Elements of chemical reaction engineering 4th edition fogler solutions manual

Elements Of Chemical Reaction Engineering 4th Edition Fogler Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/elements-of-chemical-reaction-engineering-4th-edition-fogler-solutions-manual/

Solutions for Chapter 6 – Multiple Reactions P6-1 Individualized solution P6-2 (a) Example 6-2 For PFR,

dC A = !k1 ! k 2 C A ! k 3 C A2 d" dC B = k 2C A d!

dC X = k1 d! dCY = k 3 C A2 d!

In PFR with τ = V/vo = 783 sec we get X = 0.958 and SB/XY = 0.624 also at τ = 560.24 sec SB/XY is at its maximum value of 0.644 0.4

0.7

CA

0.35

0.6

0.3 0.5

SB/XY

0.25 0.4

0.2 0.3

0.15

CY

0.2

0.1

CB

0.05

0.1

CX 0

0

0

200

400

!

600

800

0

200

400

600

!

See Polymath program P6-2-a.pol.

POLYMATH Results No Title 02-16-2006, Rev5.1.233 Calculated values of the DEQ variables Variable tau Ca Cx Cb Cy Cao X k1 k2 k3 Sbxy

initial value 0 0.4 1.0E-07 0 1.0E-06 0.4 0 1.0E-04 0.0015 0.008 0

minimal value 0 0.0166165 1.0E-07 0 1.0E-06 0.4 0 1.0E-04 0.0015 0.008 0

maximal value 783 0.4 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6437001

6-1

final value 783 0.0166165 0.0783001 0.1472919 0.1577926 0.4 0.9584587 1.0E-04 0.0015 0.008 0.6238731

800

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2 [2] d(Cx)/d(tau) = k1 [3] d(Cb)/d(tau) = k2*Ca [4] d(Cy)/d(tau) = k3*Ca^2 Explicit equations as entered by the user [1] Cao = 0.4 [2] X = 1-Ca/Cao [3] k1 = 0.0001 [4] k2 = 0.0015 [5] k3 = 0.008 [6] Sbxy = Cb/(Cx+Cy)

(2) Pressure increased by a factor of 100.

Now CA0 = P/RT = 0.4 × 100 = 40 mol/dm3 For single CSTR, CA* does not change but

"=

C ! CA C AO ! C A 40 ! 0.112 V = sec = AO = 2 vO ! rA k1 A + k 2 A C A + k 3 A C A 0.0001 + 0.00168 + 0.0001 ! = 21217 sec

P6-2 (b) Example 6-3 (a) CSTR: intense agitation is needed, good temperature control. (b) PFR: High conversion attainable, temperature control is hard – non-exothermic reactions, selectivity not an issue (c) Batch: High conversion required, expensive products (d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low, to control the conversion of a reactant. (f) and (g) Tubular with side streams: selectivity i.e. to keep a reactant concentration high, to achieve higher conversion of a reactant. (h) Series of CSTR’s: To keep a reactant concentration high, easier temperature control than single CSTR. (i) PFR with recycle: Low conversion to reuse reactants, gas reactants (j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid reactants (k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product (l) Reactive Distillation: when one product is volatile and the other is not

6-2

P6-2 (c) Example 6-4 (1) For k1 = k2, we get

) and dCd! + k C = k C d (C exp(k ! )) =kC ! (

C A = C AO exp " k1! ' B

B '

1

'

= 0 , CB = 0

Optimum yield:

dC B d!

'

vo k1

(

AO exp " k1!

1

(

)

[ (

)

)

(

= 0 = k1C AO exp " k1! ' " k1! ' exp " k1! '

)]

1 k1

and

(

)

X opt = 1 ! exp ! k1" ' = 1 ! e !1 =0.632

(2) For a CSTR:

C A0 ! C A C A0 ! C A = k1 C A ! rA' k1C A" ! + C A = C A0

#" =

C A0 (" !k1 + 1) C !C 0 ! CB # " = B0 ' B = ! [k1C A ! k 2 C B ] ! rB CA =

=

'

AO

! C B = k1C AO! ' exp " k1! '

' ! " opt =

Wopt =

1

'

d! '

Now at !

1 B

CB [k1C A ! k 2 C B ]

# !k1C A " # !k 2 C B = C B C A0 " !k1C A " !k1 CB = = (" !k 2 + 1) (" !k 2 + 1)(" !k1 + 1) k C [(# !k 2 + 1)(# !k1 + 1)]" # !k1C A0 [k1 + k 2 + 2k1 k 2# !] dC B = 0 = 1 A0 d# ! [(# !k 2 + 1)(# !k1 + 1)]2

[(! "k 2 + 1)(! "k1 + 1)]= ! "[k1 + k 2 + 2k1k 2! "] 2 2 1 + ! "k 2 + ! "k1 + k1 k 2 (! ") = ! "k1 + ! "k 2 + 2k1 k 2 (! ") 2

k 2 k1 (" !) = 1 1 ! = " OPT k1 k 2 (3) T = 310.52oC

6-3

CB =

" !k1C A0 (" !k 2 + 1)(" !k1 + 1)

POLYMATH Results No Title 02-14-2006, Rev5.1.233 Calculated values of the DEQ variables Variable t T Cao tau k1o k2o E1 E2 R k1 k2 Cb

initial value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.6944444

minimal value 0 300 5 0.5 0.4 0.01 10 20 0.001987 0.4 0.4 0.0052852

maximal value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.8040226

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 1 Explicit equations as entered by the user [1] Cao = 5 [2] tau = .5 [3] k1o = .4 [4] k2o = .01 [5] E1 = 10 [6] E2 = 20 [7] R = .001987 [8] k1 = k1o*exp((-E1/R)*(1/T-1/300)) [9] k2 = k1o*exp((-E2/R)*(1/T-1/300)) [10] Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)

6-4

final value 100 400 5 0.5 0.4 0.01 10 20 0.001987 26.513034 1757.3524 0.0052852

P6-2 (d) Example 6-5 KC = Ke = 0.25

C N CO # & 2 r2 N 2 = k 2 N 2 $C NO ' 2 2!

KC " % C N CO # & 2 .5 rNO = 'k1NO C NH 3 C 1NO ' 2k 2 N 2 $C NO ' 2 2!

KC " % C N CO $ 1 ' 2 5 .5 rN 2 = k1NO C NH 3 C 1NO + k 2 N 2 %C NO ! 2 2 " ! k 3O2 C N 2 C 022 6 KC # 2 & C N CO $ ' 2 rO2 = k 2 N 2 %C NO ! 2 2 " ! k 3O2 C N 2 C 022 KC # &

P6-2 (e) Example 6-6 For liquid phase,

F j = vo C j

Equation (E 6-6.3):

Equation (E6-6.8):

P6-2 (f) Example 6-7

dFNO dC NO .5 2 =vO = !kC NH 3 C 1NO ! 2k 2 N 2 C NO dV dV dC NO .5 2 " = !kC NH 3 C 1NO ! 2k 2 N 2 C NO d# dFNO2 dC NO 2 =vO = k 3O2 C N 2 C O22 dV dV dC NO 2 ! = k 3O2 C N 2 C O22 d"

For equal molar feed in hydrogen and mesitylene.

