Physical Chemistry for the Chemical and Biological Sciences

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Section 9 Section 9 Part III Part III
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Figure 4 Amounts of A (unabsorbed ethanol), B (absorbed ethanol) and C (ethanol oxidized by LADH) according to the integrated rate laws for first-order / zeroth-order consecutive reactions. [A] 0 = 1 mol ; k 1 = 0.00289 s -1 ; k 2 = 0.0000444 mol s -1
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Kinetics Of Alcohol Absorption k 1 k 2 A B C CH 3 CH 2 OH CH 3 CH 2 OH CH 3 CHO unabsorbed absorbed Step 1: Absorption of alcohol into body – 1st order kinetics Step 2: Metabolic oxidation of ethanol by LADH (liver alcohol dehydrogenase sometimes written as LAD) Thus: dt ] A [ d = k 1 [A] dt ] B [ d = + k 1 [A] k 2 dt ] C [ d = + k 2
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Integrated Forms: ] A [ ] A [ 0 ] A [ ] A [ d = t 0 1 dt k ln 0 ] A [ ] A [ = – k 1 t [A] = [A] 0 exp(– k 1 t) (1) dt ] B [ d = + k 1 [A] k 2 = + k 1 [A] 0 exp(– k 1 t) k 2 Thus: d[B] = { + k 1 [A] 0 exp(– k 1 t) k 2 } dt
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Therefore: ] B [ ] B [ 0 ] B [ d = + k 1 [A] 0 t 0 1 dt ) t k exp( k 2 t 0 dt = 1 1 k k [A] 0 exp(– k 1 t) k 2 t (evaluated at t = t and t = 0) = [A] 0 exp(– k 1 t) + [A] 0 k 2 t Therefore: [B] = [A] 0 [A] 0 exp(– k 1 t) k 2 t (2) assuming that: [B] 0 = 0
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Finally: ] C [ ] C [ 0 ] C [ d = t 0 2 dt k [C] = + k 2 t (3) assuming that: [C] 0 = 0 Alternatively: [A] + [B] + [C] = [A] 0 Therefore: [C] = [A] 0 [A] [B] = [A] 0 [A] 0 exp(– k 1 t) [A] 0 + [A] 0 exp(– k 1 t) + k 2 t = + k 2 t (3) Q.E.D. (quod erat demonstrandum)
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Absorbed Alcohol Maximum k 1 = 2.89 x 10 –3 s –1 k 2 = 4.44 x 10 –5 M s –1 dt ] B [ d = 0 = dt d { [A] 0 [A] 0 exp(– k 1 t) k 2 t } = + k 1 [A] 0 exp(– k 1 t) k 2 Therefore: exp(– k 1 t) = 0 1 2 ] A [ k k
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Or: – k 1 t = ln 0 1 2 ] A [ k k Or: t = 1 k 1 ln 0 1 2 ] A [ k k = 1 3 s 10 x 89 . 2 1 ln ) M 1 )( s 10 x 89 . 2 ( s M 10 x 44 . 4 1 3 1 5 = 1444.9 s = 24.1 min (3 significant figures) Legally Intoxicated 1 mole (46 g or 60 mL)
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9.7. The Effect of Temperature on Reaction Rate • the rate constants of most reactions increase as the Temperature is raised • it has been found experimentally that for many reactions a plot of ln k versus 1/T gives a straight line • this behavior is outlined by the Arrhenius equation: ln k = ln A –(E a / RT) y = b + m x ln A corresponds to the intercept of the line at 1/T = 0 A is called the pre-exponential factor or the frequency factor
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ln k = ln A –(E a / R T) y = b + m x The parameter E a is obtained from the slope which is equal to (–E a /R) E a is called the activation energy. A and E a are together referred to as the Arrhenius parameters. Atkins Text page 879
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E a is given by slope of ln k versus 1/T ……. .thus the higher the activation energy the stronger the temperature dependence of the rate constant (meaning the steeper the slope) • If a reaction has a zero activation energy , its rate is independent of temperature. • In some cases the temperature dependence of reactions is “non-Arrhenius like ” meaning a straight line is not obtained when ln k is plotted against 1/T
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Potential Energy Profile for an Exothermic Reaction How does the potential energy change in the course of a reaction ?
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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- Toronto.

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9c - Section 9 Part III Figure 4 Amounts of A (unabsorbed...

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