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, 1 st 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 rateof decomposition, degradation, biotransformation, and inactivation of a drug rate of precipitation of a drug from a formulation under specific conditions rate of releaseof a drug from a delivery system rate of uptakeof drug into cells rate of reaction between drug and biological target (or other molecules) rate of clearanceof 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 MT Cytosol nucleus 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 Liposome -encapsulated drug Free drug Drug Concentration in Plasma (ug /ml) 350 300 250 200 150 100 50 Sample blood at time points t = 15 mins, 30 mins, 1 hr, 4 hrs etc. 0 Liposome Encapsulated Drug Free Drug 0 1 2 3 4 5 Time (hours)
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 (t 2 t 1 ) may be expressed as Follows: {[R] 2 -[R] 1 } / {t 2 t 1 } = [R] / t..where [R] 1 is the concentration of R at t 1 and [R] 2 is the concentration of R at t 2 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 } / {t 2 t 1 } = [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. 2 R 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: a A + b B In general: rate [A] x [B] y = k [A] x [B] y.where k is the rate constant. c C + d D 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 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g)..the Rate Law for this reaction is known to be: Rate = k [N 2 O 5 ] reaction is first-order with respect to N 2 O 5 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] t=0 = - kdt = - kt [A] t=0 t 0 [A] t = [A] t=0 - kt [A] = [A] 0 - kt
Example: Conversion of ethanol to acetaldehyde by the enzyme LADH (liver alcohol dehydrogenase). Oxidizing agent is NAD + (nicotinamide adenine dinucleotide): LADH CH 3 CH 2 OH + NAD + CH 3 CHO + 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, k 0. The order must eventually change from being zero as [CH 3 CH 2 OH] 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
k 1 A B d[a] dt = + d[b] dt = k 1 [A] d[a] [A] = k 1 dt Integrate both sides: [ A ] d[a] 2 [ A] 1 [A] = k 1 2 dt t t 1 ln [A] = k 1 t + C
If: [A] 0 occurs at t = 0 then: ln [A] 0 = C Therefore: ln [A] [A] 0 = k 1 t Or: [A] = [A] 0 exp( k 1 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 222 218 Rn Po + 86 84 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 1 st 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] n 1 0 )) where n is the order of the reaction
Chang Text # 12.6 (a) The half-life of the first-order decay of radioactive 14 C is about 5720 years. Calculate the rate constant for the reaction. Solution: (b) The natural abundance of 14 C is 1.1. X 10-13 mol % in living matter. Radiochemical analysis of an object obtained in an archaeological excavation shows that the 14 C isotope content is 0.89 x 10 14 mol %. Calculate the age of the object. STATE ASSUMPTIONS MADE.
Solution [ 14 C] / [ 14 C] 0 = e -kt t = - (1/k) ln {[ 14 C] / [ 14 C] 0 }.the mol % of 14 C 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 14 C will decrease according to 1 st order decay kinetics...the ratio of [ 14 C] / [ 14 C] 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 10 10 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: