February 15th 2017 Chapter 16 Chapter 16 HW due Feb 24th next Friday Tutoring Wednesdays Library Help Desk 16 4 Integrated Rate Laws Concentration Changes Over Time An integrated rate law includes time as a variable Integrated rate law First order rate equation rate D A Dt k A ln A 0 A t kt Second order rate equation rate D A Dt k A 2 1 1 kt A t A 0 Zero order rate equation rate D A Dt k A 0 A t A 0 kt Sample Problem 16 5 Determining the Reactant Concentration after a Given Time At 1000 oC cyclobutane C4H8 decomposes in a first order reaction with the very high rate constant of 87 s 1 to two molecules of ethylene C2H4 a If the initial C4H8 concentration is 2 00 M what is the concentration after 0 010 s ln A 0 A t kt ln ln C4H8 0 C4H8 t 2 00 mol L 1 C4H8 t 2 00 mol L 1 C4H8 t kt 87 s 1 0 010 s 0 87 e0 87 C4H8 t 0 84 mol L 1 Sample Problem 16 5 Determining the Reactant Concentration after a Given Time At 1000 oC cyclobutane C4H8 decomposes in a first order reaction with the very high rate constant of 87 s 1 to two molecules of ethylene C2H4 b What fraction of C4H8 has decomposed in this time C4H8 0 C4H8 t C4H8 0 2 00 M 0 84 M 2 00 M 0 58 In a second order reaction the rate constant is 4 0 x 10 4 M 1 s 1 What is the concentration of reactant after 10 min if the initial concentration is 0 800 M 1 1 kt A t A 0 a 0 629 M b 0 300 M c 1 49 M d 0 671 M e 0 797 M 1 A t 1 0 800 M 1 A t 4 0 x 10 4 M 1 s 1 10 min 0 24 M 1 A t 0 671 M 1 0 800 M 60 sec 1 min Determining Rxn Orders from an Integrated Rate Law First order rate equation rate D A Dt k A ln A 0 A t kt ln A t kt ln A 0 y mx b Determining Rxn Orders from an Integrated Rate Law Second order rate equation rate D A Dt k A 2 1 1 kt A t A 0 1 1 kt A 0 A t y mx b Determining Rxn Orders from an Integrated Rate Law Zero order rate equation rate D A Dt k A 0 A t A 0 kt A t kt A 0 y mx b Figure 16 7 Graphical determination of the reaction order for the decomposition of N2O5 Reaction Half Life The half life t1 2 is the time taken for the concentration of a reactant to drop to half its initial value First order rxn ln A 0 A t kt t1 2 ln 2 0 693 k k t1 2 does not depend on the starting concentration Figure 16 8 A plot of N2O5 vs time for three reaction half lives Sample Problem 16 7 Determining the Half Life of a First Order Reaction Cyclopropane is the smallest cyclic hydrocarbon Because its 60 bond angles allow poor orbital overlap its bonds are weak As a result it is thermally unstable and rearranges to propene at 1000 C via the following first order reaction The rate constant is 9 2 s 1 a What is the half life of the reaction b How long does it take for the concentration of cyclopropane to reach one quarter of the initial value a t1 2 0 693 k 0 693 9 2 0 075 s s 1 b The time to reach of its initial value is equal to 2 half lives Time 2 0 075 s 0 15 s The thermal decomposition of N2O5 g to form NO2 g and O2 g is a first order reaction The rate constant for the reaction is 5 1 x 10 4 s 1 at 318 K What is the half life of this process a 2 0 x 103 s t1 b 3 9 x 103 s c 2 6 x 10 4 s d 1 4 x 103 s e 1 0 x 10 3 s 2 t1 2 0 693 k 0 693 1 4 103 s 4 5 1 10 s Reaction Half Life Second order rxn 1 1 kt A t A 0 1 t1 2 k A 0 Zero order rxn A t A 0 kt A 0 t1 2 2k Determining the Half Life of a 2nd Order Reaction The decomposition of NO2 to N2 and O2 is second order with a rate constant k 12 5 M 1 s 1 What is the half life for the reaction if the initial NO2 0 00260 M SOLUTION t1 2 t1 2 1 k A o 1 12 5 0 00260 M M s t 1 30 8 s 2 1 0 0325 s The decomposition of nitrosyl bromide NOBr is second order with respect to NOBr and second order overall If the initial concentration of NOBr is 0 102 M and the rate constant is 25 M 1 min 1 what is the half life of the reaction A 4 1 x 10 3 min B 4 0 x 10 2 min C 2 8 x 10 2 min D 2 6 min E 3 9 x 10 1 min t1 2 1 k A o 1 t1 25 2 0 102 M M min t 1 0 39 min 2 1 2 55 min