04 Collection and Analysis of Rate Data

8/13/2019 04 Collection and Analysis of Rate Data

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Chapter 4

Collection and Analysis of Rate Data

From Chapters 1 and 2, we have learned that

the rate of reaction ( )ir of any given chemical re-

action depends mainlyon temperature ( )T andconcentration of a specified species ( )iC in thereacting system

( ),i ir f T C (4.1)

In order to determine the effect of tempera-

ture, an experiment is carried out such that con-

centrations of all species are kept constant, whilethe rates of reaction at various temperatures are

monitored and recorded


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The rate of reaction at each temperature is

then used to calculate the corresponding rate

constant ( )k

Eventually, an Arrhenius plot (a plot between

lnkand

1

T) is employed to determine the effect

of temperature on the reaction rate; an activation

energy ( )aE and a frequency (pre-exponential)factor ( )0k are also obtained from the Arrheniusplot

On the other hand, to determine the effect of

concentration on the rate of reaction, an experi-ment is conducted such that temperature is kept

constant, while concentration of a specified spe-

cies ( )iC are varied


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3)Re-arrangethe rate equation and performan appropriate integration

4)Plota graphbetween the resulting integraland time

5)Determine the resulting plot if it is (or not)a straight line

6)Ifit is a straight line, the guessed or-dermade in 1) is correct

7)However, if it is nota straight line, anew guessis needed, and the procedure

1) 5) is to be repeated

4.1.1 Integral Method for 1st-order Reaction

Lets consider the following 1st-orderreaction:

A P (4.2)

whose rate equation can be written as follows


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AA A

dCr kC

dt- = - = (4.3)

Re-arranging Eq. 4.3 gives

A

A

dCkdt

C- = (4.4)

Integrating Eq. 4.4 and re-arranging the re-

sulting equation yields

00 0

A

A

C t t

A

AC

dCkdt k dt

C- = =

00

ln A

A

C t

A CC k t - =

0

ln lnA A

C C kt - - =

0

ln lnA A

C C kt = - (4.6a)

or

0

ln A

A

C

ktC- = (4.6b)


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Taking exponentialthroughout Eq. 4.6c and

re-arranging the resulting equation yields

( )0

exp ln expA

A

Ckt

C

= -

( )0

expA

A

Ckt

C= -

( )0

expA A

C C kt = - (4.7)

Eq. 4.7 can be presented graphically as theplot on the next Page (Figure 4.1 on Page 9)


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Figure 4.1: A plot between the concentration of

a reactant ( )AC and time ( )t for the 1st-order re-action, with

0A

C = 1.0 mol/L and k= 0.05 min-1

For a constant-volumesystem, we obtain the

fact that

( )0 1A A AC C x= - (1.91)

Combining Eq. 1.91 with Eq. 4.6c:

0

ln A

A

Ckt

C

= - (4.6c)

0.00

0.20

0.40

0.60

0.80

1.00

0 20 40 60 80 100

t (min)

CA(mol/L

)


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and re-arranging the resulting equation yields

( )0

0

1

ln

A A

A

C x

ktC

-

= -

( )ln 1 Ax kt- = - (4.8a)or

( )1

ln1

A

ktx

= -

(4.8b)

Thus, for a 1

st

-order reactionwhose systemvolumeis constant, a plot between

( )ln 1 Ax- and t(Eq. 4.8a) results in astraight linepassing through an origin

with a slope of k-

( )

1ln

1A

x

-

andt(Eq.4.8b)yields astraight

line, also passing through an origin, with a

slope of k


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Taking exponentialthroughout Eq. 4.8a and

re-arranging the resulting equation gives

( ) ( )exp ln 1 expAx kt - = - ( )1 expAx kt- = -

( )1 exp

A

x kt= - - (4.9)

which can be presented graphically as Figure 4.2

on Page 12

4.1.2 Integral Method for 2nd

-order Reaction

For 2nd-order reactions, it can be divided into

2 cases:

One-molecule 2nd-orderTwo-molecule 2nd-order


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Figure 4.2: A plot between the conversion of a re-

actant ( )Ax and time ( )t for the 1st-order reaction,with

0A

C = 1.0 mol/L and k= 0.05 min-1

Case I: One-molecule 2nd-order reactions

For an elementary one-molecule2nd-order re-

action:2A P (4.10)

the rate equation can be written as follows

2AA A

dCr kC

dt

- = - = (4.11)

0.00

0.20

0.40

0.60

0.80

1.00

0 20 40 60 80 100

t (min)

xA(-)


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Re-arranging Eq. 4.11 and integrating the re-

sulting equation yields

0

0

2

2

0

1

A

A

A

A

A

AC t

A

C A

C

AC

dCkdt

C

dCk dt

C

ktC

- =

- =

=

0

1 1A A

ktC C

- = (4.12a)

or

0

1 1

A A

kt

C C

= + (4.12b)

