IMG_0004_NEW_0025 - Experiment I[A]o:0.001 M lBlo:2 lv...

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Summary: Kinetics for Reactions of the TYPe: aA + Products First v Rate law Integrated Rate Law Plot need to give a straight line Gradient of straight Ifurc HalfJife I tv'= pfdl, '/ RaE: f Rate = klAl Rate = tlAI2 l,t!=-rt+lt\ r[z],=-tr+t'['a] k=".fr [A] venrs r ln[A] versus t l/[A] versus t Slopc:-* SloPe : -P SloPe : k '*=+ \'=Y VMfh f*t" tnl'L''s-' 6-l h.(, t.2-5-t Reaction: A+B+ products Rare = -# : -+=,rlAl,[B]' To find k, n andn = Need two experiments lsolation Method Experiment 2 lAl,):2 M [B]o = 0.001 M Example: lsolation Method Base hydrolysis of ethyl acetate in NaOH 9O "rJo""r""s *NaoH +".tXo-*"- +cHscHzoH Ethyl Acetate Sodium Hydroxide Sodium Acetate Ethanol Rate = _ d[EtoAc] - - dtNqoHl = tlEtoAcl"[NaoH]- dt dt Experiment 1: [EtOAc]o : 0.001M [r*aOH]o : 2.0 N'{ Observed Rate = f r[EtOAc] "pseudo-fi rst-order" rate law Experiment 2: [EtOAc]o:2'0M [NaOH]6 : 0.001 M
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Unformatted text preview: Experiment I [A]o:0.001 M lBlo:2 lv{ slx's*" hV*'A =l \ Time(s) goet Time(s) 'l'wt\r Combine resuls -r overall rate law What lf We Have > 1 Reactant? For more complex reactions such as: 5\^ aA + bB * cc -+ products Rate: klAl"LBl-lClP Q: How do we figure out the Integrated Rate Law for More Than One Reactant? A: Isolation Method: O Isolate the rate dependence ofone reactant (tAl) by maintaining all other reactants ([B],[C]) in great excess. Rate: l/[N t{ : k[B]i"lc)f g Use experimental data to determine l( & n O Repeat for other reactants @ Determine k In other words: deal with each reactant one at a time and combine the results at the end u Observed Rate : #2[NaOH]2 ttpseudo-second-ordertt rate law Combining the results: Overall rate law = Overall rate constant, F= QT-rjA4."7 ruunu...
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This note was uploaded on 12/27/2011 for the course CHEM 122 taught by Professor Williams during the Fall '10 term at Simon Fraser.

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