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Solutions for Chapter 8 ʹ Multiple Reactions
P8-‐1 Individualized solution
P8-‐2 (a) Example 8-‐
For PFR (gas phase with no pressure drop or liquid phase),
A k 1 k 2 C A k 3 C^2 A d
dC (^) W d^ k^1
dC (^) X W d dC W (^) B (^) k 2 C A d dCW Y (^) k 3 C A 2
In PFR with V = 1566 dm 3 we get ʏсsͬǀϬсϳϴϯ
X = 0.958 , S B/XY (instanteneous) = 0.244, and S B/XY (overall) = 0.
ĂůƐŽĂƚʏсϯϱϬ͕^ B/XY (instanteneous) is at its maximum value of 0.
See polymath problem P8-‐2-‐1.pol
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value tau 0 0 783 783 Ca 0.4 0.0166165 0.4 0. Cx Cb 1.0E 07 0 1.0E 0 07 0.07830010.1472919 0.07830010. Cy 1.0E 06 1.0E 06 0.1577926 0. Cao X 0.4 0 0.4 0 0.40.9584587 0.40. k1 1.0E 04 1.0E 04 1.0E 04 1.0E 04 k2 0.0015 0.0015 0.0015 0. k3 Sbxy_over 0.008 0 0.008 0 0.0080.6437001 0.0080. Sbxy_inst 0.4347826 0.2438609 0.8385161 0.
ODE Report (RKF45)
Differential equations as entered by the user [1] d(Ca)/d(tau) = -k1-k2Ca-k3Ca^ [2] d(Cx)/d(tau) = k [3] d(Cb)/d(tau) = k2Ca [4] d(Cy)/d(tau) = k3Ca^ Explicit equations as entered by the user [1] Cao = 0. [2] X = 1-Ca/Cao [3] k1 = 0. [4] k2 = 0. [5] k3 = 0. [6] Sbxy_overall = Cb/(Cx+Cy) [7] Sbxy_instant = k2Ca/(k1+k3Ca^2) Independent variable variable name : tau initial value : 0 final value : 783
(2) PFR -‐ Pressure increased by a factor of 100.
(a) Liquid phase: No change, as pressure does not change the liquid volume appreciably.
(b) Gas Phase:
Now CA0 = P/RT = 100 (P 0_initial)/RT = 0.4 × 100 = 40 mol/dm 3
Running with C A0 = 40 mol/dm 3 , we get:
X = 0.998 , SB/XY (instanteneous) = 0.681 , and SB/XY (overall) = 0.
As a practice, students could also try similar problem with the CSTR.
P8-‐2 (b) Example 8-‐
(a) CSTR: intense agitation is needed, good temperature control.
2 1 1 1
1 0 c c
c k k C B k^ C^ A W W
W
POLYMATH Results
See polymath problem P8-‐2-‐1.pol
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value t 0 0 100 100 T Cao (^3005 3005 4005 ) tau 0.5 0.5 0.5 0. k1o 0.4 0.4 0.4 0. k2o E1 0.01 10 0.01 10 0.01 10 0.01 10 E2 20 20 20 20 R 0.001987 0.001987 0.001987 0. k1 k2 0.40.4 0.40.4 26.5130341757.3524 26.5130341757. Cb 0.6944444 0.0052852 0.8040226 0.
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 =. [3] k1o =. [4] k2o =. [5] E1 = 10 [6] E2 = 20 [7] R =. [8] k1 = k1oexp((-E1/R)(1/T-1/300))
[9] k2 = k1oexp((-E2/R)(1/T-1/300)) [10] Cb = (tauk1Cao)/(tauk2+1)/(tauk1+1)
P8-‐2 (e) Individualized solution
P8-‐2 (f) Individualized solution
P8-‐2 (g) Individualized solution
P8-‐2 (h) Example 8-‐
SD/U Original Problem SD/U P 8 2 h
Membrane Reactor 2.58 1.
PFR 0.666 0.
Doubling the incoming flow rate of species B lowers the selectivity.
(2) The selectivity becomes 6.52 when the first reaction is changed to A+2B Æ D
P8-‐2 (i) CDROM Example
Original Case ʹ CDROM example P8-‐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 P8-2-i.pol.
