Reaction Rates Reaction Rates C 4 H 9 Cl( aq ) + H 2 O( l ) C 4 H 9 - PDF document

kinetics­presentation­student notes 2014.notebook Kinetics Chemical Kinetics kinetics­multiple­choice­free­response questions only.doc 25 Rate and Order Comp.doc AP Reaction Rates 2013.docx ChemQuest 44­42.doc Skill Practice 42­44.doc Schedule 10­7­10­28 2013.docx Reaction Rates Factors That Affect Reaction Rates 0 s 20 s 40 s A B 0.5 mol A 0.25 mol A 1 mol A 0.5 mol B 0.75 mol B Be able to describe how the rate is affected by Rates of reactions can be determined by monitoring each through Collision Theory.... the change in concentration of either reactants or products as a function of time. Reaction Rates C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) Reaction Rates Time , t(s) [C 4 H 9 Cl](M) 0.0 0.1000 50 0.0905 C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) 1 00.0 0.0820 1 50.0 0.0741 Time , t(s) [C 4 H 9 Cl](M) Average Rate (M/s) 200.0 0.0671 300.0 0.0549 400.0 0.0448 0.0 0.1000 500.0 0.0368 50 0.0905 1.9 x10 ­4 800.0 0.0200 10000 0 00.0 0.0820 1.7 x10 ­4 50. 0 0.0741 1.6 x10 ­4 In this reaction, the concentration of butyl chloride, 200. 0 0.0671 1.4 x10 ­4 C 4 H 9 Cl, was measured at various times. 300.0 0.0549 1.22 x 10 ­4 400.0 0.0448 1.01 x10 ­4 500.0 0.0368 0.88 x10 ­4 800.0 0.0200 0.560 x10 ­4 10000 0 [C 4 H 9 Cl] Average Rate = t The average rate of the reaction over each interval is the change in concentration divided by the change in time: 1

kinetics­presentation­student notes 2014.notebook Reaction Rates Reaction Rates C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) Time , t(s) [C 4 H 9 Cl](M) Average Rate (M/s) 0.0 0.1000 • A plot of [C 4 H 9 Cl] vs. 50 0.0905 1.9 x10 ­4 Instantaneous time for this reaction 00.0 0.0820 1.7 x10 ­4 rate at t=0s 50. 0 0.0741 1.6 x10 ­4 yields a curve like this. 200. 0 0.0671 1.4 x10 ­4 [C 4 H 9 Cl] (M) x 10 ­2 • The slope of a line 300.0 0.0549 1.22 x 10 ­4 400.0 0.0448 1.01 x10 ­4 tangent to the curve at 500.0 0.0368 0.88 x10 ­4 any point is the Instantaneous 800.0 0.0200 0.560 x10 ­4 rate at t=600s instantaneous rate at 10000 0 that time. • Note that the average rate decreases as the reaction proceeds. WHY? Time (s) Reaction Rates and Stoichiometry Reaction Rates C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) C 4 H 9 Cl( aq ) + H 2 O( l ) → C 4 H 9 OH( aq ) + HCl( aq ) • In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Instantaneous Instantaneous • All reactions slow • Thus, the rate of rate at t=0s rate at t=0s down over time. disappearance of C 4 H 9 Cl [C 4 H 9 Cl] (M) x 10 ­2 [C 4 H 9 Cl] (M) x 10 ­2 • Therefore, the best is the same as the rate of indicator of the rate of appearance of C 4 H 9 OH. Instantaneous a reaction is the rate at t=600s Instantaneous rate at t=600s instantaneous rate near the beginning of the reaction. ­ [C 4 H 9 Cl] [C 4 H 9 OH] Rate = Time (s) = t t Time (s) Reaction Rates and Stoichiometry Reaction Rates and Stoichiometry • To generalize, then, for the reaction • What if the ratio is not 1:1? 2HI (g) H 2 (g) + I 2 (g) aA + bB cC + dD 2