CHO = y HOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3 CMO = 0.016 lbmol/ft3 Using equations from example, solving in Polymath, we get ! opt = 0.38 hr. At ! = 0.5 hr all of the H2 is reacted and only the decomposition of X takes place..

XH CH CM CX

! ~

S X /T

Ex 6-7 0.50 0.0105 0.0027 0.00507 0.2hr 0.596

This question 0.99 0.00016 0.0042 0.0077 0.38hr 1.865

See Polymath program P6-2-f.pol.

6-5

POLYMATH Results Calculated values of the DEQ variables Variable tau CH CM CX k1 k2 r1M r2T r1H r2H r1X r2X

initial value 0 0.016 0.016 0 55.2 30.2 -0.1117169 0 -0.1117169 0 0.1117169 0

minimal value 0 1.64E-06 0.0041405 0 55.2 30.2 -0.1117169 0 -0.1117169 -0.0159818 2.927E-04 -0.0159818

maximal value 0.43 0.016 0.016 0.0077216 55.2 30.2 -2.927E-04 0.0159818 -2.927E-04 0 0.1117169 0

ODE Report (RKF45) Differential equations as entered by the user [1] d(CH)/d(tau) = r1H+r2H [2] d(CM)/d(tau) = r1M [3] d(CX)/d(tau) = r1X+r2X Explicit equations as entered by the user [1] k1 = 55.2 [2] k2 = 30.2 [3] r1M = -k1*CM*(CH^.5) [4] r2T = k2*CX*(CH^.5) [5] r1H = r1M [6] r2H = -r2T [7] r1X = -r1M [8] r2X = -r2T ~

Increasing θH decreases τopt and S X/T.

P6-2 (g) Example 6-8 Using equation from example 6-8: Polymath code: See Polymath program P6-2-g.pol.

POLYMATH Results NLES Solution Variable CH CM CX tau K1 K2 CHo CMo

Value 4.783E-05 0.0134353 0.0023222 0.5 55.2 30.2 0.016 0.016

f(x) -4.889E-11 -1.047E-11 -9.771E-12

Ini Guess 1.0E-04 0.013 0.002

6-6

final value 0.43 1.64E-06 0.0041405 0.0077207 55.2 30.2 -2.927E-04 2.986E-04 -2.927E-04 -2.986E-04 2.927E-04 -2.986E-04

NLES Report (safenewt) Nonlinear equations [1] f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0 [2] f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0 [3] f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tauCX = 0

Explicit equations [1] tau = 0.5 [2] K1 = 55.2 [3] K2 = 30.2 [4] CHo = 0.016 [5] CMo = 0.016

A plot using different values of ! is given.

For ! =0.5, the exit concentration are CH = 4.8 ×10-5 lbmol/ft3 CM =0.0134 lbmol/ft3 CX =0.00232 lbmol/ft3 The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for ! =0.5:

YMX =

FX CX 0.00232 0.89mole ! xylene ! produced = = = FMO " FM C MO " C M 0.016 " 0.0134 mole ! mesitylene ! reacted

The overall selectivity of xylene relative to toluene is:

F 8.3mole ! xylene ! produced ~ S X /T = X = FT mole ! toluene ! produced CH CM CX

!

YMX SX/T

Ex 6-8 0.0089 0.0029 0.0033 0.5 0.41 0.7

This Question 4.8 x 10-5 0.0134 0.00232 0.5 0.89 8.3

P6-2 (h) Example 6-9 (1) SD/U Original Problem Membrane Reactor 2.58 PFR 0.666 Doubling the incoming flow rate of species B lowers the selectivity.

SD/U P6-2 h 1.01 0.208

(2) The selectivity becomes 6.52 when the first reaction is changed to A+2B  D

6-7

P6-2 (i) Example 6-10 Original Case – Example 6-10

P6-2 i

The reaction does not go as far to completion when the changes are made. The exiting concentration of D, E, and F are lower, and A, B, and C are higher.

See Polymath program P6-2-i.pol.

P6-2 (j)

At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. At the same time the reactions that consume the two components begin to accelerate and the concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again. Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions.

6-8

P6-2 (k) Base case

Equal-molar feed

Increasing yM0 will increase the production of m-xylene and methane, but will result in a large amount of un-reacted mesitylene.

P6-2 (l) Individualized solution P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem ) P6-4 (a) Assume that all the bites will deliver the standard volume of venom. This means that the initial concentration increases by 5e-9 M for every bite.

After 11 bites, no amount of antivenom can keep the number of free sites above 66.7% of total sites. This means that the initial concentration of venom would be 5.5e-8 M. The best result occurs when a dose of antivenom such that the initial concentration of antivenom in the body is 5.7e-8 M, will result in a minimum of 66.48% free sites, which is below the allowable minimum.

See Polymath program P6-4-a.pol.

P6-4 (b) The victim was bitten by a harmless snake and antivenom was injected. This means that the initial concentration of venom is 0. From the program below, we see that if an amount of antivenom such that the initial concentration in the blood is 7e-9 M, the patient will die.

See Polymath program P6-4-b.pol.

6-9

POLYMATH Results Calculated values of the DEQ variables Variable t fsv fs Cv Ca fsa Cp kv ksv ka kia Cso ksa kp kov koa kop g h m j

initial value 0 0 1 0 7.0E-09 0 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -2.1E-09

minimal value 0 0 0.6655661 0 4.503E-09 0 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -2.1E-09

maximal value 0.5 0 1 0 7.0E-09 0.3344339 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -1.351E-09

ODE Report (STIFF) Differential equations as entered by the user [1] d(fsv)/d(t) = kv * fs * Cv - ksv * fsv * Ca [2] d(fs)/d(t) = -kv*fs*Cv - ka * fs * Ca + kia * fsa + g [3] d(Cv)/d(t) = Cso * (-kv * fs * Cv - ksa * fsa * Cv) + h [4] d(Ca)/d(t) = Cso*(-ka * fs * Ca + kia * fsa) + j [5] d(fsa)/d(t) = ka * fs * Ca - kia * fsa - ksa * fsa * Cv [6] d(Cp)/d(t) = Cso * (ksv * fsv * Ca + ksa * fsa * Cv) + m Explicit equations as entered by the user [1] kv = 2e8 [2] ksv = 6e8 [3] ka = 2e8 [4] kia = 1 [5] Cso = 5e-9 [6] ksa = 6e8 [7] kp = 1.2e9 [8] kov = 0 [9] koa = 0.3 [10] kop = 0.3 [11] g = ksa * fsa * Cv + ksv * fsv * Ca [12] h = -kp * Cv * Ca - kov * Cv [13] m = kp * Cv * Ca - kop * Cp [14] j = -Cso * ksv * fsv * Ca - kp * Cv * Ca - koa * Ca

