CHEMICAL KINETICS

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CHEMICAL KINETICS

Four factors affect reaction rates

Concentration of reactants

Temperature of reaction

Presence or absence of a catalyst

Surface area of liquid or solid reactants

Reaction Rates

Reaction speed defined as the change that occurs per unit

time

o

It is determined by measuring the disappearance of a reactant or the appearance of a product

o

The speed of a reaction is the reaction rate

For reaction A

B : Avg rate = change in the number of moles of B / change in time

o

B= moles of B at final time- moles of B initial

o o o o

At t= 0, there is 1 mole of A and no B

At t= 10 min, there is .74 mol A and .26 mol B

At t=20 min, there is.54 mol A and .46 mol B

Avg rate=

B/

T= Mol B at 10min- Mol B at 0 min/

o

10min –0 min= .026mol/min

Or Avg rate= -

Mol A /

t ; note the negative sign. It means reactants are disappearing

Rates in Terms of Concentration

The most useful measure is molarity

Since volume is constant, molarity and moles are directly proportional

Units for average rate are mol/L sec or M/sec

The average rate of a reaction decreases over time. Why?

The rate at any specific point in time is the instantaneous rate

If we plot the concentration of a reactant vs time, the rate at any instant is the slope of the straight tangent to the curve at that point

Stoichiometry

For the reaction :

C

4

H

9

Cl + H

2

O

C

4

H

9

OH + HCl

Rate= -

[C

4

H

9

]/

t =

[C

4

H

9

OH] /

t

What if stoichiometry is not one to one?

2HI

H

2

+ I

2

Rate = -

[HI]/ 2

t =

[H

2

]/

t =

[I

2

]/

t

Generalized : For the reaction aA + bB

cC + d D

Rate = -

[A]/a

t = -

[B]/ b

t=

[C]/c

t =

[D]/d

t

Dependence of rate on concentration

In general, rate increases with increased concentration of

Reactants and decreases with decreasing concentration

We often examine the effect of concentration by measuring the reaction rate at the beginning of a reaction

Consider A + B

C + D

We find that if we hold [B] constant and double [A], the rate doubles. Therefore , the rate is directly proportional to [A].

If we hold [A] constant and double [B], then the rate doubles. Therefore, the rate is directly proportional to [B]

The overall concentration dependence is given by the rate law. For this example the rate law is:

Rate = k [A] [B] where k is the rate constant

Reaction Order

For any reaction with rate law:

 o o

Rate= k[A] m

[B] n

, the exponents are called reaction orders. The overall reaction order is the sum of the exponents.

This reaction is first order in[A] and first order in [B]

And second order overall

Reaction orders must be determined experimentally

o

They do not necessarily correspond to stoichiometric

o

coefficients

They are typically 0,1 or 2

o

However , they can be fractional or negative

Units of rate constants

Units of the rate constant depend on the overall reaction order

For a second order reaction:

Units of rate = (Units of rate constant) (units of concentration) (units of concentration)

Therefore : Units of rate constant= Units of rate/ (Units of concentration)

2

= M/sec/M

2

= M

-1

S

-1

Using Initial Rates to determine rate laws

To determine the rate law, we observe the effect of changing the initial concentrations

o

If the reaction is zero order, changing the

o

concentration has no effect on rate

If the reaction is first order, doubling the concentration will double the rate( change is directly

o o

proportional) What if concentration is halved?

If the reaction is second order, doubling the concentration will result in a 4X increase (2

2

).

Tripling the concentration will cause a 9X increase

A reaction is nth order if doubling the concentration causes a 2 n

increase in rate

The reaction rate, NOT THE RATE CONSTANT,

o

depends on concentration.

First order Reaction

ln[A] t

= -kt + ln[A]

0 is the equation for a first order reaction. Note the similarity to y = mx + b. In this case , a plot of ln[A] t

vs t yields a straight line with a slope –k and an intercept of ln[A]

0

Half-life is the time required for the initial concentration of a reactant to decrease to one half its original value

T

1/2

= .693/k Note that the half life of a first order reaction is independent of the initial concentration of the reactant

Second order reactions

1/ [A] t

= kt + 1/[A]

0

A plot of 1/[A] t

versus t is a straight line with a slope k and an intercept of 1/[A]

0

For a second order reaction a plot of ln[A] line t is not a straight

T

1/2

= 1/k[A]

0

Half life is dependent on initial concentration

Note that a second order process can have the form : Rate= k[A] [B] . The reaction is second order overall but first order in reactant concentration

Temperature and Rate

As temperature increases, the rate increases

Since the rate law does not have temperature in it, the rate constant must be temperature dependent

The Collision Model

Rates of reaction are effected by concentration and temperature

The collision model, which is based on kinetic molecular theory, explains this

o

In order for molecules to react they must collide

o

The greater the number of collisions the faster the

o

rate

If you have more molecules the probability of collision

o

increases and the frequency of collision

Higher temperature ( higher KE ) gives the molecules higher energy and increases the number of

o

collisions(moving faster)

However , not all collisions lead to reaction. Only a small fraction of collisions lead to reaction

Activation Energy

Arrhenius: molecules must possess a minimum amount of energy to react. Why?

o

To form products , bonds in the reactants must be

o o

broken

Breaking bonds requires energy

Molecules moving too slowly, KE too low, do not react

Activation energy, E a is the minimum energy required to start a reaction

o

E a varies with the reaction

Consider the rearrangement of methyl nitrile to acetonitrile

o

Energy is required to stretch the bond between the methyl group and the nitrile group and to allow the nitrile group to rotate

o o

The carbon carbon bond forms

The energy associated with the molecule drops

o

The energy level between the starting molecule and the highest energy state found along the reaction pathway is the activation energy.

