Transcript PHT-224 Lectures 6

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+ b
Y = 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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Straight Line:
Slope = m = ∆Y / ∆X = constant
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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
e.g.,
e.g.,
y = axn
dy/dx = anxn-1
y = x2
dy/dx = 2x
y = n eax
dy/dx = an eax
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.
a
b
AUC =
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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.
aA+bB
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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C
T
Units of the rate constant K:
c = co – Kt
K = co – c /t
K = Concentration / time
= mole / liter . second
= M. sec-1
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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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lnco
C = co e –kt
Difficult to determine
slope
C
Lnc
lnc = lnco – kt
Slope = c1 – c2 / t1 – t2
Slope = -k
Log co
Log c = log co – kt / 2.303
Slope = c1 – c2 / t1 – t2
Slope = -k / 2.303
Logc
Or use semi log paper
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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
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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:
Shelf life:
t1/2 = 1 / KCo
t0.9 = 0.11 / KCo
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