6-10

final value 0.5 0 0.6655661 0 4.503E-09 0.3344339 0 2.0E+08 6.0E+08 2.0E+08 1 5.0E-09 6.0E+08 1.2E+09 0 0.3 0.3 0 0 0 -1.351E-09

P6-4 (c) The latest time after being bitten that antivenom can successfully be administerd is 27.49 minutes. See the cobra web module on the CDROM/website for a more detailed solution to this problem

P6-4 (d) Individualized Solution P6-5 (a) Plot of CA, CD and CU as a function of time (t): See Polymath program P6-5-a.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cd Cu k1 k2 K1a K2a Cao X

initial value 0 1 0 0 1 100 10 1.5 1 0

minimal value 0 0.0801802 0 0 1 100 10 1.5 1 0

maximal value 15 1 0.7995475 0.5302179 1 100 10 1.5 1 0.9198198

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a)) [2] d(Cd)/d(t) = k1*(Ca-Cd/K1a) [3] d(Cu)/d(t) = k2*(Ca-Cu/K2a)

6-11

final value 15 0.0801802 0.7995475 0.1202723 1 100 10 1.5 1 0.9198198

Explicit equations as entered by the user [1] k1 = 1.0 [2] k2 = 100 [3] K1a = 10 [4] K2a = 1.5 [5] Cao = 1 [6] X = 1-Ca/Cao

To maximize CD stop the reaction after a long time. The concentration of D only increases with time

P6-5 (b) Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)

P6-5 (c) Equilibrium concentrations: CAe = 0.08 mol/dm3 CDe = 0.8 mol/dm3 CUe = 0.12 mol/dm3

P6-5 (d) See Polymath program P6-5-d.pol.

POLYMATH Results NLES Solution Variable Ca Cd Cu Ca0 k1 k2 K1a K2a t

Value 0.0862762 0.7843289 0.1293949 1 1 100 10 1.5 100

f(x) -3.844E-14 -2.631E-14 6.478E-14

Ini Guess 1 0 0

6-12

NLES Report (safenewt) Nonlinear equations [1] f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0 [2] f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0 [3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0 !

1 min

10 min

100min

Explicit equations

CAexit

0.295

0.133

0.0862

[1] Ca0 = 1 [2] k1 = 1 [3] k2 = 100 [4] K1a = 10 [5] K2a = 1.5 [6] t = 100 [7] X = 1-Ca/Ca0

CDexit

0.2684

0.666

0.784

CUexit

0.436

0.199

0.129

X

0.705

0.867

0.914

P6-6 (a)

A! X

1/ 2 rX = k1C A

A! B

rB = k 2C A

A!Y

rY = k3C A

2

1/ 2

& mol # k1 = 0.004$ 3 ! % dm " k 2 = 0.3 min !1 k3 = 0.25

Sketch SBX, SBY and SB/XY as a function of CA See Polymath program P6-6-a.pol.

1)

SB/X =

rB k C k = 2 1A/ 2 = 2 C A1 / 2 rX k1C A k1

2)

SB/Y =

rB k C k = 2 A2 = 2 rY k 3C A k 3C A

6-13

dm 3 mol. min

min

3)

SB/XY =

rB k 2C A = 1 rX + rY k1C A / 2 + k 3C A 2

P6-6 (b) Volume of first reactor can be found as follows We have to maximize SB/XY From the graph above, maximum value of SBXY = 10 occurs at CA* = 0.040 mol/dm3 So, a CSTR should be used with exit concentration CA* Also, CA0 = PA/RT = 0.162 mol/dm3 1/ 2

And ! rA = rX + rB + rY = ( k1C A

=> V =

+ k 2 C A + k 3C A 2 )

v0 (C A0 ! C A* ) v0 (C A0 ! C A* ) = = 92.4dm 3 * * * 1 / 2 2 ! rA (k1 (C A ) + k 2 C A + k 3 (C A ) )

P6-6 (c) Effluent concentrations:

CB CB mol = ! C B * = 0.11 3 rB k 2C A dm mol mol * CY * = 0.0037 3 Similarly: C X = 0.007 and 3 dm dm We know, τ = 9.24 min => " =

P6-6 (d) 6-14

Conversion of A in the first reactor:

C A0 " C A = C A0 X ! X = 0.74

P6-6 (e) A CSTR followed by a PFR should be used.

Required conversion = 0.99 => For PFR, Mole balance:

dV FA0 = dX ! rA

0.99

" V = 10 ! 0.162 ! #

dX

1/ 2 + k 2 C A + k 3C A 2 ) 0.74 ( k1C A

= 92.8dm 3

P6-6 (f) If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B, decrease as temperature drops. So we have to compromise between high selectivity and production. To do this we need expressions for k1, k2, and k3 in terms of temperature. From the given data we know:

" ! Ei # ki = Ai exp $ % & 1.98T '

Since we have the constants given at T = 300 K, we can solve for Ai.

.004

A1 =

= 1.49e12 ! #20000 " exp $$ %% & 1.98 (300 ) ' .3 A2 = = 5.79e6 ! #10000 " exp $$ %% & 1.98 (300 ) ' .25 A3 = = 1.798e21 ! #30000 " exp $$ %% & 1.98 (300 ) ' Now we use a mole balance on species A

V= V=

FA0 ! FA !rA v (C A0 ! C A )

!rA C " CA C A0 " C A ! = A0 = 0.5 "rA k1C A + k2C A + k3C A2 A mole balance on the other species gives us:

Fi = vCi = rV i Ci = ! ri

6-15

Using these equations we can make a Polymath program and by varying the temperature, we can find a maximum value for CB at T = 306 K. At this temperature the selectivity is only 5.9. This may result in too much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature will depend on the price of B and the cost of removing X and Y, but without actual data, we can only state for certain that the optimal temperature will be equal to or less than 306 K. See Polymath program P6-6-f.pol.