Hence, for a one-molecule2ndorder reaction

taken place in a constant-volumesystem, a plot

between


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0

1 1

A AC C

- andt(Eq.4.12a)givesastraight

linepassing through an origin with a slope

of k

1A

Candt(Eq.4.12b)isastraightlinewith

oa slope of koa Y-intercept of

0

1

AC

Substituting Eq. 1.91:

( )0

1A A A

C C x= - (1.91)

intoEq.4.12aandre-arrangingtheresultingequa-

tion gives

( )( )

( )

00

0

1 1

1

1 1

1

AA A

A

A A

ktCC x

xkt

C x

- =-

- -=

-


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( )

0

1A

A A

xkt

C x=

- (4.13)

Thus, for a one-molecule2ndorder reaction

occurred in a constant-volumesystem, a plot bet-

ween ( )0

1A

A A

x

C x- and tresults in a straight line

passing through an origin with a slope of k

Case II: Two-molecule 2

nd

order reactionsFor an elementary two-molecule2nd-order re-

action:

A + B P (4.14)

the rate equation can be written as follows

A

A A B

dCr kC C

dt- = - = (4.15)


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FromEq.4.14,weobtainthefactthatif 1 mol

of A is consumed, 1 mol of B is to be consumed

Thus, when A in the amount of0

A Ax C or

0 0A A A A

C C x C - =

is consumed, B in the same amount of0A A

x C is

consumed, which means that

0 0B B A A

C C x C - = (4.16)

Re-arranging Eq. 4.16 results in

0

0

0

B

B A A

A

CC C x

C

= -

or

( )0

B A B AC C xq= - (4.17)

where 0

0

B

B

A

C

Cq =


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( )0

1A A A

C C x= - (1.91)

and Eq. 4.17 into Eq. 4.15:

A

A A B

dCr kC C

dt- = - = (4.15)

and re-arranging the resulting equation yields

( ) ( ) ( )00 0

11

A A

A A A B A

d C xk C x C x

dtq

- - = - -

( )( )( )

0 0

21

1A

A A A B A

d xC kC x x

dtq

-- = - -

( )( )0

1AA A B A

dxkC x x

dtq= - -

(4.18)


sulting equation gives

( )( )0

0 01

Ax t

A

AA B A

dxkC dt

x xq=

- -


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( )( ) 00 1

Ax

A

A

A B A

dxkC t

x xq=

- -

(4.19)

The term

( )( )0 1

Ax

A

A B A

dx

x xq- - in Eq. 4.19 can

be integrated to

( )( ) ( )( )

( )01

ln1 1 1

Ax

B AA

A B A B B A

xdx

x x x

q

q q q

-=

- - - -

(4.20)when 1

Bq

or can be integrated to

( )( ) ( )2

0 011 1

Ax xA A A

AA B AA

dx dx x

xx x xq= =

-- - -

(4.21)

when 1Bq

=


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Combining Eq. 4.20 with Eq. 4.19 and re-ar-

ranging the resulting equation yields

( )( )

( )( )

( )

( )

0

0

1ln

1 1

ln 11

B A

A

B B A

B A

B A

B A

xkC t

x

xkC t

x

q

q q

qq

q

- = - -

- = - -

( )

0

0 0

0

0 0

0

ln 1

1

B

A

A B

A

B A

AA

Cx

C CkC t

C C

xC

- = -

-

( )( )

0

0

0 0

0

0

ln

1

B

A

A

B A

B

A

A

Cx

C k C C t C

xC

- = -

-

(4.22)


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Hence, if the reaction is two-molecule2nd-order

with 1B

q , a plot between

( )

0

0

0

0

ln

1

B

AA

B

A

A

C

xC

Cx

C

-

-

and

tyields a straight linepassing through an originwith a slope of ( )

0 0B A

k C C-

Re-arranging Eq. 4.22 gives

( )

( ) ( )

0 0

0

0 0

0

0

ln1

B A A

A

B A

B A

A

C x C

Ck C C t

C x

C

- = - -

( )

( )

( )0 0 00 0

0 0

ln1

A B A A

B A

B A A

C C x C k C C t

C C x

- = - -

BC

AC


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( )00 0

0

ln A B

B A

B A

C Ck C C t

C C

= -

( )0 0

0

0

ln BB A

B

A

A

Ck C C t

C CC

= -

(4.23)

For the case where0 0

B AC C , during the pro-

gress of the reaction, the concentration of B ( )BC is nearly constant at

0B

C , or it means that

0B BC C

Thus, Eq. 4.23 becomes


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( )0

0 0

0

0

ln

B

B A

B

A

A

C

k C C t CC

C

= -

( )0 0 0ln A

B A

A

C k C C t C

= - (4.24a)

or

( )0 0

0

ln AB A

A

Ck C C t

C

- = - (4.24b)

