"=_"_'__=—‘—'——€'=_-———3‘ [HOCI]+[OCI ] 1+100 | 1+2.91110 1 2.9112 [HOCI] I“ I 10 H--(3 ~--——1-———-—097 p “ "'1+2.9x10’*‘" ‘ 104‘ 1 pH"8 " 2.9x10'8 " 0256 1+ _8 10 H" 101~ 1 —- 00034 p " ' " 2.9x10‘“ " ‘ 1+ 10—10 2.20 1—125 <::> W + 115- K :0.86x10‘7 H'* 'HS" ].=0.86x10‘7 112s] f 115 1 1 1 = ‘ _ = _—‘=_"——-:7-=————_— EHZS]+[HS] H; 1+0.86x10 1+O.86x10 7 1+[H,s] H--6 ————-—71 —092 P " “' 0.86x10‘ _ 1+——— 10-0 H 8 1 I 0104 2: :——-—?—= p 0.86x10 1+——_ 10*8 2.21 AIPO4 <2 A13+ + PO43'; Ksp:10‘22= [A13+] [130431; [A13+J:[PO43‘J 1130431: 10-11 mol/L mo1 wt PO43" : 31 +4x16: 95 g/mol [3043‘ concentration : 10‘11 mol/L x 95 g/mol x 1000 mg/g : 9.5 x 10‘7 mg/L [Es—Hmsz cm, (7714 g A. (9.001(H jig/M 7L .40?“ 3 3.75 -—1C0¢MTL zwz‘z/ 5 0400/4 = A4 ‘wQ/MMC’ 1 4H 00’ o, 4* 5?€(;gz’j : H1! 044', MAMA: [EM] ’40 t [W] Eu '1 K3455 Etc/2mm; 47A : [PM + Q41] cam/$4 3%“: [H [/5ch [M Lil/{le/ (2612/ “'47 4 Sfdywfi @H [5 5*] : [M ,3.7{ 3 /0 m4 [Mé/i’IMM fiéflL/ M/ M/fé 4/0/er £4; ., fififi, filiasug-M to; N 710]) $ 50 WW; 50/074974 7L4a/fi/m‘a balaozum/ W .7 H+ + C'7L/7/3Q’ TL 2, 0L ——17 4%;- 21420 . Iv . ‘ '1 1 ’ O ‘ V 057% 1 flaw/Md m1! v 2— M 2’ #zycfir/aé. WWW ’* 4519 + 560+ zoo») = m WW6 a W/ Wit/WK Mun/(9 __:'/ZOELEM___[_L [602 (45)] M [9/7/24 W/ 75? WW Wag/W412 WWWMW " V 750 L / , 69;" - L (/M) : 75m; M4 /0 L ‘ [W42— [533] ' Q @2- KJ = fl§39/‘7Z 4 2/672 M/Mfia 0, I MAE 2 2741/04; Mdgé : Z94 1/0 2’ (OLCJZ) ‘5 (2.4% 1/0 Z W 44 m 5%, * L IMMI-‘é i a M LCOLQZ')‘ 2’7 I ...
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Hwk 3 Solution EnvE 321 Sp09 - SOLUTIONS FOR CHAPTER 2 2 ....

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