POLYMATH Results NLE Solution Variable Ca T R k1 k2 Cao Cb k3 tau Cx Cy Sbxy

Value 0.0170239 306 1.987 0.0077215 0.4168076 0.1 0.070957 0.6707505 10 0.0100747 0.0019439 5.9039386

f(x) 3.663E-10

Ini Guess 0.05

NLE Report (safenewt) Nonlinear equations [1] f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0

Explicit equations [1] T = 306 [2] R = 1.987 [3] k1 = 1.49e12*exp(-20000/R/T) [4] k2 = 5790000*exp(-10000/R/T) [5] Cao = .1 [6] Cb = 10*k2*Ca [7] k3 = 1.798e21*exp(-30000/R/T) [8] tau = 10 [9] Cx = tau*k1*Ca^.5 [10] Cy = tau*k3*Ca^2 [11] Sbxy = Cb/(Cx+Cy)

P6-6 (g) Concentration is proportional to pressure in a gas-phase system. Therefore:

S B / XY

P which would suggest that a low pressure would be ideal. But as before the P + P2

tradeoff is lower production of B. A moderate pressure would probably be best.

P6-7 US legal limit: 0.8 g/l Sweden legal limit: 0.5 g/l k1 k2 A !! " B !!

"C

Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream 6-16

dC A = !k1C A dt dCB = k1C A ! k2 dt k1 = 10 hr !1 g k2 = 0.192 L hr Two tall martinis = 80 g of ethanol Body fluid = 40 L

C A0 =

80 g g =2 40 L L

Now we can put the equations into Polymath.

See Polymath program P6-7.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 2 0 10 0.192

minimal value 0 7.131E-44 0 10 0.192

maximal value 10 2 1.8901533 10 0.192

final value 10 7.131E-44 0.08 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

P6-7 (a) In the US the legal limit it 0.8 g/L. This occurs at t = 6.3 hours..

P6-7 (b) In Sweden CB = 0.5 g/l , t = 7.8 hrs.

P6-7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs P6-7 (d) For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L. Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half an hour.

At a half an hour CA = 0.00674 g/L and CB = 0.897 g/L. The Polymath code for after the second drink is shown below.

6-17

See Polymath program P6-7-d.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0.5 1.0067379 0.8972621 10 0.192

minimal value 0.5 5.394E-42 0.08 10 0.192

maximal value 10 1.0067379 1.8069769 10 0.192

final value 10 5.394E-42 0.08 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

for the US t = 6.2 hours Sweden: t = 7.8 hours Russia: t =10.3 hours.

P6-7 (e) The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g of ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the rate of change in concentration due to the incoming ethanol is 2 g/L/hr.

For the first hour the differential equation for CA becomes:

dC A = !k1C A + 2t after that it reverts back to the original equations. dt See Polymath program P6-7-e.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2

initial value 0 0 0 10 0.192

minimal value 0 0 -1.1120027 10 0.192

maximal value 11 0.1785514 0.7458176 10 0.192

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = if(t<1)then(-k1*Ca+2*t)else(-k1*Ca) [2] d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.192

US: CB never rises above 0.8 g/L so the is no time that it would be illegal.

Sweden: t = 2.6 hours 6-18

final value 11 6.217E-45 -1.1120027 10 0.192

Russia: t = 5.2 hours

P6-7 (f) 60 g of ethanol immediately  CA = 1.5 g/L CB = 0.8 g/L at 0.0785 hours or 4.71 minutes.

So the person has about 4 minutes and 40 seconds to get to their destination.

P6-7 (g) A heavy person will have more body fluid and so the initial concentration of CA would be lower. This means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They will take longer to reach the legal limit, as their initial concentration will be higher.

P6-8 (a)

See Polymath program P6-8-a.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb k1 k2 k4 k3

initial value 0 6.25 0 0.15 0.6 0.2 0.1

minimal value 0 0.3111692 0 0.15 0.6 0.2 0.1

maximal value 4 6.25 0.5977495 0.15 0.6 0.2 0.1

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1*Ca-k2*Ca [2] d(Cb)/d(t) = k1*Ca-k3-k4*Cb Explicit equations as entered by the user [1] k1 = 0.15 [2] k2 = 0.6

6-19

final value 4 0.3111692 0.4057018 0.15 0.6 0.2 0.1

[3] k4 = 0.2 [4] k3 = if(Cb<0)then (k1*Ca-k4*Cb) else (0.1)

P6-8 (b)

P6-8 (c) If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six hours. Doing so will result in build up of the drug in the bloodstream that can cause harmful effects.

6-20

P6-8 (d) If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also drag the drug to the intestines and may limit its effectiveness. This effect can be seen in the adsorption constant k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is slow and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the concentration profiles for k1 = 0.10 h-1 and k2 = 0.8 h-1. Note that the maximum amount of the drug in the bloodstream is reduced by two.

Concentration Profiles Ca (mg/dm3) Cb (mg/dm3)

3

0.6

6

0.4

4 0.2

2 0

CB (mg/dm3)

CA (mg/dm )

8

0 0

2

4

6

8

10

12

Time (h) Concentration profile for Tarzlon in the stomach (A) and bloodstream (B). The maximum amount of Tarzlon in the bloodstream is 0.3 mg/dm3.

P6-9 (a) Reactor selection

A+ B!D A + B !U

rD = !r1 A

r1 A = 10 exp(!8000 K / T )C AC B

rU = !r2 A

r2 A = 100 exp(!1000 K / T )C A C B

1/ 2

3/ 2

1/ 2

r 10 exp(!8000 K / T )C A C B exp(!8000 K / T )C A SDU = D = = 1/ 2 3/ 2 1/ 2 rU 100 exp(!1000 K / T )C A C B 10 exp(!1000 K / T )C B At T = 300K 1/ 2

k1 = 2.62 x 10-11 &

k2 = 3.57

SD/U =

7.35 "10 !12 C A 1/ 2 CB

At T = 1000K k1 = 3.35 x 10

-3

&

k2 =36.78

SD/U =

9.2 " 10 !5 C A CB

1/ 2

1/ 2

Hence In order to maximize SDU, use higher concentrations of A and lower concentrations of B. This can be achieved using: 1) A semibatch reactor in which B is fed slowly into a large amount of A 2) A tubular reactor with side streams of B continually fed into the reactor 3) A series of small CSTR’s with A fed only to the first reactor and small amounts of B fed to each reactor.

6-21

Also, since ED > EU, so the specific reaction rate for D increases much more rapidly with temperature.

Consequently, the reaction system should be operated at highest possible temperature to maximize SDU. 1

"C # 2 !6 Note that the selectivity is extremely low, and the only way to increase it is to keep $ B % < 10 and & CA ' add B drop by drop.

P6-9 (b) A+ B!D

and

r1 A = 100 exp(!1000 K / T )C AC B

A + B !U

and

r2 A = 10 6 exp(!8000 K / T )C AC B

SDU =

rD 100 exp(!1000 K / T )C AC B exp(!1000 K / T ) = 6 = 4 rU 10 exp(!8000 K / T )C AC B 10 exp(!8000 K / T )

At T = 300K k1 = 3.57 & k2 = 2.623 SDU = 1.14 x106 At T = 1000K k1 = 36.78 & k2 = 3354.6 SDU = 0.103 Hence we should keep the temperature low to maximize SDU but not so low that the desired reaction doesn’t proceed to a significant extent.