P8-‐2 (j)
For equal molar feed in hydrogen and mesitylene.
CHO = y HOCTO = (0.5)(0.032)lbmol/ft 3 =0.016 lbmol/ft 3
CMO = 0.016 lbmol/ft 3
Using equations from example, solving in Polymath, we get
From CD-‐ROM example
Polymath code:
See Polymath program P8-2-k.pol.
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess CH 4.783E 05 4.889E 11 1.0E 04 CM CX 0.01343530.0023222 1.047E 119.771E 12 0.0130. tau 0. K1 55. K2 CHo 30.20. CMo 0.
NLES Report (safenewt)
Nonlinear equations
[1] f(CH) = CH-CHo+K1(CMCH^.5+K2CXCH^.5)tau = 0 [2] f(CM) = CM-CMo+K1CMCH^.5tau = 0 [3] f(CX) = (K1CMCH^.5-K2CXCH^0.5)*tau-CX = 0
Explicit equations
[1] tau = 0. [2] K1 = 55. [3] K2 = 30. [4] CHo = 0. [5] CMo = 0.
A plot using different values of W is given.
For W =0.5, the exit concentration are
CH = 4.8 ×10 -‐5^ lbmol/ft 3 CM =0.0134 lbmol/ft 3
CX =0.00232 lbmol/ft 3
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for^ W^ =0.5:
m ole m esitylene reacted
m ole xylene produced C C
C F F Y^ F MO M
X MO M MX X
89
016 0. 0134
00232
The overall selectivity of xylene relative to toluene is:
m ole toluene produced
m ole xylene produced F S^ F T X T^ X
~ (^) / 8. 3
CD ROM example This Question
CH 0.0089 4.8 x 10 5
CM 0.0029 0.
CX 0.0033 0.
W 0.5 0.
YMX 0.41 0.
SX/T 0.7 8.
P8-‐2 (l)
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.
P8-‐2 (m) Individualized solution
P8-‐3 Solution is in the decoding algorithm given with the modules (ICM problem )
P8-‐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.
P8-‐4 (b)
[7] kp = 1.2e [8] kov = 0 [9] koa = 0. [10] kop = 0. [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
P8-‐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
P8-‐4 (d) Individualized Solution
P8-‐5 (a)
Plot of C A , C D and C U as a function of time (t):
See Polymath program P8-5-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t Ca (^01 0) 0.0801802 (^151 15) 0. Cd 0 0 0.7995475 0. Cu k1 01 01 0.5302179 1 0.1202723 1 k2 100 100 100 100 K1a 10 10 10 10 K2a Cao 1.5 1 1.5 1 1.5 1 1.5 1 X 0 0 0.9198198 0.
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) Explicit equations as entered by the user [1] k1 = 1. [2] k2 = 100 [3] K1a = 10 [4] K2a = 1. [5] Cao = 1 [6] X = 1-Ca/Cao
To maximize C D stop the reaction after a long time. The concentration of D only increases with time
P8-‐5 (b)
Conc. Of U is maximum at t = 0.31 min.(C A = 0.53)
P8-‐5 (c)
Equilibrium concentrations:
CAe = 0.08 mol/dm 3
CDe = 0.8 mol/dm 3
CUe = 0.12 mol/dm 3
It shows the table of SBX, S BY and SB/XY in correspondence with values of C A.
1) SB/X = 1 /^2
2 A
A
A
X
B C k
k k C
k C r
r
2) SB/Y =
A^ A
A
Y
B
k C
k k C
k C r
r 3
3) SB/XY = 2
1 1 /^23
A A
A
X Y
B
k C k C
k C r r
r
P8-‐6 (b)
Volume of first reactor can be found as follows
We have to maximize S B/XY
From the graph above, maximum value of S BXY = 10 occurs at CA*^ = 0.040 mol/dm^3
So, a CSTR should be used with exit concentration C A*
Also, C A0 = PA /RT = 0.162 mol/dm 3
And r A r X r B r Y ( k 1 C A^1 /^2 k 2 C Ak 3 C A^2 )
1 *^1 /^22 * 3
V^ v (^ C r^ C ) ( k( C )^ v (^ Ck C^ C k^ ) ( C ) ) 92. 4 d m A A A
A A
A
A A
P8-‐6 (c)
Effluent concentrations:
tĞŬŶŽǁ͕ʏсϵ͘Ϯϰ min => * 3
C r (^) k C C CB 0. (^11) d m m ol A
B
B
W^ B^
Similarly: C (^) X *^0. (^007) d^ m m^ o 3 l and CY^ *^0. (^0037) d^ m m^ o 3 l
P8-‐6 (d)
Conversion of A in the first reactor:
C A 0 C A C A 0 X X 0. 74
P8-‐6 (e)
A CSTR followed by a PFR should be used.
