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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_5320_Библиотеки_им_академика_М_И_Перельмана
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CC
kt=+
C
2
12
CC
−=
0
12
C
Ck
t
=−
00
kt=−
kt=+
C
k
693
Ck
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146
Pharmaceutical Dosage Forms and Drug Delivery
Or, the rate equation (Figure 7.3) is:
7.2.5.2 Half- Life
In a second- order reaction, the time to reach a certain fraction of the initial concentration (such as t
t
0.90
11
0
0
=
(7.39)
or
1/ 2
(7.40)
11
00
kt
−
(7.41)
/
2
/
Or,
1
kt=
(7.42)
/
half- life expression (Figure 7.3) is:
1
=
12
t
/
1/ 2
(7.43)
0
TABLE 7.1
Zero Order First Order Second Order
Denition
independent of the
reactant concentration
Rate equation
Concentration
at time t
Half- life
12
0
=
/
2
concentration
[A]
e
0
0
.
=
12
/
reactant concentration
2
[A]
11
CC
0
1
=
12
/
0
Ck
11
()
−−
ss
s
0
60 0 054
.
·
·
×
mg mL mL mg month
CC
kt=+
C
−
+×
hm
h
178
C
0 178
1
.
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Chemical Kinetics and Stability
147
1.
second- order process.
3232
M s and the initial concentration of
Solution
1
=
12
1
−−
/
21
0
=
1
×
006003
=
555 5
.=
12
/
.. ..
0063010
×
MM
2. month
, calculate:
A.
B. Shelf- life
C. Remaining drug concentration after 3 months
D.
Solution
A.
11
12
/
Ck
60 0 054
111
−−−
×
mg mL mL mg month
.·
·
=
.==
030
month
B.
0 111
%
−−−
111
90
.
=
0 034
.=
month
C.
11
11
=
mg mL
60 1
0 054 3
1
0 016 0 162 0
=+=...
==
0
.
mL mg mont
·
1
561
.mgmL
11
−−
·
ont
−
CC
kt=+
20
11
−
−
+×
−−
t
11
..
=+ ×→→= ×
−−
hm
tt
k
k
−1
d
d
[]
tt
kk
11
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148
Pharmaceutical Dosage Forms and Drug Delivery
after 3 months
D. Again, using the equation (7.39)
11
1
mg mL mg mL
50016 0 162 0050178
..
=
1160 1
mont
0 178
..month
005
.
0
.
0 054
mL mg month
··
.
onth
028
−
1
month
is 0.28 months.
7.2.6 Complex Reactions
Often, a drug undergoes more than one chemical reaction or a series of reactions in the same environ-
7.2.6.1 Reversible Reactions
k1, and the rate constant of the reverse reaction can be
k.
1
d[B]
d[A]
te
A=− ==
−
[]
B
(7.44)
−
absolute concentration.
d
]
[]
d
]
t
d
]A
(
)
[]
t
AA
kt
=
−
0
kk
12
−
[]
[]
dAt
dB
AB
[]
[]
t
kk
12
d
d
C
t
[]
[]
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Chemical Kinetics and Stability
7.2.6.2 Parallel Reactions
149
d[C]
dB
[A
=tk
1
[A
k=
2
(7.46)
d[A]
te
=− =+
[A]A [A
kk kk k
12 12
=+
[]
=
(7.47)
obs
k
is the observed rate of degradation of the reactant A.
obs
t
obs
e
(7.48)
7.2.6.3 Consecutive Reactions
product.
dA
=
=
k
1
−
[]
d
(7.49)
B
k
=
2
ERT=−
E
RT
=−
E
RT
2 303.
https://t.me/med1917
Pharmaceutical Dosage Forms and Drug Delivery
solving the above differential equations.
7.3 Factors Affecting Reaction Kinetics
7.3.1 Temperature
7.3.1.1 Arrhenius Equation
on the reaction rate constant, kArrhenius equation (Figure 7.4):
Ae
/
a
EaAR is the gas constant (1.987 calories/
degree mole); T
Figure 7.4):
lnA
k
a
Or
glogA
k
=−
A
FIGURE 7.4 Arrhenius plot. Plot of the variation of the rate constant, k, versus the reciprocal of the absolute temperature, T.
ERT
1
=
−
/
ERT
=
−
/
k
ERT
−
//
()
()
k
E
TT
21
−
k
E
TT
21
−
TT
21
LM s
jK mol
1951083145
..
×
×
−
−−−
E
600 800×
41
=−
()
−−
E
jK mol
Ea =×
−
41
.J
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Chemical Kinetics and Stability
For a straight- line plot, the equation is y mx + ck on the y
the reciprocal of the absolute temperature (1/ T) on the xEa from the slope of the straight line
(Figure 7.4
EaEa
different temperatures.
T1 and T2,
1
2
Ae
Ae
a
2
A
/
2
a
Ae
2
== ==
ERT
−
Ae
a
k
1
ERTERT
−
//
12
aa
eee
/
1
ER TT
−
// /
11
()
12
a
E
RT TTT
aa
−
2112
k
a
2
ln
=
R
1
TT
12
Or, as in Figure 7.4,
2
glog
=
k
1
2 303
.
a
R
TT
12
k
Ea, for a given reaction.
Practice problems Arrhenius equation
For the chemical reaction:
4 + 2 S22
S
2
- 8
- 1 s- 1
sEa for this reaction.
Solution
E
k
2
ln
811
27510
.
LM s
−−−−−
71111
96
.
8 3145
=
k
1
ln
=
a
11
.
−−
a
R
TT
12
41610
−
800
a
.
×
6600
K
39210
Mol
41
×
−
E
RT
E
RT
=−
×
−−
×T
×
××
..T
×
××
.K
TT
1
21
12
×
×
×
−
−−
E
×
..
E
E
RT
https://t.me/med1917
92 10
.
Pharmaceutical Dosage Forms and Drug Delivery
JMol .
2. For a certain reaction, the values of A and Ea
10 sec
-
sec
Solution
glogA
k
=−
A
2 303.
or
303 2 303..loglogA
k
A
Substituting the values in the equation:
.
1
jK mol
3
11
303 2 303 1
..
log(1.2 10 3) log(2 010)
×−=×−
80 510
2.92 10.30
=−
8 3145 2 303
.
−−
11
jK mol
8 3145
.
3
80 510
=
3. k
s
80 510
...
8 3145 2 303 13 22
=
317 98
sec
s
-
3
.
Solution
k
2
glog
=
k
E
2 303
.
−
a
R
TT
Substituting the values in the equation
31 10
g
67 10 2 303 8 3145
31
.
...
s
=
log
−− −−
41 11
s
=
19 148
a
0 00022 58314 36
()
.
067
.
a
=
jK mol
J mol
500 45
450 500
1
−
00
glogA
k
=−
A
2 303.
jK mol
2 303 8 3145
..
××
−
5500
A
×
××
−
988
E
RT
=−
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Chemical Kinetics and Stability
Substituting the values in the equation
g( slogA
67 10
.)
41
−−
×=−
58314 36 1
Jmol
.
−−
11
or
4 s
g
67 10
.
41
−−
s
=
58314 36
.
2 303 8 3145 500
..
jK mol
4 s
7.3.1.2 Shelf Life
t
(t
k, in the
0.90
1
Jmol
11
−−
=
9
5
.
) and shelf life
1/ 2
shelf life.
7.3.1.3 Thermodynamics of Reactions
temperatures.
required for them to overcome the intermolecular repulsions at close contact for effective intermolecular reactions to occur. Arrhenius equation relates the rate of a reaction, k
barrier, Ea.
G
lnA
k
a
(7.60)
∆∆ ∆GTS=−
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Pharmaceutical Dosage Forms and Drug Delivery
-
HS
the equation:
(7.61)
G
H
S and a negative TS.
7.3.2 Humidity
7.3.2.1 Water as a Reactant
reactant
7.3.2.2 Water as a Plasticizer
plasticizer
7.3.2.3 Water as a Solvent
can also act as a solvent
1.
2. Affecting the disproportionation of the salt form of the drug to its free acid or free base form,
3.
Disproportionation of the salt form of a drug in a solid dosage form to its constituent free acid or free
BR
=
()
E
=
BR ERT
=
()
−
ERTBR=+
()
[]
[]
[]
[]
https://t.me/med1917
Chemical Kinetics and Stability
7.3.2.4 Determination and Modeling the Effect of Water/ Humidity
isothermal degradation rate studies since the temperature is constant
RHk,
B, as:
H
e
(7.62)
a
−
RT
Ae
(7.63)
Ae
Ha/
(7.64)
Ae
H/
a
7.3.3 pH
7.3.3.1 Disproportionation Effect
Henderson– Hasselbalch equation.
salt
Hp log
=+
K
a
(7.66)
acid
Hp log
=+
K
base
a
salt
(7.67)
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