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14/07/1436 PHR 416 1 Principles and kinetics of drug stability (PHR 416) Nahla S Barakat, PhD Professor of Pharmaceutics

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14/07/1436 PHR 416 2 Molecularity Molecularity describes the mechanisms or the pathways of reaction (i.e the number of molecules, atoms or ions entering the reaction). Reactions may be: complex (multistep) reaction in which the reaction takes place in a series of steps the product of each step cannot be isolated and serves as starting material for the next step. Elementary reaction (single step reaction) The order is identical to molecularity in elementary reaction as it gives the number of molecules entering the reaction.

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14/07/1436 PHR 416 3 Half Life One way to gauge the speed of a reaction is to specify the amount of time required for a reactant concentration to be reduced to half its original value. The faster the reaction, the less time required for the concentration to be cut in half. We use the symbol t ½ to refer to the half life. The ½ serves as a label here. The concentration at the half life time (t ½ ) is half the starting concentration.

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Half life 14/07/1436 PHR 416 4 For the zero order, start with [A]t = [A] 0 – k t and substitute t = t½ and [A]t = ½[A] 0 to get ½[A] 0 = [A] 0 – k t½ ½[A] 0 - [A] 0 = -k t½ -½[A] 0 = -k t½ Dividing both sides of this equation by -k gives t ½ = 1/2 [A] 0 [A] 0 k = 2 k Since the initial reactant concentration ([A] 0 ) is in the numerator, the half life will increase with the initial concentration. The greater the concentration of the reactant, the longer it will take for half of it to be consumed

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14/07/1436 PHR 416 5 The half life equation also shows that the half life decreases with the rate constant. The rate constant is in the denominator, so the larger it is, the smaller (i.e., shorter) the half life will be. A larger rate constant means a faster reaction. The faster the reaction, the more quickly the reactant is used up, and therefore, the more quickly half of it is used up.

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14/07/1436 PHR 416 6 Shelf life (t90) Is the time required for 10% of the material to disappear; it is the time at which A has d decreased to 90% of its original concentration (i.e., 0.9A).

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Example 14/07/1436 PHR 416 7 A prescription of liquid aspirin containing 325mg/5ml (6.5 g/100 ml) is called for. The solubility of aspirin at 25 o C is 0.33g/100 ml; so the preparation will be suspension. The first order rate constant for aspirin degradation in this solution is 4.5 x 10 -6 sec -1. Calculate the zero order rate constant. Calculate the shelf life of the suspension k 0 = K[A] = (4.5 x 10-6 sec -1 ) x (0.33g/100 ml) K 0 = 1.5 x 10 -6 g/100ml. sec -1.

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14/07/1436 PHR 416 8 the first order reaction. We start with the integrated rate law ln [A]t = ln [A] 0 - kt and substitute t = t½ and [A]t = ½[A] 0 to get ln (½[A] 0 ) = ln [A] 0 – k t½ ln ½ + ln [A] 0 = ln [A] 0 – k t½ ln ½ = -k t½ Another way to think of ½ is 2 -1. Our equation can now be written : ln 2 -1 = -k t½ -ln 2 = -k t½ Now, dividing both sides by k, we write the equation, so that t ½ appears on the left side, we have t ½ = ln 2 k

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14/07/1436 PHR 416 9 It is an interesting result in that the initial reactant concentration, [A] 0 does not appear. This means that the half life of a first order reaction is independent of the starting concentration. The rate constant again appears in the denominator, just as it did in the zero order reaction. Therefore, once again, higher rate constant will give a shorter half life. A higher rate constant means a faster reaction, and if the reaction is faster, it should not take as long to consume half of the reactant.

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Shelf life (t 90 ) Is the time required for 10% of the material to disappear; it is the time at which A has decreased to 90% of its original concentration (i.e., 0.9A). t 0.9 = 0.105 k 1 14/07/1436 PHR 416 10

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14/07/1436 PHR 416 11 Example Consider the first order reaction A - - products which has a rate constant, k = 2.95 X 10 -3 s -1. What percent of A remains after 150 s ? Solution: For a first order reaction, the integrated rate law is: [A] t = [A] 0 e –kt The fraction remaining is the concentration divided by the initial concentration or [A]/[A] 0 thus fraction remaining = exp{-(2.95 X 10 -3 s -1 ) (150 s)} = 0.642 or 64.2%

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5/2/2015 PHR 416 12 N 2 O 5 decomposes according to 1 st order kinetics, and 10% of it decomposed in 30 s. Estimate k, t ½ and percent decomposed in 500 s. Solution: Assume [A] o = [N 2 O 5 ] o = 1.0, then [A] = 0.9 at t = 30 s or apply [A]t = [A] 0 e – k t 0.9 = 1.0 e – k t ln 0.9 = ln 1.0 – k 30 s – 0.1054 = 0 – k * 30 k = 0.00351 s – 1 t ½ = 0.693 / k = 197 s [A] = 1.0 e – 0.00351*500 = 0.173 Percent decomposed: 1.0 – 0.173 = 0.827 or 82.7 % After 2 t ½ (2*197=394 s), [A] = (½) 2 =¼, 75% decomposed. After 3 t ½ (3*197=591 s), [A] = (½) 3 =1/8, 87.5% decomposed.

