Chapter 14 -Chemical Kinetics and Stability PDF

Physical Pharmacy Chemical Kinetics and Stability Chapter 14 Chapter 14| Chemical Kinetics and Stability contents : Fundamentals and concentration effects: 4 Molecularity 4 Rate 6 Specific Rate Constant 8 Kinetic study 9 Zero-Order Reactions 9 First-Order Reactions 13 Determination of Order 19 Secon...

Physical Pharmacy Chemical Kinetics and Stability Chapter 14 Chapter 14| Chemical Kinetics and Stability contents : Fundamentals and concentration effects: 4 Molecularity 4 Rate 6 Specific Rate Constant 8 Kinetic study 9 Zero-Order Reactions 9 First-Order Reactions 13 Determination of Order 19 Second-Order Reactions 27 Factors effects on stability 35 Stability of pharmaceuticals 52 Chapter 14| Chemical Kinetics and Stability Chemical Kinetics and Stability : The stability of drug product with time is important in determination of shelf life and expired date. The stability affected by factors, such as temperature, humidity, and light. This chapter studies the rates and mechanisms of reactions specially the decomposition and stabilization of drug products. For example, thiamine hydrochloride is most stable at a pH of 2 to 3 and is unstable above pH 6, so in preparation, the pharmacist should select the buffered vehicle that prevents the degradation. Applications of chemical kinetics in pharmacy result in the production of more-stable drug preparations. Chapter 14| Chemical Kinetics and Stability FUNDAMENTALS AND CONCENTRATION EFFECTS : Rates, Order, and Molecularity of Reactions Molecularity Molecularity is the number of molecules, atoms, or ions reacting in an elementary process. Molecularity classify the reaction into unimolecular, bimolecular, and Termolecular. Molecularity cannot gives complete detail about order of reaction specially those of several steps while kinetic study gives details. Chapter 14| Chemical Kinetics and Stability Example : Br2 H2+ I 2Br unimolecular 2 2NO + O2 2HI bimolecular 2NO2Termolecular While the real detail mechanism revealed by kinetic study as follows: N2O2 (Fast) 2NO N2O 2 +O2 2NO2 (Slow) The rate of the reaction is given by the slow step. Chapter 14| Chemical Kinetics and Stability Rate : The rate, velocity, or speed of a reaction is given by the expression dc/dt, where dc is the increase or decrease of concentration over an infinitesimal time interval dt. In the reaction Reactant Rate=− 𝒅 Products Reactant 𝒅𝒕 = K Reactant a where k is the rate constant and exponent a represent the order of reaction. Chapter 14| Chemical Kinetics and Stability Chapter 14| Chemical Kinetics and Stability Specific Rate Constant The constant, k, appearing in the rate law associated with a singlestep (elementary) reaction is called the specific rate constant for that reaction. The half-life is the time required for one-half of the material to disappear; the time at which C has decreased to 1/2 C. The shelf-life is the time required for 10% of the material to disappear; it is the time at which C has decreased to 90% of its original concentration (i.e., 0.9C). Chapter 14| Chemical Kinetics and Stability Kinetic study Zero-Order Reactions Garrett found that the loss in color of a multisulfa product followed a zero order rate. The rate expression for the change of concentration, C, with time is therefore - 𝒅𝑪 𝒅𝒕 = K0 It means that the rate of reaction not depend on concentration of reactant, it is constant with time. Chapter 14| Chemical Kinetics and Stability Kinetic study Zero-Order Reactions The rate equation