Lecture 8 (2) PDF - Reaction Kinetics & Drug Stability

 Chemical (Stability) Kinetics → the study of the rate of chemical changes taking place during chemical reactions and the factors that influence such rates.  Kinetics in pharmaceutical formulations include: - Study of physical and chemical reactions in drugs and dosage forms. - Fac...

 Chemical (Stability) Kinetics → the study of the rate of chemical changes taking place during chemical reactions and the factors that influence such rates.  Kinetics in pharmaceutical formulations include: - Study of physical and chemical reactions in drugs and dosage forms. - Factors affecting rate of these chemical reactions. - Accelerated stability testing. - Prediction of formulations shelf life.  Rate of chemical reaction → the velocity with which a reactant undergo a chemical change.  Rate of chemical reaction → the change in the concentration of a reactant or product as a function of time.  Reaction rate is given by: + dc/dt - The + or - sign indicates the increase or decrease in concentration (dc) within a time interval (dt).  The rate of a chemical reaction is proportional to the product of the molar conc. of the reactants each raised to a power equal to the number of molecules of the substance undergoing reaction.  Reaction: A+B→C+D ∴ Rate α (A) (B) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 ∴ 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = − = + = + = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 - k → Reaction rate constant.  Reaction: aA + bB → Products ∴ Rate α (A)a (B)b 𝟏𝟏 𝒅𝒅𝒅𝒅 𝟏𝟏 𝒅𝒅𝒅𝒅 ∴ 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = − = 𝒌𝒌 (𝑨𝑨)𝒂𝒂 (𝑩𝑩)𝒃𝒃 𝒂𝒂 𝒅𝒅𝒅𝒅 𝒃𝒃 𝒅𝒅𝒅𝒅 - k → Reaction rate constant.  Order of reaction → Sum of the powers of concentration terms in the rate equation.  Reaction: A+B→C+D ∴ Rate α (A) (B) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 ∴ 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = − = + = + = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩) 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 - Order with respect to reactant (A) → 1st Order. - Order with respect to reactant (B) → 1st Order. - Overall reaction order = 1 + 1 → 2nd Order.   Reaction: A + 2B → Products ∴ Rate α (A)1 (B)2 𝒅𝒅𝒅𝒅 𝟏𝟏 𝒅𝒅𝒅𝒅 ∴ 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = − = 𝒌𝒌 (𝑨𝑨)(𝑩𝑩)𝟐𝟐 𝒅𝒅𝒅𝒅 𝟐𝟐 𝒅𝒅𝒅𝒅 - Order with respect to reactant (A) → 1st Order. - Order with respect to reactant (B) → 2nd Order. - Overall reaction order = 1 + 2 → 3rd Order.  Order of reaction → determine the way in which the conc. of a reactant or reactants influences the rate of reaction.  The reaction rate is independent on concentration of reactants.  The reaction has → Constant Rate.  Rate equation: 𝒅𝒅𝑪𝑪 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = 𝒌𝒌 (𝑪𝑪)𝟎𝟎 = 𝒌𝒌 𝒅𝒅𝒅𝒅 Where: o dc → change in reactant conc. with respect to change in time t. o “-” → indicate the reactant conc. is decreasing. o k → zero order rate constant. Integrated Rate Equation Graphical Plot  Plot “Ct” on y-axis and “t” on 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 x-axis → straight line (linear).  Linear equation. C0  Where: -K o Ct → Conc. remaining Ct at time t. o Co → Initial conc. o k → Zero order rate t constant. o Intercept → Co (initial conc.). o t → Time. o Slope → - k. Equation for “k” Unit of “k” calculation  Conc./time. 𝑪𝑪𝟎𝟎 − 𝑪𝑪𝒕𝒕  Ex: 𝒌𝒌 = o Mole/liter.sec (mole lit.