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Ans.: K = 0.693/ T 1/2 =0.693/5.7= 0.12hr -1 .

1. The degradation of a drug product follows zero order process. If the concentration was 300 units/ml when freshly prepared, and after 6 months, its concentration reached 240 units/ml. Calculate the constant of reaction and the shelf life of the product. Ans.:

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Ans.: K = 0.693/ T 1/2 =0.693/5.7= 0.12hr -1 .

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  1. 1. The degradation of a drug product follows zero order process. If the concentration was 300 units/ml when freshly prepared, and after 6 months, its concentration reached 240 units/ml. Calculate the constant of reaction and the shelf life of the product. Ans.: Co= 300 unit/mL Ct= 240 unit/mL t= 6 mon. Ct= Co- Kt 240=300-6K K=10 unit/mL.mon-1 t90%= Co/10K=300/10x10=3 months.

  2. 2. A drug product has an initial concentration of 125 mg/ml decays by zero order kinetics, (k=0.5 mg/ml/hr). What is the concentration of the intact drug remaining after 3 days? When the drug reaches zero concentration. • Ans.: • Co= 125 mg/mL K=0.5 mg/mL.hr-1 • t= 3 days=3x24=72 hr. • Ct= Co- Kt=125-0.5x72=89 mg/mL • Ct= Co- Kt=125-0.5t=0 • t=250 hr.

  3. 3. The degradation of a mulisulpha preparation is a zero order process. If the concentration was 0.47 mole/liter when freshly prepared & after 473 days, its concentration reached 0.225 mole/liter. Calculate k and T1/2. • Ans.: • Co= 0.47 mole/L Ct= 0.225mole/L t= 473 days. • Ct= Co- Kt • 0.225=0.47-473K • K=5.17x10-4 mole/L.day-1 • T1/2= Co/2K= 0.47/ 2 x 5. 17x10-4= 453.69 days

  4. 4. At 25 °C, the half-life period for the decomposition of N205 is 5.7 hours and is independent of the initial pressure of N205, calculate the specific rate constant for the reaction. • Ans.: • K= 0.693/T1/2=0.693/5.7= 0.12hr-1.

  5. 5. During a stability study on a new antibiotic drug A pharmacist dissolved a few mgs of the antibiotic in exactly 100 ml of distilled water and placed the solution in a refrigerator at 5°C. At various time intervals, the pharmacist removed a 10-ml aliquot from the solution and measured the amount of drug remaining in each aliquot. The following data were obtained: • Time (hr) Conc. (μg/ml) • 0.5 84.5 • 1.0 81.2 • 2.0 74.5 • 4.0 61.0 • 6.0 48.0 • 8.0 35.0 • 12.0 8.7 • a. Assuming that the decomposition is a zero order process, what is the rate constant of decomposition of this antibiotic? (Do this both graphically and mathematically). • b. What is the concentration of antibiotic were in original solution prepared by the pharmacist? • c. Give the equation for the line that best fits the experimental data? • d. Calculate the time for 50% remaining. • e. Calculate the life period for the product. • Ans.: • Co= 87.66 μg/ml • K=6.58 μg/ml. hr-1 • Equation: Ct= 87.66- 6.58t • T1/2= Co/2K= 87.66/ 2 x 6.58 = 6.66 hr. • t90%=0.1 Co/K=0.1x87.66/6.58= 1.33 hr.

  6. 6. Ampicillinwas prepared in two dosage forms for pediatrics... • a. A drop suspension where the dose is taken as 5 drops through a dropper that delivers 10 drops/ml. • b. An oral suspension where the dose is one teaspoonful. • The dose of ampicillin for pediatrics per both products is 125 mg every 6 hours, its solubility at the pH of maximum stability (the pH of both preparations) is 1.1 g /100 ml and its 1st order rate constant is 2 X 10-7 S-1. Calculate the shelf life of both products and determine how many times the drop preparation is stable than the oral suspension. • Suspension follows apparent zero order kinetics • K1 = 2 X 10-7 S-1 • K0 = K1 A =2 x 10-7 x1.1=2.2 x 10-7 g /100 ml. S-1. • Dose=125mg In drop suspension: 5 drops = 0.5 mL • 125 mg 0.5 mL • x 100 • x=25000mg/100mL=25 g/100mL • In oral suspension: 1 teaspoonful =5 mL • 125 mg 5 mL • X 100 • x=2500mg/mL=2.5g/100mL • Therefore the shelf life in drop suspension is 10 times more than that of oral suspension. t90%=0.1 C0/K =0.1x25/2.2 x 10-7 =1.1 x 107 sec. t90%=0.1 C0/K =0.1x2.5/2.2 x 10-7 =0.11 x 107 sec.

