Kinetics and Rates of reaction Reaction Rate : The speed of a reaction (usually M/s)

(1)

Chemical Kinetics

(2)

Chemical Kinetics

The area of chemistry that concerns

The area of chemistry that concerns reaction ratesreaction rates and

(3)

Kinetics and Rates of reaction

Reaction Rate: The speed of a reaction (usually M/s)

1. Can measure by the rate of increase of products or decrease of reactants

2. Rate and Time inverses (high rate, small time)

Consider the following equation:

2 HCl(aq) + Mg(s) H₂(g)+ MgCl₂(aq)

How can rate be measured? 1.Disappearance of reactants

*Decrease of HCl and Mg as time inc 2. Appearance of products

*Increase in H₂and MgCl₂ as time inc

(4)

A B

time

Can you draw a graph to represent how the concentrations (Molarity) of reactants and products change over time?

Y-axis: concentration X-axis: time

(5)

A B

(6)

Chemical Kinetics

Thermodynamics – does a reaction take place? (Unit 6) Kinetics – how fast does a reaction proceed?

Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

A B

rate = _[B]_t

_{[A] = change in concentration of A over}

time period t

_{[B] = change in concentration of B over}

time period t

*Because [A] decreases with time, [A] is negative (reactant conc. dec as time inc)

rate = - _[A]_t

(7)

Br₂ (l) + HCOOH (aq) 2Br- ₍_aq₎ + 2H+ ₍_aq₎ + CO

2 (g)

time

393 nm

light Detector

[Br₂]  Absorption

(8)

Spectrophotometry

• Spectrophotometry can be used to measure

rates of chemical rxns and predict rate laws.

• Recall that concentration is related to

absorbance

How long did it take for approximately 75 percent

of the initial Br

₂

(l) sample to react?

Br₂(l) + HCOOH(aq) 2Br-₍_aq₎ + 2H+₍_aq₎ + CO

(9)

Br₂ (l) + HCOOH (aq) 2Br- ₍_aq₎ + 2H+ ₍_aq₎ + CO

2 (g)

average rate

(time interval) = - [Br2]

t =

-[Br₂]_final – [Br₂]_initial

t_final - t_initial

slope of tangent

slope of

tangent _{slope of} tangent

instantaneous rate = rate for specific instance in time

(10)

rate _[Br₂_]

rate = k [Br₂]

k = rate

[Br₂]= rate constant = 3.50 x 10-3 s-1

Calculate the average rate

of the disappearance Br

₂

between 0 and 50

seconds.

Calculate the average rate

between 100 and 200

seconds.

Calculate the average rate

between 300 and 400

seconds.

(11)

Reaction Rates and Stoichiometry

2A B

Two moles of A disappear for each mole of B that is formed.

rate = [B]

t

rate = - [A]

t

1 2

aA + bB cC + dD

rate = - [A]

t

1

a =

-[B] t 1 b = [C] t 1 c = [D] t 1 d

(12)

Write the rate expression for the following reaction:

CH₄ (g) + 2O₂ (g) CO₂ (g) + 2H₂O (g)

rate = - [CH4]

t =

-[O₂]

t

1

2 =

[H₂O]

t

1

2

= [CO2]

t

What type of rxn is

(13)

Br₂ (l) + HCOOH (aq) 2Br- ₍_aq₎ + 2H+ ₍_aq₎ + CO

2 (g)

time

Back to bromine.

(14)

You Try: Rate Examples

2 NO

₂

(g)



2 NO

(g)

+ O

₂

(g)

If the rate of disappearance of NO

₂

is

4.2x10

-5

M/s, what is the rate of appearance

(15)

Another:

2B



P + 3Q + 4R

If the rate of appearance P is 9 x 10

-6

M/s, what

(16)

AP Sample Question:

At a certain time during the titration represented

above, the rate of appearance of O

₂

(

g

) was 1.0 x

10

-3

mol/(L s). What was the rate of

⋅

disappearance of MnO

₄-

at the same time?

A.6.0 x 10

-3

mol/(L s)

⋅

B.4.0 x 10

-3

mol/(L s)

⋅

C.6.0 x 10

-4

mol/(L s)

⋅

(17)

When the chemical reaction 2NO(g)+O

₂

(g)→2NO

₂

(g) is

carried out under certain conditions, the rate of

disappearance of NO(g) is 5.0×10

−5

Ms

−1

. What is the rate of

disappearance of O

₂

(g)under the same conditions?

A.Because two molecules of NO are consumed per molecule

of O

₂

, the rate of disappearance of O

₂

(g) is 2.5×10

−5

Ms

−1

.

B.Because two molecules of NO are consumed per molecule

of O

₂

, the rate of disappearance of O

₂

(g) is 1.0×10

−4

Ms

−1

.

C.Because the rate depends on [NO]

2

and [O

2

], the rate of

disappearance of O

₂

(g) is 2.5×10

−9

Ms

−1

.

D.Because the O

₂

(g) is under the same experimental

conditions as NO, it is consumed at the same rate

(18)

More Challenging Example:

Consider the following balanced chemical eq:

2H

₂

O

₂

(aq)



2H

₂

O

(l)

+ O

₂

(g)

1.Calculate the average rate of decomposition of

H

₂

O

₂

between 0 and 2.16 x 10

4

s.

2.31 x10

-5

M/s

2. Determine the average rate of production of

O

₂

during this same time interval.

1.16 x10

-5

M/s

Time (s) [H2O2] (M)

0 1.000

2.16 x 104 _0.500

(19)

2NO

2NO₂₂(g) (g)  2NO(g) + O 2NO(g) + O₂₂(g) (g)

Review:

In General,

Reaction Rates:

2. Can measure

appearance of

products

1. Can measure

disappearance of

reactants

(20)

2NO

2NO₂₂(g) (g)  2NO(g) + O 2NO(g) + O₂₂(g) (g)

Reaction Rates:

4. Are equal to the

slope tangent to

that point

[NO₂]

t

5. Change as the

reaction proceeds,

if the rate is

(21)

Rate Laws

A rate law is a simplified equation describing a

reaction that allows the rate of that reaction to be

calculated based on the concentration of

reactants.

aA + bB



products

Rate Law expression = k [A]

x

[B]

y

k = rate constant

(22)

•

Two key points:

1. The concentration of the

products do not

appear in the rate law

because this is an

initial rate, and initially there are NO

products.

2. The

order

must be determined

experimentally

.

It cannot be obtained from the chemical

equation nor stoichiometry.

You must be

given data to examine.

Rate Laws

(23)

Two Types of Rate Laws

•

Differential Rate Law

- describes how rate

depends on concentration.

•

Integrated Rate Law

- Describes how

concentration depends on time.

• For each type of differential rate law there is

an integrated rate law and vice versa.

(24)

• The rate

depends on the concentration

depends

of the reactants.

A Real Rate Law Expression:

• Rate =

k

[NO

2

]

x

– This is called a

rate law expression

.

–

k

is called the rate constant.

– x

is the order of the reactant -usually a

positive integer. (0, 1, or 2 for us)

(25)

Method of Initial Rates

EX: aA + bB



products

•Compare multiple sets of data.

•Write the

“

skeletal

”

rate law expression.

•Choose 2 experiments where only 1 variable

changes.

(26)

Order of the Rate Law

(What does this “order” mean?)

• If doubling the [ ] has no effect on the rate, the

reaction is

Zero Order

: Rate = k

• If doubling [ ] doubles the rate, the reaction is

first order

: Rate = k [A]

• If doubling [ ] quadruples rate, the reaction is

2

nd

(27)

Trial Initial [A]

([A]₀) Initial [B] ([B]₀) Initial Rate (v0)

1 1.00 M 1.00 M 1.25 x 10-2 M/s

2 1.00 M 2.00 M 2.5 x 10-2 M/s

3 2.00 M 2.00 M 2.5 x 10-2 M/s

What is the order with respect to A? What is the order with respect to B?

What is the overall order of the reaction? What is the rate law?

What is the rate constant, k?

