ChE 344 Winter 03 Final Exam + Solution Thursday, May, 03 Open Course Textbook Only Closed everything else (i.e., Notes, In-Class Problems and Home Problems Name Honor Code (Please sign in the space provided below) I have neither given nor received unauthorized aid on this examination, nor have I concealed any violations of the Honor Code. (Signature) The Basics ) / pts ) / 5 pts 3) / 5 pts 4) / 5 pts 5) / 5 pts 6) / 5 pts 7) / 8 pts Applications 8) /0 pts 9) /0 pts Professional 0) /0 pts ) /5 pts Total /00 pts
( pts) ) Chapter Mole Balances The reaction A + B C takes place in a membrane reactor. The feed is only A and B in equimolar proportions. Which of the following set of equations gives the correct mole balances on A, B and C. Species A and B are disappearing and Species C is being formed and C is also diffusing out the sides of a membrane reactor. Circle the correct answer where all the mole balances are correct (a) df A dv = r A df B dv = r B Ans: r C is wrong (b) df C dv = r C R C df A dv = r A df B dv = r B Ans: r C is wrong (c) df C dv = r C R C df A dv = r A df B dv = r B Ans: r C is wrong (d) df C dv = r C R C df A dv = r A df B dv = r A Ans: correct df C dv = r A R C (e) None of the above
Solution Answer is (d).
(5 pts) ) Circle the correct answer. Consider the following Levenspiel plot for a reversible reaction A Product Figure - ( pt) (a) The equilibrium conversion X e in a 3 dm 3 reactor is () X e < 0.6 () X e = 0.6 (3) X e > 0.6 (4) Can t tell from the information given ( pt) (b) The flow rate to an 8 dm 3 CSTR corresponding to Figure - where 80% conversion is achieved is () F A0 = 0.8 mol/s () F A0 = 0 mol/s (3) F A0 = mol/s (4) Can t tell from the information given ( pt) (c) If the conversion achieved in a single 8 dm 3 CSTR is 80%, what would the conversion be if the flow is equally divided into two CSTRs in parallel with each reactor having a volume of 4 dm 3 each (same total volume). " 0 " 0 " 0 " 0 vs. 8 dm 3 X=0.8 4 dm 3 4 dm 3 X=_?_ X=_?_ The total reactor volume is constant at 8 dm 3. The conversion for the two reactors in parallel is () X > 0.8 () X < 0.8 (3) X = 0.8 (4) Can t tell from the information given ( pts) (d) If the conversion achieved in a single 8 dm 3 CSTR is 80%, what would the conversion be if two CSTRs are connected in series with first reactor having a volume of approximately 3.0 dm 3 and the second reactor having a volume of 0.6 dm 3. 3
υ 0 vs. 8 dm 3 3.0 dm 3 X=0.8 0.6 dm 3 X=_?_ The conversion for the two reactors in series is () X > 0.8 () X < 0.8 (3) X = 0.8 (4) Can t tell from the information given Solution (a) Ans. (3) X e > 0.8 (b) Ans. (d) Can t tell from information given (c) Ans. (3) X = 0.8. See p6. (d) Ans. () X < 0.8. Try X = 0.6 3 dm 3 = 5 dm 3 x 0.6 = 3 dm 3 checks Try ΔV = 0. between X = 0.6 and 0.7 V = 0. x 6 dm 3 = 0.6 dm 3 checks F A0 r A (dm 3 ) 3dm 3 0 9 8 7 6 5 4 3 0. 0.4 0.6 0.8 X 4
(5 pts) 3) Consider the following reaction for parts (a), (b) and (c) A + B C Write the rate law in terms of the specific reaction rate and species concentration when ( pt) (a) The reaction is irreversible and second order in A, and independent of the concentration of C, and overall first order. r A = ( pt) (b) The reaction is elementary and reversible r A = ( pt) (c) Now consider the case when the reaction is first order in A and first order in B at high concentrations of A and B and is first order in A and second order in B at low concentrations of B. The rate law is r A = ( pt) (d) The irreversible reaction is catalyzed on a Pt surface where surface reaction limits and B is not adsorbed on the surface but reacts with adsorbed A on the surface. r A = Solution A + B C (a) r A = k A C A C B # (b) r A = k A C A C B C & C % ( $ ' (c) r A = k C A C B + k C B K C (d) A + S A S A S+ B C S C S C + S C A S = P A K A C V r B = k S C A S P B C A S = K C P C C t C V = + K A P A + K C P C k k r A = S K A P A P B + K A P A + K C P C ( ) 5
