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**Entry Task: Dec 6th 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 2 Rate = k[A]1[B]2 3 1.62 x10-5 = k[0.0100]1[0.0100]2**

1. For 2 A + B C, we’ve determined the following experimental data: 1 2 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 Rate = k[A]1[B]2 3 1.62 x10-5 = k[0.0100]1[0.0100]2 1.62 x10-5 = k= 16.2 M-1s-1 1.0 x10-6

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**2 1 Rate = k[A]2[B]1 3 2.80 x10-3 = k[0.026]2[0.015]1**

2. For 2 A + B C, we’ve determined the following experimental data: 2 1 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 Rate = k[A]2[B]1 3 2.80 x10-3 = k[0.026]2[0.015]1 2.80 x10-3 = k= 277 M-1s-1 1.01 x10-5

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3. The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H2(g) N2(g) + 2 H2O(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 [H2] = 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 H2, the rate doubled as well. This means [H2]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 H2, the rate doubled as well. This means [H2]1 Exp. 2 vs. Exp. 3, we doubled the concentration of NO and the rate quadrupled or This means that NO is 2nd order [NO]2 a) Rate = k[NO]2[H2]1

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**a) Rate = k[NO]2[H2]1 Rate 4.92 x 10-3 M/s k= = = 1.2 M-2/s-1**

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 [H2] = M. a) Rate = k[NO]2[H2]1 Rate 4.92 x 10-3 M/s k= = = 1.2 M-2/s-1 [NO]2[H2]1 [0.20]2[0.10]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 [H2] = M. a) Rate = k[NO]2[H2]1 b) k = 1.2 M1 s1 Rate = (1.2 M1 s1) (0.050 M)2(0.150) Rate = 4.5 x 10-4 M/s

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14.21 4. The following data were collected for the rate of disappearance of NO in the reaction: 2NO (g) + O2 (g) 2NO2 (g) Experiment [NO] (M) [O2] (M) Initial Rate M/s 1 0.0126 0.0125 1.41 x 10-2 2 0.0252 0.0250 1.13 x 10-1 3 5.64 x 10-2 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 [O2], if you doubled the concentration, the rate doubles so [O2]1

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14.21 4. The following data were collected for the rate of disappearance of NO in the reaction: 2NO (g) + O2 (g) 2NO2 (g) Experiment [NO] (M) [O2] (M) Initial Rate M/s 1 0.0126 0.0125 1.41 x 10-2 2 0.0252 0.0250 1.13 x 10-1 3 5.64 x 10-2 c) What is the average value of the rate constant calculated from the three data sets. SHOW YOUR WORK!! Rate 5.64 x 10-2 M/s k= = = 7.11x103 M-2/s-1 [NO]2[H2]1 [0.0252]2[0.125]1

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14.23 5. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g) Experiment [NO] (M) [Br2] (M) Initial Rate M/s 1 0.10 0.20 24 2 0.25 150 3 0.50 60 4 0.35 735 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 [Br2], if you doubled the concentration, the rate doubles so [Br2]

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**14.23 Rate 150 M/s k= = = 1.2 x104 M-2/s-1 [NO]2[Br2]1 [0.25]2[0.20]1**

Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g) Experiment [NO] (M) [Br2] (M) Initial Rate M/s 1 0.10 0.20 24 2 0.25 150 3 0.50 60 4 0.35 735 b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets. Rate 150 M/s k= = = 1.2 x104 M-2/s-1 [NO]2[Br2]1 [0.25]2[0.20]1

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**14.23 1 [Br2] t 1 2 [NOBr] t Rate = − =**

5. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g) Experiment [NO] (M) [Br2] (M) Initial Rate M/s 1 0.10 0.20 24 2 0.25 150 3 0.50 60 4 0.35 735 c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2? 1 [Br2] t 1 2 [NOBr] t Rate = − =

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**14.23 Rate = k [NO]2[Br2]1 = 2(1.2 x104 M-2/s-1) [0.075]2[0.185]1**

5. Consider the gas-phase reaction between nitric acid oxide ad bromine at 273°C: 2NO (g) + Br2 (g) 2NOBr (g) Experiment [NO] (M) [Br2] (M) Initial Rate M/s 1 0.10 0.20 24 2 0.25 150 3 0.50 60 4 0.35 735 d) What is the rate of disappearance of Br2 when [NO] = 0.075M and [Br2] = M? Rate = k [NO]2[Br2]1 = 2(1.2 x104 M-2/s-1) [0.075]2[0.185]1 = 6.1 M/s

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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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**Equation Sheet Under thermochemistry and kinetics**