0 1

1 Rate = k [B]

(28)

DETERMING AND WRITING a

(differential) Rate Law – the

mathematical way

2 NO(g) + Cl

₂

(g)



2 NOCl(g)

Problem

- Write the rate law, determine the

value of the rate constant, k, and the overall

order for the following reaction:

Experiment

Experiment [NO] [NO] (mol/L) (mol/L)

[Cl [Cl₂₂] ] (mol/L) (mol/L) Rate Rate Mol/L·s Mol/L·s 1

1 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

2

2 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6

3

3 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6

4

(29)

Writing a Rate Law

Part 1

– Determine the values for the exponents

in the rate law:

Experiment

1 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

2

2 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6

3

3 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6

4

4 0.5000.500 0.5000.500 1.14 x 101.14 x 10-5-5

In experiment 1 and 2, [Cl

₂

] is constant

while [NO] doubles.

R = k[NO]

x

[Cl

2

]

y

The rate quadruples, so the reaction is

second order with respect to [NO]

R = k[NO]

2

[Cl

2

]

y

In experiment 2 and 4, [NO] is constant

while [Cl

₂

] doubles.

The rate doubles, so the reaction is

first order with respect to [Cl

₂

]

R = k[NO]

2

[Cl

(30)

Writing a Rate Law

Part 2

– Determine the value for k, the rate

constant, by using any set of experimental data:

Experiment

1 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

Rate = k[NO]

2

[Cl

2

]

2 6

1.43 10

x

mol

k

0.250

mol

0.250

mol

L s

L





 



 

 







 



6 3 2

5

3 3 2

1.43 10

9.15 10 0.250

x mol L L

k x

L s mol mol s

(31)

Writing a Rate Law

Part 3

– Determine the overall order for the

reaction.

Rate = k[NO]

2

[Cl

2

]

Overall order is the sum of the exponents,

or orders, of the reactants

2 + 1 = 3

(32)

Easy way to find units of k?

Formula:

1

s*M

(overall order – 1)

(33)

YOU TRY: Determine based on the following info: 1.The rate law

2.The rate constant, k, with proper units 3.The overall reaction order

S₂O₈2- ₍_aq₎ + 3I- ₍_aq₎ 2SO

42- (aq) + I3- (aq)

rate = k [S₂O₈2-]x[I-]y

Double [I-], rate doubles (experiment 1 & 2) y = 1

Double [S₂O₈2-], rate doubles (experiment 2 & 3) x = 1

k = rate

[S₂O₈2-][I-] =

2.2 x 10-4 M/s

(0.08 M)(0.034 M) = 0.08/M•s

rate = k [S₂O₈2-][I-]

Experiment [S₂O₈2-] _[I-] Initial Rate

(M/s) 1 0.08 0.034 2.2 x 10-4

2 0.08 0.017 1.1 x 10-4

(34)

Expressing the answer to rate

laws and rate constant, k.

rate =

k

[S

₂

O

₈2-

][I

-

]

rate constant, k = 0.08/

M

•

s

*Do not substitute the rate constant into the

rate law equation, write it separately.

(Reminder: Easy way to find units of k)

Formula: 1

(35)

AP Sample Question:

According to the rate law for the reaction, an

increase in the concentration of hydronium ion

has what effect on this reaction?

A.The rate of reaction increases.

B.The rate of reaction decreases.

(36)

REVIEW of the RATE LAW

The rate law expresses the relationship of the rate of a

reaction to the rate constant and the concentrations of the reactants raised to some powers.

aA + bB cC + dD Rate = k [A]x[B]y

reaction is xth order in A reaction is yth order in B

(37)

F₂ (g) + 2ClO₂ (g) 2FClO₂ (g)

rate = k [F₂][ClO₂]

Review of Rate Laws

• Rate laws are always determined experimentally.

• Reaction order is always defined in terms of reactant (not product) concentrations.

• The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical

equation and must be determined experimentally

1

(38)

The Integrated Rate Law

• The rate laws we have considered,

express the rate as a function of the

reactant concentrations. (Differential

Rate Laws)

• Not useful b/c cannot determine conc

easily at any time.