can be integrated between the initial concentration, C0, at t = 0, and Ct , the absorbance after time (t): Ct = Co - kot t =Co−Ct ko Chapter 14| Chemical Kinetics and Stability Kinetic study Zero-Order Reactions The rate equation can be integrated between the initial concentration, C0, at t = 0, and Ct , the absorbance after time (t): Co−1/2Co t1/2 = ko 1/2Co t1/2 = ko Chapter 14| Chemical Kinetics and Stability For shelf life Co−0.9Co 0.1Co T90% = = ko ko The unit of zero order rate constant is: K0= - 𝒅𝑪 𝒅𝒕 𝒎𝒐𝒍𝒆/𝒍𝒊𝒕𝒆𝒓 = 𝒔𝒆𝒄𝒐𝒏𝒅 = 𝒎𝒐𝒍𝒆 𝒍𝒊𝒕𝒆𝒓 𝒔𝒆𝒄𝒐𝒏𝒅 = mole liter −1 second −1 Chapter 14| Chemical Kinetics and Stability Kinetic study First-Order Reactions Harried showed that the decomposition rate of hydrogen peroxide catalyzed by 0.02 M KI was proportional to the concentration of hydrogen peroxide remaining in the reaction mixture at any time. The data for the reaction 2H202 = 2H2O + O2 Although two molecules of hydrogen peroxide appear in the equation, the reaction was found to be first order. Chapter 14| Chemical Kinetics and Stability The rate equation is written - 𝒅𝑪 𝒅𝒕 = KC where c is the concentration of hydrogen peroxide remaining undecomposed at time t and k is the first-order rate constant Chapter 14| Chemical Kinetics and Stability Integrating above equation, we have ln C = ln C0 – Kt (Common logarithm) log C =logC0 – 𝑲 t 𝟐.𝟑𝟎𝟑 For calculation of half life of reaction follows first order; C log 0 = C 𝑲 t 𝟐.𝟑𝟎𝟑 Chapter 14| Chemical Kinetics and Stability at half life; t= t1/2, Ct= ½ C0 log C0 = 1/2 C 𝑲𝒕𝟏/𝟐 𝟐.𝟑𝟎𝟑 , t ½ = 0.693/k For shelf life t90; 𝑲𝒕𝟏𝟎% C 0 log = 𝟐.𝟑𝟎𝟑 0.9 C , t 90 = 0.105/k The unit of first order rate constant is: K0= - 𝒅𝑪 𝟏 𝒎𝒐𝒍𝒆/𝒍𝒊𝒕𝒆𝒓 = 𝒅𝒕 𝒄 𝒔𝒆𝒄𝒐𝒏𝒅.𝒎𝒐𝒍𝒆/𝒍𝒊𝒕𝒆𝒓 = 𝟏 𝒔𝒆𝒄𝒐𝒏𝒅 = second −1 Chapter 14| Chemical Kinetics and Stability Determination of order Chapter 14| Chemical Kinetics and Stability Determination of Order The order of a reaction can be determined by several methods. 1. Substitution Method. The data from kinetic study can be substituted in the equations for each order. When the calculated k values remain constant for different Ct, the reaction is considered to be of that order. Chapter 14| Chemical Kinetics and Stability 2. Graphic Method. A plot of the data in the form of a graph can also be used to ascertain the order. If a straight line results when concentration is plotted against t, the reaction is zero order. The reaction is first order if log (concentration) versus t yields a straight line, and it is second-order if 1/( concentration) versus t gives a straight line. Chapter 14| Chemical Kinetics and Stability Chapter 14| Chemical Kinetics and Stability 3. Half-Life Method. In a zero-order reaction, the half-life is proportional to the initial concentration, a, as observed in Table 15-2. The half-life of a first order reaction is independent of initial concentration a; t1/2 for a second-order reaction, in which a = b, is proportional to 1 /a. Chapter 14| Chemical Kinetics and Stability Chapter 14| Chemical Kinetics and Stability Suspensions, Apparent Zero-Order Kinetics Suspensions are another case of zero-order kinetics, in which the concentration in solution depends on the drug's solubility. As the drug decomposes in solution, more drug is released from the suspended particles, so that the concentration remains constant. The equation for an ordinary solution, with no reservoir of drug to replace that depleted, is the first-order expression, equation : Chapter 14| Chemical Kinetics and Stability -d[C]/dt = kf[C] where