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SECTION 9:
Kinetics
Chapter 12 in Chang Text
Outline for Section 9 – Part I
9.1. Kinetics in Pharmaceutical Science
9.2. Rates of Reactions
9.3. Reaction Order:
Zero-Order, 1st Order
9.1. Chemical Kinetics in Pharmaceutical Science
Why do we, as pharmaceutical scientists, care about Chemical
Kinetics and Kinetics in General ?
We use rate equations and rate laws to define and describe processes
involved in everything from …………..
“Making the Drug”
to
Formulation
to
Drug Administration
in Animals
Rates are important for the following:
• rate of reaction to synthesize a drug molecule
…..…..rate of side reactions under specific conditions
• rate of decomposition, degradation, biotransformation,
and inactivation of a drug
• rate of precipitation of a drug from a formulation under
specific conditions
• rate of release of a drug from a delivery system
• rate of uptake of drug into cells
• rate of reaction between drug and biological target
(or other molecules)
• rate of clearance of drug from bloodstream
• rate of elimination of drug from body
Rate of Decomposition of Drug
• characterization of the decomposition or stability of drugs is
of critical importance
• decomposition may proceed by hydrolysis, oxidation, isomerization,
epimerization, photolysis etc.
• many groups study the effect of ingredients of dosage forms or
formulations and environmental factors (i.e. temperature, pH )
on the chemical and physical stability of drugs
Examples:
1.
Higuchi studied the decomposition of Chloramphenicol.
Chloramphenicol decomposes via hydrolytic cleavage of the amide
bond (shown above).
• rate of degradation is low and independent of pH for pH
range 2 – 7
• hydrolysis is catalyzed by presence of acids
• maximum stability occurs at pH 6 at room temperature……….the
half-life under these conditions is 3 years
2. Stability of Doxorubicin (anti-cancer drug)
• very difficult drug to study as it chelates with metal ions,
self-associates in solution (DOX-DOX), absorbs to plastic
and glass, undergoes oxidative and photolytic decomposition
• Beijnen et al. studied the kinetics of degradation of DOX as a
function of pH, ionic strength, temperature and concentration of
drug.
• decomposition followed first order kinetics at constant
temperature
• pH – rate profile demonstrated that maximum stability was
achieved at about pH = 4.5
• DOX is an anthracycline……includes
chromophoric anthraquinone moiety
and a charged sugar within structure
Doxorubicin
• anthracyclines are generally light
sensitive (many have investigated
photodegradation of DOX when
exposed to room light)
• exposure to light, increase in pH and
and adsorption to container known to
result in DOX loss (decrease in concentration
of intact DOX in solution)
• kinetics of photodegradation of DOX has been studied in buffer,
and biologically relevant media
• incorporation of DOX into a delivery system, termed liposomes,
has been found to reduce rate of photodegradation of drug
can make
liposomes
such that
internal
pH = 4.5
DOXIL (Caelyx in Canada)
Liposome encapsulated
Doxorubicin
• Doxil is approved for refractory ovarian cancer and AIDs-related
Kaposi’s sarcoma
Bandak et al. studied the UV-induced degradation of DOX as a
free agent and as a liposome-encapsulated agent………
(Pharm. Res. 1999, 16, 841)
….very important study as administration of DOXIL
(liposome-encapsulated DOX) is known to result in localization of
high concentrations of DOX under the skin……………
………….actually one of the dose-limiting toxicities of DOXIL is
called “hand-foot syndrome”.
Photodegradation of Encapsulated DOX is Reduced
(Pharm. Res. 1999, 16, 841)
Why is Encapsulated DOX More Stable ????
• protection by lipid bilayer
• high concentration of DOX inside liposome leads to aggregate
formation (DOX-DOX aggregates)
• low pH of the intra-liposomal aqueous phase
Rate of Reaction of Drugs with Biological Components
• Cisplatin (CDDP) is an
anti-cancer drug that
exerts anti-cancer activity
by binding to cellular DNA
• drug enters the cell, passes
through cytosol and enters
nucleus where it binds DNA
• CDDP in cytosol may also
bind metallothionein (MT)
and other endogenous thiols
mitochondria
nucleus
MT
Cytosol
Figure 1: Schematic of CDDP entry into cell, passage through cytosol
and binding to DNA in nucleus. (not drawn to scale)
• binding of CDDP to MT limits amount available for binding to DNA
therefore reducing anti-cancer activity
• MT is a small cellular protein (6 kDa) that binds strongly to metal ions
(CDDP molecule contains platinum within chemical structure)
• Hagrman et al. (Drug Metabolism and Disposition 2003)
studied the kinetics of the reaction of CDDP with MT
• their studies included characterization
of the rate of reaction between CDDP
and MT under specific conditions
Cisplatin
(CDDP)
also dependence of reaction rate on
concentration of CDDP and MT was
examined.
……..this data is critical as it provides understanding of ability
of MT to trap CDDP……….and thus alter the therapeutic
effect of this drug.
Rate of Clearance of Drug from Circulation
Free drug
350
Drug Concentration in Plasma
(ug /mL)
Liposome
-encapsulated
drug
Liposome Encapsulated Drug
Free Drug
300
250
200
150
100
50
0
0
Sample blood at time points
t = 15 mins, 30 mins, 1 hr, 4 hrs etc.
1
2
3
Time (hours)
4
5
9.2. Rates of Reactions
Rate of a reaction is expressed as change in reactant concn. with time
R  P
Rate of reaction over time interval (t2 – t1) may be expressed as
Follows:
{[R]2 - [R]1} / {t2 – t1} =  [R] /  t
………..where [R]1 is the concentration of R at t1
and [R]2 is the concentration of R at t2
Since [R]2  [R]1 we introduce a minus sign so rate has a positive value:
Rate = -  [R] /  t
Rate can also be expressed in terms of concentration of product:
Rate
= {[P]2 - [P]1} / {t2 – t1} =  [P] /  t
since [P]2  [P]1 we don’t need minus sign in equation.
……actually we are not usually interested in rate over a time interval
because this is an AVERAGE……..rather we are interested in
instantaneous rate………
The rate of a reaction at a specific time may be given by:
rate = - d [R] / dt
= d [P] / dt
Units of reaction rate are usually M s-1 or M min-1
For reactions where we have coefficients in front of the reactants or
products…………….
2R  P
……..in this case the reactant disappears twice as fast as the product
appears…………
Rate = - ½ ( d [R] / dt )
= d [P] / dt
The rate for reactions is then:
a A + b B

