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Entry Task: Dec 6 th Thursday Question : For the general rate law, Rate = k[A] [B] 2, what will happen to the rate of reaction if the concentration of A is tripled? You have 5 minutes!

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Agenda: Go through the answers to Rate Law ws #1 Walkthrough NOTES Ch. 14 sec 3 – The change in concentration with time (integrated – graph) Rate Law ws #2

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1. For 2 A + B C, weve determined the following experimental data: a. Rate order for A is _____and B is______ b. The rate law for this reaction is: c. The overall reaction order is_______. d. Provide the rate constant for this reaction 12 Rate = k[A] 1 [B] x10 -5 = k[0.0100] 1 [0.0100] x10 -5 = k= 16.2 M -1 s x10 -6

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2. For 2 A + B C, weve determined the following experimental data: a. Rate order for A is _____and B is______ b. The rate law for this reaction is: c. The overall reaction order is_______. d. Provide the rate constant for this reaction 21 Rate = k[A] 2 [B] x10 -3 = k[0.026] 2 [0.015] x10 -3 = k= 277 M -1 s x10 -5

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3. The following data were measured for the reaction of nitric oxide with hydrogen:2 NO(g) + 2 H 2 (g) N 2 (g) + 2 H 2 O(g) Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = M and [H 2 ] = M.

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3. Using these data, determine (a) the rate law for the reaction Exp. 1 vs. Exp. 2, we doubled the concentration of H 2, the rate doubled as well. This means [H 2 ] 1

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3. Using these data, determine (a) the rate law for the reaction Exp. 1 vs. Exp. 2, we doubled the concentration of H 2, the rate doubled as well. This means [H 2 ] 1 Exp. 2 vs. Exp. 3, we doubled the concentration of NO and the rate quadrupled or 2 2. This means that NO is 2 nd order [NO] 2 a) Rate = k[NO] 2 [H 2 ] 1

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3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, c) the rate of the reaction when [NO] = M and [H 2 ] = M. a) Rate = k[NO] 2 [H 2 ] 1 k= Rate [NO] 2 [H 2 ] 1 = 4.92 x M/s [0.20] 2 [0.10] 1 = 1.2 M -2 /s -1

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3. Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [NO] = M and [H 2 ] = M. a) Rate = k[NO] 2 [H 2 ] 1 b) k = 1.2 M 1 s 1 Rate = (1.2 M 1 s 1 ) (0.050 M) 2 (0.150) Rate = 4.5 x M/s

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The following data were collected for the rate of disappearance of NO in the reaction: 2NO (g) + O 2 (g) 2NO 2 (g) Experiment[NO] (M)[O 2 ] (M)Initial Rate M/s x x x a) What is the rate law for the reaction? For [NO], if you doubled the concentration, the rate goes up by a factor of 4 so [NO] 2 For [O 2 ], if you doubled the concentration, the rate doubles so [O 2 ] 1

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The following data were collected for the rate of disappearance of NO in the reaction: 2NO (g) + O 2 (g) 2NO 2 (g) Experiment[NO] (M)[O 2 ] (M)Initial Rate M/s x x x c) What is the average value of the rate constant calculated from the three data sets. SHOW YOUR WORK!! k= Rate [NO] 2 [H 2 ] 1 = 5.64 x M/s [0.0252] 2 [0.125] 1 = 7.11x10 3 M -2 /s -1

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Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br 2 (g) 2NOBr (g) Experiment[NO] (M)[Br 2 ] (M)Initial Rate M/s a) Determine the rate law? For [NO], if you doubled the concentration, the rate goes up by a factor of 4 so [NO] 2 For [Br 2 ], if you doubled the concentration, the rate doubles so [Br 2 ]

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14.23 Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br 2 (g) 2NOBr (g) Experiment[NO] (M)[Br 2 ] (M)Initial Rate M/s b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets. k= Rate [NO] 2 [Br 2 ] 1 = 150 M/s [0.25] 2 [0.20] 1 = 1.2 x10 4 M -2 /s -1

