# DRUG STABILITY

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1 STABILITY CALCULATION AND RATE EQUATION SRAVANKUMAR NARLA M PHARMACY ,PHARMACEUTICS VIKAS COLLEGE OF PHARMACY

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2 2 General Outline Definition of drug stability and drug kinetics Influence of temperature on drug stability Importance of studying kinetics Shelf life and half life Basic math principles Drug kinetics reaction orders Determination of reaction orders

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3 Definition of drug stability and drug kinetics Stability Drug Stability refers to the capacity of a drug substance or product to remain within established specifications of identity, strength quality, and purity in a specified period of time. factors that affect drug stability include temperature condition, moisture, light, microbes, packaging materials, transportation, components of drug composition and the nature of the active ingredient . .

### Arrhenius Equation:

Arrhenius Equation Arrhenius developed a mathematical relationship between k and E a : where A is the frequency factor , a number that represents the likelihood that collisions would occur with the proper orientation for reaction. 4

### Arrhenius Equation:

Arrhenius Equation 5 Taking the natural logarithm of both sides, the equation becomes y = mx + b When k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. 1/T

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6 Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy , E a Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier .

### Accelerated stability studies:

7 Accelerated stability studies Stability study is to predict the shelf life of a product by accelerating therate of decomposition, preferably by increasing the temperature of reaction conditions . With the advancement in branch of kinetics shelf life of dosage form can be predicted with in months based on accelerated stability reports

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8 Storage condition of 40 0 C and relative humidity of 75% has been recommended for all the four zones for drug substances and drug products. Studies carried out for 6 months . Accelerated storage conditions must be at least 15 0 C above the expected actual storage temperature and appropriate relative humidity

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9 the rapid detection of deterioration in different initial formulations of the same product - this is used in selecting the best formulation from a series of possible choices the prediction of shelf life, which is the time a product will remain satisfactory when stored under expected or directed storage condition; Objectives of accelerated stability tests :

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10 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 The rate, velocity or speed of a reaction is given by + or – dc/dt Here dc is small changes in the concentration with in a given time interval dt .

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11 Importance of studying kinetics It determines: Stability of drugs (t 1/2 ) Shelf life ( (t 0.9 ) Expiration date Stability of drugs (t 1/2 ) The half life (t 1/2 ) 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 (t 0.9 ) It is defined as the time necessary for the drug to decay to 90% of its original concentration.

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12 Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0 Half-Life

### SHELF LIFE:

SHELF LIFE 13 It is defined as the time required for the concentration of the reactant To reduce to 90% of its initial concentration . As per defination the term in equation change to C =90a/100 ; t =t 90 Shelf life is represented as t 90

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14 14 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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15 15 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) Logarithms: (a) Common log (base10) log 100 = log 10 2 = 2 log 1000 = log 10 3 = 3 (a) Natural log (base e = 2.72) In 100 = In e x In 100 = In 2.72 x = 4.61 16 16

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17 17 Slope = m = ∆Y / ∆X = constant Straight Line:

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18 18 Curve: Slope is not constant but function of X Slope = 1 st derivative of y with respect to X

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19 Integration Determination of area under the curve i.e., sum or amount. AUC = a b Where; Y is the function of the graph b = upper limit a = Lower limit

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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 20

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Example: The reaction of acetic anhydride with ethyl alcohol to form ethyl acetate and water (CH 3 CO) 2 + 2 C 2 H 5 OH 2 CH 3 CO 2 C 2 H 5 + H 2 O Rate = K [(CH 3 CO) 2 O] . [C 2 H 5 OH] 2 Order for (CH 3 CO) 2 O is 1 st order Order for [C 2 H 5 OH] 2 is 2 nd order Overall order of reaction is 3 rd Order 21

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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 , 22 - dc/dt =Ko Oxidation of vitamin A in an oily solution etc ….

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23 23 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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24 24 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 t 1/2 = t substitute in equation; c = co – k t t 1/2 = co / 2K Note: Rate constant (k) and t 1/2 depend on co Determination of t 0.9 Let c = 0.9 co and t= t 0.9 substitute in equation; c = co –k t t 90% = t 0.9 = 0.1 co / k 25 25

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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: 26 26

### C = co e –kt Difficult to determine slope:

C = co e –kt Difficult to determine slope 27 27 lnc = lnco – kt Slope = c 1 – c 2 / t 1 – t 2 Slope = -k lnco Log co Log c = log co – kt / 2.303 Slope = c 1 – c 2 / t 1 – t 2 Slope = -k / 2.303 C Lnc Logc

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Determination of t 1/2 Let t = t 1/2 and c = co /2 substitute in ln c = ln co – Kt t 1/2 = ln 2/ K = 0.693 / K K units = 0.693 / t 1/2 = time -1 Determination of t 0.9 Let t = t 0.9 c = 0.9 co substitute in ln c = ln co – Kt t 0.9 = 0.105 / K and K = 0.105/ t 0.9 28 28

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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 29 29

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30 30 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 t 1/2 = 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 1 st 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 t 1/2 is dependent on concentration 31 31

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2 nd Order reaction When you have two components reacting with each other or one component reacting with itself . 32

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2 nd order graph 33 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: t 1/2 = 1 / KCo Shelf life: t 0.9 = 0.11 / KCo

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