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 gasphase 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=InOut+Gen
Steady state PPR so no accumulation. Q is constant 0 : QCAIV _QCAV+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—arrFtrfn 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 countercurrent 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 NewtonRaphsah technique, attempttoﬁnd the root of the equation: x — 58—“ : 1
State your initial guess and show two iterations, Has the NewtonRaphson 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.
 Spring '10
 Gilman

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