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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5320_Библиотеки_им_академика_М_И_Перельмана
.pdf
=
[]
k
3232 332
+→ +
d
d
ttt
=
k
32
https://t.me/med1917
136
Pharmaceutical Dosage Forms and Drug Delivery
te CH COOC HNaOH
325
(7.6)
equation.
molecularity
molecularity of a reaction. In the
7.2.1 Pseudo- nth Order Reactions
pseudo- nth- order
a pseudo- 1st- order reaction.
COOCHCHHOCHCOOH CH CH OH
dCHCOOC H
te
=−
3253 32
dCHCOOHddCHCHOH
=
=
(7.7)
te CH COOC H
(7.8)
5
7.2.2 Determining the Order of a Reaction
1.
−=
dCt
C
t
Ck
∫∫
dd
t
−=−−
()
=−
00 0
0
https://t.me/med1917
Chemical Kinetics and Stability
137
concentration of reactant(s).
2.
infer reaction order.
3.
4.
to reach half of the measured initial concentration, on the initial concentration of the reactant is
7.2.3 Zero- Order Reactions
of light for photochemical reactions or the interfacial surface area for heterogeneous reactions (i.e.,
order reactions.
7.2.3.1 Rate Equation
Ct
(Figure 7.1
Or
k0
k0.
Integrating this equation from concentration C0Ctt,
Figure 7.1) is:
d
k0 (7.9)
0
t
=−
0
t
(7.11)
0
0
(7.10)
(7.12)
t
=−
00
C
t12 0012//
=−
C
Ck
0012
2
C
k
0
https://t.me/med1917
138
Pharmaceutical Dosage Forms and Drug Delivery
FIGURE 7.1 C, versus time, t.
(7.13)
y mx + cCt, on the y
time, t, on the xk 0
yC0.
7.2.3.2 Half- Life
t
1/ 2
(C
C0); that is:
t1/ 2
t12
0
=
(7.14)
/
2
0
=−/ (7.16)
t
Figure 7.1
1
0
=× (7.17)
12
/
2
1.
120 seconds the concentration of
238
238
A. Determine the rate constant k of the reaction.
B.
C. Calculate the amount of
238
238
t
=−
00
0
MMk=−
)
07
./
ec
−
)
= ./
ec
C
0
./
t
=−
00
t
=−
()
=
(.
cM
M
c
C
⋅
ys
https://t.me/med1917
Chemical Kinetics and Stability
139
D. If the original concentration is reduced to 1.0 M in the previous problem, does the half- life
Solution
A.
Inserting the given values in the equation:
515
..
MM
120
(
sec
00 00625
515 120
..(
=− =k
0
0 00625
Ms
sec
Ms
B.
1
order half- life is:
12
/
Inserting the values in the equation:
=×
2
0
k
=×
12
/
1
2150 00625
.
M
Msec
C.
15 0 00625 150 0 5625.. /sec)
MMse
238
D. If the initial concentration is reduced to 1.0 M the half- life is calculated as:
1
=× =
12
/
2100 00625
.
Msec
./
80
se
2. k
Solution
1
0
=×
12
/
12
/
k
2
0
1
2800 180
./
mg mL days
/
mg mL
=
222 2
.=×
da
−=
ddC
t
kC
−=
dC
C
Ct
C
kt
0
∫∫
CC kt kt−=−−
()
=−
https://t.me/med1917
140
7.2.4 First- Order Reactions
Pharmaceutical Dosage Forms and Drug Delivery
7.2.4.1 Rate Equation
Figure 7.2) is:
(7.18)
C is the reactant concentration at time t, and k
kdt (7.19)
Integrating this equation from concentration C0Ctt,
=−dd (7.20)
C
0
Solving this integral,
ln
0
0 (7.21)
:
FIGURE 7.2 C, against time, t (A), and plot of natural logarithm of the concen-
tration, C, against time, t.
kt
−=
−
2 303.
CCkt=−
kt=−
0
kt
303
kt
−=
−
2 303.