-1 sec-1). 𝒕𝒕 o Gram/liter.hr (gram lit-1 hr-1).  Half life (t1/2 or t50%) Shelf life or Expiry date (t90%)  Time required for initial drug  Time required for 10% of drug to conc. to be decreased to be degraded (90% of its original half → ∴ Ct = ½ Co. conc. remaining) → ∴ Ct = 0.9 Co.  Substitute in integrated  Substitute in integrated equation equation & rearrange: & rearrange: 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝟏𝟏 ∴ 𝑪𝑪 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒕𝒕𝟏𝟏 ∴ 𝟎𝟎. 𝟗𝟗 𝑪𝑪𝒐𝒐 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒕𝒕𝟗𝟗𝟗𝟗% 𝟐𝟐 𝒐𝒐 𝟐𝟐 𝟏𝟏 𝑪𝑪𝟎𝟎 − 𝑪𝑪 𝑪𝑪𝟎𝟎 − 𝟎𝟎. 𝟗𝟗 𝑪𝑪𝒐𝒐 ∴ 𝒕𝒕𝟏𝟏 = 𝟐𝟐 𝒐𝒐 ∴ 𝒕𝒕𝟗𝟗𝟗𝟗% = 𝟐𝟐 𝒌𝒌 𝒌𝒌 𝟏𝟏 𝟎𝟎. 𝟏𝟏𝑪𝑪𝒐𝒐 𝑪𝑪𝒐𝒐 𝑪𝑪𝒐𝒐 𝑪𝑪𝒐𝒐 ∴ 𝒕𝒕𝟏𝟏 = 𝟐𝟐 = ∴ 𝒕𝒕𝟗𝟗𝟗𝟗% = = 𝒌𝒌 𝟐𝟐𝟐𝟐 𝒌𝒌 𝟏𝟏𝟏𝟏𝒌𝒌 𝟐𝟐 o Both half life & shelf life are directly proportional to initial conc.  Fading of color of some drugs (Ex: Multi-Sulpha Drugs) → follow a zero order rate.  Color loss is measured by the decrease in spectrophotometric absorbance at wavelength 500 nm.  Rate of color fading → is Constant & Independent on drug conc.  Decomposition of drugs in solution → follow 1st order kinetics. 𝒅𝒅𝒅𝒅 − = 𝒌𝒌𝟏𝟏 (𝑪𝑪) 𝒅𝒅𝒅𝒅 - k1 → 1st order degradation rate constant. - C → Conc. of remaining drug at time (t).  In suspension → drug conc. exceeds its maximum solubility.  As long as there is excess suspended solid particles → the conc. of drug in solution remains constant → WHY??? o Because when the drug in solution decompose → more drug is released from the suspended drug particles reservoir. o Drug conc. in solution remains constant. o Degradation become an Apparent (Pseudo) zero order reaction.  N.B: Drug conc. in solution (C) = Saturation solubility of drug (S) in this solvent.  Since the conc. (C) become constant: 𝒅𝒅𝒅𝒅 ∴ − = 𝒌𝒌𝟎𝟎 𝒅𝒅𝒅𝒅 o Where: 𝒌𝒌𝟎𝟎 = 𝒌𝒌𝟏𝟏 (𝑪𝑪) = 𝒌𝒌𝟏𝟏 (𝑺𝑺) - k1 → 1st order degradation rate constant. - S → Saturation solubility of drug in solvent.  Once all the suspended drug particles are dissolved and converted to solution → the system changes to 1st order degradation again. 1) In a zero order reaction, if the initial conc. was 10 mole/100 ml and after 10 hrs the conc. was 8 mole /100 ml. What was the conc. after 5 hrs? Solution: - Co = 10 mole/ 100 ml. - Ct = 8 mole/ 100 ml. 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝟖𝟖 = 𝟏𝟏𝟏𝟏 − (𝒌𝒌 𝒙𝒙 𝟏𝟏𝟏𝟏) ∴ 𝒌𝒌 = 𝟎𝟎. 𝟐𝟐 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎⁄𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎. 𝒉𝒉𝒓𝒓 - Ct after 5 hrs: 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝑪𝑪𝒕𝒕 = 𝟏𝟏𝟏𝟏 − (𝟎𝟎. 𝟐𝟐 𝒙𝒙 𝟓𝟓) = 𝟗𝟗 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎⁄ 𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎 2) The decomposition of a multi-sulfa compound, was found to follow zero order kinetics. If its concentration was 0.47 mole/liter, when freshly prepared, and after 473 days its concentration reached 0.225 mole/liter, calculate its degradation rate constant and t1/2? Solution: - Co = 0.47 mole/L. - Ct = 0.225 mole/L. 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐 = 𝟎𝟎. 𝟒𝟒𝟒𝟒 − (𝒌𝒌 𝒙𝒙 𝟒𝟒𝟒𝟒𝟒𝟒) ∴ 𝒌𝒌 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝑳𝑳−𝟏𝟏 𝒅𝒅𝒅𝒅𝒅𝒅−𝟏𝟏 - Half life: 𝑪𝑪𝒐𝒐 𝟎𝟎. 