c= concentration of drug in solution =solubility Kf= first order rate constant of solution Chapter 14| Chemical Kinetics and Stability When the concentration [C ] is rendered constant, as in the case of a suspension, we can write Kf [C] = K0 where K0= zero order rate constant of suspension Thus, K0 = Kf x Solubility -d[C]/dt = K0 Chapter 14| Chemical Kinetics and Stability Second-Order Reactions The rates of bimolecular reactions, which occur when two molecules come together, are frequently described by the second-order equation. A+B Products When the speed of the reaction depends on the concentrations of A and B with each term raised to the first power, the rate of decomposition of A is equal to the rate of decomposition of B, and both are proportional to the product of the concentrations of the reactants: Chapter 14| Chemical Kinetics and Stability Second-Order Reactions 𝒅*A+ 𝒅*B+ =− = K *A+*B+ 𝒅𝒕 𝒅𝒕 If a and b are the initial concentrations of A and B respectively, and x is the concentration of each species reacting in time t, the rate law can be integrated and written according to 3 possibilities: Chapter 14| Chemical Kinetics and Stability 1. When, in the simplest case, both A and B are present in the same concentration so that a = b, x =Kt 𝒂(𝒂 − 𝒙) The rate constant, k, in a second-order reaction therefore has the dimensions liter/ (mole sec) or liter mole -1 sec -1. The half-life of a second-order reaction is t1/2= 1/ak Chapter 14| Chemical Kinetics and Stability 2. When, in the general case, A and B are not present in equal concentrations but the difference in concentration is not big, integration of equation yields: 2. 𝟑𝟎𝟑 𝒃(𝒂 − x) 𝒍𝒐𝒈 =Kt 𝒂−𝒃 𝒂(𝒃 − 𝒙) Chapter 14| Chemical Kinetics and Stability 3. When A and B are not present in equal concentrations and the difference in concentration is very big (Pseudo-first-order reaction): Suppose that in this reaction, A was in great excess and B was in a relatively low concentration. As the reaction proceeded, B would change appreciably from its original concentration, whereas the concentrations of A would remain essentially unchanged because they are present in great excess. Chapter 14| Chemical Kinetics and Stability Chapter 14| Chemical Kinetics and Stability In this case the contribution of A to the rate expression is considered constant and the reaction rate can be written as A+B Products 𝒅*B+ − = K *A+*B+ 𝒅𝒕 K is constant and [A] is constant 𝒅*B+ − = K′*B+ 𝒅𝒕 where K'= K [A], K'= Pseudo-first-order constant Factors effects on stability Chapter 14| Chemical Kinetics and Stability Factors effects on stability A number of factors other than concentration may affect the reaction velocity. Among these are temperature, solvents, catalysts, and light. 1. Temperature effect Collision Theory Reaction rates are expected to be proportional to the number of collisions per unit time. Chapter 14| Chemical Kinetics and Stability Because the number of collisions increases as the temperature increases, the reaction rate is expected to increase with increasing temperature. Chapter 14| Chemical Kinetics and Stability The effect of temperature on reaction rate is given by the equation, first suggested by Arrhenius, k = Ae-Ea/RT or log K = log A – 𝑬𝒂 𝟏 𝟐.𝟑𝟎𝟑𝑹𝑻 𝑻 where k is the specific reaction rate, A is a constant known as the Arrhenius factor or the frequency factor, Ea is the energy of activation. R is the gas constant, 1.987 calories/deg mole, and T is the absolute temperature. Chapter 14| Chemical Kinetics and Stability In case at 2 temperatures t1 and t2, the equation becomes: or log 𝑲𝟐 𝑲𝟏 = 𝑬𝒂 𝑻𝟐−𝑻𝟏 ( ) 𝟐.𝟑𝟎𝟑𝑹𝑻 𝑻𝟐𝑻𝟏 Chapter 14| Chemical Kinetics and Stability Accelerated stability testing The k values for the decomposition of drug at various elevated temperatures are obtained by plotting log of concentration against time as shown in this figure: Chapter 14| Chemical Kinetics and Stability Then the logarithm of rate constants (k) at various temperatures are plotted against reciprocal of absolute temperature and the resulting line extrapolated to the room temperature to get K25°C as shown in this figure: The shelf life t90% can be calculated from equation t90% = 0.105/ K25°C Chapter 14| Chemical Kinetics and Stability Factors effects on stability 1. Medium Effects: Solvent, Ionic Strength, Dielectric Constant a. Effect of the Solvent In summary, it can be said that the polarity of solvents affect the rate of reactions depending on the polarity of reactant. b. Effect of the ionic strength For ionic compound, the ionic strength affects the rate of reaction while for neutral molecule, the rate of reaction independent on ionic strength. Chapter 14| Chemical Kinetics and Stability a. Effect of the Dielectric Constant The dielectric constant affects the rate constant of an ionic reaction. For a reaction between ions of opposite sign, an increase in dielectric constant of the solvent results in a decrease in the rate constant. For ions of like charge, on the other hand, an increase in dielectric constant results in an increase in the rate of the reaction. Chapter 14| Chemical Kinetics and Stability 3. Catalysis effect: Specific Acid-Base Catalysis Effects The rate of a reaction is frequently influenced by the presence of a catalyst. A catalyst is defined as a substance that influences the speed of a reaction without itself being altered chemically. Solutions of a number of drugs undergo accelerated decomposition on the addition of acids or bases. If the drug solution is buffered, the decomposition may not be accompanied by an appreciable change in the concentration of acid or base, so that the reaction can be considered to be catalyzed by hydrogen or hydroxyl ions. Chapter 14| Chemical Kinetics and Stability Best example of specific acid-base catalysis, is the hydrolysis of esters. Chapter 14| Chemical Kinetics and Stability The general formula for hydrolysis of ester which affected by both H+ and OH is - 𝒅𝑪 𝒅𝒕 = K observed [ester] K observed = K H + H + +KOH - OH− K observed = total rate constant of the system K H + = rate constant for acid catalysis reaction KOH - =rate constant for base catalysis reaction KOH H + =hydrogen ion concentration OH− =hydroxide ion concentration Chapter 14| Chemical Kinetics and Stability Note: K H + and KOH- are second order rate constants. The pH- Rate profile for the specific acid-base-catalyzed hydrolysis of ester as shown in this figure: Chapter 14| Chemical Kinetics and Stability Explanation of pH- Rate profile 1. At low PH K observed = K H + H + +KOH - OH− Since [OH-] concentration value is very low, thus the part (KOH - OH− ) is neglected from equation at low pH. So, K observed = K H + H + log K observed = log KH ++ log H + log K observed = log KH + - pH log K observed = log KH + -1XpH Chapter 14| Chemical Kinetics and Stability Chapter 14| Chemical Kinetics and Stability Explanation of pH- Rate profile 1. At high PH K observed = K H + H + +KOH - OH− Since H + concentration value is very low, thus the part (K H + H + ) is neglected from equation at low pH. So, K observed = KOH - OH− Kw= [H+ ] [OH- ] K observed = KOH - OH− [OH- ] = Kw/ [H+ ] Chapter 14| Chemical Kinetics and Stability K observed = KOH - kw H+ log K observed = ( log KOH - +log kw)- log H + log K observed = ( log KOH - +log kw)+pH log K observed = ( log KOH - +log kw)+1XpH STABILITY OF PHARMACEUTICALS Chapter 14| Chemical Kinetics and Stability STABILITY OF PHARMACEUTICALS Decomposition and Stabilization of Medicinal Agents Pharmaceutical decomposition can be classified as hydrolysis, oxidation, isomerization, epimerization, and photolysis, and these processes may affect the stability of drugs in liquid, solid, and semisolid products. Chapter 14| Chemical Kinetics and Stability Examples of Martin physical pharmacy text book Examples 1: In the reaction of acetic anhydride with ethyl alcohol to form ethyl acetate and water, (CH3CO)2O + 2C2H5OH = 2CH3CO2C2H5 + H2O The rate of reaction is Rate= − 𝒅 (CH3CO)2O 𝒅𝒕 =k [(CH3CO)2O] [C2H5OH]2 Chapter 14| Chemical Kinetics and Stability What is the order of the reaction with respect to acetic anhydride? With respect to ethyl alcohol? What is the overall order of the reaction? If the alcohol, which serves here as the solvent for acetic anhydride, is in large excess such that a small amount of ethyl alcohol is used up in the reaction, write the rate equation for the process and state the order. Answer: The reaction appears to be first order with respect to acetic anhydride, second order with respect to ethyl alcohol, and overall third order. Chapter 14| Chemical Kinetics and Stability However, because alcohol is the solvent, its concentration remains essentially constant, and the rate expression can be written -d [(CH3C0)2O]/dt = k'[(CH3CO)2O] Kinetically the reaction is therefore a pseudo-first-order reaction. Chapter 14| Chemical Kinetics and Stability Example.2 : Shelf Life of an Aspirin Suspension. A prescription for a liquid aspirin preparation is called for. It is to contain 325mg/5 mL or 6.5 g/100 mL. The solubility of aspirin at 25°C is 0.33 g/100 mL; therefore, the preparation will definitely be a suspension. The other ingredients in the prescription cause the product to have a pH of.6.0 The first-order rate constant for aspirin degradation in this solution is 4.5 x 10-6 sec-.1 Chapter 14| Chemical Kinetics and Stability Calculate the zero-order rate constant. Determine the shelf life, t90, for the liquid prescription, assuming that the product is satisfactory until the time at which it has decomposed to 90% of its original concentration (i.e., 10% decomposition) at 25°C. Chapter 14| Chemical Kinetics and Stability Answer : ko = k x [Aspirin in solution], from equation (k [A] =k.)0 Thus , Ko = ( 4.5x 10-6 sec-1) x (0.33 g/100 mL) ko = 1.5 x 10-6 g/100 mL sec- 1 t90 = 0.10[A]o /k0 = [(0.10X6.5 g/100 mL)]/ [(1.5 x 10-6 g/100 mLsec-1) ] = 4.3x 105 sec = 5.0 days Chapter 14| Chemical Kinetics and Stability EXAMPLE.3 Decomposition of Hydrogen Peroxide : The catalytic decomposition of hydrogen peroxide can be followed by measuring the volume of oxygen liberated in a gas burette. From such an experiment, It was found that the concentration of hydrogen peroxide remaining after 65 min, expressed as the volume in milliliters of gas evolved, was 9.60 from an initial concentration of 57.90. Chapter 14| Chemical Kinetics and Stability a) Calculate k using equation k = 2.303 𝐶0 log 𝑡 𝐶. b) How much hydrogen peroxide remained undecomposed after 25 min ? 