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Lecture 2: Rate Laws (1/6)

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Lecture 2: Rate Laws (1/6) Chem 31B

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Notes from Lecture 2: Rate Laws, on January 6, 2016
Principles of Chemistry II
Jennifer Poehlmann
Class Notes
Chemistry, Rate Laws, Rates, reactions, reaction rate, integrated rate law, chem 31B, Half Life, first order, second order, zero order
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This 2 page Class Notes was uploaded by it's lit notes on Wednesday January 6, 2016. The Class Notes belongs to Chem 31B at Stanford University taught by Jennifer Poehlmann in Fall 2015. Since its upload, it has received 42 views. For similar materials see Principles of Chemistry II in Chemistry at Stanford University.


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Date Created: 01/06/16
Wednesday  January 6, 2016  1  Important Information:  ­Pset #1 due Monday  ­Lab Write Up #1 due Friday @ 2:15pm    Lecture 2: Rate Laws  ● net reaction­​  only depicts the start and end result → does not include the reaction  mechanism  ○ the stoichiometry of the net rxn does not give us the rxn mechanism  ○ the rxn mechanism is composed of ​ elementary steps​  (cannot be decomposed to  smaller steps)  ● rate = k[A]​ x[B]​[C]​  ○ x = reaction order with respect to A  ○ y = reaction order with respect to B  ○ z = reaction order with respect to C  ○ overall order = x + y + z  ○ x,y,z are determined experimentally → have nothing to do with the stoichiometry  of the reaction    Clicker Question #1  The following reaction is first order in A, second order in B, and zero order in C. What is  the basic form of the rate law for this rxn?  2A + B + C → products  a.) rate = k[A]​[B][C]  b.) rate = k[A][B][C]  c.) rate = k[A][B]​2  d.) rate = k[A][B]  e.) rate = k[A][B]​  solution: do not look at coefficients    Clicker Question #2  The following reaction is first order in A, second order in B, and zero order in C. How will  the rate change if the concentration of C is doubled?  2A + B + C → products  a.) the rate will not change  b.) the rate will double  c.) the rate will halve  d.) there is insufficient information to determine this  e.) the rate will increase by a factor of 3  solution: for a zero order reaction, the concentration is independent of the rate  Wednesday  January 6, 2016  2    Clicker Question #3  The following reaction is first order in A, second order in B, and zero order in C. By what  factor will the rate increase if the concentrations of A and B are both doubled?  2A + B + C → products  a.) 0  b.) 2  c.) 4  d.) 8  e.) 16  solution: doubling the concentration of A will increase the rate 2x because the concentration is  directly proportional to the rate for a first order rxn; doubling the concentration of B will 4x the  reaction rate because the rate is directly proportional to the square of the concentration. 4 x 2 = 8.    Rate Laws  ● rate =  ­Δ[A] /Δt   ● for all rate laws, for any order reaction, and for all half­life calculations, we are really just  using the same equation over and over again  ● half life­​ the time required for the initial concentration to halve    Zero­Order Reaction  ● differential rate law: ​ rate = k = ­Δ[A] /Δt  ● integrated rate law: ​ [A]t​= ­kt + [A]0 ● half life expression:​  1/2​ [A]0​/ 2k  ○ t​ 1/2​nds on [A]​ 0​→ will have different half lives depending on the initial  amount you start with at each time  ● if only a certain amount of the reactable surface is exposed, adding more reactant to the  part that is unexposed (increasing the concentration) will have no effect on the reaction  rate    First­Order Reaction  ● differential rate law: ​ rate = k [A] = ­Δ[A] /Δt  ● integrated rate law: l n ( [A]t​[A]0 ​= ­kt  ● half life expression: t​1/2​.693/k  ○ rate depends on [A]​ 0​but 1/2​oes not depend on [A]​0  ○ the half life is not dependent on the initial concentration because as the reaction  proceeds the half life remains constant   


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