• Now we will look at the reactant

(39)

Integrated Rate Laws

• Can all be derived from integrating the

differential rate laws (calculus – yuck!)

• Do not need to memorize all of the steps

of derivation, just small pieces of info

•

Recognize the formula for 0, 1, 2 order

plot that will produce a straight line

(y = mx +b)

(40)

1

st

Order Integrated Rate Law

• Most common type of rate law • Describes all radioactive

decay processes

• Graphing ln [A] vs. time is straight line plot

• Equations:

• Units of k 1/s, negative slope

ln[A]

t

- ln[A]

0

= -

kt

t

_½ =

0.693

(41)

First-Order Reactions

A product rate = - [A]

t rate = k [A] k = rate

[A] = 1/s or s-1

M/s M

=

[A]

t = k [A]

-[A] is the concentration of A at any time t

[A]₀ is the concentration of A at time t=0

(42)

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take

for A to decrease from 0.88 M to 0.14 M ?

ln[A] - ln[A]₀ = - kt kt = ln[A]₀ – ln[A]

t = ln[A]0 – ln[A]

k = 66 s

[A]₀ = 0.88 M

[A] = 0.14 M

ln [A]0 [A]

k

=

ln 0.88 M 0.14 M 2.8 x 10-2 s-1

(43)

First-Order Reactions

The half-life, t_½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t_½ = t when [A] = [A]₀/2

ln [A]0 [A]₀/2

k

=

t_½ ln2

k

= 0.693

k

=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t_½ ln2 k

= 0.693

5.7 x 10-4 s-1

= = 1200 s = 20 minutes

How do you know decomposition is first order?

(44)

For 1

st

Rxn order only:

# of half

lives

1

2

3

4

5

Fraction

remaining

1/2

1/4

1/8

1/16 1/32

Example:

The half-life of 226Ra is 1600 years. How much of an

80 gram sample of 226Ra would remain after a period of

4800 years?

(45)

(46)

2

nd

order Integrated Rate Law

• Graphing

1 / [A] vs.

time

is straight line

plot

• Equation:

• Units of k = 1/M*s

* Positive slope

1 [A]_t

-1

(47)

0 order Integrated Rate Law

• Graphing

[A] vs. time

is straight line plot

• Equation:

(48)

Rate Laws Summary

Zero Order

Zero Order First OrderFirst Order Second OrderSecond Order

Rate Law

Rate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Integrated Rate Law

Rate Law [A] = -kt + [A]0 ln[A] = -kt + ln[A]0 Plot that Plot that produces a produces a straight line straight line

[A] versus t

*Memorize

ln[A] versus t

these!

Relationship of

Relationship of rate constant rate constant to slope of to slope of straight line straight line

Slope = -k Units = M/s

Slope = -k Units = 1/s

Slope = k Units = 1/M*s

Half-Life Half-Life

1

[ ]A versus t

0

1 1

[ ]A  kt [ ]A

0 1/ 2 [ ] 2 A t k

 t_{1/ 2} 0.693

k

 _{1/ 2}

(49)

Eq. Given on AP Exam

(50)

Graphing data to determine Rxn

order

*If given [ ] vs. time data, constructing a

graph of [ ] vs time, ln[ ] vs time and 1/[ ] vs.

time will allow you to determine the order of

the reaction.

*The set of data that gives the straight line

plot will determine order.

(51)

Zero Order Second Order

(52)

Example #1

The following data was obtained for the

decomposition of NO

₂

(g)

at 300°C. Is the

reaction first or second order in NO

₂

?

Time (s) [NO₂] (M)

0.0 5

20.0 4.44

40.0 3.92

60.0 3.46

80.0 3.06

(53)

EXAMPLE #2:

Two graphs were constructed for a reaction at

25°C. Is the reaction first or second order with

(54)

Example #3

Given the equation below, what is the

concentration of H

₂

SO

₄

after 30 min? The

initial concentration of H

₂

SO

₄

is 0.01 M, rate

constant, k = 3 x 10

-4

s

-1

.

