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Chapter 12 Chemical Kinetics. The study of reactions rates Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed?. Chemical Kinetics. The study of reaction rates

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chapter 12 chemical kinetics
Chapter 12Chemical Kinetics

The study of reactions rates

Thermodynamics – does a reaction take place?

Kinetics – how fast does a reaction proceed?

chemical kinetics
Chemical Kinetics

The study of reaction rates

Spontaneous reactions---is the tendency for the reaction to occur but it does not imply how fast.

Ex. Diamond will spontaneously turn into graphite --- eventually

reaction mechanism
Reaction Mechanism

A goal of chemical kinetics is to understand the steps by which a reaction takes place.

The series of steps is called the reaction mechanism.

If we understand the mechanism, we can find ways to facilitate the reaction

let s review collision theory
Let’s review collision theory

Particles have to collide to react

They have to hit hard enough

Things that increase this increase rate

High temps – faster reaction

High concentration – faster reaction

Small particles = greater surface area – faster reaction

reaction rate
Reaction Rate

The brackets mean concentration (Molarity)

  • Change in concentration of a reactant or product per unit time
slide6

A B

rate =

D[A]

D[B]

rate = -

Dt

Dt

slide8

Concentration

N2 + 3H2→ 2NH3

  • As the reaction progresses the concentration H2 goes down

[H2]

Time

slide9

Concentration

N2 + 3H2→ 2NH3

  • As the reaction progresses the concentration N2 goes down 1/3 as fast

[N2]

[H2]

Time

slide10

Concentration

N2 + 3H2→ 2NH3

  • As the reaction progresses the concentration NH3 goes up 2/3 times

[N2]

[H2]

[NH3]

Time

calculating rates
Calculating Rates
  • Average rates are taken over long intervals
  • Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time
  • Derivative.
slide12

Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g)

slope of

tangent

slope of

tangent

slope of

tangent

[Br2]final – [Br2]initial

D[Br2]

average rate = -

= -

Dt

tfinal - tinitial

instantaneous rate = rate for specific instance in time

defining rate
Defining Rate
  • We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.
  • In our example
  • N2 + 3H2 2NH3
  • D[H2] = 3D[N2] DtDt
  • D[NH3] = -2D[N2] DtDt
rate of no 2
Rate of NO2
  • Since the concentration of the reactant will typically be decreasing over time, the rate expression will include negative; however, the slope of the line will be negative which will then yield a positive rate.
compare the three rates 2no 2 g 2no g o 2 g
Compare the three rates 2NO2 (g)  2NO(g) + O2 (g)

Rate of NO2 = 2.4 x 10-5 mol/L•s

Rate of NO = 8.6 x 10-6 mol/L•s

Rate of O2 = 4.3 x 10-6 mol/L•s

Notice the rate of NO is twice that of O2. This makes sense since twice as much NO is produced according to the reaction.

12 2 rate laws an introduction
12.2 Rate Laws: An Introduction

We looked at the reaction

2NO2 (g) 2NO(g) + O2 (g)

But we need to understand that the reverse happens too.

2NO(g) + O2 (g)  2NO2 (g)

Chemical reactions ARE reversible

chemical equilibrium
Chemical equilibrium

When forward and reverse reaction rates are equal, it is called chemical equilibrium. When this occurs, there will be NO changes in the concentrations of reactants and products.

2no 2 g 2no g o 2 g
2NO2 (g) 2NO(g) + O2 (g)

in the reaction depends on the difference in the rates of the forward and reverse reactions.

If you choose conditions where the reverse reaction can be neglected. Then the reaction rate will depend only on the concentrations of the reactants. The rate for the decomprxn is

slide22

This shows that the rate depends on the concentration of the reactants…this is the rate law expression.

k = the rate constant

n = order of the reactant

both must be determined by experiment

(although n can be an integer or a fraction, we will typically look at positive integer orders in this book.)

slide23

Two important things:

    • The concentration of the products is not considered here because the reverse reaction does not contribute to the rate
    • The value of the exponent n must be determined from experimentation
so what do we choose
So what do we choose?