Since0 0

B AC C , it results in the fact that

( )0 0 0

B A Bk C C kC k - =

Hence, Eq. 4.24b becomes

0

ln A

A

Ck t

C- = (4.25a)


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or

0

ln A

A

Ck t

C

= - (4.25b)

or

( )ln 1 Ax k t- = - (4.26)

Eqs. 4.25a & b and 4.26 are similar, or almost

identical, to Eqs. 4.6b & c and 4.8a, respectively

Thus, for a two-molecule2nd-order reaction in

which0 0

B AC C , the resulting equation is similar

to that of the 1st-orderreaction

We call such two-molecule2nd-order reaction

a pseudo1st-order Rxn.


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Combining Eq. 4.21:

( )( ) ( )20 0 11 1

Ax x

A A A

AA B AA

dx dx x

xx x xq = = -- - - (4.21)

with Eq. 4.19:

( )( ) 00 1

Ax

AA

A B A

dx kC tx xq

=- -

(4.19)

results in

01A

A

A

x kC tx

=-

(4.27)

Hence, for the case of two-molecule2nd-order

reaction with 1B

q = , a plot between1

A

A

x

x-and t

yields a straight linepassing through an origin

with a slope of0

AkC


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4.1.3 Integral Method for 0th-order Reaction

We have learned from Chapter 1 that the

rate equation for a 0th-order reaction can be writ-

ten as follows

AdC

kdt- = (1.103)


sulting equation gives

0

0

0

A

A

A

A

C t

AC

A A

dCk

dtdC kt

C k t

C C kt

- =

- =

- =

- - =

0A A

C C kt - =


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0

A AC C kt - = - (4.28a)

or

0A A

C C kt = - (4.28b)

Accordingly, for a 0th-order reaction, a plot

between

0

A AC C- and t(Eq. 4.28a) results in a

straight linepassing through an origin

with a slope of k-

AC and t(Eq. 4.28b) gives a straight linewith

oa slope of k- oa Y-intercept of

0A

C


( )0

1A A A

C C x= - (1.91)

into Eq. 4.28a and re-arranging the resulting

equation yields


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( )( )

0 0

0

0

1

1 1

A A A

A A

A A

C C x kt

C x kt

x C kt

- - = - - = =

0

A

A

kx t

C= (4.29)

Hence, for a 0th-order reaction, a plot between

Ax and tresults in a straight linepassing through

an origin with a slope of0

A

kC


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4.2 Differential Method

In the integralmethod, as we have to take a

guess of the order of the reaction, it is almost al-

ways the case that the answer is an integer(-

), e.g., 0th, 1st, or 2ndorder [actually, we

can guess an order with a non-integernumber

(e.g., 0.5, 1.5), but it is unlikely that we shall do

so]

Isthereanyothermethodthatwedonothave

to take a guess of an order of reaction in priori?

A differentialmethodis the answer

Taking lnthroughout the rate equation for

an nth-order reaction:


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nAA A

dCr kC

dt- = - = (4.30)

yields

ln ln

ln ln

nA

A

n

A

dCkC

dt

k C

- =

= +

ln ln lnAA

dCk n C

dt

- = +

(4.31)

Hence, when plotting a graph between

ln AdC

dt

-

(Y-axis) and lnA

C (X-axis), it results

in a straight line, with

a slope of n(order of reaction)a Y-intercept of lnk


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The calculated slope at each point (time) is

summarised in the following Table

t[s] CA[mol/L]Slope

- A

dC

dt

0 10.0 -(10.0-0)/(0-90) = -(-0.11) = 0.1110

20 8.0 -(10.0-0)/(0-105) = -(-0.0952) = 0.0952

40 6.0 -(8.0-0)/(0-165) = -(-0.0484) = 0.0484

60 5.0 -(7.0-0)/(0-200) = -(-0.0350) = 0.0350

120 3.0 -(6.0-0)/(0-245) = -(-0.0245) = 0.0245

180 2.0 -(4.4-0)/(0-320) = -(-0.0138) = 0.0138

300 1.0 -(3.5-0)/(0-420) = -(-0.0083) = 0.0083

0

2

4

6

8

10

12

0 100 200 300 400 500

CA[mol/L

]

Time [s]

1

23

4

56

7


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Plotting a graph between ln AdC

dt

- and ln

AC

yields

t[s] CA[mol/L] -dCA/dt lnCA ln(-dCA/dt)

0 10.0 0.1110 2.303 -2.198

20 8.0 0.0952 2.079 -2.352

40 6.0 0.0484 1.792 -3.02860 5.0 0.0350 1.609 -3.352

120 3.0 0.0245 1.099 -3.709

180 2.0 0.0138 0.693 -4.283

300 1.0 0.0083 0 -4.791

y = 1.1503x - 4.9612

R2= 0.9675

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

0.0 0.5 1.0 1.5 2.0 2.5

ln CA

ln[-dC

A/dt]