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Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br 2 (g) 2NOBr (g) Experiment[NO] (M)[Br 2 ] (M)Initial Rate M/s c) How is the rate of appearance of NOBr related to the rate of disappearance of Br 2 ? Rate = 1111 [Br 2 ] t = 1212 [NOBr] t

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Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br 2 (g) 2NOBr (g) Experiment[NO] (M)[Br 2 ] (M)Initial Rate M/s d) What is the rate of disappearance of Br 2 when [NO] = 0.075M and [Br 2 ] = M? = k Rate [NO] 2 [Br 2 ] 1 [0.075] 2 [0.185] 1 = 2(1.2 x10 4 M -2 /s -1 ) = 6.1 M/s

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Chemical Kinetics I can… Graph the relationship of time with amount of reactant concentrations and integrate this with rates of reactions. Determine the graphical relationship between time and the rate order.

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Chemical Kinetics Equation Sheet Under thermochemistry and kinetics 1 st order 2 nd order Arrhenius Equation

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Chemical Kinetics Chapter 14 section 3 Notes

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Chemical Kinetics Two Types of Rate Laws 1.Differential- Data table contains RATE AND CONCENTRATION DATA. Uses table logic or algebra to find the order of reaction and rate law 2.Integrated- Data table contains TIME AND CONCENTRATION DATA. Uses graphical methods to determine the order of the given reactant. K=slope of best fit line found through linear regressions

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Chemical Kinetics Integrated Rate Law Can be used when we want to know how long a reaction has to proceed to reach a predetermined concentration of some reagent

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Chemical Kinetics Graphing Integrated Rate Law Time is always on x axis Plot concentration on y axis of 1 st graph Plot ln [A] on the y axis of the second graph Plot 1/[A] on the y axis of third graph You are in search of a linear graph

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Chemical Kinetics Zero order Reactions- Use A B as an example. What happens when we double [A], what happens to the rate of reaction that is zero order? So does the concentration affect rate? Y/N____ What would the rate law be for a zero order? The rate of reaction does not change No Rate = k

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Chemical Kinetics Sketch a graph with rate on Y and concentration on X axis- Label axis!! As concentration increases, the rate of reaction remains the same.

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! Integrated Rate laws We look for straight lines. This provides a clean visual about the relationship of concentration and time.

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! So when we plot our data table and get a negative straight line it is ____________order! Slope is negative (-k)

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Chemical Kinetics First Order Reactions A B in a reaction. Write the rate expression for reactant A. (sec. 1stuff) Rate = - [A] t

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Chemical Kinetics The Change of Concentration with Time Write the rate law for reactant A. (sec 2 stuff) Rate = k[A] 1

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Chemical Kinetics The Change of Concentration with Time Describe a First Order reaction. Double the concentration the reaction doubles. Low amount of reactant = low rate of reaction

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Chemical Kinetics Sketch a graph with rate on Y and concentration on X axis- Label axis!! As we double our concentration, the rate doubles. Its a direct relationship.

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! Integrated Rate laws We look for straight lines. This provides a clean visual about the relationship of concentration and time. This does not provide a straight line

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! Integrated Rate laws We can manipulate the data to provide a straight line plot. Change how we plot concentration. Natural log x Concentration In [A] Slope is negative (-k)

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Chemical Kinetics The Change of Concentration with Time Take equations and and smash them together. Rate = - [A] t = k[A]

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Chemical Kinetics The Change of Concentration with Time How do you use this equation to solve for concentration? Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 = kt Where [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t, during the course of the reaction.

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Chemical Kinetics Integrated Rate Laws Manipulating this equation produces… ln [A] t [A] 0 = kt ln [A] t ln [A] 0 = kt ln [A] t = kt + ln [A] 0 …which is in the form y = mx + b

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Chemical Kinetics First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A] t = -kt + ln [A] 0 Relate this equation to the slope.