𝟒𝟒𝟒𝟒 𝒕𝒕𝟏𝟏 = = = 𝟒𝟒𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 = 𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝟐𝟐 𝟐𝟐𝟐𝟐 𝟐𝟐 𝒙𝒙 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 3) A prescription for a liquid aspirin preparation is called for 325 mg/5ml. The solubility of aspirin at 25oC is 0.33g /100ml; the 1st order rate constant for aspirin degradation in the solution is 4.5 x 10-6 sec-1. Determine the shelf life (t90%) for the liquid prescription at 25oC? Solution: - 325 mg aspirin → 5 ml ∴ 6500 mg → 100 ml. ∴ Aspirin conc. = 6.5 g / 100 ml. - Since aspirin conc. >>> solubility (0.33 g/ 100 ml) → ∴ it is a suspension follow zero order. - So, 𝒌𝒌𝟎𝟎 = 𝒌𝒌𝟏𝟏 (𝑺𝑺) = 𝟒𝟒. 𝟓𝟓 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟔𝟔 𝒙𝒙 𝟎𝟎. 𝟑𝟑𝟑𝟑 = 𝟏𝟏. 𝟓𝟓 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟔𝟔 𝒈𝒈⁄𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎. 𝒔𝒔𝒔𝒔𝒔𝒔. - Shelf life: 𝑪𝑪𝒐𝒐 𝟔𝟔. 𝟓𝟓 𝟓𝟓 𝒔𝒔𝒔𝒔𝒔𝒔. = 𝒕𝒕𝟗𝟗𝟗𝟗% = = −𝟔𝟔 = 𝟒𝟒. 𝟑𝟑 𝒙𝒙 𝟏𝟏𝟏𝟏 𝟓𝟓 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 𝟏𝟏𝟏𝟏𝒌𝒌𝟎𝟎 𝟏𝟏𝟏𝟏 𝒙𝒙 𝟏𝟏. 𝟓𝟓 𝒙𝒙 𝟏𝟏𝟏𝟏  The reaction rate is Directly proportional (Dependent) on the remaining concentration of single reactant.  The reaction has → Decreasing Rate with time.  Rate equation: 𝒅𝒅𝒅𝒅 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 = − = 𝒌𝒌 (𝑪𝑪) 𝒅𝒅𝒅𝒅 Where: o dc → change in reactant conc. with respect to change in time t. o “-” → indicate the reactant conc. is decreasing. o k → first order rate constant. o C → remaining conc. of reactants at time t.  Integrated Rate Equation & Graphical Plot 𝒌𝒌𝒌𝒌 𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝒌𝒌𝒌𝒌 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑  Plot “Ln Ct” Vs “t” → straight  Plot “Log Ct” Vs “t” → straight line (linear). line (linear). o Intercept → Ln Co. o Intercept → Log Co. o Slope → - k. o Slope → - k/2.303.  Relation between Conc. & Time in 1st order reaction  Relation between “Conc.” and “Time” → Exponential (Not linear). 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 𝒆𝒆−𝒌𝒌𝒌𝒌 𝑪𝑪𝒕𝒕 = 𝑪𝑪𝟎𝟎 𝟏𝟏𝟏𝟏−𝒌𝒌𝒌𝒌⁄𝟐𝟐.𝟑𝟑𝟑𝟑𝟑𝟑  Plot “Ct” Vs “t” → Curve → Indicate exponential relationship.  Conc. starts at Co with highest decomposition rate (seen from line slope).  Conc. decreases exponentially → reaction rate decreases → reaction become very slow due to conc. dependence.  Equation for “k” Unit of “k” calculation  Time-1. 𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕  Ex: 𝒌𝒌 = o Sec-1. 𝒕𝒕 o Min-1. 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝟎𝟎 𝒌𝒌 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝒕𝒕 𝑪𝑪𝒕𝒕  Half life (t1/2 or t50%) Shelf life or Expiry date (t90%)  ∴ Ct = ½ Co.  ∴ Ct = 0.9 Co.  Substitute in integrated  Substitute in integrated equation & rearrange: equation & rearrange: 𝒌𝒌𝒌𝒌 𝒌𝒌𝒌𝒌 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏 𝒌𝒌𝒕𝒕𝟏𝟏⁄𝟐𝟐 𝒌𝒌𝒕𝒕𝟗𝟗𝟗𝟗% ∴ 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒐𝒐 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − ∴ 𝑳𝑳𝑳𝑳𝑳𝑳 𝟎𝟎. 𝟗𝟗 𝑪𝑪𝒐𝒐 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝟐𝟐 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝟎𝟎 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝟎𝟎 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟏𝟏 ∴ 𝒕𝒕𝟏𝟏 = 𝑳𝑳𝑳𝑳𝑳𝑳 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟐𝟐 ∴ 𝒕𝒕𝟗𝟗𝟗𝟗% = 𝑳𝑳𝑳𝑳𝑳𝑳 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟐𝟐 𝒌𝒌 𝟏𝟏 𝟐𝟐 𝑪𝑪𝟎𝟎 𝒌𝒌 𝒌𝒌 𝟎𝟎. 𝟗𝟗𝑪𝑪𝟎𝟎 𝒌𝒌 𝟎𝟎. 𝟗𝟗 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏 ∴ 𝒕𝒕𝟏𝟏 = ∴ 𝒕𝒕𝟗𝟗𝟗𝟗% = 𝟐𝟐 𝒌𝒌 𝒌𝒌 o Both half life & shelf life are Independent on initial conc. (constant).  