Lecture 3: Rate Laws Continued (1/8)
Lecture 3: Rate Laws Continued (1/8) Chem 31B
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This 3 page Class Notes was uploaded by it's lit notes on Saturday January 9, 2016. The Class Notes belongs to Chem 31B at Stanford University taught by Jennifer Poehlmann in Fall 2015. Since its upload, it has received 18 views. For similar materials see Principles of Chemistry II in Chemistry at Stanford University.
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Date Created: 01/09/16
Friday January 8, 2016 Important Information: Pset #1 due Monday @ 2:30pm Lecture 3: Rate Laws Continued Clicker Question #1 You suspect that the reaction 2NO → 4NO + O is first order with respect toO]. How 2 5 2 2 2 5 should you plot [N2 5versus time (t) to confirm your suspicion? a.) [NO] vs. t 2 5 b.) [2 5vs. 1/t c.) ln[NO] vs t 2 5 d.) ln[2 5vs. 1/t e.) 1/[NO] vs. t 2 5 solution for a first order reaction, the equation2 5= kt + ln2 50 This equation is in the form y = mx + b (the equation of a line), so in order to see if this reaction follows the trend, you would plot y (ln[2 5 vs. x (time). Radioactive Decay ● obeys first order kinetics and therefore the first order rate law ● the general first order reaction rate can be applied to the rate of radioactive decay ● rate = kN ○ N = number of radioactive nuclei Clicker Question #2 The radioactive element J has a half life of 3.5 years. How many years must pass before a mole of J decays to 1/1000th of a mole? a.) cannot be determined becaus1/2 of a firstorder reaction is independent of concentration b.) 7.5 years c.) 35 years d.) 3500 years e.) 38 years solution t = 3.5 years /[N] = 1/1000 t = 3.5 = .693/k → k = (.693/3.5) 1/2 t 0 1/2 ln ([Nt[N0 = kt ln (1/1000) = (.693/3.5)t t = 34.9 years 5 years Second Order Reactions ● rate: rate = k[A] ● differential rate law:Δ[A]/Δ[t] = k[A]2 ● integrated rate law:1/[A]t= 1/[A0+ kt Friday January 8, 2016 ● halflife expression: t1/21/k[A] 0 ○ half life increases with time Clicker Question #3 The following results were obtained for the reaction below at 25 oC. 2 2 2I (aq) + S2 8aq) → I 2(aq) + 2SO 4q) Trial # (I) M (S2 8M Initial Rate M/s 6 1 .080 .040 12.5 x 10 2 .040 .040 6.25 x 106 3 .080 .020 6.25 x 106 6 4 .032 .040 5.00 x 10 5 .060 .030 7.00 x 106 a.) Which 2 trials should be chosen to find the order of the reaction with respect to S 2 8 a.) 1 and 2 b.) 2 and 3 c.) 2 and 5 d.) 1 and 3 e.) 1 and 5 solution: look for the trials where I is held constant because we are looking to see how the rate 2 depends on changing S 2 8ncentration b.) What are the units of the rate constant? a.) M/s b.) 1/(M x s) 2 c.) (M/s) d.) M2/s e.) 1/s solution: looking at trials 1 and 3, we see that the concentration of S 2 8d the rate of the reaction are directly proportional, thus the order with respect to that reactant is 1. Looking at trials 1 and 2, we can see that the rate of the reaction is also directly proportional to the concentration of I. Thus, the order with respect to I is also 1. Overall, the order of the reaction is 2, and the units for k for a 2nd order reaction are 1/(M x s). Friday January 8, 2016 Summary: Rate Laws ● the reaction order (n) is experimentally determined and cannot be deduced from the net reaction ● molecules must collide before a reaction can occur between then ● in an elementary step → the reaction order describes how many molecules must meet at once ● if the elementary step isA → products ○ the reaction is first order because the coefficient of A is one ○ no collisions are required ● if the elementary step isA → products ○ the reaction is second order because the coefficient is 2 ○ 2 molecules must collide ● ratedetermining step the slowest step is a reaction mechanism determines the overall rate