The question is which reactant or product do we choose for defining the “rate” in the rate equation:

The value of the rate constant, k, depends on how the rate is defined.

two types of rate laws
Two types of rate laws

One type:

Differential rate laws – once you know the value of n – the rate law expresses how the rate depends on concentration. Often just called “the rate law” – rate as a function of concentration.

the other type
The other type:

Integrated rate law – expresses how concentration depends on time

Once you find one experimentally, you know the other. The one you find experimentally is based on which is easier to collect data.

slide27
Why?
  • What is the purpose of knowing the rate?
    • Infer the steps by which a reaction occurs
    • Understand the reaction by learning what the steps are in the reaction
      • Where can I interrupt the reaction?
      • How can I speed up the reaction?
  • Rate laws are not used for the purpose of knowing the rate, but what the rate reveals about the STEPS of the reaction. (Reaction mechanism)
12 3 determining the form of the rate law
12.3 Determining the Form of the Rate Law

aA + bBcC + dD

Rate = k [A]x[B]y

Reaction is xth order in A

Reaction is yth order in B

Reaction is (x +y)th order overall

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 power.

lets look at the decomp of n 2 o 5 first differential rate
Lets look at the decomp of N2O5 first. (differential rate)

2N2O5 4NO2 + O2

Rate = k [N2O5]n

Here’s some data

slide30

Notice that the concentration of N2O5 is halved, the rate is also halved. This means that the rate is completely dependent on concentration of N2O5 to the FIRST power.

Notice: It is NOT based on the coefficient from the reaction.

initial rates
Initial rates.

The initial rate is the instantaneous rate found immediately after the reaction has begun.

The experiment must be run several times with various initial concentrations

The initial rate is found for EACH “run”

Then compare rates to determine the form of the rate law.

the reaction nh 4 no 2 n 2 2h 2 o
The reaction:NH4+ + NO2- N2 + 2H2O

NO2-

The general form of the rate law

We need experimental data to determine the values of n and m

slide33

Simplify

NO2-

Compare: [NO2-] doubles as the rate doubles there fore m = 1

Here’s the math as a ratio to find m

This means that “m” must be 1

Do this again for “n” with rates 2 & 3

slide34

Simplify

NO2-

Again, when the [NH4+] doubles so does the rate n = 1

Which mean “n” must be 1 and the rate law is

slide35

The rate law is first order for both reactants with an overall reaction order of 2

The reaction order is the sum of the orders for the various reactants

With all this info you can now calculate the rate constant for this reaction.

slide36

NO2-

Pick ANY one of the experiments and substitute and solve for k.

We’ll do the same one as the book. But you could choose either!

use the reaction and data below to find
Use the reaction and data below to find:

The orders of each reactant, the overall reaction order, and the value of the rate constant.

12 4 integrated rate law
12.4 Integrated rate law

So far we’ve looked at rate as a function of reactant concentrations.

Integrated rate laws relate concentration to reaction time.

Let’s look at reactions with a single reactant.

aa products
aA products

What is the basic rate law?

First Order - double the concentration… double the rate

rate law integrated if n=1

slide40

Shows how concentration of A depends on time. If [A]o &k are known [A] at any time can be calculated.

If plotting ln[A] vs. time gives a straight line where k is the slope, then it proves the reaction is first order.

Often shown as a ratio.

half life
Half Life

Half life is the time required for a reactant to reach half its original concentration.

For a first order reaction

second order
Second order

Starting with the basic rate law

Double the concentration…quadruple the rat OR triple the concentration …nine times the rate.

Integrated law when n=2

slide43

If you plot versus t and get a straight line with a slope of k…the reaction is proven to be second order.

half life for second order
Half life for second order

The equation for 2nd order half life

For second order reactions, t1/2 is dependent on [A]0. For first order reactions t1/2 is independent on [A]0.

For a 2nd order reaction each successive ½ life is double the preceding one.(ex. The 1st ½ life of C4H6 reacting to form a dimer is 1630 sec, but the 2nd ½ life takes 3570 seconds.)

zero order rate laws
Zero order rate laws

The reaction has a constant rate.

When n=0 Rate=k

Integrated rate law when n=0

zero order half life
Zero order half life

Most zero order reactions are when a substance such as a metal or an enzyme is required for the reaction to occur.

integrated rate laws for reactions with more than one reactant
Integrated rate laws for reactions with more than one reactant

Most often the kinetics of a complicated reaction can be studied by opbserving the behavior of one reactant at a time.

If one reactant has a concentration that is much smaller than the others, then it is used to determine the order of the reaction in that component.

summary1
Summary
  • We only consider only the forward reaction so that rate laws contain only reactant concentrations.
  • There are two types of rate laws.
    • Differential – rate depends on concentrations.
    • Integrated – rate is a function of concentration with respect to time.
summary2
Summary

The type of rate law depends on the type of data that is most conveniently collected. Once we have one type of rate law we can determine the other.

The most common method for differential rate law is the method of initial rates.

For integrated rate laws concentrations are measured at various values of t.

12 6 reaction mechanisms
12.6 Reaction Mechanisms

Most chemical reactions occur by a series of steps called reaction mechanisms.