(5 pts) 4) The following figure shows the energy distribution function at 300 K for the reaction A + B C f(e,t) (kcal) 0.5 0. 0.5 0. 0.05 0 3 4 5 6 7 8 E (kcal) (a) What fraction of the collisions have energies between 3 and 5 kcal? (b) What fraction of collisions have energies greater than 5 kcal? Solution (a) Between 0 and 4 k cal Between 4 and 8 kcal " f( E,T) = 0.5 % $ ' E f(e, T) = 0.5 0.5C # 4 & 4 Graphical (0.5) () + 0(0.98)() = 0.448, i.e., 45% f(e,t) (kcal) 0.5 0. 0.5 0. 0.05 0.88 Algebraic f(e,t) (kcal) 0.5 0. 0.5 0. 0.05 0 3 4 5 6 7 8 E (kcal) 0 3 4 5 6 7 8 E (kcal) at E = 3 f ( E, T) = 3.5 = 0.88 4 at E = 5 f ( E, T) = 3.5 = 0.88 4 Area = ( 0.5) + ( ) 3 4.5 = 0.448 = 44.8% 6
(b) " 3% $ '.5 # 4& ( ) f(e,t) (kcal) 0.5 0. 0.5 0. 0.05 " Area = 3 % $ '.5 # 4& = 0.8 = 8% ( ) 3 = 0 3 4 5 6 7 8 E (kcal) 7
(5 pts) 5) For elementary reaction A B the equilibrium conversion is 0.8 at 7 C and 0.5 at 7 C. What is the heat of reaction? ΔH Rx = cal/mole A Solution X e X e = K C At 7 C, T = 400 K 0.8 0.8 = 4 = K C At 7 C, T = 500 K 0.5 0.5 == K C ln K C = ΔH Rx K C R ΔH Rx = T T T T # & % ( = ΔH # Rx % $ T T ' R $ ( ) R ln K C K C ( )( 400) = 500 500 400.987ln 4 T T T T = (, 000K).987 cal ( molk.39 ) = 5, 509 cal mola & ( ' 8
(5 pts) 6) A Hanes-Woolf plot is shown below for the different types of enzyme inhibition. Match the line with the type of inhibition. A B C C S r S None (a) Inhibition Mechanism C S (b) Inhibition Mechanism Ans: (c) Inhibition Mechanism Inactive Ans: Ans: Solution (a) B (b) A (c) C 9
(8 pts) 7) (pt) (a) Keeping in mind the explosions we discussed this term, suggest at least one possible cause of the West, Texas fertilizer plant explosion on April 5, 03 that resulted in 4 fatalities. In the following, circle the correct answer below. ( pt) (b) Which of the following figures best represents the relationship between the amount of down time, t d, and the time the heat exchanger failed after start up, t s for which the ONCB/ammonia reactor would not explode () () (3) (4) t d t d t d Can t tell from information given t s t s t s ( pt) (c) Currently 50% conversion is being achieved in an endothermic liquid phase reaction A + B C + D in a CSTR when the reaction is carried out adiabatically and the feed is stoichiometric. If the reaction molar flow rate of B is doubled with everything else held constant, the exit temperature will Increase Decrease Remain the same Insufficient information to tell ( pt) (d) A co-current heat exchanger with a variable ambient temperature will always have a greater equilibrium conversion than a heat exchanger with a constant ambient temperature when an exothermic reversible reaction is taking place. True False Insufficent informtion to tell 0
Solution (a) (b) (c) Ans. Heat Exchange Failure or Ans. Cold feed to the reactor interrupted. Either answer acceptable (3) The later the heat exchanger fails. Since the start of the reaction the more reactant will have been conserved. Consequently when the heat exchanger fails at a later time, t s, the rate will be slower allowing one down time, t a before Q g > Q R Increase (d) Excess species B acts in the same way as an inert does. False
(0 pts) 8) Experimental data for the gas phase catalytic reaction A+B C is shown below. The limiting step in the reaction is known to be irreversible, so that the overall reaction is irreversible. The reaction was carried out in a differential reactor (i.e., virtually no concentration gradient down the reactor) to which A, B, and C were all fed. Run Number 3 4 5 6 7 PA (atm) 0 0 0. PB (atm) 0 0 0 PC (atm) 0 Reaction rate (mol)/(gcat s) 0.4.40 0.80.73 0.96 0.86 0.043 (a) Suggest a rate law consistent with the experimental data. (Hint: Sketch ( ra" ) as a function of PA, as a function of PB, and as a function of PC.) (b) From your rate expression, which species can you conclude are adsorbed on the surface? (c) Suggest a mechanism that is consistent with the rate law in part (a).
