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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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Substituting Eq. 1.91:
( )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)
Re-arranging Eq. 4.18 and integrating the re-
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
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)
Re-arranging Eq. 1.103 and integrating the re-
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
Substituting Eq. 1.91:
( )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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43
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- =
-