The whole purpose of kinetics is to learn about the steps.

2no g o 2 g 2no 2 g

Elementary step:

NO + NO N2O2

+

Elementary step:

N2O2 + O2 2NO2

Overall reaction:

2NO + O2 2NO2

2NO (g) + O2 (g) 2NO2 (g)

N2O2 is detected during the reaction!

Intermediate – something formed in one step and used up in a subsequent step. It is never seen as a product.

elementary steps molecularity
Elementary steps & Molecularity
  • Molecularity of a reaction is the number of molecules reacting in an elementary step.
    • Unimolecular – elementary step with 1 molecule – overall rate is 1st order
    • Bimolecular – elementary step with 2 molecules – overall rate is 2nd order
    • Termolecular – elementary step with 3 molecules - – overall rate is 3rd order
slide54

Termolecular are rare. The probability of 3 molecules colliding at the same time is extremely small.

reaction mechanisms are
Reaction mechanisms are…
  • A series of steps that must satisfy two requirements:
    • Sum of the elementary rate steps must give the overall balanced equation for the reaction.
    • The mechanism must agree with the experimentally determined rate law.
doest this reaction follow the rules
Doest this reaction follow the rules?

First requirement is met. Overall reaction is obtained from adding the elementary steps.

to test for the second requirement
To test for the second requirement…

The rate determining step must be found.

Multistep reactions always have one step that is much slower.

The rate determining step is the slowest step in the sequence of the steps leading to product formation.

The RDS should predict the same rate law that is determined experimentally.

assuming
Assuming…

if you assume the first step is the rate determining step the rate law would be:

This is the experimentally determined rate therefore, we can say this mechanism satisfies both requirements.

given
Given:

Slow

Fast

The experimentally determined rate law is

Is the mechanism below acceptable?

collision model
Collision Model

Based on the idea that molecules must collide to react

Concentration and temperature affect collisions.

Collisions must be hard enough to cause a reaction.

only a small fraction of collisions produce a reaction
Only a small fraction of collisions produce a reaction
  • There are two reasons for this.
    • Activation energy
    • Molecular orientation
activation energy
Activation Energy

The minimum energy required to initiate a chemical reaction.

Rate is NOT determined by E but rather by the size of Ea

this means
This means…

The higher the activation energy, the slower the reaction at a given temperature.

molecular orientations
Molecular orientations

The observed reaction rate is considerably smaller than the rate of collisions with enough energy to overcome the barrier.

This is because even though the collision has enough energy it still does not produce a reaction.

the correction factor for the rate constant
The correction factor for the rate constant

Arrhenius said that the number of collisions with the necessary energy increases exponentially with temperature.

z=collision frequency

p= steric factor (the fraction of collisions with efffective orientations)

= the fraction of collisions with sufficient energy to produce a reaction.

the arrhenius equation
The Arrhenius equation

A, the frequency factor, replaces zp

And taking the natural log of each side gives:

Ln(k) versus 1/T yields a straight line.

The slope is Ea/R and y-intercept is ln(A)

slide68

The fact that most rate constants obey the Arrhenius equation indicates that the collision model is physically reasonable.

using the arrhenius equation
Using the Arrhenius equation.

Given the reaction and data below, calculate the value of Ea for the reaction.

to solve
To solve…

Plot the ln(k) versus 1/T & find slope

another way
Another way…

If you find the rate constant at only two temperatures and subtract the first from the second equation, you’d get the formula:

OR

12 8 catalysts
12.8 Catalysts

Sometimes raising the temperature to speed up a reaction is not feasible.

Catalysts are often used. Catalysts speed up a reaction without being used up by the reaction.

Catalysts lower the reaction energy, but do not affect the E.

two types
Two types
  • There are two types of catalysts
    • Homogeneous – present in the same phase as the reactants.
    • Heterogeneous – exist in a different phase (typically solid)
  • Catalysts show up as a reactant in one step and a product in a later step.
how they work heterogeneous
How they work: heterogeneous
  • Adsorption and activation of reactants.
  • Adsorption is the collection of one substance on the surface of another.
  • Migration of the adsorbed reactants on the surface (H-H bonds broken,
  • π bond broken, metal-carbon
  • bond formed.)

Reaction of the adsorbed substances. New C-H

bonds are formed.

Escape or desorption of the products. New

molecule escapes.

catalysts and rate
Catalysts and rate

Catalysts will speed up a reaction, but only to a certain point.

Past a certain point, adding more reactants won’t change the rate. Rate increases until the active sites of catalyst are filled.

Then rate is independent of concentration.