Fractional times required for the decomposition of certain drug fraction (t1/2 and t90%) is more appreciated than the time for a substance to decompose completely (Life time) → Why?? o Theoretically, it takes an infinite period of time for a process to subside completely. o So, time required for complete disintegration has no meaning. o Except → zero order reaction → process reach completion (reactants reach zero conc.).  Decomposition of hydrogen peroxide (catalyzed by KI) → is directly proportional to remaining conc. of H2O2 at any time → ∴ 1st Order. 2 H2O2 → 2 H2O + O2  Although 2 molecules of H2O2 appear in equation → the reaction is 1st order. 1) A solution of a drug contained 500 units per ml when prepared. It was analyzed after a period of 40 days and was found to contain 300 units per ml. Assuming that the decomposition is first order, at what time will the drug have decomposed to one half of its original concentration? Solution: - Co = 500 unit/ ml. - Ct = 300 Unit/ ml. 𝒌𝒌𝒌𝒌 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝒌𝒌 𝒙𝒙 𝟒𝟒𝟒𝟒 𝑳𝑳𝑳𝑳𝑳𝑳 𝟑𝟑𝟑𝟑𝟑𝟑 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟓𝟓𝟓𝟓𝟓𝟓 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 ∴ 𝒌𝒌 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒅𝒅𝒅𝒅𝒅𝒅−𝟏𝟏 - T1/2: 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝟎𝟎. 𝟔𝟔𝟔𝟔𝟔𝟔 𝒕𝒕𝟏𝟏 = = = 𝟓𝟓𝟓𝟓 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 𝟐𝟐 𝒌𝒌 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 2) The first order catalytic decomposition of H2O2 may be followed by measuring the volume of O2 liberated in a gas burette. From such an experiment it was found that the concentration of H2O2 remaining after 65 minutes, expressed as the volume in ml of gas evolved was 9.6 from an initial concentration of 57.9. - Calculate K. - How much H2O2 remains un-decomposed after 25 minutes? Solution: - Co = 57.9 ml & Ct = 9.6 ml. 𝒌𝒌𝒌𝒌 𝒌𝒌 𝒙𝒙 𝟔𝟔𝟔𝟔 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝑳𝑳𝑳𝑳𝑳𝑳 𝟗𝟗. 𝟔𝟔 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟓𝟓𝟓𝟓. 𝟗𝟗 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 ∴ 𝒌𝒌 = 𝟎𝟎. 𝟎𝟎𝟐𝟐𝟐𝟐𝟐𝟐 𝒎𝒎𝒎𝒎𝒎𝒎−𝟏𝟏 - H2O2 remaining after 25 min: 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒙𝒙 𝟐𝟐𝟐𝟐 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟓𝟓𝟓𝟓. 𝟗𝟗 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝑪𝑪𝒕𝒕 = 𝟐𝟐𝟐𝟐. 𝟎𝟎𝟎𝟎 𝒎𝒎𝒎𝒎 3) A drug product is known to be ineffective after it has decomposed 30%. The original concentration of one sample was 50 mg/ml. When assayed 20 months later the concentration was found to be 40 mg/ml, assuming that the decomposition is first order, what should be the expiration time on the label? Solution: - Co = 50 mg/ml & Ct = 40 mg/ml. 𝒌𝒌𝒌𝒌 𝒌𝒌 𝒙𝒙 𝟐𝟐𝟐𝟐 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝒕𝒕 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝑪𝑪𝟎𝟎 − 𝑳𝑳𝑳𝑳𝑳𝑳 𝟒𝟒𝟒𝟒 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟓𝟓𝟓𝟓 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 ∴ 𝒌𝒌 = 𝟎𝟎. 𝟎𝟎𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎−𝟏𝟏 - Product ineffective when decompose 30% → ∴ Expiry date when 70% remaining: 𝟕𝟕𝟕𝟕 𝑪𝑪𝒕𝒕 = 𝒙𝒙 𝟓𝟓𝟓𝟓 = 𝟑𝟑𝟓𝟓 𝒎𝒎𝒎𝒎/𝒎𝒎𝒎𝒎 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎. 𝟎𝟎𝟏𝟏𝟏𝟏 𝒙𝒙 𝒕𝒕𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝑳𝑳𝑳𝑳𝑳𝑳 𝟑𝟑𝟑𝟑 = 𝑳𝑳𝑳𝑳𝑳𝑳 𝟓𝟓𝟓𝟓 − 𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑 𝒕𝒕𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 = 𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒 𝒎𝒎𝒐𝒐𝒐𝒐𝒐𝒐𝒐𝒐