chapter_13_Feb_5_2010 - 16-1Important….New office hours...

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Unformatted text preview: 16-1Important….New office hours: Mon and Fri 11 – 12; Wed. 4 – 5http://www.nytimes.com/2009/02/18/education/18college.html?scp=1&sq=max%20roosevelt&st=cse Chemistry Tutor Room – Fleming 23616-2Chapter 16Kinetics: Rates and Mechanisms of Chemical Reactions16-3Average, Instantaneous, and Initial Reaction RatesFor example: C2H4(g) + O3(g) C2H4O(g) + O2(g)We will only consider reactant concentrations. So….Rate = - ∆[C2H4]/∆t = - ∆[O3]/∆tC2H4(g) + O3(g) C2H4O(g) + O2(g)Time (s)Concentration of O3(mol/L)0.020.030.040.050.060.010.03.20x10-52.42x10-51.95x10-51.63x10-51.40x10-51.23x10-51.10x10-5Rate = - ∆[O3]/∆t =[(1.10 x 10-5mol/L) – (3.20 x 10-5mol/L)]/(60s – 0s) =3.50 x 10-7mol/L.sThis is the average rate – during 60s of the reaction,O3 decreases an average of 3.50 x 10-7mol/L eachsecond.Over the entire 60s -16-4C2H4(g) + O3(g) C2H4O(g) + O2(g)Time (s)Concentration of O3(mol/L)0.020.030.040.050.060.010.03.20x10-52.42x10-51.95x10-51.63x10-51.40x10-51.23x10-51.10x10-5Average rate over two shorter times- 1st10 secondsRate = - ∆[O3]/∆t =[(2.42 x 10-5mol/L) – (3.20 x 10-5mol/L)]/(10s – 0s) =7.80 x 10-7mol/L.s-Last 10 secondsRate = - ∆[O3]/∆t =[(1.10 x 10-5mol/L) – (1.23 x 10-5mol/L)]/(60s – 50s) =1.30 x 10-7mol/L.sThe earlier rate is 6x as fast as the later rate. Rate decreases during the course of the reaction.16-5The concentration of O3vs. time during its reaction with C2H4....
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chapter_13_Feb_5_2010 - 16-1Important….New office hours...

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