P6-9 (c) A + B ! D and

r1 A = 10 exp(!8000 K / T )C ACB

B + D !U

r2 A = 109 exp(!10, 000 K / T )CB CD

and

10 exp (!8000 K / T )C ACB r S DU = 1 A = 9 r2 A 10 exp (!10000 K / T )CB CD S DU =

exp (!8000 / T )C A 8

10 exp (!10000 / T )CD

Therefore the reaction should be run at a low temperature to maximize SDU, but not too low to limit the production of desired product. The reaction should also take place in high concentration of A and the concentration of D should be limited by removing through a membrane or reactive distillation.

P6-9 (d) A ! D and D ! U 1 and A ! U 2 and

r1 A = 4280 exp(!12000 K / T )C A r2 D = 10,100 exp(!15000 K / T )C D

r3 A = 26 exp(!10800 K / T )C A

6-22

S D / U 1U 2 =

rD 4280 exp(!12000 K / T )C A ! 10,100 exp(!15000 K / T )C D = rU 1 + rU 2 10,100 exp(!15000 K / T )C D + 26 exp(!10800 K / T )C A

At T = 300K k1 = 1.18 x 10-14 &

k2 = 1.94 X 10-18 & k3 = 6.03 x 10-15

If we keep CA > 1000CD

S D / U 1U 2 =

1.18 " 10 !14 C A ! 1.94 " 10 !18 C D 1.18 ! = 1.96 1.94 " 10 !18 C D + 6.03 " 10 !15 C A .603

At T = 1000K k1 = 0.026 &

k2 = 3.1 X 10-3 & k3 = 5.3 x 10-4

If we keep CA > 1000CD

S D / U 1U 2 =

0.026C A ! 3.1 " 10 !3 C D .026 ! = 49 !3 !4 3.1 " 10 C D + 5.3 " 10 C A .00053

Here, in order to lower U1 use low temperature and high concentration of A But low temperature and high concentration of A favours U2 So, well have to optimize the temperature and concentration of A. Membrane reactor in which D is diffusing out can be used.

P6-9 (e) A+ B!D D! A+ B A + B !U

and

r1 A = 10 9 exp(!10000 K / T )C A C B

and

r2 D = 20 exp(!2000 K / T )C D

and

r3 A = 10 3 exp(!3000 K / T )C AC B

rD 10 9 exp(!10000 K / T )C A C B ! 20 exp(!2000 K / T )C D S D /U = = rU 10 3 exp(!3000 K / T )C A C B At T = 300K k1 = 3.34 x 10-6 &

k2 = 0.025

& k3 = 0.045

The desired reaction lies very far to the left and CD is probably present at very low concentrations so that:

3.34 " 10 !6 C A C B ! 0.025C D 0.000334 ! = 0.000074 4.5 0.045C A C B S D /U ! 0 S D /U =

At T = 1000K k1 = 45399.9 & k2 = 2.7 & k3 = 49.7 If we assume that CACB > 0.001CD then,

S D /U =

45399.9C A C B ! 2.7C D 45399 ! = 913 49.7 49.7C A C B

Here we need a high temperature for a lower reverse reaction of D and lower formation of U Also we need to remove D as soon as it is formed so as to avoid the decomposition.

6-23

P6-9 (f) A+ B!D

and

!r1 A = 10 exp(!8000 K / T )C ACB

A !U

and

!r2 A = 26 exp(!10,800 K / T )C A

!r3U = 1000 exp(!15, 000 K / T )CU and U!A r 10 exp(!8000 K / T )C ACB S D /U = D = rU 26 exp(!10,800 K / T )C A ! 1000 exp (!15, 000 K / T )CU We want high concentrations of B and U in the reactor. Also low temperatures will help keep the selectivity high.

If we use pseudo equilibrium and set –rA = 0.

" !8000 # " !10800 # " !15000 # !rA = 10 exp $ % C ACB + 26 exp $ % C A ! 1000 exp $ % CU = 0 & T ' & T ' & T ' " !8000 # " !10800 # 10 exp $ CB + 26 exp $ % % CA T ' T & & ' = CU " !15000 # 1000 exp $ % & T ' CA 1 26 ! 7000 " ! 4200 " = exp # exp # $ CB + $ CU 100 1000 % T & % T &

P6-9 (g) A+ B!D

and

!r1 A = 800 exp(!8000 K / T )C A0.5CB

A + B ! U1

and

!r2 B = 10 exp(!300 K / T )C ACB

D + B ! U2

and

!r3 D = 106 exp(!8000 K / T )CD CB

(1)

S D / U1 =

800 exp (!8000 / T )C A0.5CB 10 exp (!300 / T )C ACB

=

80 exp (!8000 / T ) exp (!300 / T )C A0.5

At T = 300

S D / U1 =

2.098*10!10 0.368C A0.5

At T = 1000

S D / U1 =

29.43 0.7408C A0.5

To keep this selectivity high, low concentrations of A, and high temperatures should be used.

800 exp (!8000 / T )C A0.5CB

800C A0.5 S D /U 2 = 6 = 6 10 exp (!8000 / T )CD CB 10 CD To keep this selectivity high, high concentrations of A and low concentrations of D should be used. Try to remove D with a membrane reactor or reactive distillation. The selectivity is not dependant on temperature.

6-24

To keep optimize the reaction, run it at a low temperature to maximized SD/U1 in a membrane reactor that allows only D to diffuse out.

(2)

S D / U 1U 2 = S D / U 1U 2 =

800 exp (!8000 / T )C A0.5CB 10 exp (!300 / T )C ACB + 106 exp (!8000 / T )CD CB 800 exp (!8000 / T )C A0.5 10 exp (!300 / T )C A + 106 exp (!8000 / T )CD

At T = 300

2.09*10!9 C A0.5 S D / U 1U 2 = "0 3.67C A + 2.62*10!6 CD At T = 1000 and very low concentrations of D

S D / U 1U 2 =

0.268C A0.5 .03617 = 7.408C A + 335CD C A0.5

If temperature is the only parameter that can be varied, then the highest temperature possible will result in the highest selectivity. Also removing D will help keep selectivity high.

P6-9 (h) No solution will be given P6-9 (i) rD exp(!7000 K / T )C A1/ 2 = rU 10CB1/ 2 dFA dFB = rA = rB + RB dV dV dFU dFD = rU = rD dV dV F RB = A0 VT rD = !r1 A A+ B!D

A + B !U rA = rB = r1 A + r2 A F C A = CT 0 A FT mol dm3 FA0 = CT 0 v0

CT 0 = 0.4

r2 A = !100 exp(!1000 K / T )C A1/ 2CB 3/ 2

rU = !r2 A

CB = CT 0

v0 = 10

r1 A = !10 exp(!8000 K / T )C ACB

FB FT

dm3 s

These equations are entered into Polymath and the plots below are for the membrane reactor. The code can be modified to compare with the PFR results.

