HW3_solution - «c P3-7(a Ifa reaction rate doubles for an...

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Unformatted text preview: «c P3-7 (a) Ifa reaction rate doubles for an increase in 10°C, at T = Tl let k = k1 and at T = T2 = T1+10 let k 2 k2 : 2k.. Then with k = Ae‘m in general, k1 .. Ae-E/RTI and k2 _ Ae—E/RT; or Therefore: k2 111(79— )G (T‘ +10» (1112)(T1(T1 +10» E-R -R——-——-———-- (Tz—Tl) 10 10E T :- whi hc nbe roximatedb T EEO—‘5- c a app y R1112 P3-7 (b) E Equation 3-18 is k - Ae RT ~E/RT, From the data, at T. = 0°C, k1 - Ae'E/RT‘ , and at T2 = 100°C, k2 - Ae k E 1 I) . . . . Ri T DiViding gives —2 =- e T’ ' or ’ le‘l E__ in __ RT1T2 111(5) k1 (1 1) (Tl—T2) T2 T1 cal 1.99 [273K1373K] EJ moIK| In .050 ”960335 ~. 383 g 100K .001 mol MA . _E_ 7960—031— A-kleRT' —10'3min"exp _———l——"’—"L———— -2100min" (1.99 03K)(273K) m0 P3-7 (C) Individualized solution ____________________________________._____._———————————————————— J mmin WM - P3-14 C6H1206 + 2102 + bNHJ "’ C(C4AH7JN036012) + deo + eCO2 T . . o calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients , , C, , ' a I) (l and e 0 W a e . 3-16 proceeds to completion and calculating the ending mass of the cells. P3-14 (a) Apply mass balance ForC 6=4.4c + e ForO For N b = 0.86c For H Also for C, 6(2/3) = 4.4c which gives c = 0.909 Next we solve for e using the other carbon balance 6 = 4.4 (0.909) + e e = 2 We can solve for b using the nitrogen balance b = 0.86c = 0.86“ (0.909) b = 0.78 Next we use the hydrogen balance to solve for d 12 + 3b =7.3c+ 2d 12 + 3(0.78) = 7.3(0.909) + 2d d = 3.85 Finally we solve for a using the oxygen balance 6+2a=1.2c+d+2e 6 + 2a = 1.2(0.909) + 3.85 + 2(2) a = 1.47 P3-14 (b) Assume 1 mole of glucose (180 g) reacts: 6+2a=1.2c+d+2e 12+3b=7.3c+2d Yo“: mass of cells / mass of glucose = mass of cells / 180 g mass of cells = c*(molecular weight) = 0.909 mol* (91.34g/mol) mass of cells = 83.12 g Yd. = 83.12 g/180 g ng=0.46 Yam: mass of cells / mass of 02 If we assume 1 mole of glucose reacted, then 1.47 moles of 02 are needed and 83.12 g of cells are produced. mass of 02 = 1.47 mol * (32 g/mol) mass of 02 = 47.04 g Yd.2 = 83.12 g /47.04 g Yc/oz "—1.77 W «4.. P3-15 (a) Isothermal gas phase reaction. 3-17 7; éN, +§H2 —+ NH, (1 X) MamgszdwbwmohmkMMMm -m,“L X 1____. H2+%N2->%NH3 ( 3) A 1 B 2 C (2) For batch system, constant volume. +— —> — 3 3 3 m1c~z1y21cm14 Stoichiometric table: X )6 / -- fl __ 32 “(C/40)} ‘ 3 [(14)] W FEM-m y 2 0x13 —rN1-1.6[1—1:—] [1-24/2 P3-15 (b) 5 .. 3-1-1) _ -3 3 3 3 w P3-l6 (a) 2 1 Liquid phase reaction -) assume constant volume anyw6n05x(—§)-—§ lkwLwdmwmmkmmmm: C 1 .4 tm -' - k C C --C- CA0 - 0.5———(—£—E—3-l————=0.2 mol/dm3 A [ A B KC (0.082 atm'dm )(SOOK) Stoichiometry: mol.K CA - C,o(1—X), C,5 - CA0(1—X), CC - CAOX C C C A 0 (1— X) 0,2(1— X) 0 lmol / dm3 To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe. H II A - -—-—————- . ————--—-— - . 1 1 + 8X X ( ) (1—3— CACBKC - CC ' C2 (1 —X )2 K - C X C l—X 0.2 X A0 a 0 A0 2 CNHJ nCC-Engg—J-EX—jfl-BOJMOI/dm3 1 3 (HEX) 3 1-... X3— 2+ Xe+1-0 3 C ,1ch LY; -0.80 P3-15 (c) km 2 40 dm3/mol s To find the equilibrium concentrations, substitute the equilibrium conversion into the stiochiometric . relations. (I) For Flow system: 3 mol mol _er _ k”: [C11, 1% [C11, ]A C, - C,0 (1 —X)- 2 dm, (1 —0.so)— 0.4 dm3 3 1 mol mol 1_£ % A CB-CA0(1—X)-2dm3(1—0.80)-0.4dm3 =’=4(’(CAO)2 U; (1—? C -C X—2m"l*030-16’"°l (1—?) (1—?) A A0 dm3 . . dm3 P3-16 (b) Stoichiometry: 3-19 3-18 e - oncS - (1)(3—1)- 2 and 0C -0 NA NA0(1_X) (I'X) CA'_'_—_'CAo‘—— V Vo(1+eX) (1+2x) C..._A:C____§1V_AP_’¥__C 3X C V Vo(l+eX) "°(1+2x) Combine and solve for Xe. (1 —Xe) 3X _______ - C _____e__ ch’” (1+ 2X9) A" (1+ 2X9) KC (1 — X,)(1+ 2X9)2 - 27cfmxe3 2 —(4+ 27C’“’)Xe3 +3Xe +l-0 KC Xe - 0.58 Equilibrium concentrations: -i-—————————10“”" -030530—5- CA0 3 d R73 (400K)(0.082 ‘1’" at“) m molK (1—0.58) moI C - 0.305 - 0.059— " (1+ 2(0.58)) dm3 3(0.58)(0.305) mol C - - 0246—— C l1+2(0.58)i dm3 P3-16 (c) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction. Stoichiometry: N N l—X CA-fi-J—(I-lZ—J-CAOO—X) CC__ALC__M-3CAOX V V0 Combine and solve for Xe KCCAO (1' Xe )' (361101“)3 Xe =- 0.39 Equilibrium concentrations 3-20 moI CA - (0.305)(1_0.39)- 0195;3- CC - (0.305)(0.39)- 0.36fl0—5- dm P3-l6 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction): 3 —rA -k[CA —%— C Stoichiometry: 5-onc5-(l)(3—l)-2andHC-O ___ALL_NA0(1_X)_C (1_X) V Vo(l+eX) "°(1+2x) N 3NA0X 3 X CC a-——C—III V Vo(1+eX)- GATE—275 CA Combine and solve for Xe: 1+2Xe 1+2Xe Xe —0.58 KCCAO (1' Xe ) a 3CA0Xe J3 Equilibrium concentrations: 0.305 (1 —0.58) 0 m0] "' 1+2(0.58) ' ”059dm3 3(O.305)(O.58) mg] C -—-—-—-—-——-—n _ —— C 1+2(0.58) 0246dm3 ...
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