Lecture 2: Rate Laws (1/6)
Lecture 2: Rate Laws (1/6) Chem 31B
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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 1/2nds 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: t1/2.693/k ○ rate depends on [A] 0but 1/2oes 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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