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- DRUG STABILITY KINETICS
- General Outline
- Definition of drug stability and drug kinetics
- Importance of studying kinetics
- Basic math principles
- Drug kinetics reaction orders
- Determination of reaction orders
- Shelf life and half life
- Overage
- Degradation pathways
- Influence of packaging on drug stability
- Influence of temperature on drug stability
- Influence of catalysts on drug stability

1

- Definition of drug stability and drug kinetics
- Stability
- It is defined as the study of the extent

to which the properties - of a drug substance or drug product remain within

specified limits at - certain temperature. Properties may be physical,

chemical, - microbiological, toxicological or performance

properties such as - disintegration and dissolution.
- Drug Kinetics
- It is defined as how drug changes with time

i.e., study of rate - of change. Many drugs are not chemically stable

and the - principles of chemical kinetics are used to

predict the time span for - which a drug (pure or formulation) will maintain

its therapeutic - effectiveness or efficacy at a specified

temperature.

2

- Importance of studying kinetics
- It determines
- Stability of drugs (t1/2)
- Shelf life ((t0.9)
- Expiration date
- Stability of drugs (t1/2)
- The half life (t1/2) is defined as the time

necessary for a - drug to decay by 50 (e.g., From 100 to 50, 50

to 25, 20 to - 10)
- Shelf life (t0.9)
- It is defined as the time necessary for the

drug to decay to - 90 of its original concentration.

3

- Basic Math principles
- i) The straight Line
- General equation Y mx b
- Y dependent variable
- m slope
- X independent variable
- b intercept
- also
- Ordinate dependent variable axis
- abscissa independent variable axis

4

m slope ?Y / ?X

5

- Advantages of use of straight line
- Easier to determine parameters (slope and

intercept) - Simultaneous determination of two parameters (m

b) - Logarithms
- (a) Common log (base10)
- log 100 log 102 2
- log 1000 log 103 3
- (a) Natural log (base e 2.72)
- In 100 In ex
- In 100 In 2.72x 4.61

6

- Relation between Log and Ln
- Ln X 2.303 Log X
- Rules for calculating with Log
- log (a . b) log a log b
- log (a / b) log a - log b
- log an n log a
- log ex X
- (iii) Differentiation
- Determination of the rate of change ( slope

in graph)

7

Straight Line

Slope m ?Y / ?X constant

8

Curve

Slope is not constant but function of X Slope

1st derivative of y with respect to X

9

Rules of differentiation y axn

dy/dx anxn-1 e.g., y x2

dy/dx 2x y n eax dy/dx an

eax e.g., y 3e-2x dy/dx

-6e-2x y ln x dy/dx

1/x y 1/x dy/dx -

1/x2 y ex dy/dx ex

Example y 10 x3 2 x2 5x 5

dy/dx 30 x2 4 x 5

10

(iv) Integration Determination of area under

the curve i.e., sum or amount.

a

b

AUC

Where Y is the function of the graph b

upper limit a Lower limit Rules of

Integration

- Example
- Determine the area under the curve for the

relationship - y mx b, upper limit a and Lower limit 0

(No Transcript)

- If you do not know the equation of the line you

can use the trapezoidal - rule to calculate the area under curve (AUC)
- 4) Order of Reactions
- Law of mass action
- The rate of a reaction is proportional to the

molar concentrations of the reactants each raised

to power equal to the number of molecules

undergoing reaction. - a A b B

Product - Rate a Aa .Bb
- Rate K Aa .Bb
- Order of reaction sum of exponents
- Order of A a and B b
- Then Overall order a b

- Example
- The reaction of acetic anhydride with ethyl

alcohol to form ethyl acetate and water - (CH3 CO)2 2 C2H5OH 2 CH3

CO2 C2H5 H2O - Rate K (CH3 CO)2 O . C2H5OH2
- Order for (CH3 CO)2 O is 1st order
- Order for C2H5OH2 is 2nd order
- Overall order of reaction is 3rd Order

- Types of reaction orders
- Zero order reaction
- It is a reaction where reaction rate is not

dependent on the concentration of material i.e

concentration is not changing (i.e. negligible

amount of change). - Example Fading of dyes

Equation for zero order a A k

Product (P) Rate - dc/dt K c0 - dc/dt

k dc - k dt co Initial

concentration ct Concentration at time

t

18

C

T

Units of the rate constant K c co

Kt K co c /t K Concentration / time

mole / liter . second M. sec-1

19

- Determination of t1/2
- Let c co /2 and t1/2 t
- substitute in equation
- c co k t
- t1/2 co / 2K
- Note Rate constant (k) and t1/2 depend on co
- Determination of t0.9
- Let c 0.9 co and t t0.9
- substitute in equation
- c co k t
- t90 t0.9 0.1 co / k

20

- (b) First order reaction
- The most common pharmaceutical reactions e.g

drug absorption -

drug degradation - The reaction rate of change is proportional to

drug concentration i.e. - drug conc. is not constant.
- a A

k Product (P) - Rate - dc/dt K c1
- Equation

21

C co e ktDifficult to determine slope

lnco

lnc lnco kt Slope c1 c2 / t1 t2 Slope

-k

C

Lnc

Log co

Log c log co kt / 2.303 Slope c1 c2 / t1

t2 Slope -k / 2.303

Logc

Or use semi log paper

22

Semi log paper Slope -K / 2.303

Slope log c1 log c2 / t1 t2 NOT c1 c2 /

t1 t2 Units of K lnc lnco Kt K ( lnco

lnc ) / t Unit time-1

23

- Determination of t1/2
- Let t t1/2 and c co /2
- substitute in ln c ln co Kt
- t1/2 ln 2/ K 0.693 / K
- K units 0.693 / t1/2 time-1
- Determination of t0.9
- Let t t0.9 c 0.9 co
- substitute in ln c ln co Kt
- t0.9 0.105 / K and K 0.105/ t0.9

24

- Example A drug degrades according to the

following - Time (min.) Conc. ()
- 0 100
- 1 65.6
- 2 43.0
- 3 28.19
- 4 18.49
- 10 1.50
- Plot c against t on semi log paper and determine

slope, K and t1/2

25

Solution log 28.195 1.45 and log 1.5

0.176 slope 1.45 0.176 / 3 10 1.27 /

-7 - 0.181 Equation log c log co Kt /

2.303 slope -K/ 2.303 - 0.181 - K /

2.303 K 0.417 min-1 t1/2 0.693 / K t1/2

0.693 / 0.417 1.66 minute

26

- Special Case
- Apparent zero order of reaction
- In aqueous suspensions of drugs, as the dissolved
- drug decomposes more drug dissolve to maintain

drugconcentration - i.e. drug concentration kept constant, once all

undissolved drug is - dissolved, rate becomes first order.
- Another special case Pseudo 1st order
- When we have two components, one of which is

changing appreciably from its initial

concentration and the other is present in excess

that it is considered constant or nearly

constant. - Note In first order reactions, neither K or nor

t1/2 is dependent on - concentration

27

- (c) 2nd Order reaction
- When you have two components reacting with

each - other or one component reacting with itself.

- 2nd order graph

Units of K 1/C 1/Co Kt K (1/C - 1/Co)

/ t K M-1. sec -1

i.e, K is dependent on initial drug

concentration. Half life t1/2 1 /

KCo Shelf life t0.9 0.11 / KCo

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