Assin2 Solution

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ChE 321 Washington State University

Kinetics and Reactor Design Department of Chemical Engineering

Spring, 21 Richard !" #ollars

Assignment #2

Due: January 29, 2001

We are going to work one of the problems from the D that goes along with !ogler"s tet

but in two $ifferent %ariations&

A rather sinister'looking gentleman si$les up to you one night an$ in a sibilant

whisper asks you to make him some methyl per(hlorate& )ou *uestion his moti%es

be(ause the pro$u(t of the rea(tion between soli$ sil%er per(hlorate an$ methylio$i$e +eplo$es %iolently when stru(k+ -& !& .a$ies an$ /& re$ale,J. Phys.

Chem., 48, 22 1934& 5e respon$s by telling you the truth: he owns a tree

stump remo%al business, an$ he nee$s (heap eplosi%es& /he legal route haso((urre$ to him, but his fun$s are *ui(kly $epleting&

)ou6re not too (omfortable with this situation, but times are har$, an$ you nee$

the money& What you $on6t nee$, howe%er, is to $estroy your laboratory&/herefore, you $e(i$e to make the material in bat(h in a ben7ene solution, gi%e

the sinister stranger the pro$u(t still in the ben7ene, an$ let him figure out how to

get the methyl per(hlorate out&

)ou use a %essel (ontaining 80 $m8of solution, starting with 0&M58 an$ 0&

MAgl;(on(entrations& f you let the rea(tion pro(ee$ for 10 hours, how mu(hmethyl per(hlorate will be pro$u(e$< 5ow mu(h will be pro$u(e$ in 20 hours Agl; 58l;> Ag

where

2?8

188 AgClO

CICH

CkICH

r =

at 29@ in ben7ene, kB 0&0002 $m8?mol38?2?s&


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C;E/;F

f we let A B 58, G B Agl;, B 58l;an$ D B Ag then the only thing we ha%eleft to $o is to epress Gin terms of A& /his relationship is gi%en by

( ) ( )oAoAAAoo CCCCCCC ==

Esing this in the rate epression we now (an sol%e the problem, first using -athAD&


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!inal2 H Ao I 100 1, 3:=

I .ka$apt A 0, 80, 100, D,( ):=

!inal1 H Ao I 100 1, 3:=

I .ka$apt A 0, 20, 100, D,( ):=

Now repeat this for final times of 20, 30, 40, 50, and 100 hours.

!inal0 H Ao I 100 1, 3:=

Compute the amount of C and save it for further calculations

I .ka$apt A 0, 10, 100, D,( ):=

Now integrate

D t A,( ) k A A Go+ Ao( )1&:

:=

Now define the derivative

A

Ao

:=

Now give the initial condition

H 80:=

Go 0&::=

Ao 0&:=

The factor of 300 converts the time unit to hours rather than minutesk 0&0002 8J00:=

!ssignment "2

#et ! $ C%3&, ' $ !gCl(4, C $ C%3Cl(4, and ) $ !g&. Now define the parameters in the pro*lem.


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!inal8 H Ao I 100 1, 3:=

I .ka$apt A 0, :0, 100, D,( ):=

!inal H Ao I 100 1, 3:=

I .ka$apt A 0, 100, 100, D,( ):=

!inal H Ao I 100 1, 3:=

)efine a time vector that contains the final times for each

time

10

2080

0

:0

100

:=

Thus the results are

where time is in hours and +inal is moles

time

10

20

80

0

:0

100

= !inal

18&11

1&181

1&:

1&J

1&1

1&918

=

(r in graphical form

i 0 :..:=

0 0 10018

1

1

!inali

timei


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An$ now in -atlab

/his %ersion is for e$u(ational (lassroom use only&

/o get starte$, type one of these: helpwin, help$esk, or $emo&

!or pro$u(t information, type tour or %isit www&mathworks&(om&

K L Assignment #2

K L .i(har$ & IollarsK L January 2, 2001

K L /he m'file (ontaining the $eri%ati%e has been (reate$ so start the integrator

K t,A4Bo$e6assin26,0:0&1:104,0&3 M L /his will integrate to 10 hours

K L Fow (ompute the amount of methyl (hlorate 3 pro$u(e$ an$ store this an$ thefinal time in an array

K time13Bt1013M

K final13B0&'A10133N80M

K L Fow repeat these last three steps for the other final timesK t,A4Bo$e6assin26,0:0&2:204,0&3 M L /his will integrate to 20 hours

K time23Bt1013MK final23B0&'A10133N80M

K t,A4Bo$e6assin26,0:0&8:804,0&3 M L /his will integrate to 80 hours

K time83Bt1013MK final83B0&'A10133N80M

K t,A4Bo$e6assin26,0:0&:04,0&3 M L /his will integrate to 0 hours

K time3Bt1013M

K final3B0&'A10133N80MK t,A4Bo$e6assin26,0:0&:04,0&3 M L /his will integrate to 0 hours

K time3Bt1013M

K final3B0&'A10133N80MK t,A4Bo$e6assin26,0:1:1004,0&3 M L /his will integrate to 100 hours

K time3Bt1013M

K final3B0&'A10133N80MK L thus the tabulate resutls are

K time4

ans B

10 20 80 0 0 100

K final4

ans B

18&100 1&180 1& 1&8 1&10 1&9128

K L or gi%en in a graphi(al form


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K plottime:3,final:33

K

function dCA=assin2(t,CA)

dCA=zeros(1,1); %create a 1x1 array

k=0.00042*!00; % define t"e rate constant #it" units of "ours

CAo=0.$;Co=0.& % define initia' concentrations

dCA(1)=k*CA*(CACoCAo)1.& ; % define t"e deri+ati+e of CA

10 20 30 40 50 60 70 80 90 10013

13.2

13.4

13.6

13.8

14

14.2

14.4

14.6

14.8

15

ssgnmen

Amount of Methyl Perchlorate Produced as a Function of Reaction Time

Reaction Time (hours)

MolesofMethylPerchlorate

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Assin2 Solution