1st order 2nd order Arrhenius Equation

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Chapter 14 section 3 Notes

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Two Types of Rate Laws Differential- Data table contains RATE AND CONCENTRATION DATA. Uses “table logic” or algebra to find the order of reaction and rate law 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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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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**Graphing Integrated Rate Law**

Time is always on x axis Plot concentration on y axis of 1st 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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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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**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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**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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**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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First Order Reactions AB in a reaction. ① Write the rate expression for reactant A. (sec. 1stuff) Rate = - ∆[A] ∆t

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**14.3- The Change of Concentration with Time**

② Write the rate law for reactant A. (sec 2 stuff) Rate = k[A]1

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**14.3- 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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**Sketch a graph with rate on Y and concentration on X axis- Label axis!!**

As we double our concentration , the rate doubles. It’s a direct relationship.

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**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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**Natural log x Concentration**

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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**14.3- The Change of Concentration with Time**

Take equations ① and ② and smash them together. Rate = - ∆[A] ∆t = k[A]

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**14.3- 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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**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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**First-Order Processes**

Relate this equation to the slope. ln [A]t = -kt + ln [A]0 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.

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**Sample Exercise 14.5 Using the Integrated First-Order Rate Law**

The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 107 g/cm3. 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 3.0 10–7 g/cm3? PLUG & CHUG k = 1.45 yr-1 ln [A]0 = [5.0 x 10-7g/cm3] SET IT UP ln [insecticide]t-1yr = [X] t = 1 year ln [insectacide]t-1yr = -(1.45 yr-1) (1 year) + ln [5.0 x 10-7g/cm3] ln[insecticide]t -1 yr = (14.51) ln[insecticide]t - 1 yr = 15.96 Get rid of ln by ex on both sides [insecticide]t = 1 yr = e15.96 = 1.2 107 g/cm3

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**Sample Exercise 14.5 Using the Integrated First-Order Rate Law**

The decomposition of a certain insecticide in water at 12 C follows first-order kinetics with a rate constant of 1.45 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 107 g/cm3. 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 3.0 10–7 g/cm3? PLUG & CHUG k = 1.45 yr-1 ln [A]0 = [5.0 x 10-7g/cm3] SET IT UP ln [ ]t = ln [ ]t = -(1.45 yr-1) X + ln [5.0 x 10-7g/cm3] t = X Get X by itself- move to left side 15.02 ln [ ]t - ln [5.0 x 10-7g/cm3] =X =0.35 years 1.45 yr 1.45 yr-1

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**Sample Exercise 14.5 Using the Integrated First-Order Rate Law**

Continued Practice Exercise The decomposition of dimethyl ether, (CH3)2O, at 510 ºC is a first-order process with a rate constant of 6.8 10–4 s–1: (CH3)2O(g) CH4(g) + H2(g) + CO(g) If the initial pressure of (CH3)2O is 135 torr, what is its pressure after 1420 s? k = 6.8 x 10-4-s-1 ln [A]0 = [135 torr] SET IT UP ln [X torr]t = [X] t = s ln [X torr]t = -(6.8 x 10-4) (1420 s) + ln [135 torr] ln[torr]t = (4.91) ln[torr]t = 3.94 Get rid of ln by ex on both sides [torr]t = e3.94= 51.6 torr

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**14.3- 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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**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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**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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**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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**Second-Order Processes**

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

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**Second-Order Processes**

1 [A]t = kt + [A]0 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.

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**14.3- 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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**Second-Order Processes**

The decomposition of NO2 at 300°C is described by the equation NO2 (g) NO (g) + 1/2 O2 (g) and yields data comparable to this: Time (s) [NO2], M 0.0 50.0 100.0 200.0 300.0

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**Second-Order Processes**

Graphing ln [NO2] vs. t yields: The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 −4.610 50.0 −4.845 100.0 −5.038 200.0 −5.337 300.0 −5.573

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**Second-Order Processes**

Graphing ln 1/[NO2] vs. t, however, gives this plot. Because this is a straight line, the process is second-order in [A]. Time (s) [NO2], M 1/[NO2] 0.0 100 50.0 127 100.0 154 200.0 208 300.0 263

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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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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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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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NOTES 14-3 obj 14.4. 1.) Using calculus to integrate the rate law for a first-order process gives us A.) INTEGRATED RATE LAWS ln [A] t [A] 0 = −kt Where.

NOTES 14-3 obj 14.4. 1.) Using calculus to integrate the rate law for a first-order process gives us A.) INTEGRATED RATE LAWS ln [A] t [A] 0 = −kt Where.

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