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Chemical Kinetics The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr 1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of g/cm 3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to –7 g/cm 3 ? PLUG & CHUG Sample Exercise 14.5 Using the Integrated First-Order Rate Law ln[insecticide] t -1 yr = ( 14.51) k = 1.45 yr -1 ln [A] 0 = [5.0 x g/cm 3 ] t = 1 year ln [insecticide] t-1yr = [X] -(1.45 yr -1 ) SET IT UP (1 year) ln [insectacide] t-1yr = + ln [5.0 x g/cm 3 ] Get rid of ln by e x on both sides ln[insecticide] t - 1 yr = [insecticide] t = 1 yr = e = g/cm 3

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Chemical Kinetics The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr 1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of g/cm 3. Assume that the average temperature of the lake is 12 ºC. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to –7 g/cm 3 ? PLUG & CHUG Sample Exercise 14.5 Using the Integrated First-Order Rate Law k = 1.45 yr -1 ln [A] 0 = [5.0 x g/cm 3 ] t = X ln [ ] t = -(1.45 yr -1 ) SET IT UP X ln [ ] t = + ln [5.0 x g/cm 3 ] Get X by itself- move to left side 1.45 yr -1 =X ln [ ] t - ln [5.0 x g/cm 3 ] 1.45 yr =0.35 years

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Chemical Kinetics Practice Exercise The decomposition of dimethyl ether, (CH 3 ) 2 O, at 510 ºC is a first-order process with a rate constant of –4 s –1 : (CH 3 ) 2 O(g) CH 4 (g) + H 2 (g) + CO(g) If the initial pressure of (CH 3 ) 2 O is 135 torr, what is its pressure after 1420 s? Continued Sample Exercise 14.5 Using the Integrated First-Order Rate Law ln[torr] t = (4.91) k = 6.8 x s -1 ln [A] 0 = [135 torr] t = 1420 s ln [X torr] t = [X] -(6.8 x ) SET IT UP (1420 s)ln [X torr] t = + ln [135 torr] Get rid of ln by e x on both sides ln[torr] t = 3.94 [torr] t = e 3.94 = 51.6 torr

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Chemical Kinetics The Change of Concentration with Time Describe a second-order reaction. When you double the reactant the rate increases by a power of 2, to quadruple the rate

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Chemical Kinetics Sketch a graph with rate on Y and concentration on X axis- Label axis!! The relationship is more pronounced. Double your concentration and the rate goes up by the power of 2. Hence- second order.

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! Integrated Rate laws We look for straight lines. This provides a clean visual about the relationship of concentration and time. This does not provide a straight line

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Chemical Kinetics Sketch a graph with concentration on Y and time on X axis- Label axis!! Integrated Rate laws We can manipulate the data to provide a straight line plot. Change how we plot concentration. 1 divided by Concentration 1/[A] And the slope is positive (k)

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Chemical Kinetics Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b Provide the second order equation.

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Chemical Kinetics Second-Order Processes So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k. 1 [A] t = kt + 1 [A] 0

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Chemical Kinetics The Change of Concentration with Time What does second order reactions depend on? A second order reaction is one whose rate depends on the initial reactant concentration

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Chemical Kinetics Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields data comparable to this: Time (s)[NO 2 ], M

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Chemical Kinetics Second-Order Processes Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] The plot is not a straight line, so the process is not first-order in [A].

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Chemical Kinetics Second-Order Processes Graphing ln 1/[NO 2 ] vs. t, however, gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] Because this is a straight line, the process is second- order in [A].

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Chemical Kinetics

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Chemical Kinetics Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph. What order is this reaction and what formula would I use to calculate various times/concentrations? First order and

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Chemical Kinetics Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph. What order is this reaction and what formula would I use to calculate various times/concentrations? Zero order and

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Chemical Kinetics Practice with graphs- After creating regression graphs of various reactions, provide the rate order for each graph. What order is this reaction and what formula would I use to calculate various times/concentrations? Second order and

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Chemical Kinetics

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