Question

Pharmacokinetics

1-An analgesic was given in a dose of 300,

Vd=20L, Kel = 0.15hr-1

a- calculate Cp 2 Hr after injection.

b-Calculate the half-life of the drug

c- Calculate time for 80% to be completely eliminated

d- How many half-lives will the drug take when 50 mg of it remain in the body?

Answer 1

om the question, I understand that t Information is POL PAR 165 Half life of the drug Half life of the drug is the time when concentration of the injected drug remains half of its initial concen tration le Ct = co , then t-tyzie half life from the above derived formula luct = -kt, we have In Co2) - - Kty In(y) = -KtYz -4.303 x 0.3010 = -K tz Half life, ty = 0.693 An analgesic was given ima dose of 300 mg, Rato constant, K = 0.15 HY? om the rate constant we know that this is a order reaction is the amalge sie is reacting with pst Order reaction For aist order reaction, let CP rate law is given by , Y = 0[] = d [P] |dt at - ve sign is there because is consuming in this reaction now, rate expression, r = -066) = K[c] If take concentration of cle. (C) in g and time in hours., we get k= hourl] in this question, unit of given rate constant is hourt hence, given process is a 1st order kinetics. (a) Calculate Cp le concentration of poodust, 2 Hafter o Injection To calculate concentration of product first we howe derive a expression for progression of the reaction let say A - Product is a storder reaction Initial Q Co= 0 dt Here, half life of the drug = 0.693 = 4.62hr - olshit Half life of drug is 4.62 hours = 4 houts 37 min 12 seconds (c) Calculate time for 80% to be completely eleminated When 80% of drug is completely eleminated, concen tration of drug remained in. Ce is given by- Ct = 60-80.. = Co(1-018) = 0.2C. now ln 0.2% = -kt after some (=Q-X fime't' (tot) Here Co is the intral count ration the reactant 4 and ct is its coucentration at time t x is the concentration of product formed arome F Rate of reaction, Y = K[NJ - - d[n] - dre I dt at - Ct = KC+ dt On rearranging we get, dei - kot → t = - 0.2 - 1.60944 olshit (d) How many half-lives will the drug take when some. of il remain in the body. For a first order reaction, Co trz, cu EYE L Wars Eve Concentration remaining after n half lives is given by con Here somg drug is remained in body Hems, somg = 300m =) ? = 300 mg = 6 507 =) <n<3 Hence the drug takes & half lives when some tof it remain in the body. On integrating both sides from to ating both sides from t=0 to tot, we have e [lucile - [t] - lu Ct = -kt = ctacoekt now, concentration of product formed is given by [P] = Cocct Here, Co= 300 mg = 15 m 1' = POL After 2 hours of Section, Reactant concentration remained, CL = Co en combe written as, en c = -0.15 hotxahr = -0.3 0.015 81303 logo (os) = -0.3 Oors = Antilog (- 030) - Antilog (-0.1303) enors = Antilog (1.0.8697) = 7.408 xv^= 0.741 now, C4 = 0.015 X0.741 764 = 0.01111592" Hence, concentration of product after 2 hours of injection, Cp or [P] = C.-C4 = 0.01597'-0.01111571? = 0.00388576' = 3.885 mg 14 * 3.9 mg i' [cp = 3.9 mg/L