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DRUG STABILITY & KINETICS

General Outline

1) Definition of drug stability and drug kinetics

2) Importance of studying kinetics

3) Basic math principles

4) Drug kinetics reaction orders

5) Determination of reaction orders

6) Shelf life and half life

7) Overage

8) Degradation pathways

9) Influence of packaging on drug stability

10) Influence of temperature on drug stability

11) Influence of catalysts on drug stability

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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.

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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.

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3) Basic Math principles

i) The straight Line:

General equation: Y = mx+ bY = dependent variable

m = slope

X = independent variable

b = intercept

also

Ordinate = dependent variable axis

abscissa = independent variable axis

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m = slope = ∆Y / ∆X

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Advantages of use of straight line

Easier to determine parameters (slope and intercept)

Simultaneous determination of two parameters (m + b)

ii) 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

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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)

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Slope = m = ∆Y / ∆X = constant

Straight Line:

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Curve:

Slope is not constant but function of X

Slope = 1st derivative of y with respect to X

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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

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(iv) Integration

Determination of area under the curve i.e., sum or amount.

AUC =

a b

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Where; Y is the function of the graph

b = upper limit

a = Lower limit

Rules of Integration:

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Example

Determine the area under the curve for the relationship

y = mx + b, upper limit = a and Lower limit = 0

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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 α [A]a .[B]b

Rate = K [A]a .[B]b

Order of reaction = sum of exponents

Order of A = a and B = b

Then Overall order = a + b

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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] . [C2H5OH]2

Order for (CH3 CO)2 O is 1st order

Order for [C2H5OH]2 is 2nd order

Overall order of reaction is 3rd Order

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Types of reaction orders

(a)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

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Equation for zero order:

a [A] k Product (P)

Rate = - dc/dt = K [c]0

- dc/dt = k dc = - k dt

co = Initial concentration

ct = Concentration at time t

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Units of the rate constant K:

c = co – Kt

K = co – c /t

K = Concentration / time

= mole / liter . second

= M. sec-1

C

T

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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

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(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 [c]1

Equation:

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C = co e –kt

Difficult to determine slope lnc = lnco – kt

Slope = c1 – c2 / t1 – t2

Slope = -k

lnco

Log co

Log c = log co – kt / 2.303Slope = c1 – c2 / t1 – t2

Slope = -k / 2.303

Or use semi log paper

C Lnc

Logc

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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

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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

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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

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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

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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

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(c) 2nd Order reaction

When you have two components reacting with each

other or one component reacting with itself.

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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