c C + d D
Rate = - (1/a) ( d [A] / dt ) = - (1/b) ( d [B] / dt )
= + (1/c) ( d [C] / dt ) = + (1/d) ( d[D] / dt )
(t is time after start of reaction)
9.3. Reaction Order:
…relationship between rate of a chemical reaction and the concns.
of reactants and products is complicated and must be determined
experimentally……..
For reaction:
In general:
a A + b B

rate  [A]x [B]
= k
[A]x
c C + d D
y
[B]
…….where k is the rate constant.
y
The Rate Law :
rate is proportional to conc.
of reactants raised to some
power
The rate constant does not depend on concentrations it is only
dependent on Temperature.
How do we define the order of a reaction ?
….for example……
Rate = k [A] x [B] y
…….in this case the reaction is x order with respect to A and
y order with respect to B.
………..the reaction has an overall order of x + y
IMPORTANT: in general there is NO relationship between
order of reaction and stoichiometric coefficients in reaction.
Example:
2 N2O5 (g)  4 NO2 (g) + O2 (g)
……..the Rate Law for this reaction is known to be:
Rate = k [N2O5 ]
……reaction is first-order with respect to N2O5
ORDER of REACTION……..gives dependence of Rate on concns.
9.3.1.
Zero – Order Reactions
Rate law for a zero-order reaction is given by………
A

Products
rate = - d [A] / dt = = k [A] 0 = k
k (in units of M s-1) is the zero- order rate constant.
……for a zero-order reaction the rate is independent of reactant
concentration.
d [A] = - k dt
Integration between t = 0 and t = t at concentrations of [A]t=0
and [A]t gives the following expression:

[A] t
d [A] = [A] t – [A]
[A] t=0
[A] t = [A]
t=0
- kt
[A] = [A]0 - kt
t=0
=-
t
0
kdt = - kt
Example:
Conversion of ethanol to acetaldehyde by the enzyme
LADH (liver alcohol dehydrogenase). Oxidizing agent is
NAD+ (nicotinamide adenine dinucleotide):
LADH
CH3CH2OH + NAD+ →
CH3CHO + NADH + H+
In the presence of an excess of alcohol over the enzyme
and with the NAD+ buffered via metabolic reactions that
rapidly restore it, the rate of this reaction in the liver is
zero order over most of its course.
The reaction cannot be of zero order for all times; because,
obviously, the reactant concentration cannot become less than
zero and the product concentration must also reach a limit.
For the oxidation of alcohol by LADH, the reaction is zero
order only while alcohol is in excess !!
Plot of concentration versus time for a zero order reaction.
The magnitude of the slope of each straight line is equal to
the rate constant, k0 . The order must eventually change from
being zero as [CH3CH2OH] approaches zero.
For a zero-order
reaction the Rate
is independent of
concentration of the
reactant.
9.3.2.
First - Order Reactions
rate = - d [A] / dt = = k [A]
…rate of reaction depends on concn. of reactant raised to the
first power.
….in this case the units of k are s-1
A
–
k1
→
B
d[A ]
d[B]
= +
= k1 [A]
dt
dt
d[A ]
= – k1 dt
[A]
Integrate both sides:
t
d[A]
= – k1  dt

t
[ A ] [A]
[ A ]2
2
1
1
ln [A] = – k1 t + C
If:
[A]0 occurs at t = 0
then:
ln [A]0 = C
Therefore:
 [A] 
 = – k1 t
ln 
 [A]0 
Or:
[A] = [A]0 exp(– k1 t)
[A] = [A]0 e
–kt
……this equation shows us that in first – order reactions there
is an exponential decrease in reactant concentration with time.
Plot of ln { [A] / [A]0 } versus t gives a straight line with a slope
that is given by – k .
Exponential Decay of [A] with Time
Chang Text. Page 449
Example (1). The Kinetics of Radioactive Decay is First – Order
222
86
Rn

218
84
Po
+
 is the helium nucleus (He

2 +)
Table 12.1 in your text gives examples of other radioactive
decay reactions.
HALF – LIFE (t
1/2)
of a Reaction
….half –life of a reaction is defined as “the time it takes for the
concentration of the reactant to decrease by half of its original
value”.
Therefore for a 1st order reaction the concentration of A would
be [A] = [A]0/2 at t = t 1/2
ln { [A] / [A] 0 } = - k t 1/2
When t = t 1/2 the equation becomes:
ln { ([A]0/2 )/ [A] 0 } = - k t 1/2
t
1/2
=
(ln 2) / k =
0.693 / k
t
1/2
=
(ln 2) / k =
….from this equation we see that t
concentration of the reactant.
1/2
0.693 / k
is independent of the initial
Thus for A to decrease from 2 M to 1 M will take just as much
time as decrease from 0.1 M to 0.05 M.
For other types of reactions the half-lives do depend on the
initial concentration of reactant.
In general the expression describing relationship between
[reactant] and half-life is as follows:
t
1/2
 ( 1 / ( [ A]0
……where n is the order of the reaction
n – 1))
Chang Text # 12.6
(a) The half-life of the first-order decay of radioactive 14C is about
5720 years. Calculate the rate constant for the reaction.
Solution:
(b) The natural abundance of 14C is 1.1. X 10-13 mol % in living matter.
Radiochemical analysis of an object obtained in an archaeological
excavation shows that the 14C isotope content is 0.89 x 10–14 mol %.
Calculate the age of the object. STATE ASSUMPTIONS MADE.
Solution
[14C] / [14C]0 = e - kt
t = - (1/k) ln {[14C] / [14C]0 }
…….the mol % of 14C for all living matter is assumed to be the same
……when matter dies, it no longer exchanges material with the
environment and the mol % of 14C will decrease according to 1st order
decay kinetics.
…..the ratio of [14C] / [14C]0 depends on time elapsed since death….
Chang Text # 12.5
A certain first-order reaction is 34.5 % complete in 49 minutes
at 298 K. What is the rate constant ?
Chang TEXT # 12.49: The activity of a radioactive sample is the
number of nuclear disintegrations per second, which is equal to the
first order rate constant times the number of radioactive nuclei
present. The fundamental unit of radioactivity is curie (Ci), where
1 Ci corresponds to exactly 3.70 x 1010 disintegrations per second.
This decay rate is equivalent to that of 1 g of Ra-226. Calculate the
rate constant and the half life for the radium decay. Starting with
1.0 g of the radium sample, what is the activity after 500 years ?
The molar mass of radium-226 is 226.03 g/mol.
Solution: