PHA 111 Chemical Kinetics PDF

PHA 111 Chemical Kinetics: Application Dr Stephen Childs Senior Lecturer in Pharmaceutical Chemistry [email protected] Slide 1 PHA 111: Chemical Kinetics  Integ...

PHA 111 Chemical Kinetics: Application Dr Stephen Childs Senior Lecturer in Pharmaceutical Chemistry [email protected] Slide 1 PHA 111: Chemical Kinetics  Integrated Rate Laws How can we find the concentration of A at any given time (t)? Or how long to fall to a given concentration of A? Reaction Order Rate Integrated Rate Law 0 K[A]⁰ [A]t = -kt + [A]0 1 K[A]1 ln[A]t = -kt + ln[A]0 2 K[A]2 1/[A]t = kt + 1/[A]0 Each integrated rate law can be incorporated in y=mx+c. The x term is always time. We can now draw a straight line graph for each reaction order! Slide 2 PHA 111: Chemical Kinetics  Reaction Order Zero Order 1st Order 2nd Order 1/[A] [A] ln[A] time time time We can determine the order of the reaction simply by plotting the data in each model and seeing which gives a straight line. Slide 3 PHA 111: Chemical Kinetics  A side note… The natural logarithm (ln) is used here where we are concerned with change over time Slide 4 PHA 111: Chemical Kinetics  Second Order Reactions (i) A + A → Product Rate = k[A]2 time = 0 [A]0 , time = t [A]t We know that: −Δ[𝐴] 2 𝑅𝑎𝑡𝑒 = 𝑎𝑛𝑑 𝑅𝑎𝑡𝑒 = 𝑘 𝐴 Δ[𝑡] Rate of change w.r.t time: −𝑑[𝐴] = 𝑘 𝑑𝑡 𝐴2 Then integrating: [𝐴]𝑡 𝑡 න [𝐴]−2 𝑑 𝐴 = 𝑘 න 𝑑𝑡 [𝐴]0 0 We end up with: 1 1 1 1 − = 𝑘𝑡 or = + 𝑘𝑡 [𝐴]𝑡 𝐴0 [𝐴]𝑡 𝐴0 Slide 5 PHA 111: Chemical Kinetics  Second Order Reactions (i) 1 1 = + 𝑘𝑡 [𝐴]𝑡 𝐴0 Fractional loss depends on [A] 0 (initial concentration) Consider environmentally harmful products which persist due to long half lives at low concentrations! -Time to fall to 50% initial (half life): −𝑡1/2 = 1/[𝐴]0 𝑘 -Time to fall to 90% initial (shelf life?): −𝑡90 = 1/9[𝐴]0 𝑘 Slide 6 PHA 111: Chemical Kinetics  Second Order Reactions (ii) A+B→C+D If [A]0 ≠ [B] 0 (different concentrations) The second order integrated rate law: 1 [𝐵]𝑡 [𝐴]𝑡 𝑙𝑛 − 𝑙𝑛 − 𝑘𝑡 [𝐴]0 −[𝐵]0 [𝐵]0 [𝐴]0 No simple equation for 𝑡1/2 or 𝑡90 ! Time it takes for [A] = ½ might not equal the time taken for [B] = ½ Slide 7 PHA 111: Chemical Kinetics  Pseudo First Order Reactions Eg. Base hydrolysis of benzocaine: Rate =k[benzocaine][OH-] Second order reaction, unless pH is constant = [OH-] is constant Rate = k1[benzocaine] (where k1 = k.[OH-]) Referred to as a pseudo first order reaction Slide 8 PHA 111: Chemical Kinetics  First Order Reactions (i) A+B→C Integrated rate law: ln[A]t = -kt + ln[A]0 Fractional loss takes a constant time, it is independent of [A] 0 𝐴 𝑡 𝐴 0 −𝑙𝑛 = 𝑙𝑛 = 𝑘𝑡 𝐴 0 𝐴 𝑡 -Time to fall to 50% initial: 𝑡1/2 = ln(100/50)/𝑘 = 0.693/𝑘 -Time to fall to 90% initial: 𝑡90 = ln 100/90 /𝑘 = 0.105/𝑘 Slide 9 PHA 111: Chemical Kinetics  First Order Reactions (ii) Hydrolysis of ampicillin – Pseudo first order (excess water) If k = 2.4 x10-4 hour-1 at 5°C Lowest permitted concentration is 95% label strength What is the shelf life of ampicillin?? 𝒕𝟗𝟓 = 𝒍𝒏 𝟏𝟎𝟎/𝟗𝟓 / 2.4 x10-4 𝒕𝟗𝟓 = 𝟐𝟏𝟒 𝒉𝒐𝒖𝒓𝒔 = 𝟗 𝒅𝒂𝒚𝒔 𝒂𝒕 5°C Slide 10 PHA 111: Chemical Kinetics  Zero Order Integrated Rate Law A -> Product k Rate Rate = k [A]⁰ Therefore, Rate = k Time (t) We also know that: -Δ[A] / Δt = k [ A ]t t As a differential equation: -d[A] / dt = k  d [ A] = −k  dt [ A ]0 0 Integrating gives: [A]t - [A]0 = -k t Rearranging gives: [A]t = -kt + [A]0 -k [A] Now it looks a bit like y=mx+b (and so we have a straight line graph) Time (t) Slide 11 PHA 111: Chemical Kinetics  Zero Order Half Life Fractional loss is dependent of [A]0 Integrated rate law: [A]t = -kt + [A]0 Half life is time taken for fall in [A]0 from 100 % to 50 % We can call this time t50 So t50 is when [A]t = [A] 0 / 2 (100/50 =2) Therefore, [A] 0 / 2 = -kt + [A]0 kt = [A]0 - [A] 0 / 2 (the initial conc. minus half the initial conc.) kt = [A] 0 / 2 t50 = [A] 0 / 2k Or… t50 = 0.5 [A] 0 / k Slide 12 PHA 111: Chemical Kinetics  Zero Order Shelf Life Integrated rate law: [A]t = -kt + [A]0 or, [A]t = [A]0 – kt [ A]0 − [ A]t (1 − x)[ A]0 t= = k k For half life; x = 0.5 t½=0.5[A]0/k For 90% shelf life; x = 0.9 t90=0.1[A]0/k Shelf life increases as [A]0 increases Slide 13 PHA 111: Chemical Kinetics  Zero Order Shelf Life Example Example: Hydrolysis of ampicillin suspension (125mg/5mL) Ampicillin solubility: 12 mg/mL Zero order k = (12.0 x 2.4 x 10-4) mg mL-1 hour-1 Zero order k = 2.88 x 10-3 mg mL-1 hour-1 What is the shelf life of ampicillin suspension?? 𝑡95 = 0.05[A]0 /𝑘 [A]0 = 25 mg/mL Shelf life = (0.05 x 25.0)/(2.88 x 10-3) = 434 hours (18 days) Reaction will follow zero order kinetics until conc. fall to 12 mg/mL Change in solubility indirectly affects shelf life! Shelf life increases as [A]0 increases Suspensions often follow zero order kinetics-only dissolved drug degrades appreciably Drugs in solid phase or enzyme reactions (high substrate conc.) may also follow zero order kinetics Slide 14 PHA 111: Chemical Kinetics  Using the Arrhenius Equation Taking natural logs: 𝐸𝑎 ln 𝑘𝑟 = ln 𝐴 − 𝑅𝑇 Hence plotting ln(k) versus 1/T gives a straight line Slope = -Ea/R Intercept = ln(A) (1/T=0 is at infinite temperature) Ea can be calculated from the slope Slide 15 PHA 111: Chemical Kinetics  Hydrolysis of Bupivacaine Data presented as % initial value Time (months) 10 °C 25° C 35° C 50° C 0 100 100 100 100 1 99.6 99.3 98.3 96.1 2 99.2 98.8 94.9 93.1 4 99 97.6 91.2 86.8 6 98.3 96.3 88.2 80.2 8 97.9 94.1 86.6 74.9 10 97.4 93.5 83.1 69.4 12 97 91.8 80.3 65.9 Slide 16 PHA 111: Chemical Kinetics  First Order Plots 4.65 4.6 4.55 4.5 4.45 ln[A] 4.4 4.35 4.3 4.25 4.2 4.15 0 2 4 6 8 10 12 14 Time (months) Slide 17 PHA 111: Chemical Kinetics  Hydrolysis of Bupivacaine Temp (°C) k 1/T(°K) ln(k) 10 0.0025 0.003534 -5.99146 25 0.0071 0.003356 -4.94766 35 0.0178 0.003247 -4.02856 50 0.0354 0.003096 -3.34104 Arrhenius plot y = -6208.2x + 15.96 -2 -2.50.003 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 -3 -3.5 -4 ln(k) -4.5 -5 -5.5 -6 -6.5 1/T Slide 18 PHA 111: Chemical Kinetics  Arrhenius Calculations Intercept = 15.96 (= lnA) A = e15.96 = 8.54x 106 months-1 – Note units Slope = -6208 °K Ea = -slope x R = -(-6208) x 8.3145 J mol-1 Ea = 51.7 kJ mol-1 What will the rate constant (k) be at 15 °C? 15 °C = 288 °K 1/T = 0.003472 From graph ln(k) at this temperature = -5.63 k = 0.0036 months-1 Slide 19 PHA 111: Chemical Kinetics  Prediction of Degradation What is the half life (t50) at 20°C? From the graph, at this temperature k = 0.00536 months-1 t50 = 0.693/k = 0.693 / 0.00536 = 129.3 months What is the shelf life (t90) at 20°C? k = 0.00536 months-1 t90 = 0.105/k = 0.105 / 0.00536 = 19.6 months Remember – Ln (100/50) = 0.693 – Ln (100/90) = 0.150 Slide 20 PHA 111: Chemical Kinetics