Chapter 13
This presentation is the property of its rightful owner.
Sponsored Links
1 / 81

Chapter 13 Chemical Kinetics PowerPoint PPT Presentation


  • 151 Views
  • Uploaded on
  • Presentation posted in: General

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,.

Download Presentation

Chapter 13 Chemical Kinetics

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


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. [NO2]: molarity of NO2

reactants  products,

[reactant]↓ and [product]↑

Unit:

mol·L−1·s−1 ≡ M·s−1


2NO2 (g)  2NO (g) + O2 (g)

= r(NO2)

rate is a function of time

0 s → 50 s:

50 s → 100 s:


2NO2 (g)  2NO (g) + O2 (g)

= r(NO2)

rate is a function of time

50 s → 100 s:

50 s → 100 s:


2NO2(g)  2NO(g) + O2(g)


a A + b B  c C + d D

r =

r does not depend upon the choice of species


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 ?


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.


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.


= r(NO2)

All the rates in this table are average rates.


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


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 =


Reaction rate is a function of time


Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst


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.


(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]


aA + bB  cC +dD

r = k [A]m [B]n

Units

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


(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]


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


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?


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


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


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


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


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


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


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


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.


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


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.


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


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


Read a similar Example 13.3 on page 575


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

at 150 s after the start of the reaction.

0.0354 M


Practice on Example 13.4 on page 576 and

check your answer


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

reactant to reach one-half of its initial concentration.


The half-life for first order reaction:

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

concentration.


A Plot of [N2O5] versus Time for the Decomposition Reaction of N2O5


  • 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


Try Example 13.6 and For Practice 13.6

on page 579 and check your answers


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.


Second order reaction integrated rate law:

y = mx + b

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

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


The half-life for second order reaction:

The half-life for second order reaction depends on initial

concentration.


  • 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 ?


(show your work, do not copy the question)

For first order reaction, show that

from the integrated rate law


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.


Zero order reaction integrated rate law:

y = mx + b

Plot [A] vs. t gives a straight line

Slope = −k, intercept = [A]0


The half-life for zero order reaction:

The half-life for zero order reaction depends on initial

concentration.


The Decomposition Reaction 2N2O(g)  2N2 (g) + O2 (g) takes place on a Platinum Surface


  • 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 ?


Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst


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

Collision model: molecules must collide to react.

Not all collisions lead to products

T ↑ v ↑ kinetic energy = ½ mv2↑


activation energy

Fraction of molecules whose Ek > Ea is

T ↑ f↑

Ea↑ f↓

Chapter 14


Another factor needs to be taken into account

molecular orientation


Several Possible Orientations for a Collision Between Two BrNO MoleculesBrNO + BrNO  2NO +Br2


steric factor p ≤ 1

Arrhenius equation

A: frequency factor

How k depends on T

T ↑ k↑

Ea↑ k↓


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


Try Example 13.8 on page 585 and

check your answers


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.


(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


Reaction Mechanism

Chemical Equation

Reactants  Products


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


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.


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).


1 ― unimolecular

2 ― bimolecular

3 ― termolecular


What determines the rate of an overall reaction?


Team A

Team B

Slowest step: rate determining step


Which step is the rate determining step?

2nd

exothermic or

endothermic?

Ep

intermediate

Ea, 2

Ea, 1

Reactants

∆E

Products

Reaction progress


Factors that affect reaction rates

State of the reactants

Concentrations of the reactants

Temperature

Catalyst


Catalyst is a substance that speeds up a reaction without

being consumed itself.

Catalyst changes the reaction mechanism through a lower

activation energy pathway.


Energy Plots for a Given Reaction

Ea

Ea


homogeneous

catalyst

heterogeneous


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 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

C = C

+ H2

− C − C −


milk sugar = lactose


  • Login