cheme140-03-mt2-Radke-soln (1)

cheme140-03-mt2-Radke-soln (1) - Chemical Engineering 140...

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Unformatted text preview: Chemical Engineering 140 Midterm Examination #2 Solutions 1. (100 points) Air (nitrogen and oxygen) is fed as a reactant in marry chemical processes. However, we know that oxygen and nitrogen can also react to form NO by the following gas-phase reaction: N2 + O] A 2ND Consider a reactor into which air is fed and the only reaction occurring is the conversion of oxygen and nitrogen to NO. Assume the reactor is long enough that the outlet stream is in equilibrium. Figure 15.3 shows equilibrium constants as a function of temperature for some common reactions. \ 3. Assume the reactor is run at 2000K and 1 arm. i. What is the Equilibrium constant (K) at 2000K? K = e'4 = 0.0183 for the reaction VzNz + 5/20; a NO or K = 3.35 x 10" for the reaction N2 + 02 m) 2N0 5 pts ii. Write K in terms of equilibrium mole fractions of productsfreactants. K z m% : % for the reaction VzNz+ 1/20; —> NO (x NEP) ttxoz P) ‘ XN‘zxo; : (X HOP): Z Xiao for the reaction N2+ O; —> 2N0 (X NIP)(XOZ P) anxog 5 pts for either of these expressions 10 pts for being consistent with part {ii iii, Write species balances for the system in terms of conversion of N2. We know that for each mol of N; fed, 21/79 = 0.266 mol 0; are fed; no NO is fed to the reactor. At steady state we write 0 = IN — OUT + GEN. Expressing the outlet molar flow rates (m) in terms of conversion of N; (Km) we have N2: 11m "—‘ Dwain - fluamxnz : antin (1‘XN2) —L10 t5 0;: 1101 = 1102;“ _- HNZmXNz = Hsz - X312} 5 913 NO! 11910 : anmxm 5 [its 10 pts off if the species balances are not written in terms of conversion of N3 iv. What is the equilibrium conversion [Jsz '? “N0 Xxo ,_ “mu: 2 END _ ZERXN: xii Kg: an} H noI (l; 1113:1151: winii,(1— XML!”in _ XM) it): E 23in2 . . . . . . 00183: —__.T 10 Hts 5 pts off if K is moonsmtent wtth expressxon (r — XE )5 (0.266— X1045 2 3353.110“t = Aim—— is also acceptable (1 - XN2 )(0266 — XNZ) Xm = 4.66’7n10'3 (KN; = 0.032 is the answer ime is used instead of K) 5 pts b. Assume the reactor is run at room temperature (25°C) and pressure. i. Is the reaction exothermic or endothermic? Since the slope of the plot of 111K vs 1/T is negative, the reaction is endothermic Spts ii. What is the equilibrium constant at room temperature? _4 — —6 — AH" Slope of line is ——_-4(T(—))_71—_1 = ~1x104K = “mm—3”“ 5 pts 5x10 K H 7X10 K R AHmD = 33.1 lemol Assuming AH”: is independent of temperature K — AH” 1 1 in —— = —”‘” — — —-- 10 ts (K1) R (T T!) _p_ K 1 1 rim—29%) = —1x10“K(—_—— — —) 0.0183 298K 2000K 5 pts Km e 7.25x10‘” iii, Would the equilibrium conversion of N; at 25°C be higher or lower than at 2000K? Lower. K is smaller} the reaction is endothermic. 10 pts c. How would die equilibrium conversion change if the system pressure was 100 atm? There would be no change. There is no net change in moles. p—u |: Ft- m 2) a) Consider the following gas phase reaca‘ons.‘ A +B —)2D {EA—rt} where Dis the desired product and U is undesired. Write expressions for the equilibrium constants fin each reaction in terms of the motefi‘acttons and totat pressure. Wat are the units of each equitibiurn constant? 2 (xDPF x5 1 fl (xAPXxBP) # xAxB /._ (35911)) _ xUP% (xAP}% — x? The equilibrium constants don’t have unitsl 5 points for each K, and 5 points for no units. K2: h) The graph to the right shows the temperature dependence of the equitihriurn constants in part (a). What reactor conditions (T and P) wilt increase set’ect‘tvity to the destredproduot these reactions are run simultaneoust in a very long PPR? As temperature increaeee, K1 becomes larger than K;. As P 111K increases, the conversion to the undesired product is lowered by LeChatlier’s principle, but conversion to the desired product is unaffected. So run at high T and high P. 5 points each for temperature and pressure dependance c) Derive a a’tjfi’erentia! equation for the change in concentration ofethytene oxide along the length of a PPR as it reacts with water to form ethylene glycol, Assume elementary kinetics. Start with a difierenttat mass balance and Show your work! 0 OH i: + H20 —-—-l-— HON (3} E (C) V V‘HJV I‘A:-kCACB Acc=In-Out+Gen Steady state PPR so no accumulation. Q is constant 0 : QCAIV _QCA|V+uV + "AAV QCA1V+3V — QCA IV 2 FA AV take the limit as Nit-+0 to find dCA _ rA —kCACB dV F Q Q 3 points for 13;, 5 for mass balance, 2 for limit, 5 for final equation. a?) How are the concentrations ofthe reactants and products retatea‘ for the following general homogeneous tiquid phase reaction? Assume that C Am) = C An, (33(0) 2 C30, Ccl’G) = Cm (13(0) =‘Cm nd—LhR—arrF-t-rfn CAD—CAHCBD-‘CB Coo—Cc Ccc‘cc —a —b C d 5 points for the correct format, 5 for the correct signs e) Draw the temperature prafilesfii‘ a art—current and a counter-current doubfe pipe heat exchanger running at steady state without reaction. Co -current 5 points for each of the 4 profiles. fl Draw the iemperatureprofite for an adiabatic PPR running with an endothermic reaction at steaaj» State. Adiabatic PPR with endothermic rxn 10 points all or none gJUsing the Newton-Raphsah technique, attempttofind the root of the equation: x — 58—“ : 1 State your initial guess and show two iterations, Has the Newton-Raphson technique converged? fix): 0 2X - sex—1 fix) = o = 1+5e'x Taylor expansion of {(x) around x0 fix) = flxo) + f‘lxoxxwxo) Xm = Xi— Ext)! 1’09) . xi” = xi — (Xi — Se'x‘ - I)f(1 + 56’“) guess = 1 X1 = 0352 X2 = 1274 technique has not converged 5 for f(x)=D, 5 for fix), 5 for taylor series expansion, 1 for guess, 3 for iterations, 1 for convergence it) When calculatihg reactor Size, what is the “design tradeof’fir an exothermic reaction? The design tradeofl" is the balance between equilibrium conversion and kinetic rate of reaction. As temperature goes up the equilibrium will go down (for an exothermic reaction) but the rate will go up. 10 points ail or none ...
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This note was uploaded on 02/18/2010 for the course CHEME 120A taught by Professor Gilman during the Spring '10 term at Berkeley.

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