2.303 𝑐0 k= log 𝑡 𝑐 a) 2.303 57.90 k= log = 0.0277 min−1 𝑡 9.60 (b) 0.0277 = 2.303 57.90 log 25 𝑐 𝑐 = 29.01 Chapter 14| Chemical Kinetics and Stability EXAMPLE.4 : First-Order Half-Life : A solution of a drug contained 500 units/mL when prepared. It was analyzed after 40 days and was found to contain 300 units/mL. Assuming the decomposition is first order, at what time will the drug have decomposed to one-half of its original concentration? Chapter 14| Chemical Kinetics and Stability We have 𝟐. 𝟑𝟎𝟑 𝟓𝟎𝟎 𝐤= 𝐥𝐨𝐠 = 𝟎. 𝟎𝟏𝟐𝟖𝒅𝒂𝒚−𝟏 𝟒𝟎 𝟑𝟎𝟎 𝟐. 𝟑𝟎𝟑 𝟓𝟎𝟎 𝐭= 𝐥𝐨𝐠 = 𝟓𝟒. 𝟑 days 𝟎. 𝟎𝟏𝟐𝟖 𝟐𝟓𝟎 Chapter 14| Chemical Kinetics and Stability Example.5 : Saponification of Ethyl Acetate Walker investigated the saponification of ethyl acetate at25°C: CH3COOC2H5 + NaOH CH3COONa + C2H5OH The initial concentrations of both ethyl acetate and sodium hydroxide in the mixture were 0.01000 M. The change in concentration, x, of alkali during 20 min was 0.000566 mole/liter, therefore, (a - x) = 0.01000 - 0.00566 =0.00434 Compute (a) the rate constant and (b) the half-life of the reaction. (a) Using equation Chapter 14| Chemical Kinetics and Stability 𝒙 = 𝐊𝐭 𝒂(𝒂 − 𝒙) or 𝟏 𝒙 𝐤= 𝒂𝒕 𝒂 − 𝒙 , we obtain 𝒂 = 𝟎. 𝟎𝟏 𝐱 = 𝟎. 𝟎𝟎𝟎𝟓𝟔𝟔 𝒂 − 𝒙 = 𝟎. 𝟎𝟎𝟗𝟒𝟑𝟒 𝑲= 𝟏 𝟎.𝟎𝟎𝟎𝟓𝟔𝟔 𝟎.𝟎𝟏×𝟐𝟎 𝟎.𝟎𝟎𝟗𝟒𝟑𝟒 = 𝟎. 𝟐𝟗𝟗 liter 𝒎𝒐𝒍𝒆−𝟏 𝒎𝒊𝒏−𝟏 Chapter 14| Chemical Kinetics and Stability (b) The half-life of a second-order reaction is 𝐭𝟏/𝟐 = 𝟏/𝐚𝐤 It can be computed for the reaction only when the initial concentrations of the reactants are identical. In the present example, 𝒕𝟏/𝟐 𝟏 = = 𝟑𝟑𝟒. 𝟒𝟒 𝐦𝐢𝐧 𝟎. 𝟎𝟏 × 𝟎. 𝟐𝟗𝟗 Chapter 14| Chemical Kinetics and Stability Example.6 : The rate constant k1 for the decomposition of 5-HMF (5hydroxymethylfurfural ) at 120°C (393 K) is 1.173 hr-1 or 3.258 x 10-4 sec -1 , and k2 at 140°C (413 K) is 4.860 hr-1. What is the activation energy, Ea, in kcal/mole and the frequency factor, A, in sec-1 for the breakdown of 5-HMF within this temperature range? Chapter 14| Chemical Kinetics and Stability We have 𝐥𝐨𝐠 𝐥𝐨𝐠 𝑲𝟐 𝑬𝒂 𝑻𝟐 − 𝑻𝟏 = 𝑲𝟏 𝟐. 𝟑𝟎𝟑𝑹 𝑻𝟐𝑻𝟏 𝟒. 𝟖𝟔𝟎 𝑬𝒂 𝟒𝟏𝟑 − 𝟑𝟗𝟑 = 𝟏. 𝟏𝟕𝟑 𝟐. 𝟑𝟎𝟑 × 𝟏. 𝟗𝟖𝟕 𝟒𝟏𝟑 × 𝟑𝟗𝟑 𝐄𝐚 = 𝟐𝟑𝟎𝟎𝟎𝐜𝐚𝐥/𝐦𝐨𝐥𝐞 𝐄𝐚 = 𝟐𝟑𝐤𝐜𝐚𝐥/𝐦𝐨𝐥𝐞 At 𝟏𝟐𝟎∘ 𝐂, using equation 𝑬𝒂 𝟏 𝐥𝐨𝐠 𝐊 = 𝐥𝐨𝐠 𝑨 − 𝟐. 𝟑𝟎𝟑𝑹 𝑻 , we obtain 𝐥𝐨𝐠 𝟑. 𝟐𝟓𝟖 × 𝟏𝟎−𝟒 𝒔𝒆𝒄−𝟏 = 𝐥𝐨𝐠 𝑨 − 𝐀 = 𝟐 × 𝟏𝟎𝟗 𝒔𝒆𝒄−𝟏 𝟐𝟑, 𝟎𝟎𝟎𝒄𝒂𝒍 𝟏 𝟐. 𝟑𝟎𝟑 × 𝟏. 𝟗𝟖𝟕 𝟑𝟗𝟑 Chapter 14| Chemical Kinetics and Stability Example.7 : Increased Shelf Life of Aspirin : Aspirin is most stable at pH 2.5. At this pH the apparent fist-order rate constant is 5 x 10-7 sec-1 at 25°C. The shelf life of aspirin in solution under these conditions can be calculated as follows: t= 90 0.105 /(5 x 10-7 )= 2.1 x 105 sec 2 =days As one can see, aspirin is very unstable in aqueous solution. Chapter 14| Chemical Kinetics and Stability Would making a suspension increase the shelf life of aspirin? The solubility of aspirin is 0.33g/100mL. At pH 2.5, the apparent zero- order rate constant for an aspirin suspension is ko = 5 x 10-7 sec -1 x 0.33 g/100 mL = 1.65 x 10-7 g/mL sec If one dose of aspirin at 650 mg per teaspoonful is administered, then one has 650 mg/5 mL = 13 g/100 mL. For this aspirin suspension , t 90= [(0.1) (13)]/ [1.65 x 10-7] = 7.9 x 106 sec = 91 days Chapter 14| Chemical Kinetics and Stability The increase in the shelf-life of suspensions as compared to solutions is a result of the interplay between the solubility and the stability of the drug. In the case of aspirin, the solid form of the drug is stable, whereas when aspirin is in solution it is unstable. As aspirin in solution decomposes, the solution concentration is maintained as additional aspirin dissolves up to the limit of its aqueous solubility.