H

₂

SO

₄



H

₂

O + SO

₃

(55)

A.The decomposition of N

₂

O

₅

is a zero-order

reaction.

B.The decomposition of N

₂

O

₅

is a first-order reaction.

C.The decomposition of N

₂

O

₅

is a second-order

reaction.

(56)

Based on the information, which of the following is the rate

law expression for the reaction?

2HO

₂

(g)→H

₂

O

₂

(g)+O

₂

(g)

A. Rate=k

B. Rate=k[HO

₂

]

C. Rate=k[HO

₂

]

2

(57)

AP Sample Question

Gaseous cyclobutene undergoes a first-order reaction to

form gaseous butadiene. At a particular temperature, the

partial pressure of cyclobutene in the reaction vessel drops

to one-eighth its original value in 124 seconds. What is the

half-life for this reaction at this temperature?

(58)

The rate constant (k) for the decay of the

radioactive isotope I-131 is 3.6×10

−3

hours

−1

. The

slope of which of the following graphs is correct

for the decay and could be used to confirm the

value of k ?

A.

B.

C.

(59)

A reaction was observed for 20 days and the

(60)

Rate =

k

[M][N]

2

The rate of a certain chemical reaction between

substances M and N obeys the rate law above. The

reaction is first studied with [M] and [N] each 1 x 10

-3

molar. If a new experiment is conducted with [M]

and [N] each 2 x 10

-3

molar, the reaction rate will

increase by a factor of

A.2

(61)

How does a rxn start?

• Mix reactants together and…

• Collisions necessary to initiate or start

rxns, but not all collisions are considered

effective.

• Why? Collisions?

(62)

Collision Theory

1.

Collisions must have sufficient energy

Collisions must have

sufficient energy

to produce the reaction (must equal or

exceed the activation energy).

Activation Energy (Ea)

looked at as barrier,

always need energy

input to start rxn

(63)

Collision Theory

2.

Colliding particles must be correctly

Colliding particles must be

correctly

oriented

to one another in order to

produce a reaction.

Which diagram shows proper orientation of reactant

molecules to

produce CO₂ and NO from a

(64)

What can be done to increase

RATE?

• Need to increase effective collisions

• Need two important factors to make

collisions effective

– Energy

– Orientation

(65)

Factors Affecting Rate

Increasing temperature

always increases

the rate of a reaction.



Particles collide more frequently



Particles collide more energetically

Increasing surface area

increases the

rate of a reaction

Increasing Concentration

USUALLY increases

the rate of a reaction

(66)

Graphical Representation?

Two types of energy

profile diagrams

(67)

Endothermic Reactions

(68)

Exothermic Reactions

(69)

Activation Energy (Ea)

• Minimum amount of

energy required to

overcome energy

barrier and start a

chemical reaction

• Rate can increase if

Ea is lowered or get

to the top of the

(70)

How would this E Profile

(71)

(72)

Recall Maxwell-Boltzmann

Distribution Graphs:

• What effect will

increasing

temperature or

adding a catalyst

have on this

diagram?

• Draw?

Threshold energy

: minimum

amount of energy to break

bonds (activation E)

• The

KE

distribution

(73)

(74)

Add a Catalyst

• Causes a shift of

the threshold or

activation energy

to the left

• More particles

have the

necessary

activation energy

• Area under

(75)

Increasing Temperature

•

Temperature

is

the

average

KE

• Increasing

temperature

spreads out curve

to the right,

increases average

KE

(76)

Catalysis (Different Types)

•

Catalyst

: A substance that speeds up a

reaction without being consumed by lowering

activation energy (Ea)

•

Enzyme

: A large molecule (usually a

protein) that catalyzes biological reactions.

•

Homogeneous catalyst

: Present in the same

phase as the reacting molecules.

•

Heterogeneous catalyst

: Present in a

different phase than the reacting molecules

(77)

Catalysts Increase the Number of

Effective Collisions

(78)

(79)

Heterogeneous Catalysis

Step #1:

Adsorption and

activation of the

reactants.