Required conversion = 0.
=> For PFR, Mole balance:
A
A r
F dX
dV
Using these equations we can make a Polymath program and by varying the temperature, we can find a
maximum value for C B 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 P8-6-f.pol.
POLYMATH Results
NLE Solution
Variable Ca 0.0170239Value (^) 3.663Ef(x) 10 Ini Guess0. T 306 R 1. k1 k2 0.00772150. Cao 0. Cb 0. k3 tau 0.6707505 10 Cx 0. Cy Sbxy 0.00194395.
NLE Report (safenewt)
Nonlinear equations
[1] f(Ca) = (Cao-Ca)/(k1Ca^.5+k2Ca+k3*Ca^2)-10 = 0
Explicit equations
[1] T = 306
[2] R = 1.
[3] k1 = 1.49e12exp(-20000/R/T) [4] k2 = 5790000exp(-10000/R/T) [5] Cao =. [6] Cb = 10k2Ca [7] k3 = 1.798e21exp(-30000/R/T) [8] tau = 10 [9] Cx = tauk1Ca^. [10] Cy = tauk3*Ca^ [11] Sbxy = Cb/(Cx+Cy)
P8-‐6 (g)
Concentration is proportional to pressure in a gas-‐phase system. Therefore:
B / XY ~^ A 2
A A
S^ P
P P which would suggest that a low pressure would be ideal. But as before the
tradeoff is lower production of B. A moderate pressure would probably be best.
We know that :
SB/XY = 2
1 1 /^23
A A
A
X Y
B
k C k C
k C r r
r
Substituting P A = C A R T ;
in the above co ʹ relation we get
SB/XY = 2 1/2 2
1 3
( / ) ( / ) ( / )
B A X Y A A
r k P RT r r k P RT k P RT
See Polymath program P8-6-g.pol.
Thus we find that the maximum is obtained at P= 1 atm.
P8-‐7 (b)
In Sweden C B = 0.5 g/l , t = 7.8 hrs.
P8-‐7 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs
P8-‐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 C A = 0.00674 g/L and C B = 0.897 g/L. The Polymath code for after the second drink is
shown below.
See Polymath program P8-7-d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value t 0.5 0.5 10 10 Ca Cb 1.00673790.8972621 5.394E0.08 42 1.00673791.8069769 5.394E0.08 42 k1 10 10 10 10 k2 0.192 0.192 0.192 0.
ODE Report (RKF45)
Differential equations as entered by the user [1] d(Ca)/d(t) = -k1Ca [2] d(Cb)/d(t) = -k2+k1Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.
for the US t = 6.2 hours
Sweden: t = 7.8 hours
Russia: t =10.3 hours.
P8-‐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 C A becomes:
d dCt (^) A (^) k 1 C (^) A 2 tafter that it reverts back to the original equations.
See Polymath program P8-7-e.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value t 0 0 11 11 Ca 0 0 0.1785514 6.217E 45 Cb k1 010 1.1120027 10 0.7458176 10 1.1120027 10 k2 0.192 0.192 0.192 0.
ODE Report (RKF45)
Differential equations as entered by the user [1] d(Ca)/d(t) = if(t<1)then(-k1Ca+2t)else(-k1Ca) [2] d(Cb)/d(t) = -k2+k1Ca Explicit equations as entered by the user [1] k1 = 10 [2] k2 = 0.
US: C B never rises above 0.8 g/L so there is no time that it would be illegal.
Sweden: t = 2.6 hours
Russia: t = 5.2 hours
P8-‐7 (f)
60 g of ethanol immediately Æ 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.
P8-‐7 (g)
A heavy person will have more body fluid and so the initial concentration of C A 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.