Solution From runs, 3, & 6 r A # Suggests P A is in both numerator and denominator of the rate law r A # ~ ( P A )(? ) ( ) + K A P A +? (A) P A From runs,, & 4 r A # Suggests P B is only in the numerator of the rate law r A # ~ P B (B) P B From runs 4, 5 r A # Suggests P C is in the denominator of the rate law r A # ~ (C) + K C P C +... P C Combining Equations (A), (B) and (C) above k P (a) r A " = A P B (b) A and C are on the surface + K A P A + K C P C (c) A + S AS A S+ B C S C S C + S 3
(0 pts) 9) The reversible liquid phase reaction A B is carried out in a dm 3 CSTR with heat exchange. Both the entering temperature, T 0, and the heat exchange fluid, T a, are at 330 K. An equal molar mixture of inerts and A enter the reactor. (6 pt) (a) What product of the heat transfer coefficient and heat exchange area would give the maximum conversion? Ans: UA = cal/h/k (4 pt) (b) Using UA from part (a), what is the maximum conversion that can be achieved in this reactor? Ans: X max = Additional Information The G(T) curve for this reaction is shown below C PA = C PB =00 cal mol K, C PI =50 cal mol K F A0 =0 mol h, C A0 = mol dm 3, υ 0 =0 dm 3 h ΔH Rx = 4,000 cal mol k = 0.00 h at 300K with E = 30,000 cal mol K C = 5,000,000 at 300K 4
Solution a) T C = T a = T 0 R( t) = C P0 ( + κ) ( T T C ) Slope = C P0 ( + κ), Slope = 36,000 cal mol 366 ( )K = 36,000 cal 36 mol K ( ) 330 =,000 cal mol K = 50 cal mol K + κ κ = 3 C P0 = C PA + C PI = 50 " UA = 3 $ # ( ) 0 mol h = 7,500 cal hk %" ' 50 cal % $ ' &# molk& Ans: UA = 7,500 cal/h/k b) X max = G = 36,000 ΔH Rx 4,000 = 0.86 Ans: X max = 0.86 ( ) 5
360 6
(0 pts) 0) The irreversible elementary gas phase reaction A + B " C+ D is carried out isothermally at 305 K in a packed bed reactor with 00 kg of catalyst. The entering pressure was 0 atm and the exit pressure is atm. The feed is equal molar in A and B and the flow is in the turbulent flow regime, with F A0 = 0 mol/min and C A0 = 0.4 mol/dm 3. Currently 80% conversion is achieved. What would be the conversion if the catalyst particle size were doubled and everything else remained the same.? X = Solution (a) C A0 = y A0 P AT 0 RT 0 = 0.5 ( 0) ( ) ( 0.08) ( 305) = 0 ( 305) ( 0.08) C A0 = 0.4mol dm 3 (b) dx dw = r A " = kc A0 X F A0 ( ) y F A0 = kc A0 ( X) αw F A0 ( ) 0.8 0.8 = X X = kc # A0 W αw & F A0 $ % '( y α = W y = 0 = 0. y = ( αw) 0.0 = = 0.99 00 00 α = 9.9 0 3 kg " k 0.4 mol % # $ dm 3 & ' " 0 mol 00 9.9 0 3 0 $ 4 $ # min 4 = k 0.6 ( ) [ ][ 00 49.5] 0 dm 6 k = 4.95 kg mol min % kg ' ' & 7
k = ( 9 0) 50.5 ( )( 0.059) = 30. dm 3 mol min For the turbulent flow X X = α ~ D P D α = α P = α D P = 9.9 0 3 kg = 4.95 0 3 kg $ dm 6 ' $ mol ' & 4.95 % kg mol min )& 0.4 (% dm 3 ) ( * 0 mol 00 4.95 0 3 00, +, min = 0.08 00 4.7 [ ] = 5.95 X = 5.95 6.95 = 0.86 ( ) - kg/. / 8