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Assignment 3. 5/2/2015 PHR 416 13 The decomposition of A is first order, and [A] is monitored. The following data are recorded: t, min 0 2 4 8 [A], [M]0.1000.09050.08190.0670 Calculate k (What is the rate constant? k = 0.0499) Calculate the half life (What is the half life? Half life = 13.89) Calculate [A] when t = 5 min. ( What is the concentration when t = 5 min? ) Calculate t when [A] = 0.0100 M (Estimate the time required for 90% of A to decompose.)

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14/07/1436 PHR 416 14 Now let's consider the reaction that is second order in a single reactant. We take the integrated rate law multiplying the numerator and denominator by the same number (both get multiplied by 2),

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14/07/1436 PHR 416 15 Dividing both sides by k (or multiplying by 1/k and we writing the equation so that t ½ appears on the left As usual, the rate constant is in the denominator, meaning that the larger the rate constant is, the shorter the half life will be. As we have seen, this is because a larger rate constant means a faster reaction, and the faster the reaction, and less time it should take to consume half of the reactant.

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14/07/1436 PHR 416 16 Shelf life (t 90 ) Is the time required for 10% of the material to disappear; it is the time at which A has d decreased to 90% of its original concentration (i.e., 0.9A). t 0.9 = 0.11 / A 0 k

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14/07/1436 PHR 416 17 In summary then, the following features pertain to half lives: Half life always decreases as the rate constant increases. This is true regardless of the reaction order. For zero order reactions, the greater the reactant concentration, the longer the half life. For first order reactions, the half life is independent of reactant concentration For second order reactions, the greater the reactant concentration, the shorter the half life.

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14/07/1436 PHR 416 18 Example: The decomposition of sulfuryl chloride, SO 2 Cl 2 is described by the following equations: SO 2 Cl 2 (g) ----------> SO 2 (g) + Cl 2 (g) Rate = k [SO 2 Cl 2 ] At 320 o C, the rate constant k has the value 2.20 x 10 -5 s -1. If the reaction begins with SO 2 Cl 2 at an initial concentration of 1.00 x 10 -4 mol L -1, how long will it take for the SO 2 Cl 2 concentration to be reduced to 2.50 x 10 -5 mol L -1 ?

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14/07/1436 PHR 416 19 Since the half life of a first order reaction does not depend on concentration, the second half life will be the same duration as the first. In fact, every half life in a first order reaction will be the same duration as the first. So if we need an integer number of half lives (i.e., 2 half lives, 3 half lives, 4 half lives, etc), we can calculate the half life just once and then multiply by the number of half lives we need We had to wait for 2 half lives and they were each 31507 seconds in duration. Our total waiting time has been 2 (31507 s) = 63014 s 6.30 x 10 4 s

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14/07/1436 PHR 416 20 Zero-Order Reaction For a zero-order reaction, the rate of reaction is a constant. When the limiting reactant is completely consumed, the reaction abrupts stops. Differential Rate Law: r = k The rate constant, k, has units of mole L -1 sec -1. First-Order Reaction For a first-order reaction, the rate of reaction is directly proportional to the concentration of one of the reactants. Differential Rate Law: r = k [A] The rate constant, k, has units of sec -1. Second-Order Reaction For a second-order reaction, the rate of reaction is directly proportional to the square of the concentration of one of the reactants. Differential Rate Law: r = k [A] 2 The rate constant, k, has units of L mole -1 sec -1.

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Summary 14/07/1436 PHR 416 21

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example 14/07/1436 PHR 416 22

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14/07/1436 PHR 416 23 2HI(g) ----------> H2(g) + I2(g) Rate = k [HI]2 At 427 C, the rate constant k has the value 1.20 x 10-3 mol -1 L s -1 If the initial concentration of HI is 0.560 mol L-1, what will the HI concentration be after 2.00 hours of reaction time? Example: The decomposition of hydrogen iodide is described by the following equations:

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14/07/1436 PHR 416 24 Solution: Since the rate constant has the time units in seconds, we must convert the 2.00 hour of reaction time to seconds: 60 min 60 s 2.00 h x --------- x -------- = 7200 s 1 h 1 min [HI]0 = 0.560 mol L -1 and k = 1.20 x 10-3 mol -1 L s -1 From these, we can calculate [HI]t The rate law indicates that this is a second order reaction, so we use the second order integrated rate law. Substituting HI for the generic substance A in that equation, we have

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14/07/1436 PHR 416 25 [HI]t = 0.09592 mol L -1

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Summary 14/07/1436 PHR 416 26

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