Carbon monoxide and

nitrogen monoxide

(80)

Step #2:

Migration of the

adsorbed reactants

on the surface.

Carbon monoxide and

nitrogen monoxide

(81)

Step #3:

Reaction of the

adsorbed

substances.

Carbon dioxide and

nitrogen form from

(82)

Heterogeneous Catalysis

Step #4:

Escape, or

desorption, of

the products

.

Carbon dioxide and

nitrogen gases escape

(83)

(84)

Homogeneous Catalyst?

• Destruction of Ozone, O

₃

by CFCs

• Heterogeneous Catalyst example

•

https://www.youtube.com/watch?v=OfP5h

(85)

AP Sample Question

Relatively slow rates of chemical reaction are

associated with which of the following?

A.The presence of a catalyst

B.High temperature

C.High concentration of reactants

(86)

What effect will increasing the [H

+

] at constant

temperature have on the reaction represented

above?

A.The activation energy will increase.

B.The activation energy will decrease.

C.The frequency of collisions between H

+

(aq)

and

ClO

-

(aq)

ions will increase.

(87)

The energy diagram for the reaction

X + Y → Z is shown below. The addition

of a catalyst to this reaction would cause

a change in which of the indicated energy

differences?

(88)

Reaction Mechanisms

(89)

Reaction Mechs

• Many reactions

proceed by a series

of elementary steps.

• E profile diagram

showing a reaction

proceeding by two

steps

• Each step has its

own Ea.

(90)

Reaction Mechanisms

The overall progress of a chemical reaction can be

represented at the molecular level by a series of simple

elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

(91)

Elementary step: NO + NO N₂O₂ Elementary step: N₂O₂ + O₂ 2NO₂ Overall reaction: 2NO + O₂ 2NO₂ +

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.

An intermediate is always formed in an early elementary step and consumed in a later elementary step,

intermediates cannot appear in the rate law.

(92)

Overall Rxn: 2NO (

g

) + O

₂

(

g

) 2NO

₂

(

g

)

(93)

Elementary step: Cl + O₃ ClO + O₂ Elementary step: ClO + O O₂ + Cl Overall reaction: O₃ + O 2O₂ +

Catalysts are species that speed up a chemical reaction by lowering the activation energy required for a reaction to take place.

A catalyst is always present in the beginning of the

reaction and the end. Catalysts might appear in the rate law if part of the RDS.

(94)

Rate Laws and Rate Determining Steps Rules:

Writing plausible reaction mechanisms:

• The sum of the elementary steps must give the overall balanced equation for the reaction.

(95)

Unimolecular

reaction A products rate = k [A] Bimolecular reaction A + B products rate = k [A][B] Bimolecular reaction A + A products rate = k [A]2

Rate Laws and Elementary Steps

(96)

The experimental rate law for the reaction between H₂ and ICl to produce I₂ and HCl is rate = k[H₂][ICl]. The reaction is believed to occur via two steps:

Step 1: H₂ + ICl HI+ HCl

Step 2: HI + ICl I₂ + HCl What is the equation for the overall reaction?

H₂+ 2 ICl I₂ + 2 HCl

What is the intermediate?

HI

What can you say about the relative rates of steps 1 and 2?

(97)

Identifying the Rate-Determining Step

For the reaction:

2H

₂

(g) + 2NO(g)



N

₂

(g) + 2H

₂

O(g)

The experimental rate law is:

R = k[NO]

2

[H

2

]

Which step in the reaction mechanism is the

rate-determining (slowest) step?

Step #1

H

₂

(g) + 2NO(g)



N

₂

O(g) + H

₂

O(g)

Step #2

N

₂

O(g) + H

₂

(g)



N

₂

(g) + H

₂

O(g)

(98)

Identifying Intermediates

For the reaction:

2H

₂

(g) + 2NO(g)



N

₂

(g) + 2H

₂

O(g)

Which species in the reaction mechanism are

intermediates (do not show up in the final,

balanced equation?)