(5 pts) ) The following reactions are taking place in a,000 dm 3 liquid phase batch reactor under a pressure of 400 psig k A A + B " " C ΔH RxB = 5,000 cal mol r A = k A C A C B k A 3C + A " " D ΔH RxC = +0,000 cal mol r A = k A C A C C B+ 3C k 3C " " E ΔH Rx3B = 50,000 cal mol r 3C = k 3C C B C C The initial temperature is 450 K and the initial concentrations of A, B and C are.0, 0.5 and 0. mol/dm 3 respectively. The coolant flow rate was at its maximum value so that T a = T a = T a = 400 K so that the product the exchange area and overall heat transfer coefficient, UA, is UA = 00 cal/s K. ( pt) (a) If Q r > Q g at time t = 0, and there is no failure of the heat exchange system, is there any possibility that reactor will run away? Explain (4 pt) (b) What is Q r at t = 0? (0 pt) (c) What is Q g at t = 0? (3 pt) (d) What is the initial rate of increase in temperature, (dt/dt) at t = 0? dt dt = (3 pt) (e) Suppose that the ambient temperature T a is lowered from 400 K to 350 K, what is the initial rate of reactor temperature change? dt dt = (3 pt) (f) A suggestion was made to add 50 moles of inerts at a temperature of 450 K. Will the addition of the inerts make runaway more likely or less likely? How? Show quantitatively. Additional information As a first approximation, assume all heats of reaction are constant (i.e., ΔC Pij 0 ) Specific reaction rates at 450 K are k A = 0 3 ( dm 3 mol) s k A = ( 3 0 3 dm 3 mol) s k 3C = 0.6 0 3 ( dm 3 mol) s C PA =0cal mol K C PB =0cal mol K C PC = 50cal mol K C PD = 80cal mol K C PE = 50cal mol K 9
Solution Part (a) Part (b) Part (c) dt dt = Q g Q r N A C PA + N B C PB + N C C PC If Q r > Q g then the temperature can only decrease causing the specific reaction rates k i to decrease, hence runaway is unlikely. Q r = UA( T T a ) =00 cal [ 450 400]K = 5, 000 cal s K s [ ] Q g = V r B ΔH RxB + r C ΔH RxC + r 3B ΔH Rx3B Initially T = 350 K Reaction : Reaction : r A = r B = r C r A = r C 3 = r D r B = r A r C = 3 r A Part (d) Reaction 3: r 3B = r 3C 3 = r 3E Q g = V[ k A C A C B ] ΔH RxB $ % & 0 r 3B = 3 r 3C [ ] + V 3 k A C A C C ' ( ) ΔH RxC $ % & [ ] + V 3 k 3C C B C C $ ' = (,000) [( ) ( 0 3 ) ( ) ( 0.5 ) 5,000] +,000 & 3 +. ) & -, 3 0 3 0 ( ) ( 0.) [ 0,000] ) + & / ) 5,000 %& () ( ) 0.5 $ +,000 ' 0.6 0 3 ( )( 0.) % & 3 ( ) [ 50,000] = 5,000 Q g = 5,000cal s dt dt = +,000 Q r Q g N A0 C PA + N B0 C PB + N C0 C PC = N A0 C PA = C A0 VC PA = ( ) (,000) ( 0) = 0,000 N B0 C PB = C B0 VC PB = ( 0.5) (,000) ( 0) =0,000 N C0 C PC = C C0 VC PC = ( 0.) (,000) ( 50) = 0,000,000 5,000 5,000 N A0 C PA + N B0 C PB + N C0 C PC = 0 ' ( ) ΔH Rx3C [ ]
Part (e) Part (f) dt dt = Q g Q r 50,000 Drop T a by 50 = 5,000 5,000 50,000 = 0 Q r = UA( T T a ) =00( 450 350) =0,000 dt dt = 5,000 0,000 50,000 = 0. [ ] UA( T T a ) dt dt = Q g Q r = r B VΔH RxB + r C VΔH RxC + r 3A VΔH Rx3A N i C Pi N A C PA + N B C PB + N C C PC + N D C PD + N E C PE + N Inerts C PInerts Inerts (N Inerts ) will not change Q g or Q r, they will only slow the rate of temperature increase or decrease.