# Physical Chemistry for the Chemical and Biological Sciences

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Section 9 – Part II

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9.3.2. First Order Kinetics Continued……. Rate = - d [A] / dt = k [A] [A] = [A] 0 e -kt ln ([A]/[A] 0 ) = - k t Plot of ln { [A] / [A] 0 } vs. t gives a straight line with a slope that is given by –k
Exponential decay of [A] with time [A] = [A] 0 e -kt

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For a 1 st order reaction the concentration of A would be [A] = [A] 0 /2 at t = t 1/2 When t = t 1/2 the equation becomes: ln { ([A] 0 /2 )/ [A] 0 } = - k t 1/2 t 1/2 = (ln 2) / k = 0.693 / k t 1/2 is independent of [A] 0
For example, assume the elimination of a drug follows 1 st order kinetics: If we start with an initial plasma concentration of 100 g/mL: [Cp] 0 = 100 g/ml if k = 0.35 hr –1 and t 1/2 = 0.693/0.35 then t 1/2 = 2 hours how much remains after 6 hours ? 100 g/ml 50 g/ml 25 g/ml 12.5 g/ml 6.25 g/ml Following: 2 hrs. 4hrs. 6 hrs. 8hrs.

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What Happens if Initial Concentration of Drug in Plasma is Different ? k el = 0.347 hr –1 then t 1/2 = 0.693/0.35 = 2 hours Initial Conc. 200 g/mL 100 g/mL 50 g/mL 100 g/mL 50 g/mL 25 g/mL 50 g/mL 25 g/mL 12.5 g/mL 25 g/mL 12.5 g/mL 6.25 g/mL 12.5 g/mL 6.25 g/mL 3.13 g/mL t= 0 t = 2 t = 4 = 6 t= 8
How Does Rate Constant For Elimination Effect Concentration of Drug in Plasma ? t 1/2 = 0.693/ k el = ? [Cp] 0 = 100 g/mL k el 0.17 hr -1 0.347 hr -1 0.7 hr -1 t 1/2 4.08 hrs. 2 hrs. 1 hour In each case what is the concentration of drug in plasma following 8 hours ?

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Atkins Phys. Chem 7 th Edn. Page 871 “The larger the rate constant the more rapid the decay”.
Radioactive Decays Follow First Order Kinetics: • we can use this as a way to determine age of something……… as in problem #12.6 in Chang Text.

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Radiocarbon Age Determination continued……… • all plants and animals contain carbon • most carbon is 12 C (stable) , but a small fraction is 14 C (radioactive) 14 C is generated by interaction in upper atmosphere of cosmic rays with gas molecules, resulting in neutrons which bombard 14 N to produce 14 C ( 14 N 7 + 1 n 0 14 C 6 + 1 p 1 ) 14 C is radioactive and emits a beta particle according to the following reaction: 14 14 0 6 7 -1 C N + • the radioactive half-life for this reaction is 5730 years
…….the natural abundance of 14 C is 1.1. X 10 -13 mol % in living matter . ……. the mol % of 14 C for all living matter is assumed to be the same ……when matter dies, it no longer exchanges material with the environment and the mol % of 14 C will decrease according to 1 st order decay kinetics . ……when matter dies [ 14 C] becomes [ 14 C] 0 and decays with t 1/2 = 5730 years If we measure the [ 14 C] at some time post-death, we can estimate elapsed t since t 0 (its “ age ”)****………. . ****for this determination we are limited by the minimum amount of material we can measure reliably …….limit is approximately 10 half lives (57,000 years old)

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9.3.3. Second Order Reactions 1) 2 nd Order Reaction with One Reactant : A Products Rate = - d [A] / dt = k [A] 2 Rate is proportional to [A] raised to second power •k u n i t s a r e M -1 s -1 d [A]/[A] 2 = - kdt (1 / [A]) – (1 / [A] 0 ) = kt or (1/[A]) = kt + (1/[A] 0 ) [A] t [A] 0 t 0 Second order in [A] and Second order overall
( 1/[A] ) = kt + ( 1/[A] 0 ) Plot of 1/[A] versus t gives a straight line with slope equal to k To obtain t 1/2

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## This note was uploaded on 05/09/2010 for the course CHEMISTRY CHM223H taught by Professor Macdonald during the Spring '09 term at University of Toronto.

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9b - Section 9 Part II 9.3.2 First Order Kinetics Continued...

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