SECTION 9: Kinetics

Chapter 12 in Chang Text

Outline for Section 9 Outline for Section 9 –– Part IPart 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 processesinvolved 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 underspecific 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 isof critical importance

• decomposition may proceed by hydrolysis, oxidation, isomerization,epimerization, photolysis etc.

• many groups study the effect of ingredients of dosage forms orformulations 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 plasticand 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 ofdrug.

• decomposition followed first order kinetics at constanttemperature

• pH – rate profile demonstrated that maximum stability wasachieved at about pH = 4.5

Doxorubicin• DOX is an anthracycline……includeschromophoric anthraquinone moietyand a charged sugar within structure

• anthracyclines are generally lightsensitive (many have investigatedphotodegradation of DOX when exposed to room light)

• exposure to light, increase in pH andand adsorption to container known toresult in DOX loss (decrease in concentrationof 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

DOXIL (Caelyx in Canada)

Liposome -encapsulated Doxorubicin

can make liposomessuch thatinternal pH = 4.5

• Doxil is approved for refractory ovarian cancer and AIDs-relatedKaposi’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 ofhigh concentrations of DOX under the skin……………

………….actually one of the dose-limiting toxicities of DOXIL iscalled “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 aggregateformation (DOX-DOX aggregates)

• low pH of the intra-liposomal aqueous phase

Rate of Reaction of Drugs with Biological Components

Cytosol

MT

• Cisplatin (CDDP) is an anti-cancer drug thatexerts anti-cancer activityby binding to cellular DNA

• drug enters the cell, passesthrough cytosol and entersnucleus where it binds DNA

• CDDP in cytosol may also bind metallothionein (MT) and other endogenous thiols

• binding of CDDP to MT limits amount available for binding to DNAtherefore 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)

Figure 1: Schematic of CDDP entry into cell, passage through cytosoland binding to DNA in nucleus. (not drawn to scale)

nucleus

mitochondria

Cisplatin(CDDP)

• Hagrman et al. (Drug Metabolism and Disposition 2003)studied the kinetics of the reaction of CDDP with MT

• their studies included characterizationof the rate of reaction between CDDPand MT under specific conditions

also dependence of reaction rate on concentration of CDDP and MT wasexamined.

……..this data is critical as it provides understanding of ability of MT to trap CDDP……….and thus alter the therapeutic effect of this drug.

Time (hours)0 1 2 3 4 5

Dru

g C

once

ntra

tion

in P

lasm

a (u

g /m

L)

0

50

100

150

200

250

300

350Liposome Encapsulated DrugFree Drug

Liposome-encapsulated drug

Free drug

Sample blood at time points t = 15 mins, 30 mins, 1 hr, 4 hrs etc.

Rate of Clearance of Drug from Circulation

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 intervalbecause 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…………….

2 R P

……..in this case the reactant disappears twice as fast as the productappears…………

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: a A + b B c C + d D

In general:rate [A]x [B] y

= k [A]x [B] y

…….where k is the rate constant.

The rate constant does not depend on concentrations it is only dependent on Temperature.

The Rate Law :rate is proportional to conc.of reactants raised to somepower

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 andy order with respect to B.

………..the reaction has an overall order of x + y

IMPORTANT: in general there is NO relationship betweenorder 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 reactantconcentration.

d [A] = - k dt

Integration between t = 0 and t = t at concentrations of [A]t=0

and [A]t gives the following expression:

d [A] = [A] t – [A] t=0 = - kdt = - kt

[A] t = [A] t=0 - kt

[A] = [A]0 - kt

[A] t

[A] t=0

t

0

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 frombeing zero as [CH3CH2OH] approaches zero.

For a zero-orderreaction 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 thefirst power.

….in this case the units of k are s-1

k1 A → B

– dt

]A[d = + dt

]B[d = k1 [A]

]A[]A[d = – k1 dt

Integrate both sides:

2

1

]A[

]A[ ]A[]A[d = – k1

2

1

t

t

dt

ln [A] = – k1 t + C

If: [A]0 occurs at t = 0 then:

ln [A]0 = C

Therefore: ln

0]A[]A[ = – k1 t

Or:

[A] = [A]0 exp(– k1 t)

[A] = [A]0 e –kt

……this equation shows us that in first – order reactions thereis an exponential decrease in reactant concentration with time.

Plot of ln { [A] / [A]0 } versus t gives a straight line with a slopethat is given by – k .

Exponential Decay of [A] with Time

Chang Text. Page 449

Example (1). The Kinetics of Radioactive Decay is First – Order

Rn Po + 222 218

86 84

is the helium nucleus (He 2 +)

Table 12.1 in your text gives examples of other radioactivedecay reactions.

HALF – LIFE (t 1/2) of a Reaction

….half –life of a reaction is defined as “the time it takes for theconcentration of the reactant to decrease by half of its originalvalue”.

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 = 0.693 / k

….from this equation we see that t 1/2 is independent of the initial concentration of the reactant.

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 n – 1))

……where n is the order of the reaction

Chang Text # 12.6

(a) The half-life of the first-order decay of radioactive 14C is about5720 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 orderdecay 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 minutesat 298 K. What is the rate constant ?

Chang TEXT # 12.49: The activity of a radioactive sample is thenumber of nuclear disintegrations per second, which is equal to thefirst order rate constant times the number of radioactive nucleipresent. The fundamental unit of radioactivity is curie (Ci), where1 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 therate constant and the half life for the radium decay. Starting with1.0 g of the radium sample, what is the activity after 500 years ?The molar mass of radium-226 is 226.03 g/mol.

Solution: