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Chapter 13 Chemical Kinetics. The study of reaction rate is called chemical kinetics. Reaction rate is measured by the change of concentration (molarity) of reactants or products per unit time. Molarity of A: [A], e.g. [NO 2 ]: molarity of NO 2. −. −. reactants  products,.

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slide1

Chapter 13

Chemical Kinetics

slide2

The study of reaction rate is called chemical kinetics.

Reaction rate is measured by the change of

concentration (molarity) of reactants or products

per unit time.

slide3

Molarity of A: [A], e.g. [NO2]: molarity of NO2

reactants  products,

[reactant]↓ and [product]↑

Unit:

mol·L−1·s−1 ≡ M·s−1

slide4

2NO2 (g)  2NO (g) + O2 (g)

= r(NO2)

rate is a function of time

0 s → 50 s:

50 s → 100 s:

slide5

2NO2 (g)  2NO (g) + O2 (g)

= r(NO2)

rate is a function of time

50 s → 100 s:

50 s → 100 s:

slide7

a A + b B  c C + d D

r =

r does not depend upon the choice of species

slide8

2N2O5(g)  4NO2(g) + O2(g)

r(N2O5) = 4.2 x 10−7 M·s−1

What are the rates of appearance of NO2 and O2 ?

slide9

Example 13.1. page 567

  • Consider the following balanced chemical equation:
  • H2O2(aq) + 3 I–(aq) + 2 H+(aq)  I3–(aq) + 2 H2O(l)
  • In the first 10.0 seconds of the reaction, the concentration of I– dropped from 1.000 M to 0.868 M.
  • Calculate the average rate of this reaction in this time interval.
  • (b) Predict the rate of change in the concentration of
  • H+ (that is, [H+]/t) during this time interval.
slide10

Consider the general reaction aA + bB cC and the

following average rate data over some time period Δt:

Determine a set of possible coefficients to balance this

general reaction.

slide11

= r(NO2)

All the rates in this table are average rates.

slide12

2:00 pm

2:16 pm

3:16 pm

Δt

0 mile

Barnesville

16 miles

Griffin

56 miles

Atlanta

Δl

Average Speed from B to G = 16 miles ÷ 16 min = 1.0 mile/min

Average Speed from G to A = 40 miles ÷ 60 min = 0.7 mile/min

Instantaneous speed at green spot

Instantaneous speed contains more information

slide13

l

Atlanta

56 miles

Δl

Griffin

20 miles

Δt

Barnesville

0 mile

t

16 min

0 min

76 min

Instantaneous speed at the red point = slope of the red solid line =

slide15

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

slide16

a A + b B  c C + d D

r = k [A]m [B]n

Differential rate law: how r depends on concentrations

m, n: reaction order, mth order for A, nth order for B

(m+n): overall reaction order

k: rate constant: depends on temperature, but not

concentrations

m and n must be measured from experiments. They can

be different from the stoichiometry.

slide17

(1) 2N2O5(g)  4NO2(g) + O2(g) r = k[N2O5]

(2) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2

(3) H2(g) + I2(g)  2HI(g) r = k[H2][I2]

slide18

aA + bB  cC +dD

r = k [A]m [B]n

Units

overall reaction order (m+n)⇌ unit of k

slide19

(1) 2N2O5(g)  4NO2(g) + O2(g) r = k[N2O5]

(2) CHCl3(g) + Cl2(g)  CCl4(g) + HCl(g) r = k[CHCl3][Cl2]1/2

(3) H2(g) + I2(g)  2HI(g) r = k[H2][I2]

slide20

How to find the rate law by experiment: method of initial rates

NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)

r = k [NH4+]m [NO2−]n

slide21

A very common method to investigate how each factor

affects the whole system:

Change one thing at a time while keep the others constant.

z = f (x,y)

How does the change of x affect z?

How does the change of y affect z?

slide22

How to find the rate law by experiment: method of initial rates

NH4+ (aq) + NO2− (aq)  N2(g) + 2H2O(l)

r = k [NH4+]m [NO2−]n

slide23

2NO(g) + 2H2(g)  N2(g) + 2H2O(g)

  • Determine the differential rate law
  • Calculate the rate constant
  • Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
slide24

2NO(g) + 2H2(g)  N2(g) + 2H2O(g)

again

  • Determine the differential rate law
  • Calculate the rate constant
  • Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M
slide25

A + B  C

  • Determine the differential rate law
  • Calculate the rate constant
  • Calculate the rate when [A] = 0.050 M and [B] = 0.100 M
slide26

EXAMPLE 13.2 Determining the Order and Rate Constant of a Reaction

NO2(g) + CO(g)  NO(g) + CO2(g)

From the data, determine:

(a) the rate law for the reaction

(b) the rate constant (k) for the reaction

slide27

Use the data in table to determine

1) The orders for all three reactants

2) The overall reaction order

3) The value of the rate constant

slide28

r = k [BrO3−]m [Br−]n [H+]p

r1 = k (0.10 M)m (0.10 M)n (0.10 M)p = 8.0 x 10−4 M · s−1

r2 = k (0.20 M)m (0.10 M)n (0.10 M)p = 1.6 x 10−3 M · s−1

r3 = k (0.20 M)m (0.20 M)n (0.10 M)p = 3.2 x 10−3 M · s−1

r4 = k (0.10 M)m (0.10 M)n (0.20 M)p = 3.2 x 10−3 M · s−1

slide29

one quiz after lab

Relationship among reaction rates as expressed

by different species.

r =

overall reaction order (m+n)⇌ unit of k

Method of initial rates:

table of experimental data  rate order, k, rate at other

concentrations.

slide30

a A + b B  c C + d D

r = k [A]m [B]n

Differential rate law: how r depends on concentrations

Differential rate law: differential equation

How concentration changes as a function of time 

integrated rate law

slide31

A  Products

First order reaction differential rate law:

First order reaction integrated rate law:

or

integrated rate law: how concentration changes as a

function of time.

slide32

First order reaction integrated rate law:

[A] is the molarity of A at t

y = mx + b

Plot ln[A] vs. t gives a straight line

Slope = −k, intercept = ln[A]0

slide33

N2O5(g)  2NO2(g) + ½ O2(g)

Use these data, verify that the rate law is first order in N2O5,

and calculate the rate constant.

k = 6.93 x 10−3 s−1

slide35

Using the data given in the previous example, calculate [N2O5]

at 150 s after the start of the reaction.

0.0354 M

slide37

The half-life of a reaction, t1/2, is the time required for a

reactant to reach one-half of its initial concentration.

slide38

The half-life for first order reaction:

The half-life for first order reaction does NOT depend on

concentration.

slide41

A certain first order reaction has a half-life of 20.0 s.

  • Calculate the rate constant for this reaction.
  • How much time is required for this reaction to be 75 %
  • complete?

a) k = 0.0347 s−1

b) k = 40.0 s

slide42

Try Example 13.6 and For Practice 13.6

on page 579 and check your answers

slide43

A  Products

Second order reaction differential rate law:

Second order reaction integrated rate law:

integrated rate law: how concentration changes as a

function of time.

slide44

Second order reaction integrated rate law:

y = mx + b

Plot 1/[A] vs. t gives a straight line

Slope = k, intercept = 1/[A]0

slide45

The half-life for second order reaction:

The half-life for second order reaction depends on initial

concentration.

slide46

A certain reaction has the following general form: A  B

  • At a particular temperature [A]0 = 2.80 x 10−3 M,
  • concentration versus time data were collected for this reaction,
  • and a plot of 1/[A] versus time resulted in a straight line with a
  • slope value of 3.60 x 10−2 M−1·s−1
  • Determine the (differential) rate law, the integrated rate law,
  • and the value of the rate constant.
  • b) Calculate the half-life for this reaction.
  • c) How much time is required for the concentration of A to
  • decrease to 7.00 x 10−4 M ?
slide47

(show your work, do not copy the question)

For first order reaction, show that

from the integrated rate law

slide48

A  Products

Zero order reaction differential rate law:

Zero order reaction integrated rate law:

integrated rate law: how concentration changes as a

function of time.

slide49

Zero order reaction integrated rate law:

y = mx + b

Plot [A] vs. t gives a straight line

Slope = −k, intercept = [A]0

slide50

The half-life for zero order reaction:

The half-life for zero order reaction depends on initial

concentration.

slide52

The decomposition of ethanol on alumina surface

  • C2H5OH(g)  C2H4(g) + H2O(g) was studied at 600 K.
  • Concentration versus time data were collected for this reaction, and a plot of [C2H5OH] versus time resulted in a straight line with a slope value of −4.00 x 10−5 M·s−1
  • Determine the (differential) rate law, the integrated rate law,
  • and the value of the rate constant.
  • b) If the initial concentration of C2H5OH was 1.25 x 10−2 M,
  • calculate the half-life of this reaction.
  • c) How much time is required for all the 1.25 x 10−2 M C2H5OH
  • to decompose ?
slide54

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

slide55

aA + bB  cC +dD

r = k [A]m [B]n

k: rate constant: depends on temperature, but not

concentrations

a plot showing the exponential dependence of the rate constant on absolute temperature
A Plot Showing the Exponential Dependence of the Rate Constant on Absolute Temperature

Collision model: molecules must collide to react.

Not all collisions lead to products

T ↑ v ↑ kinetic energy = ½ mv2↑

slide57

activation energy

Fraction of molecules whose Ek > Ea is

T ↑ f↑

Ea↑ f↓

Chapter 14

several possible orientations for a collision between two brno molecules brno brno 2no br 2
Several Possible Orientations for a Collision Between Two BrNO Molecules BrNO + BrNO  2NO +Br2
slide60

steric factor p ≤ 1

Arrhenius equation

A: frequency factor

How k depends on T

T ↑ k↑

Ea↑ k↓

slide61

At 550 °C the rate constant for

CH4(g) + 2S2(g)  CS2(g) + 2H2S(g)

is 1.1 M·s−1, and at 625 °C the rate constant is

6.4 M·s−1. Using these value, calculate Ea for this reaction.

Ea = 1.4 x 105 J/mol

slide63

h

The ball can climb over the hill only if its kinetic energy is greater

than Ep = mgh, where m is the mass of the ball, h is the height

of the hill, and g is gravitational acceleration.

slide64
(a) The Change in Potential as a Function of Reaction Progress (b) A Molecular Representation of the Reaction

BrNO + BrNO  2NO +Br2

O N Br

exothermic reaction

slide65

Reaction Mechanism

Chemical Equation

Reactants  Products

slide66

NO2 + CO  NO + CO2

Step 1: NO2 + NO2 NO3 + NO

Step 2: NO3 + CO  NO2 + CO2

NO2 + CO  NO + CO2

+

NO3: intermediate, does not

appear in overall reaction

slide67

NO2 NO2 NO3 NO

NO3 CO NO2 CO2

Whole process is called the reaction mechanism.

Each single step is called an elementary reaction/step.

An elementary reaction is a single collision.

slide68

For an elementary reaction

aA + bB  cC + dD

r = k[A]a[B]b

Not for overall reaction!

Overall reaction: r = k[A]m[B]n , find m and n by experiments

The number of molecules that react in an elementary reaction is called the molecularity of that that elementary reaction (not applicable to overall reaction).

slide69

1 ― unimolecular

2 ― bimolecular

3 ― termolecular

slide71

Team A

Team B

Slowest step: rate determining step

slide72

Which step is the rate determining step?

2nd

exothermic or

endothermic?

Ep

intermediate

Ea, 2

Ea, 1

Reactants

∆E

Products

Reaction progress

slide73

Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst

slide74

Catalyst is a substance that speeds up a reaction without

being consumed itself.

Catalyst changes the reaction mechanism through a lower

activation energy pathway.

slide76

homogeneous

catalyst

heterogeneous

slide77

Homogeneous catalyst: same phase as reactants.

3O2(g)  2O3(g)

2NO(g) + O2(g)  2NO2(g)

light

2NO2(g)  2NO(g) + 2O(g)

2O2(g) + 2O(g)  2O3(g)

+

3O2(g)  2O3(g)

NO(g): catalyst;

NO2(g), O(g): intermediates

the decomposition reaction 2n 2 o g 2n 2 g o 2 g takes place on a platinum surface1
The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g) takes place on a Platinum Surface

Heterogeneous catalyst: different phase from reactants.

these cookies contain partially hydrogenated vegetable oil
These Cookies Contain Partially Hydrogenated Vegetable Oil

C = C

+ H2

− C − C −

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