entry task: dec 6 th thursday question : for the general rate law, rate = k[a] [b] 2, what will...

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!

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. For 2 A + B C, we’ve 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

1 2

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. For 2 A + B C, we’ve 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

2 1

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] = 0.050 M and [H2] = 0.150 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 22 . This means that NO is 2nd order [NO]2

a) Rate = k[NO]2[H2]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] = 0.050 M and [H2] = 0.150 M. a) Rate = k[NO]2[H2]1

k=Rate

[NO]2[H2]1=

4.92 x 10-3 M/s[0.20]2[0.10]1

= 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] = 0.050 M and [H2] = 0.150 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

14.214. 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 0.0252 0.0125 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

14.214. 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 0.0252 0.0125 5.64 x 10-2

c) What is the average value of the rate constant calculated from the three data sets. SHOW YOUR WORK!!

k=Rate

[NO]2[H2]1=

5.64 x 10-2 M/s[0.0252]2[0.125]1

= 7.11x103 M-2/s-1

14.235. 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 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735

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]

14.23Consider 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 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735

b) Calculate the average value of the rate constant for the appearance of NOBr from our four data sets.

k=Rate

[NO]2[Br2]1=

150 M/s[0.25]2[0.20]1

= 1.2 x104 M-2/s-1

14.235. 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 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735

c) How is the rate of appearance of NOBr related to the rate of disappearance of Br2?

Rate = −11

[Br2]t

= 12

[NOBr]t

14.235. 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 0.20 242 0.25 0.20 1503 0.10 0.50 604 0.35 0.50 735

d) What is the rate of disappearance of Br2 when [NO] = 0.075M and [Br2] = 0.185 M?

= kRate [NO]2[Br2]1

[0.075]2[0.185]1= 2(1.2 x104 M-2/s-1)

= 6.1 M/s

ChemicalKinetics

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.

ChemicalKinetics

Equation SheetUnder thermochemistry and kinetics

1st order

2nd order

Arrhenius Equation

ChemicalKinetics

Chapter 14 section 3 Notes

ChemicalKinetics

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

ChemicalKinetics

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

ChemicalKinetics

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

ChemicalKinetics

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

ChemicalKinetics

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

As concentration increases, the rate of reaction remains the same.

ChemicalKinetics

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.

ChemicalKinetics

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)

ChemicalKinetics

First Order Reactions

AB in a reaction.

① Write the rate expression for reactant A. (sec. 1stuff)

Rate = - ∆[A]∆t

ChemicalKinetics

14.3- The Change of Concentration with Time

② Write the rate law for reactant A. (sec 2 stuff)

Rate = k[A]1

ChemicalKinetics

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

ChemicalKinetics

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.

ChemicalKinetics

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

ChemicalKinetics

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 ConcentrationIn [A]

Slope is negative (-k)

ChemicalKinetics

14.3- The Change of Concentration with Time

Take equations and and smash ① ②them together.

Rate = - ∆[A]∆t

= k[A]

ChemicalKinetics

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.

ChemicalKinetics

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

ChemicalKinetics

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.

ChemicalKinetics

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

Sample Exercise 14.5 Using the Integrated First-Order Rate Law

ln[insecticide]t -1 yr = 1.45 + (14.51)

k = 1.45 yr-1

ln [A]0 = [5.0 x 10-7g/cm3]

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 10-7g/cm3]

Get rid of ln by ex on both sides

ln[insecticide]t - 1 yr = 15.96

[insecticide]t = 1 yr = e15.96 = 1.2 107 g/cm3

ChemicalKinetics

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

Sample Exercise 14.5 Using the Integrated First-Order Rate Law

k = 1.45 yr-1

ln [A]0 = [5.0 x 10-7g/cm3]

t = X

ln [3.0 10-7]t = -(1.45 yr-1)

SET IT UPXln [3.0 10-7]t = + ln [5.0 x 10-7g/cm3]

Get X by itself- move to left side

1.45 yr-1

=Xln [3.0 10-7]t - ln [5.0 x 10-7g/cm3]1.45 yr

=0.35 years15.02 - -14.51

ChemicalKinetics

Practice ExerciseThe 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?

Continued

Sample Exercise 14.5 Using the Integrated First-Order Rate Law

ln[torr]t = 0.9656 + (4.91)

k = 6.8 x 10-4-s-1

ln [A]0 = [135 torr]

t = 1420 s

ln [X torr]t = [X]

-(6.8 x 10-4)

SET IT UP

(1420 s)ln [X torr]t = + ln [135 torr]

Get rid of ln by ex on both sides

ln[torr]t = 3.94

[torr]t = e3.94= 51.6 torr

ChemicalKinetics

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

ChemicalKinetics

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.

ChemicalKinetics

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

ChemicalKinetics

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 Concentration1/[A]

And the slope is positive (k)

ChemicalKinetics

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]0also in the form

y = mx + b

Provide the second order equation.

ChemicalKinetics

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

ChemicalKinetics

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

ChemicalKinetics

Second-Order ProcessesThe 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 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

ChemicalKinetics

Second-Order Processes• Graphing ln [NO2] vs. t

yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 −4.610

50.0 0.00787 −4.845

100.0 0.00649 −5.038

200.0 0.00481 −5.337

300.0 0.00380 −5.573

• The plot is not a straight line, so the process is not first-order in [A].

ChemicalKinetics

Second-Order Processes• Graphing ln

1/[NO2] vs. t, however, gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• Because this is a straight line, the process is second-order in [A].

ChemicalKinetics

kt[A][A] 0 ln[A]

0[A]

1kt

[A]

1

0ln[A]kt

s

M

s

1sM

1

t vs.[A] t vs.ln[A] t vs.[A]

1

ChemicalKinetics

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 ln[A] 0ln[A]kt

ChemicalKinetics

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 kt[A][A] 0

ChemicalKinetics

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

0[A]

1kt

[A]

1

ChemicalKinetics