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CH 102 Chapter 15 Notes from 2/15/16 Lecture

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by: Jess Snider

CH 102 Chapter 15 Notes from 2/15/16 Lecture CH 102

Marketplace > University of Alabama - Tuscaloosa > Chemistry > CH 102 > CH 102 Chapter 15 Notes from 2 15 16 Lecture
Jess Snider
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Class notes from lecture given on 2/15/16
General Chemistry
Dr. Bakker
Class Notes
CH102, Chapter 15




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Paula Ramirez

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This 6 page Class Notes was uploaded by Jess Snider on Monday February 15, 2016. The Class Notes belongs to CH 102 at University of Alabama - Tuscaloosa taught by Dr. Bakker in Winter 2016. Since its upload, it has received 43 views. For similar materials see General Chemistry in Chemistry at University of Alabama - Tuscaloosa.


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Date Created: 02/15/16
Chapter 15: Chemical Kinetics A. Chemical Kinetics a) Thermodynamics i) Does a reaction take place? b) Kinetics i) How fast does a reaction proceed? c) Reaction Rate: change in concentration of reactant or product with time (M/s) A B rate= ∆[A] ∆t Δ[A]= change in concentration of A over time period Δt ∆[B ] rate= Δ[B]= change in concentration of B over time ∆t period Δt Because [A] decreases with time, Δ[A] is negative Br (aq)+HCOOH (aq)−→2Br− aq +2H+ (aq)+CO (g) d) 2 2 Br 2 initial ¿ Br2¿ final Average rate= ¿ −∆ B[ 2] =−¿ ∆t i) Instanteaous rate= rate for specific instance in time rate=∝ [r 2] rate=k[Br 2 k= rate =rateconstant [Br 2 −3 −1 ¿3.50∗10 s ii) 2H 2 2(aq)−→2H O 2 (l+O 2(g) PV=nRT n P= RT=[O ]2 RT V [ 2 1 P Measure∆ P RT time 1 ∗∆P ∆[O 2 RT rate= = ∆ t ∆t B. Reaction Rates and Stoichiometry 2A B a) Two moles of A disappear for each mole of B that is formed −1 ∗∆[A] ∆[B] 2 rate= rate= ∆ t ∆ t aA+bB−→cC+dD −1 −1 1 1 a ∗∆ A[ ] b ∗∆ [ ] c∗∆ [ ] d ∗∆[D] rate= = = = ∆t ∆t ∆t ∆t b) Example: i) Write the rate expression for the following reaction: C H4(g)+2O 2g −→C O 2g +2 H 2(g) −1 1 −∆ C[ 4] 2 ∗∆ [ 2 ∆ [ O 2 2∗∆[H O2 rate= = = = ∆t ∆ t ∆t ∆t ii) The rate of disappearance of NO in the reaction 2NO+ O2=¿2 N O 2 is 0.066 Ms −1 . The rate of the reaction is _____−1 ? (a) 0.066 (b) 0.132 (c) 0.033 (d) 0.66 iii) The rate of disappearance of NO in the reaction 2NO+O2 => 2N 0 2 is 0.066 M s−1 . The rate of disappearance of O 2 is ____ M s−1 ? (a) 0.066 (b) 0.132 (c) 0.033 (d) 0.66 O 2 2NO 2 Ms−1 iv) If rate of the reaction 2NO+ => is 0.033 . The rate of −1 disappearance of NO is ___ M s ? 1 − 2 N[ ] (a) 0.066 rate=0.033= ∆t ∆[NO] (b) 0.132 rate of disappearance of NO = ∆ t = 2-rate=0.006 (c) 0.033 (d) 0.66 C. The Rate law a) Rate law: relationship of reaction rate to rate constant & concentrations of reactans aA+bB cC+dD y B¿ A¿ ¿ rate=k¿ i) Reactions is xth order in A ii) Reaction is yth order in B iii) Reaction is (x+y)th order overall (1) X and y are NOT related to a,b,c, or d D. Initial Rate Method a) Can determine order of reactant from effect on initial rate of the reaction when the initial concentration of that reactant is changed i) For multiple reactants, keep initial concentration of all reactants constant except one ii) Zero order= changing the concentration has no effect on the rate iii) First order= rate changes by the same factor as the concentration (1) Doubling initial concentration will double the rate iv) Second order= rate changes by square of factor by which concentration changes (1) Doubling initial concentration will quadruple the rate b) Example i) F2(g)+2ClO 2g −→2FClO (g)2 ClO ¿ y 2 F 2 ∗¿ Rate=k¿ Double [F 2 with [Cl O2] constant Rate doubles X=1 rate= k [ 2ClO ] 2 Quadruple [ClO2] with [F2] constant Rate quadruples y=1 E. Rate Laws a) Rate laws are always determined experimentally b) Reaction order is always defined in terms of reactant(not product) concentrations c) The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation F (g +3Cl O (g −→2FClO (g) 2 2 2 1 ClO ¿2 rate=k F ¿ [ ]2 d) Example: i) Determine the rate law & calculate the rate constant for the following −¿(aq) 2−¿ (aq +I ¿ 3 ¿ reaction from the following data: −¿ ( )−→2SO 4 ¿ 2−¿ aq )¿3I S 2 8 r.f exp. 1+2 (S .0172O82-) is constant −4 initialrate2=2∗2∗10 [I-] change term .034 0.017 [1∗10 −4 .036/.107 =2  [I-] S ¿ 2−¿ C.f. 2+3 [I-] is constant double => can double initial ¿ ¿ rate 4 2−¿¿ S O ¿ => 2 8 ¿ I−¿ 2−¿¿ ¿4 Rate= ¿ S2O 8 k¿ 4 .034¿ ¿ (08)¿ −1 1 2,2∗10 1 exp.= 0.034¿ =¿k= ¿ 1 .08¿ ¿ 2.2∗10 −4 −1 =k∗¿ M s ii) Determine the rate law & calculate the rate constant for the following −¿(aq) 2−¿ aq +I 3 ¿ reaction from the following data: −¿ ( )−→2SO 4 2−¿ (aq)+3I ¿ ¿ S2O 8 B¿ =3.20∗10 −1 x [A][B] A¿ ¿ initialrate:1=k∗¿ 1.50¿ =3.2∗10 −1 ¿ y −1 2.50¿ =3.2∗10 y A¿ 1 1.50¿ ¿ 1.50 y=1 rate=k ¿ k¿ 2.50 1.50¿ ¿ k¿ ¿ F. First-Order Reactions rate= −∆[A] a) Aproduct ∆ t rate=k[A] rate MIs 1 K= = = ∨s −1 −∆ [A ]=K[ A] [A] M s ∆t [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 A¿ 0xp? (−kt) A¿ okt A =¿ ln A −ln? ¿ [ ] [ ]


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