C
kt
0
−
C
C
kt
2 303=.
t
C
C
k
C
C
C
0
k
2
https://t.me/med1917
Chemical Kinetics and Stability
glogCC
0
141
Figure 7.2) is:
ln
(7.22)
0
e (7.23)
10),
=
og C
−
2
.
(7.24)
0
Figure 7.2).
constant, k.
7.2.4.2 Half- Life
t
1/ 2
(C0); that is, Ct C0
glogCC
=
g
g
=
C
0
.
2 303
.
2 303
=
0
2 303
.
log
log
(7.26)
(7.27)
0
(7.28)
0
(7.29)
2 303
.
= lo g
12
/
k
0
(7.30)
C
2
/
2 303
= lo g
12
/
.
(7.31)
k
kt
0
303
t
A
A
25
./
gMh
70.%
gM
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142
Pharmaceutical Dosage Forms and Drug Delivery
Figure 7.2) is:
t
in concentration (e.g., t
, i.e., time to 90% of initial concentration), is a constant number and independent
0.9
12
0 693/.
=
(7.32)
1/ 2
of the initial reactant concentration, C0.
1. 42
4222O (l)
A. Calculate the rate constant k
B.
C. 42
Solution
=−log
2
.
Rearranging equation for the rate constant
.
2 303
=
log
0
A
t t
t90
90%
2 303
.
k
B. At time t t
42
70%
=
10 0 092
lo
()
=
2 303 100
=
lo
t
30
Mh
0 092
./
()
gM
..
Mh
t
A
A
A
0
2 303
.
gM
×
A
t
A
A
540
25
sec
gm
c
k
https://t.me/med1917
Chemical Kinetics and Stability
2 303
t
70
%
=
70
%
0 092
2 303
.
Mh
./
70%
.
333
.=
lo
0 522 13 06
()
=
143
.
2 303
092 35
./
=
092
./
Mh
Mh h
=
2 303
35
.
log
lo
0
100
log=
h
100
399
=
.
42
2.
Calculate the rate constant and the half- life for this reaction.
Solution:
.
2 303
=
log
0
2 303
.
=
100
logm
=
0 00256
.
g/se
0 693 0 693
12
/
0 0025
.
..
277 2
.sec== =
3.
A. Determine the rate constant k.
B.
C.
CC kt kt−=−−
()
=−
CCkt
−=
−=−
()
k
k
0 00447
.
n
MM
0115
00
CC
0
..
MM
e
M
https://t.me/med1917
144
Pharmaceutical Dosage Forms and Drug Delivery
Solution:
A.
ln
0
0
Rearranging the equation
(0.045) ln(0.023)
ln
0
450 300
kk
k
B.
0 693 0 693
..
== = mi
12
/
155
C.
ln lnCCkt
−=−→
0 023
.
0
0 00447 450
.
=− ×=
C
C
0
0 023
.
=−
kt
2
.
=−
023 0 023
C
0
0
2 0115
.
−
ee==
0 023
.
M
−
.
2 0115
2 0115
.
−
0 1719==
.
7.2.5 Second- Order Reactions
molecules.
[]=−[]
[]
d
d
AB
A
tt
−
[]=−[]
[]
AB
A
t
B
t
kk
22
−=
dCt
2
dC
C
Ct
C
kt
∫∫
CC
kt
https://t.me/med1917
Chemical Kinetics and Stability
change in the concentrations of products and reactants in second- order reactions is proportional either to
A is equal to the rate of decomposition of B, and both are pro-
−
7.2.5.1 Rate Equation
d
dB
=
k (7.33)
[]
Assuming that the initial concentrations of A and B are the same, that is, C0, and their concentration after
time, t, is C
d
d
d
=
d
=
[]
(7.34)
Or, using their concentration value, C, the rate expression (Figure 7.3) is:
d
2
kC
kdt
=−
(7.36)
Integrating,
=−dd
2
C
0
11
−=−
0
0
(7.37)
(7.38)
FIGURE 7.3 C, against time, t.
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