Question

Can I please get help on how to workout this problem step by step?

Summary of the Kinetics of Zero-Order, First-order and Second-Order Reactions Concentration Time En JA-ALAI 0 rate --- rate - RIAL IHAJ-HAI- 2 mateRIAT - 2.The decomposition of a certain antibiotic in water has a rate constant at 20 °C of 2.45 yr'. Determine the molarity of an antibiotic solution that site for 2 months if its original concentration is 0.0043 M. (hint - be sure units of time match/cancel, also unit tells you order). (20 pts) a. Set up equation and define variables with units b. insert values with correct units that can cancel C Solve the problem (STP)! d. Calculate the half-life.

Answer 1

decomposition nate constant of a wtain antibiotic Hence it follows 2.45 yrt first order kinetics. Explanation ; A can be written as Simple r=k (conej order ú к 'Lone) onda now unit of ra mole / lt time unit of lone let order =n > mole lt mole et time k mole Inn time -0 раах et او of order = 1 order = 1 i n=1 ten timet mole lt timet which is independent of concentration, for the reaction A HB Ist oder knietis following we have B t=o A [A]. tot Alt rate law ها rak [Alt _d[A K[A]t dt d [A =kt ü

oh integration, we have -bge [AJE kt +C (mme (dhe me + c) At t=0 c=. -loge [A]. kt: Loge [A), - loge [Alt = log (1) I loge of Ixl. 303 logro ZASE [A] [A]. [A]t mw ear=t I year a -yr quin wien problem k=2.45 yrt t= 2 months 2 / [A] =.0043 M [A]. Now wing ka t loc t ť hoge TALE 2.45 = (Al. to loge (A) + (Alo =.41 ve .ul i =ė4 = 1.5 loge [Alt (AJ. [Alt [Alta [A]. (concertation after 2 months) 1 0043 = 2.878103 M 1.5 A Half life At t= tiz (Alt [A]. 2 0.693 (A). :: - loge (AIO th Loge Cal. = 14 loge 2 0.693 tila .693 2.45 2.28 year K