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Assignment 4 Solutions

# Assignment 4 Solutions - 0%5 2W M 5mm‘m(91 We WW WM Wm...

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Unformatted text preview: 0%5 2W , M 5mm‘m (91) We WW WM Wm /; 0/711, £201 : 40/2 +WQO W WW M W 97%;»ng Wyﬂdgf W fi/M‘@,WM/'s (Q1. ch=‘————k . =§i§l=06 ”Ant 100 “EVBkEk g = J”..— = ——1— _ 0 25 M nAm 100 Solving, we ﬁnd El = 25 52 =10 Yield of C from A is “VAl EVCkEk 2 E1 -E2 2(25 -10) _O 30 yA—»c = *— = — = “'— - - VCI "A, fed 1 "A, fed 100 Selectivity of converting A to C is SA... = "VA: EVCkEk = 2(_§1_:§_2~)= 2(25 ’1°)= 0.50 VCl EVAkEk 1 251*‘52 50+“) Check if yield = conversion times selectivity: 0.6 x 0.5 = 0.3 — yes! .a . way—wwwwm . Wm . .. ' 63%, H H Li First We’ll check the total mass walker OUT=27g+52g+121g+68 =268 7 WW ”£9,199! mass is O . g g We can proceed by either (1) checking element mass balances or (2) converting to moles, and checking stoichiometry. We will take approach 2 here. IN: 134 gIBBxM=1gm011BB, 134 gAAnxw =l.314 gmol AAn 134 g IBB 102 g AAn 1 gmol IBB OUT: 27 g IBBx 134g IBB = 0.20 gmol IBB Therefore, 0.80 gmol IBB are consumed by reaction. The balanced chemical reaction is C10H14 + C4H603 —> C12H16O + CH3COOH Balances on other compounds: In (moles) Consumed or Molar generated mass (grams) . [- CH COOH [- The last column is calculated from material balances. Agreement with the data for acetic anhydride is good, but the calculated masses of IBA and acetic acid are wrong by 20 g. Thus, the Chemist’s data does not satisfy material balance reauirements. and must be incorrect G4" A diagram of the system is shown, assuming a basis of 100 gmoles total initially charged to the reactor. Gases 5.1% E 10 E Reactor/separator E0 12 O N 78 N C02 W This is a batch reactor. Since there are 2 chemical reactions of known stoichiometry, we will choose compounds as components, which we indicate as E for ethylene, E0 for ethylene oxide, N for N2, 0 for 02, and C fOr C02. The material balance equation for all components is: ni,sys, f - "imp = V1151 + W252 'f where 5k = fékdt. to For each compound, using the known initial charge to the reactor, the equations become: 0-12:"51-362 OI' EI+3E2=12 "E,sys,f “ 10 = -2§1—§2 or "Egysj =10 " 251'" 52 nN,syS,f -' 78 = O or nNJys,f = 78 "50:”. f = 251 "C,sys,f = 352 We need one more equation, which comes from the stream composition speciﬁcation: 0.051 = "Em”: = “10 _ 2&1 — E2 "E,sys, f + "N ,yys, f + nCJst + "50,:st 88 + £2 , and from that calculate fractional conversron and yield 51 =O.8 gmol £2 = 3.74 gmol _ _ 2 0.8 + 3.74 M = -335. = ( > =0534 n E ”3,0 10 —2 2 0.8 yE—sgo = —(—) 251 = —(—) =O.16 +2 nEJySD 10 ”ﬂu—um,wewmmwmﬁmmﬂmamw-uwWVWMWWWMWWWWM mmw ,, _ _ evlaemwmmqqm.wmym IN OUT A 100 A reactor 3 100 B C D The reaction is 2A + B <—> C + D Since this is a gas-phase reaction, we say that The mole fractions of each are related to the moles of each component and the total moles, as well as the extent of reaction at equilibrium: "1' ,out Mole fraction y, = "out Substituting in the expressions for mole fraction into the K, equation and simplifying, we ﬁnd: K M: “ (loo-atom)” 1 52(200—é) K, is a constant; If P increases, then — decreases, and . 2 . (100 —25) (100 — £3) keep Ka constant. This happens if E increases. (You can check this by plugging in two different must increase to values for E and calculating this term. The maximum value of E is 50.) Thus, increasing P will increase fractional conversion of A. ...
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