# Entry Task: Dec 6th Thursday

## Presentation on theme: "Entry Task: Dec 6th Thursday"— Presentation transcript:

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!

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

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

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

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.

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

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

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

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 M1 s1 Rate = (1.2 M1 s1) (0.050 M)2(0.150) Rate = 4.5 x 10-4 M/s

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

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

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]

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

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 = − =

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

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.

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

Chapter 14 section 3 Notes

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

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

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

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

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

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.

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)

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

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

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

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.

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

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)

14.3- The Change of Concentration with Time
Take equations ① and ② and smash them together. Rate = - ∆[A] ∆t = k[A]

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.

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

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.

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 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  107 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 = e15.96 = 1.2  107 g/cm3

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 yr1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0  107 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

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 = e3.94= 51.6 torr

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

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.

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

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)

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

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.

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

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

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

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

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

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

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