Step #1

H

₂

(g) + 2NO(g)



N

₂

O(g) + H

₂

O(g)

Step #2

N

₂

O(g) + H

₂

(g)



N

₂

(g) + H

₂

O(g)

2H

₂

(g) + 2NO(g)



N

₂

(g) + 2H

₂

O(g)

(99)

Write the rate law for this reaction. Rate = k [HBr] [O₂] List all intermediates in this reaction.

List all catalysts in this reaction.

HOOBr, HOBr

(100)

Fast Initial Step?

Step #1

O

₂

(g) + NO(g)



NO

₃

(g) Fast

Step #2

NO

₃

(g)+ NO(g)



2NO

₂

(g) Slow

• If first step fast, sometimes, reaction goes reverse and equilibrium is achieved.

• Steady state assumption addresses this problem. • Cannot use intermediates in rate laws.

• Must use substitution to eliminate the intermediate from the rate law.

(101)

The rate law for the reaction represented by the equation

above is rate =

k

[NO

₂

][F

₂

]. Which of the following could be

the first elementary step of a two-step mechanism for the

reaction if the first step is slow and the second step is fast?

A.

B.

C.

(102)

2 NO

(g)

+ O

₂

(g)

→ 2 NO

₂

(g)

Consider the following mechanism for the reaction

represented above.

Step 1: 2 NO N

⇄

₂

O

₂

(

fast reversible

)

Step 2: N

₂

O

₂

+ O

₂

→ 2 NO

₂

(

slow

)

Which of the following statements is true?

A.Step 1 represents a unimolecular reaction.

B.Increasing the concentration of NO will decrease the overall rate of the reaction.

C.Raising the temperature will have no effect on the numerical value of the rate constant.

D.The rate law that is consistent with the mechanism is rate = k[NO]2[O

(103)

The particle models shown above represent a proposed two-step mechanism for the destruction of ozone (O3) in the upper atmosphere.

(104)

Answers:

(105)

Based on the models, which of the following

represents a species that acts as a catalyst for

the reaction?

(106)

Based on the proposed mechanism, what is the

balanced chemical equation for the overall

reaction?

(107)

Step 1: Cl

(g)

+ O

₃

(g)

→ ClO

(g)

+ O

₂

(g)

Step 2: ClO

(g) +

O

(g)

→ Cl

(g) +

O

₂

(g)

A proposed mechanism for destruction of ozone gas in the

stratosphere is represented above. Which of the following is

evidence that the mechanism is occurring?

A.The presence of Cl(

g

) increases the rate of the overall

reaction.

B.The presence of Cl(

g

) decreases the rate of the overall

reaction.

C.The presence of Cl(

g

) increases the equilibrium constant of

the overall reaction.

(108)

Temperature Dependent Rate

Constant, k

• If graph k vs. Temp,

find exponential

growth curve.

(109)

The Arrhenius Equation

/

a

E RT

k Ae







k

= rate constant at temperature T



A

= frequency factor



E

_a_a

= activation energy

(110)

The Arrhenius Equation, Rearranged

1

ln( )

k

E

a

ln( )

A

R T

 

 

 



 

Simplifies solving for

E

_a

-E

_a

/ R

is the slope when

(

1/T

)

is plotted

against

ln(

k

)

ln(

A

)

is the y-intercept

Linear regression analysis of a table of

(1/

(

T)

vs.

ln(

k

)

can quickly yield a slope

(111)

Overall Review Types Qs

2 A

(g)

+ B

(g)

→ 2 C

(g)

When the concentration of substance B in the reaction above

is doubled, all other factors being held constant, it is found that

the rate of the reaction remains unchanged. The most

probable explanation for this observation is that

A.the order of the reaction with respect to substance B is 1

B.substance B is not involved in any of the steps in the

mechanism of the reaction

C.substance B is not involved in the rate–determined step of

the mechanism, but is involved in subsequent steps

(112)

(113)

Important Numbers:

•

What does each mean?

(114)

Important Numbers:

•

119 days until AP Chem Exam

•

36 “A” days until Exam

•

31 days instruction remain (review inc)

•

6 “B” Day labs remain

•

1 “A” Day Lab

•Homogeneous catalyst

•Heterogeneous catalyst