kinetics­presentation­student notes 2014.notebook Concentration and Rate Concentration and Rate + ] Initial [NO 2 ­ ] Observed initial Experiment Initial[ NH 4 number (M) [M] Rate (M/s) 1 0.0100 0.200 5.4 x10 ­7 2 0.0200 0.200 10.8 x10 ­7 One can gain information about the rate of a 3 0.0400 0.200 21.5 x10 ­7 reaction by seeing how the rate changes 4 0.200 0.0202 10.8 x10 ­7 5 0.200 0.0404 21.6 x10 ­7 with changes in concentration. 6 0.200 0.0808 43.3 x10 ­7 + (aq) + NO 2− (aq) NH 4 N 2 (g) + 2 H 2 O(l) Concentration and Rate Concentration and Rate + ] Initial [NO 2 ­ ] Observed initial Experiment Initial[ NH 4 number (M) [M] Rate (M/s) 1 0.0100 0.200 5.4 x10 ­7 2 0.0200 0.200 10.8 x10 ­7 3 0.0400 0.200 21.5 x10 ­7 4 0.200 0.0202 10.8 x10 ­7 5 0.200 0.0404 21.6 x10 ­7 6 0.200 0.0808 43.3 x10 ­7 + (aq) + NO 2− (aq) NH 4 N 2 (g) + 2 H 2 O(l) Likewise, when we compare Experiments 5 and 6, we see ______________________________________ Rate Laws Order of a reaction Order of the reaction can be determined only from experimental data. Concentration changes → Rate independent → zero order Concentration doubles → Rate doubles → first order Concentration doubles → Rate quadruples → second order 3

kinetics­presentation­student notes 2014.notebook AP Questions 2002 & 2003.doc Integrated Rate Laws AP Chemistry Exam 2003 5Br­ (aq) + BrO 3 ­ (aq) + 6H+ (aq) ­­­> 3Br 2 (l) + 3H 2 O (l) Integrated rate law equations can be used to determine the In a study of the kinetics of the reaction represented above, the following data were concentration of either a reactant or a product at any time. obtained at 298 K. In a typical problem, you may be asked to find either: 1) concentration of a reactant at a given time, t, after the start of a rxn 2) the time it takes for a certain amount of reactant to be consumed a. From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning. 3) the time it takes for the concentration of a Br ­ i. reactant or product to reach a certain level ­ ii. BrO 3 H + iii. b. Write the rate law for the overall reaction. c. Determine the value for the specific rate constant for the reaction at 298 K. Include the correct unit. Integrated Rate Laws Integrated Rate Laws Manipulating this equation produces… Using calculus to integrate the rate law [A] t ln [A] 0 =­kt for a first­order process gives us [A] t R= k[A] time 0 to t [A] 0 =­kt ln ln[A] t ­ ln[A] 0 = ­kt ln[A] t = ­kt + ln[A] 0 …which is in the form Where [A] 0 is the initial concentration of A, and y = mx + b [A] t is the concentration of A at some time, t, during the course of the reaction. First­Order Processes First­Order Processes CH 3 NC ln [A] t = ­kt + ln [A] 0 Consider the process in which methyl Therefore, if a reaction is first­order, a plot isonitrile is converted to acetonitrile. of ln [A] vs. t will yield a straight line, and the slope of the line will be ­k. CH 3 CN CH 3 NC CH 3 CN 4

kinetics­presentation­student notes 2014.notebook First­Order Processes First­Order Processes CH 3 NC CH 3 CN • When ln P is plotted as a function of time, a straight line results. • Therefore, This data was collected for this reaction at 198.9 °C. • The process is first­order. • k is the negative of the slope: 5.1 × 10­5 s ­1 . Second­Order Processes Second­Order Processes 1 [A]t = kt + 1 Similarly, integrating the rate law for a process that is second­order in reactant A, R= k[A] 2 , we get [A]0 So if a process is second­order in A, a plot of 1/[A] vs. t 1 1 will yield a straight line, and the slope of that line is k . [A] t = kt + also in the form [A] 0 y = mx + b Second­Order Processes Second­Order Processes Plotting ln [NO 2 ] vs. t yields the graph below The decomposition of NO 2 at 300°C is described by The plot is not a straight the equation line, so the process is not first­order in [A]. and yields data comparable to this: 1 NO 2 (g) NO (g) + O 2 (g) 1 2 2 Time (s) [NO 2 ], M 0.0 0.01000 Time (s) [NO 2 ], M ln [NO 2 ] 50.0 0.00787 0.0 0.01000 ­4.610 100.0 0.00649 50.0 0.00787 ­4.845 200.0 0.00481 100.0 0.00649 ­5.038 300.0 00.0380 200.0 0.00481 ­5.337 300.0 0.00380 ­5.573 5