See Polymath program P6-9-i.pol.

6-25

POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fd Fu Cto T Ft Cb Ca r1a rd r2a ra rb ru Vt Fao Rb Sdu

initial value 0 4 0 0 0 0.4 600 4 0 0.4 0 0 0 0 0 0 5 4 0.8 5.423E+05

minimal value 0 0.5141833 0 0 0 0.4 600 4 0 0.0241566 -4.575E-07 0 -0.4461944 -0.4461948 -0.4461948 0 5 4 0.8 2.894E-07

maximal value 10 4 4.5141833 3.034E-06 3.4858137 0.4 600 8.5141833 0.2120783 0.4 0 4.575E-07 0 0 0 0.4461944 5 4 0.8 5.423E+05

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb+Rb [3] d(Fd)/d(V) = rd [4] d(Fu)/d(V) = ru Explicit equations as entered by the user [1] Cto = .4 [2] T = 600 [3] Ft = Fa+Fb+Fd+Fu [4] Cb = Cto*Fb/Ft [5] Ca = Cto*Fa/Ft [6] r1a = -10*exp(-8000/T)*Ca*Cb [7] rd = -r1a [8] r2a = -100*exp(-1000/T)*Ca^.5*Cb^1.5 [9] ra = r1a+r2a [10] rb = ra [11] ru = -r2a [12] Vt = 5 [13] Fao = 4 [14] Rb = Fao/Vt [15] Sdu = exp(-7000/T)*Ca^.5/(10*Cb^.5+.00000000001)

6-26

final value 10 0.5141833 4.5141833 3.034E-06 3.4858137 0.4 600 8.5141833 0.2120783 0.0241566 -8.297E-08 8.297E-08 -0.2867066 -0.2867066 -0.2867066 0.2867066 5 4 0.8 2.894E-07

P6-9 (j) No solution will be given P6-9 (k) No solution will be given P6-9 (l) No solution will be given P6-10 (a)

6-27

P6-10 (b) For CSTR, ! = 0.5h First calculate k1 and k2 :

! E ! 1 1 "" k = k0 exp $ $ # % % $R T T % '' & & 0 ! 10, 000 ! 1 1 "" k1 = 0.4 exp $ # %% $ & R & 373 T ' '

! 20, 000 ! 1 1 "" k2 = 0.01exp $ # %% $ & R & 373 T ' ' See Polymath program P6-10-b-nle.pol or P6-10-b-DEQ.pol.

(1) Non-linear equation solver method POLYMATH Results No Title 03-06-2006, Rev5.1.233 NLES Solution Variable Ca Cb Cc tau T E1 R k1 E2 k2 ra rb rc

Value 0.6236527 0.7473796 0.2289677 0.5 440.12 10000 1.987 3.1310605 2.0E+04 0.6127213 -1.9526945 1.4947592 0.4579354

f(x) 4.693E-10 -2.086E-09 0

Ini Guess 1.5 0 0

NLES Report (safenewt) Nonlinear equations [1] f(Ca) = tau*(-ra)-(1.6-Ca) = 0 [2] f(Cb) = tau*(rb)-Cb = 0 [3] f(Cc) = tau*(rc)-Cc = 0

Explicit equations [1] tau = .5 [2] T = 440.12 [3] E1 = 10000

6-28

[4] R = 1.987 [5] k1 = 0.4*exp(E1/R*(1/373-1/T)) [6] E2 = 20000 [7] k2 = 0.01*exp(E2/R*(1/373-1/T)) [8] ra = -k1*Ca [9] rb = k1*Ca-k2*Cb [10] rc = k2*Cb

(2) Differential equation solver method POLYMATH Results No Title 03-06-2006, Rev5.1.233 Calculated values of the DEQ variables Variable t T tau R E2 k2 E1 k1 cao ca ra cb rc rb cc

initial value 0 200 0.5 1.987 2.0E+04 7.289E-13 10000 3.415E-06 1.6 1.5999973 -5.464E-06 2.732E-06 1.991E-18 5.464E-06 9.957E-19

minimal value 0 200 0.5 1.987 2.0E+04 7.289E-13 10000 3.415E-06 1.6 0.0145145 -3.1709711 2.732E-06 1.991E-18 5.464E-06 9.957E-19

maximal value 500 700 0.5 1.987 2.0E+04 2983.0609 10000 218.46962 1.6 1.5999973 -5.464E-06 0.7495133 3.1688465 1.4990265 1.5844232

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 1 Explicit equations as entered by the user [1] tau = 0.5 [2] R = 1.987 [3] E2 = 20000 [4] k2 = 0.01*exp(E2/R*(1/373-1/T)) [5] E1 = 10000 [6] k1 = 0.4*exp(E1/R*(1/373-1/T)) [7] cao = 1.6 [8] ca = cao/(1+tau*k1) [9] ra = -k1*ca [10] cb = tau*k1*ca/(1+tau*k2) [11] rc = k2*cb [12] rb = k1*ca-k2*cb [13] cc = tau*rc

6-29

final value 500 700 0.5 1.987 2.0E+04 2983.0609 10000 218.46962 1.6 0.0145145 -3.1709711 0.0010623 3.1688465 0.0021246 1.5844232

0.8 0.7

CB [mol/dm 3]

0.6

Therefore, CB is maximum at T=440K.

0.5 0.4 0.3 0.2 0.1 0.0 200

250

300

350

400

450

T [K]

P6-10 (c)

6-30

500

550

600

650

700

P6-10 (d)

P6-10 (e)

6-31

P6-11 (a) Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCl2 catalyst reacted with K-terephthalate in an autocatalytic reaction step: k1 k2 Series A !! " R !!

"S k3 Autocatalytic R + S !! " 2S P 110kPa C AO = = = 0.02mol / dm3 3 RT ! kPa.dm " # 8.314 $ (683K ) mol.K & %

Maximum in R occurs at t = 880 sec.

See Polymath program P6-11-a.pol.

POLYMATH Results Variable t A R S k1 k2 k3

initial value 0 0.02 0 0 0.00108 0.00119 0.00159

minimal value 0 0.003958 0 0 0.00108 0.00119 0.00159

maximal value 1500 0.02 0.0069892 0.0100382 0.00108 0.00119 0.00159

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 1.08e-3 [2] k2 = 1.19e-3 [3] k3 = 1.59e-3

P6-11 (b) 1) T = 703 K CAO = 0.019 mol/dm3

! E ! 1 1 "" k1' = k1 exp $ 1 $ # ' % % & R & T T ''

6-32

final value 1500 0.003958 0.005868 0.0100382 0.00108 0.00119 0.00159

" (42600cal / mol ) " 1 1 ## !3 !1 k1' = (1.08 $10!3 s !1 ) exp % ! % & & = 2.64 $10 s ' (1.987cal / mol.K ) ' 683K 703K ( (

Similarly,

k2' = 3.3 "10!3 s !1 '

And, k3 = 3.1" 10

!3

dm3 / mol.s

Maxima in R occurs at around t =320 sec.