  7. 7. 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? What is the half-life of this product? Ans.: Co= 50mg/mL Ct= 40mg/mL t= 20 mon. log Ct= log Co- Kt/2.303 log 40= log 50- 20K/2.303 K= 0.0111mon-1 log Ct= log Co- Kt/2.303 Decomposed 30% Remaining 70% (50x70/100=35 mg/mL) log 35= log 50- 0.0111t/2.303 t= 32.13 mon. expiration date on the label. T1/2= 0.693/K= 0.693/ 0.0111= 62.43 mon.

  8. 8. An ophthalmic solution dispensed at 5 mg/ml conc. exhibits a first order reaction, with a rate of 0.0005 day-1. How much drug remaining after 120 days? & how long will it take for the drug to degrade to 90% of its initial concentration? Ans.: Co= 5mg/mL K= 0.0005 day-1 t= 120 days. log Ct= log Co- Kt/2.303 log Ct = log 5- 120x0.0005/2.303=0.67 Ct=4.7 mg/mL t90% =0.105/K= 0.105/0.0005= 210 days.

  9. 9. The initial concentration of both ethyl acetate and sodium hydroxide in a reaction mixture was 0.01 M each. The change in the concentration "x" of alkali during 20 minutes was 0.00566 mole/liter. Compute * The second order rate constant. * The half life of the reaction. • Ans.: • a= 0.01 x= 0.00566 mole/litre t= 20 min. • K= 6.521 M-1 min-1. • T1/2 = 1/ak = 15.33 min.

  10. 10. The rate of decomposition of 5.6 molar glucose at 140 °c in an aqueous solution containing 0.35 HCI was found to be: What is the order and the specific reaction rate of this decomposition? determine the half-life period. (Graphically and Mathematically).

  11. 11. The following are the concentrations of Salphadiazien remaining at various times during the thermal decomposition at 40o C. Using the above data determine: a. The order of the decomposition reaction using two different methods. b. The rate constant of the reaction. c. The shelf life of the product.

  12. 12. The rate constant K, for the decomposition of hydroxymethyl furfural at 120ºC (393ºK) is 1.173 hr -1, and K2 at 140ºC (413ºC) is 4.860 hr -1. What is the activation energy "Ea" in Kcal / mol and the frequency factor A sec-1 for the breakdown of the considered substance within this temperature range? • Log K2/K1=Ea/2.303R (T2-T1/T2T1) • Log 4.86/1.173 = (Ea/2.303x1.987) (413-393/413 x393) • 0.617 = (Ea/2.303x1.987) x 0.00012 • Ea = 23255.8 cal. = 23.255 Kcal.

  13. 13.The initial stage of chain reaction was found to be first order. The initial concentration of the solution was 0.050 moles/L and after 10 hours at 40ºC, the concentration was 0.015 moles/L, a: compute the specific rate constant at this temperature. b: What is the remaining concentration after 2 hours? c: If the k value for this reaction at 20ºC is 0.002 hr-1 what is the activation energy and the frequency factor for the reaction? • Ans.: • Co= 0.05 mole/L Ct= 0.015 mole/litre t= 10 hr. • log Ct= log Co- Kt/2.303 • log 0.015= log 0.05- 10K/2.303 • K= 0.12hr-1 • Log K2/K1=Ea/2.303R (T2-T1/T2T1) • Log 0.12/0.002 = (Ea/2.303x1.987) (413-393/413 x393) • 1.778 = (Ea/2.303x1.987) ( 0.0004) • Ea = 18181.81 cal. = 18.181 Kcal.

  14. 14. Garrett determined the apparent first order rate constant for the degradation of streptozotocin. The results obtained at approximately pH 7 and at various temperature are oC 40 oC 50 oC 60 oC K 0.00018 0.00073 0.00268 a. Plot the results according to the Arrhenious relationship and compute the activation energy Ea. b. Extrapolate the results to 25oC to obtain K at room temperature.

  15. 15. The specific reaction rates at 50oC and 70oC for the decomposition of dibromosuccinicacid in solution are 1.80 x 10-6 and 2.08 x 10-4 min-1 respectively. Calculate the energy of activation for the reaction. Ans.: Log K2/K1=Ea/2.303R (T2-T1/T2T1) Log 2.08 x 10-4 /1.8 x 10-6= Ea/2.303x1.987 (343-323/343x323) 2.06 = (Ea/2.303x1.987) (0.00018) Ea= 52631 cal. = 52.631 Kilocalorie. .

  16. 16. The accelerated stability data for a solution of a drug was found to be as follows:- ________________________________________________________ Time (mo.) Assay (mg/ml) ________________________________________________________ 35oC 45oC 55oC ________________________________________________________ 0.0 100 100 100 0.5 ----- ----- 97.5 1.0 99.5 98.4 95.1 2.0 99.0 96.7 90.4 3.0 98.5 95.2 86.1 4.0 98.0 93.6 81.9 5.0 97.5 92.1 77.9 6.0 97.0 90.6 74.1 ________________________________________________________ If the decomposition was proved to follow first order (t # ln amount remaining), determine: a. The rate constant at each temperature. b. The rate constant at 25oC c. How long the preparation will retain 90% of the labelled quantity at 25oC.

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