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From the resulting graph above, we obtain the

following information:

Slope = n(reaction order) = 1.15Y-intercept = ln k= 4.9612

Thus,

exp( 4.9612) 0.007k= - =

Accordingly, the rate equation of this reaction

can be written as follows

1.150.007AA AdC

r Cdt- = - =


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When the system volume ( )V is constant, thefollowing equation (Eq. 1.91):

( )0 1A A AC C x= - (1.91)is applicable

Substituting Eq. 1.91 into Eq. 4.30:

nA

A A

dCr kC

dt- = - = (4.30)

and re-arranging the resulting equation gives

( )( )

( )( )

0

0

0

0

11

11

nA A

A A A

nA A n

A A

d C x

r k C x dt

C d xkC x

dt

- - = - = - - - = -

( )0

1 1n

nA

A A

dxkC x

dt

-= - (4.32)


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Taking lnthroughout Eq. 4.32 yields

( )0

1ln ln 1 n

nA

A A

dxkC x

dt

- = -

( )0

1ln ln ln 1nAA A

dxkC n x

dt

- = + -

(4.33)

Thus, when plotting a graph between ln Adx

dt

and ( )ln 1 Ax- , it results ina slope of na Y-intercept of -

0

1ln nA

kC

Hence,( )

0

1 exp Y-interceptnA

kC - =

( )

0

1

exp Y-intercept

n

A

kC -

= (4.34)


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Example A decomposition of species A taken

place at a given temperature, and the data of

conversion of A at various times are summarised

in the following Table

Time [min] 13 34 59 120

xA

[] 0.112 0.257 0.367 0.552

Determine a rate equation of this reaction;

given0

AC = 0.2 mol/L

To determine a rate equation of this reaction,

we follow the following procedure:

1) Plot a graph betweenA

x and t, as shown

on the next Page

2) Draw a tangential line at each point ofdata and determine its slope (try doing it

yourself)


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3) Plot a graph between ln Adxdt

and

( )ln 1 Ax-

0

0.1

0.2

0.3

0.4

0.5

0.6

0 50 100 150

t [min]

xA[-]


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t[min] xA dxA/dt ln(dxA/dt) (1-xA) ln(1-xA)

13 0.112 0.00734 -4.914 0.888 -0.119

34 0.257 0.00551 -5.201 0.743 -0.29759 0.367 0.00362 -5.621 0.633 -0.457

120 0.552 0.00289 -5.846 0.448 -0.803

From the plot above, we obtain the following

information:

y = 1.3783x - 4.8183

R2= 0.9212

-6.20

-5.80

-5.40

-5.00

-4.60

-1.00 -0.75 -0.50 -0.25 0.00

ln (1-xA)

ln

(dxA/dt)


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Slope = n(order of reaction) = 1.38Y-intercept = 4.8183

Thus,

( )

( )

( )

0

1

1.38 1

exp Y-intercept

exp 4.8183

0.2

0.0149

n

A

kC

k

-

-

=

-

=

=

Hence, the rate equation of this reaction is as

follows

1.380.0149A A

r C- =


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2

1 1

1 1 A

A A A

k C

r k C k C = +

-

2

1 1

1 1

A A

k

r k C k = +

- (4.36)

Plotting a graph between1

Ar-

and1

AC

gives

a straight linewith

a slope of1

1k

a Y-intercept = 21

k

k


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Example The rate equation for a reaction:

A P

can be expressed as the following equation:

1

21

A A

A

A

dC k C r

dt k C - = - =

+

The data of0

AC and the corresponding

0A

r-

are summarised in the following Table

CA0[mol/L] 0.23 0.90 0.68 0.45

-rA0[mol/(L-min)] 0.00176 0.0108 0.00819 0.00447

Determine the values of1

kand2

k

Plotting a graph between0

1A

r- and0

1A

Cyields


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1/CA0 1/(-rA0)

4.348 568.2

1.111 92.61.471 122.1

2.222 223.7

From the resulting plot above, we obtain the

following information:

y = 150.3x - 92.219

R2= 0.9949

0

200

400

600

0.0 2.0 4.0 6.0

1/CAo

1/(-rA)


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slope = 150.3 =1

1

k; thus,

11 0.00665

150.3k = =

Y-intercept = 92.219 = 21

k

k; hence,

( )( )2 1

2

Y-intercept

0.00665 92.219

0.613

k k

k

=

= -= -

Accordingly, the rate equation for this reac-

tion can be written as follows

0.00665

1 0.613A

A

A

Cr

C- =

-

04 Collection and Analysis of Rate Data

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