See Polymath program P6-11-b1.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 0.00264 0.0033 0.0031

minimal value 0 3.622E-04 0 0 0.00264 0.0033 0.0031

maximal value 1500 0.019 0.0062169 0.0174625 0.00264 0.0033 0.0031

final value 1500 3.622E-04 8.856E-04 0.0174625 0.00264 0.0033 0.0031

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 2.64e-3 [2] k2 = 3.3e-3 [3] k3 = 3.1e-3

2) T = 663 K CAO = 0.19 mol/dm3

" (42600cal / mol ) " 1 1 ## !3 !1 k1' = (1.08 $10!3 s !1 ) exp % ! % & & = 0.42 $10 s ' (1.987cal / mol.K ) ' 683K 663K ( ( ' !3 !1 k2 = 0.4 "10 s k3' = 0.78 "10!3 dm3 / mol.s See Polymath program P6-11-b2.pol.

POLYMATH Results Calculated values of the DEQ variables Variable t A R S k1 k2 k3

initial value 0 0.019 0 0 4.2E-04 4.0E-04 7.8E-04

minimal value 0 2.849E-04 0 0 4.2E-04 4.0E-04 7.8E-04

maximal value 10000 0.019 0.0071414 0.016889 4.2E-04 4.0E-04 7.8E-04

6-33

final value 10000 2.849E-04 0.0012573 0.016889 4.2E-04 4.0E-04 7.8E-04

ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1*A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 0.42e-3 [2] k2 = 0.4e-3 [3] k3 = 0.78e-3 Independent variable variable name : t initial value : 0 final value : 10000

Maxima in R occurs around t = 2500 sec.

P6-11 (c) Use the Polymath program from part (a) and change the limits of integration to 0 to 1200. We get: CAexit = 0.0055 mol/dm3 CRexit = 0.0066 mol/dm3 CSexit = 0.0078 mol/dm3

P6-12 (a)

P6-12 (b)

P6-12 (c)

P6-12 (d)

P6-12 (e)

P6-12 (f)

P6-12 (g)

6-34

P6-12 (h) Mole balance:

C AO " C A = (" rA )!

C C = (rC )!

C D = (rD )!

Rate law:

Solving in polymath:

1 & # rA = ' $k1 A C A + k 2 D C A C C2 !

3 % " 2 &1 # rC = $ k1 A C A ' k 2 D C A C C2 ' k 3 E C C C D !

3 %3 " 4 & # rD = $k 2 D C A C C2 ' k 3 E C C C D !

3 % " C A = 0.0069 M C B = 0.96 M C C = 0.51M C D = 0.004 M SB/D = rB/r D = 247

SB/C = 1.88

See Polymath program P6-12-h.pol.

POLYMATH Results NLES Solution Variable Ca Cb Cc Cd Ce kd ka rb ra ke rc rd re tau Cao

Value 0.0068715 0.9620058 0.5097027 0.0038925 0.2380808 3 7 0.0160334 -0.0498855 2 0.008495 6.488E-05 0.003968 60 3

f(x) -2.904E-10 -1.332E-15 -1.67E-08 -2.391E-08 1.728E-08

Ini Guess 3 0 0 0 0

NLES Report (safenewt) Nonlinear equations [1] f(Ca) = Cao-Ca+ra*tau = 0 [2] f(Cb) = Cb - rb*tau = 0 [3] f(Cc) = Cc-rc*tau = 0

6-35

[4] f(Cd) = Cd-rd*tau = 0 [5] f(Ce) = Ce - re*tau = 0

Explicit equations [1] kd = 3 [2] ka = 7 [3] rb = ka*Ca/3 [4] ra = -(ka*Ca+kd/3*Ca*Cc^2) [5] ke = 2 [6] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc [7] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc [8] re = ke*Cd*Cc [9] tau = 60 [10] Cao = 3

P6-12 (i) For PFR and gas phase: Mole balance: Rate law:

Stoichiometry:

dFC dFA dFB = rA = rB = rC dV dV dV 1 & # rA = ' $k1 A C A + k 2 D C A C C2 !

3 % " &1 # rB = $ k1 A C A ! %3 " 1 2 & # rC = $ k1 A C A ' k 2 D C A C C2 ' K 3 E C C C D !

3 %3 " 4 & # rD = $k 2 D C A C C2 ' k 3 E C C C D !

3 % " rE = [k 3 E C C C D ]

F FA y C C = CTO C y FT FT FT = FA + FB + FC + FD + FE

C A = CTO

dFD = rD dV

C D = CTO

dFE = rE dV

FD y FT

dy " ! FT = dV 2 y FTO Plot of CB and CC are overlapping.

See Polymath program P6-12-i.pol.

POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fc Fd Fe y Ft Cto

initial value 0 20 0 0 0 0 1 20 0.2

minimal value 0 9.147E-04 0 0 0 0 0.9964621 13.330407 0.2

maximal value 100 20 6.6638171 6.6442656 0.0201258 0.0043322 1 20 0.2

6-36

final value 100 9.147E-04 6.6638171 6.6442167 0.0171261 0.0043322 0.9964621 13.330407 0.2

Cc ka kd ke Ca rb ra Cd Fto rc rd re alfa X

0 7 3 2 0.2 0.4666667 -1.4 0 20 0.4666667 0 0 1.0E-04 0

0 7 3 2 1.367E-05 3.191E-05 -1.4 0 20 -1.923E-05 -7.012E-05 0 1.0E-04 0

0.0993605 7 3 2 0.2 0.4666667 -9.586E-05 3.0E-04 20 0.4666667 8.653E-04 5.908E-05 1.0E-04 0.9999543

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc [4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(y)/d(V) = -alfa*Ft/(2*y*Fto) Explicit equations as entered by the user [1] Ft = Fa+Fb+Fc+Fd+Fe [2] Cto = 0.2 [3] Cc = Cto*Fc/Ft*y [4] ka = 7 [5] kd = 3 [6] ke = 2 [7] Ca = Cto*Fa/Ft*y [8] rb = ka*Ca/3 [9] ra = -(ka*Ca+kd/3*Ca*Cc^2) [10] Cd = Cto*Fd/Ft*y [11] Fto = 0.2*100 [12] rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc [13] rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc [14] re = ke*Cd*Cc [15] alfa = 0.0001 [16] X = 1-Fa/20

6-37

0.0993325 7 3 2 1.367E-05 3.191E-05 -9.586E-05 2.56E-04 20 -1.923E-05 -6.742E-05 5.087E-05 1.0E-04 0.9999543

P6-12 (j) Changes in equation from part (i): dFC = rC ! RC dV

RC = k diffuse C C

k diffuse = 2 min !1

See Polymath program P6-12-j.pol.

