1. Kinetics By Pharmacist Muhanad S. Al Ani

2. Rate and orders of reactions: The rate, velocity, or speed of a reaction is given by the expression dc/dt dc: is the increase or derease in the conc. dt: is the time interval of this increase or derease. so for the reaction A B reactant product Rate of disappearance = - dA / dt Rate of appearance = dB / dt Also aA + bB +….. Products a : is the no. of molecules of substance A b: is the no. of molecules of substance B

3. The rate of reaction = K[A] [B] K : is the rate constant. If we have a reaction A + B product rate = K[A] [B] so the reaction is said to be 1 st order in relation to substance A, and also 1 st order in relation to substance B and the over all reaction is 2 nd order. a b 11

4. Molecularity: is the no. of molecules, atoms or ions that react to give the product. Ex. Br 2 2Br is unimolecular since single molecule of Br 2 is decomposes to form two bromine atoms. Ex. H 2 + I 2 2HI this reaction is bimolecular since 2 molecules react to give the product. Complex reaction: is the reaction that proceed through more than one step.

5. Zero – order kinetics In this case the rate of the reaction is constant with time and not depend on the conc. of the reactant. A B K = - dA / dt = K o the initial conc. That is A o is a The conc. at time t is c At = A o – K o t or Ct = a - K o t so if we plot this equation in which C on the vertical axis and t on the horizontal axis rate time Slop = - K o Ct time AoAo

6. The units of the rate constant in this case: K = - = = = mole liter second Half life : is the time required for one half of the material to disappear so it is the time at which the conc. A is 0.5A t 1/2 = shelf life : time at 90% of the initial conc. remain i.e. 10% of the drug disappear. t 90% = 0.1 C o / K o dA dt Mole / liter second - 1 Liter second Mole - 1 0.5 C o KoKo

7. Suspensions: are apparent zero – order kinetic ?? suspension are an example on zero order kinetic in which the conc. of the drug in solution depends on the drug solubility, in that as the drug decomposes in solution more drug is released from the suspension particles so that the conc. remains constant. The important point is that the conc. of the drug in solution remains constant despite its decomposition with time. The reservoir of the solid drug in suspension is responsible for this constancy.

8. For any solution with no reservoir of drug to replace that decomposed drug is 1 st order kinetic. [A] : is the conc. of the drug remaining undecomposed at time t. k : is the 1 st order rate constant. When the conc. remain constant i.e. A is constant K o = K [A] Or K o = - d[A] / dt this equation appear as a zero – order kinetic. so it is referred to as an apparent zero – order equation because the suspended drug reservoir that ensures constant conc. - d[A] dt = K [A]

9. Example: a prescription for a liquid aspirin preparation is called for to contain 6.5g/100 ml, the solubility of aspirin at 25C˚ is 0.33g/100 ml therefore the preparation will be definitely be a suspension. The 1 st order rate constant for aspirin degradation in this solution is 4.5 x 10 sec Calculate the zero order rate constant and the shelf life for this liquid preparation. -6

10. First – Order Reaction In this case the rate of the reaction is proportional to the concentration of the substance remaining in the reaction mixture at time t. K c K : is the first –order rate constant. C: is the concentration of drug remain undecomposed at time t. So by integrating the equation 1:1 C 0 at time t = 0 and C t at time t

11. In C – In C 0 = - k (t - 0) In C = In C 0 - kt By converting to the logarithem Log C = Log C 0 – k = log Half – life in first order kinetic t ½ = shelf –life t 90 % = Slope = log conc. Time

12. The unit of rate constant K [A] = - K= - K= = sec ̄ Example: A solution of a drug contained 500 unit/ml when prepared. It was analyzed after a period of 40 days and was found to contain 300 unit/ml. assuming the decomposition is 1st order. At what time will the drug have decomposed to one half its original concentration.

13. second – order kinetic A + B → product so in this case we have bimolecular reaction. if a : is initial concentration of drug A b : is initial concentration of drug B x : is the concentration of each drug reacting at time t conc.=conc. a= b

14. Half – life = Unit for second order kinetic = liter mole ̄ ¹ second ̄ ¹ Example : The initial concentration of both ethylacetate and Sodium hydroxide in the mixture were 0.01M. the change in concentration of the alkali during 20 min. was 0.00566 mole/L. Calculate (a) rate constant (b) t1/2

15. Determination of order We have several methods: 1) Substitution Method Data accumulated from kinetic study substituted in the equation that describe the various order. 2) Graphic Method We plot the data of kinetic study in the from of a graph. a) if the concentration plotted against (t) results in straight line, its zero order. b) if (a – x) plotted against (t) results in straight line, its 1 st order. c) if plotted against (t) results in straight line, its second order if (a=b).

16. 3) Half – life Method in zero order t1/2 is proportional to a in 1 st order t1/2 is independent on a in second order t1/2 is proportional to 1/a if a=b

17. Factors that affect on the rate of reaction: 1) Temperature The speed of many reactions increases about 2-3 times with each 10 C ̊ rise in temperature. The effect of temperature on reaction rate is given by Arrhenius equation K = A e or log K = log A A: Arrhenius factor or frequency factor Ea: Energy of activation R: Gas constant = 1.987 calorie/deg mole T: Absolute temperature - Ea/ RT

18. Log K Log A Slope = So at any temperature we have a rate constant, if we have 2 temperature mean 2 rate constant log Example: The rate constant for the decomposition of 5- hydroxymethyle furfural at 120 C ̊ is 1.173 hr ̄ ¹ and at 140 C ̊ is 4.860 hr ̄ ¹ what is the activation energy in Kcal/mole and the frequency factor in sec ̄ ¹ for this breakdown.

19. 2) Specific Acid – Base catalysis Solution of a number of drugs undergo accelerated decomposition upon the addition of acids or bases. When the rate law for such an accelerated decomposition is found to contain a term involving the concentration of hydrogen or hydroxyl in the reaction is said to be subject to specific acid – base balance.

20. 3) Influence of light (photodegradation) Light is not classified as a catalyst, light energy, like heat may provide the activation necessary for a reaction to occur.