The chemical species at the top of the barrier is the activated complex or the transition state

The energy change for the reaction is the difference in energy between the products and the reactants

o

E rxn

has no effect on reaction rate

The activation energy is the difference in energy between the reactant and the activated complex.

o

Rate depends on activation energy. The lower E a

, the faster the reaction

How does this relate to temperature

o

At any particular temperature, molecules present have an average kinetic energy associated with the population

o

In the same population , some molecules have more energy than average , some have less.

The fraction of molecules with energy equal to or greater than E a is given by f = e

-E a

/RT

As we increase temperature, f becomes larger

Orientation Factor

Orientation has a major effect on whether or not a reaction will occur

Consider the reaction between NOCl and Cl

Cl + NOCl

NO + Cl

2

If the Cl collides with the Cl the reaction proceeds

If the Cl collides with the O, no reaction occurs

Arrhenius Equation

Arrhenius discovered that most reaction rate data obeys an equation based on 3 factors

o

Number of collisions per unit time

o

Fraction of collisions with the correct orientation

o

Fraction of molecules that have energy greater than or equal to E a

lnk = -E a

/RT + ln A where k is the rate constant, E a

is the activation energy, R is the gas constant(8.314J/Kmol) T is the temperature in K

A is the frequency factor. It is related to the frequency of collision and the probability that the orientation will be correct

Both A and E a

are specific to a reaction

Rearranging the equation we get lnk = -E a

/RT + lnA. Once again , if you graph the lnk vs 1/T you get a straight line with a slope –E a

/R

Reaction Mechanisms

Balanced chemical equations provide information about substances present at the beginning and end of a reaction

The reaction mechanism is the process by which the reaction occurs

Elementary Steps

Any process that occurs in a single step

The number of molecules present in an elementary step gives the molecularity

o

Unimolecular-one molecule

o o

Bimolecular- two molecules

Termolecular- three molecules

A multistep mechanism consists of a series of elementary steps

o o

Elementary step must give a balanced reaction

Some multistep reactions include intermediates

Species that appear in an elementary step but

not as a product or reactant

Intermediates are formed in one elementary

step and consumed in another

They are not found in the balanced equation for the reaction

Rate Laws for Multistep Reactions

Rate laws of the elementary steps determine the overall rate law

Rate law of the elementary step is determined by molecularity

o

Uni- processes are first order

o o

Bi- processes are second order

Ter- processes are third order

Most reactions have more than one elementary step

o

Often one step is much slower

o

This step limits the reaction rate and governs the rate law

Consider: NO

2

+ CO

NO + CO

2

The experimentally derived rate law is : rate = k[NO

2

]

2

A proposed mechanism is

o

Step 1 : NO

2

+ NO

2

NO

3

+ NO slow

o

Step 2 : NO

3

+ CO

NO

2

+ CO

2

fast

Note NO

3

is an intermediate

The rate law for the slow step is rate = k[NO

2

]

2

This theoretical rate law is in agreement with the experimental

This supports(does not prove) a proposed mechanism

Mechanisms with an initial fast step

Consider the reaction 2NO + Br

2

2NOBr

The experimentally determined rate is : Rate = k[NO]

2

[Br

2

]

Proposed Mechanism

o o

Step 1 : NO + Br

2

NOBr

2

fast

Step 2 : NOBr

2

+ NO

2NOBr slow

o

The theoretical rate law for the slow step is : rate = k

2

[NOBr

2

] [NO]

Problem: this rate law depends on the concentration of an intermediate

o

Intermediates are usually unstable and have

o

low/unknown concentrations

We can express [NOBr

2

] in terms of NOBr and Br

2

o

by assuming an equilibrium in Step 1

In a dynamic equilibrium the e forward rate equals the reverse rate

Therefore we get k

1

[NO] [Br

2

] = k

-1

[NOBr

2

]

Rearranging : [NOBr

2

] = k

1

/k

-1

[NO] [Br

Substituting in the rate law : rate = k

2

k

1

2

]

/k

-

1

[NO] [Br

2

] [NO] = k[NO]

2

[Br

2

]

This is consistent with the experimental rate law

Catalysis

A catalyst is a substance that changes the rate of a reaction without undergoing a permanent change

There are 2 types of catalysts : homogeneous and heterogeneous

Homogeneous Catalysts

A catalyst that is present in the same phase as the reactants

How do catalysts increase rates?

o o o

By lowering the activation energy

Increasing the number of collisions

Catalysts provide a completely different mechanism for reaction

Heterogeneous Catalysts

A catalyst that exists in a different phase from the reactants.

Ie. Solid catalyst and gas reactants: the catalytic convertor on a car

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