P6-13 (a) m-xylene → benzene + methane A → B + M m-xylene → p-xylene A → P See Polymath program P6-13-a.pol.

POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fp Fm Fi Ft k1 k2

initial value 0 75 0 0 0 25 100 0.22 0.71

minimal value 0 6.1072984 0 0 0 25 100 0.22 0.71

maximal value 6000 75 16.297198 52.595503 16.297198 25 116.2972 0.22 0.71

6-38

final value 6000 6.1072984 16.297198 52.595503 16.297198 25 116.2972 0.22 0.71

Cto Ca r1 Cb r2 Cp Spb tau Y X

0.05 0.0375 -0.00825 0 -0.026625 0 0 0 0 0

0.05 0.0026257 -0.00825 0 -0.026625 0 0 0 0 0

0.05 0.0375 -5.777E-04 0.0070067 -0.0018643 0.0226125 3.2272267 3 0.7634409 0.9185694

ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = r1+r2 [2] d(Fb)/d(V) = -r1 [3] d(Fp)/d(V) = -r2 Explicit equations as entered by the user [1] Fm = Fb [2] Fi = 25 [3] Ft = Fa+Fb+Fp+Fm+Fi [4] k1 = .22 [5] k2 = .71 [6] Cto = .05 [7] Ca = Cto*Fa/Ft [8] r1 = -k1*Ca [9] Cb = Cto*Fb/Ft [10] r2 = -k2*Ca [11] Cp = Cto*Fp/Ft [12] Spb = Cp/(Cb+.0000001) [13] tau = V/2000 [14] Y = Fp/(75.00000001-Fa) [15] X = (75-Fa)/75

a τ = 2.8 is necessary to achieve 90% conversion

P6-13 (b) CSTR Mole Balances:

FA = FA0 + rAV

FB = rBV

FM = rM V

FP = rPV 6-39

0.05 0.0026257 -5.777E-04 0.0070067 -0.0018643 0.0226125 3.2272267 3 0.7634409 0.9185694

C A = C A0 + rA!

C B = rB!

C M = rM !

C P = rP!

Rate Laws:

rA = !(k1 + k 2 )C A

rB = rM = k1C A

rP = k 2 C A

&E , 1 1 )# k1 = k1@ 946 K exp $ 1 * - '! % R + 946 T (" &E , 1 1 )# k 2 = k 2 @ 946 K exp $ 2 * - '! % R + 946 T (" Combining:

C A = C A0 " (k1 + k 2 )!C A CA =

C A0 1 + (k1 + k 2 )!

C P = k 2!C A =

k 2!C A0 1 + (k1 + k 2 )!

Using these equations and Polymath we find the optimal temperature is 878.6 K. The maximum concentration of p-xylene is 0.0264 mol/dm3 See Polymath program P6-13-b.pol.

POLYMATH Results No Title 03-06-2006, Rev5.1.233 Calculated values of the DEQ variables Variable t T E2 E1 R k2 tau cao k1 ca cb cm cp

initial value 0 500 10000 2.0E+04 1.987 0.0061708 10 0.0375 1.662E-05 0.0353149 5.869E-06 5.869E-06 0.0021792

minimal value 0 500 10000 2.0E+04 1.987 0.0061708 10 0.0375 1.662E-05 3.403E-04 5.869E-06 5.869E-06 0.0021792

maximal value 9000 1400 10000 2.0E+04 1.987 3.9856769 10 0.0375 6.9328239 0.0353149 0.0235949 0.0235949 0.0264436

6-40

final value 9000 1400 10000 2.0E+04 1.987 3.9856769 10 0.0375 6.9328239 3.403E-04 0.0235949 0.0235949 0.0135647

ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 0.1 Explicit equations as entered by the user [1] E2 = 10000 [2] E1 = 20000 [3] R = 1.987 [4] k2 = 0.71*exp(E2/R*(1/946-1/T)) [5] tau = 10 [6] cao = 0.05*0.75 [7] k1 = 0.22*exp(E1/R*(1/946-1/T)) [8] ca = cao/(1+tau*(k1+k2)) [9] cb = tau*k1*ca [10] cm = tau*k1*ca [11] cp = tau*k2*ca 0.03

0.025

3

CP [mol/dm ]

0.02

0.015

0.01

0.005

0 400

600

800

1000

T [K]

P6-14(a)

6-41

1200

1400

1600

See Polymath program P6-14-a.pol.

Calculated values of the DEQ variables Variable V Ca Cb Cc Cd Ce Cf vo kf3 ke2 kd1 rf3 rd1 re2 rf re rd ra rb rc

initial value 0 1.5 2 0 0 0 0 10 5 0.1 0.25 0 1.5 0 0 0 1.5 -1.5 -3 1.5

minimal value 0 0.3100061 0.3384816 0 0 0 0 10 5 0.1 0.25 0 0.0088793 0 0 0 -0.0128612 -1.5 -3 -0.0490933

maximal value 50 1.5 2 0.1970387 0.6751281 0.1801018 0.3621412 10 5 0.1 0.25 0.2569483 1.5 0.0523431 0.2569483 0.0523431 1.5 -0.0694818 -0.0365984 1.5

ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(V) = ra/vo [2] d(Cb)/d(V) = rb/vo [3] d(Cc)/d(V) = rc/vo [4] d(Cd)/d(V) = rd/vo [5] d(Ce)/d(V) = re/vo [6] d(Cf)/d(V) = rf/vo Explicit equations as entered by the user [1] vo = 10

6-42

final value 50 0.3100061 0.3384816 0.105508 0.6516263 0.1801018 0.3621412 10 5 0.1 0.25 0.0188398 0.0088793 0.0202008 0.0188398 0.0202008 -0.0126825 -0.0694818 -0.0365984 -0.0085994

[2] kf3 = 5 [3] ke2 = .1 [4] kd1 = 0.25 [5] rf3 = kf3*Cb*Cc^2 [6] rd1 = kd1*Ca*Cb^2 [7] re2 = ke2*Ca*Cd [8] rf = rf3 [9] re = re2 [10] rd = rd1-2*re2+rf3 [11] ra = -rd1-3*re2 [12] rb = -2*rd1-rf3 [13] rc = rd1+re2-2*rf3

Concentration [mol/dm 3]

2.5

Ca Cb Cc Cd Ce Cf

2

1.5

1

0.5

0 0

5

10

15

20

25

3

V [dm ]

P6-14 (b)

6-43

30

35

40

45

50

P6-14 (c)

6-44