Collision Rate ModelThree conditions must be met at the nano-scale level if a reaction is to occur:

• the molecules must collide;

• they must be positioned so that the reacting groups are together in a transition state between reactants and products;

• and the collision must have enough energy to form the transition state and convert it into products.

Transition State: Activated Complex or Reaction Intermediates

• an unstable arrangement of atoms that has the highest energy reached during the rearrangement of the reactant atoms to give products of a reaction

Activation Energy

the minimum energy required to start a reaction

In fact, usually only a VERY SMALL fraction of the total number of

collisions are successful……..resulting in product formation.

The 3 conditions must be fulfilled for a successful collision resulting in product formation.

In reality we are mainly limited by the overall kinetic energy of the molecules (1st condition).

……..molecules are often moving too slowly which prevents a close enough approach for rxn. to occur….

Most collisions DON’T result in a

reaction……………..so even

though molecules are constantly

in motion and colliding…….only

a few of these collisions are

actually successful.

Why ?????

…..because there are THREE

CONDITIONS that must be met

for a reaction to occur. 

Only some collisions are Only some collisions are SUCCESSFUL………..SUCCESSFUL………..

Unsuccessful Collision

THREETHREE conditions conditions must be met for a reaction to occur : must be met for a reaction to occur :

1st Condition: Two molecules must have sufficient kinetic energy(meaning they must be moving fast enough)

2nd Condition: The sum of the molecular momentum vectors must be near zero(meaning molecules have to hit each other directly)

3rd Condition: There must be overlap of the involved electronic orbitals of the molecules (meaning the molecules must be properly oriented)

Product

Successful Collision

What affect does What affect does

TEMPERATURETEMPERATURE

have on the rate have on the rate

of a reaction?of a reaction?

Temperature & Reaction Rate

What is a Catalyst?

Catalyst• substance which speeds up the rate of a

reaction while not being consumed

What are the two types of catalysts?

Catalyst• substance which speeds up the rate of a

reaction while not being consumed

Homogeneous Catalysis - a catalyst which is in the same phase as the reactants

Heterogeneous Catalysis - a catalyst which is in the different phase as the reactants

catalytic converter– solid catalyst working on gaseous materials

Enzymes

any one of many specialized organic substances, composed of polymers of amino acids, that act as catalysts to regulate the speed of the many chemical reactions involved in the metabolism of living organisms.

ln k = ln A – (Ea / R T)

y = B + m x The parameter Ea is obtained from the slope which is equal to (–Ea/R) Ea is called the activation energy.

A and Ea are together referred to as the Arrhenius parameters.

• Ea is given by slope of ln k versus 1/T ……..thus the higher the activation energy the stronger the temperature dependence of of the rate constant (meaning the steeper the slope)

• If a reaction has a zero activation energy, its rate is independent of temperature.

• In some cases the temperature dependence of reactions is “non-Arrhenius like” meaning a straight line is not obtained when ln k is plotted against 1/T

Potential Energy Profile for an Exothermic Reaction

How does the potential energy changein the course of a reaction ?

…..consider a reaction where reactantsA and B react to form products

……A and B collide, come into contactand distort, begin to exchange ordiscard atoms………………..

…….potential energy rises to maximumand cluster of atoms at this point is termed activated complex…………..

……potential energy then falls and reaches characteristic value for products

Activated Complex

Arrhenius Parameters

ln k = ln A – (Ea / R T)

or k = A e – Ea/RT

Pre-exponential Factor

……the pre-exponential factoris a measure of the rate at whichcollisions occur irrespective of their energy.

Activation Energy

Fraction of collisionswith a kinetic energyin excess of an Ea is given by the Boltzmanndistribution as e – Ea/RT

….this is fraction of collisions with enough kinetic energy to lead to reaction

What fraction of the molecules have the minimum amount of kinetic energy required for a successful collision ?

………to define number of molecules with sufficient KE to react

we use the Boltzmann distribution with Ea as the energy

difference……………

(*Remember we discussed Boltzmann Distribution in Section 2)

The ratio of the populations of molecules in any two energy levels (E1 and E2) is given by the Boltzmann Distribution Law:

N2 = e –(E)/(kT)

N1

………where E = Ea kB = Boltzmann Constant = 1.381 x 10-23 JK-1

N2 = e –(Ea)/(kB

T)

N1

……so probability of molecules being in higher energy state is:

probability e – (ΔE/RT)

..….the rate of our reaction (successful collisions per unit time) is also proportional to this fraction:

RATE e – (Ea/RT)

Arrhenius Parameters

ln k = ln A – (Ea / R T)

or k = A e – Ea/RT

Pre-exponential Factor

……the pre-exponential factoror frequency factor is a measure of the rate at whichcollisions occur irrespective of their energy.

Activation Energy

Fraction of collisionswith a kinetic energyin excess of an Ea is given by the Boltzmanndistribution as e –Ea/RT

….this is fraction of collisions with enough kinetic energy to lead to reaction

• activation energy (Ea) is the minimum amount of energy required to initiate the chemical reaction.

• factor e-Ea/RT resembles Boltzmann distribution law….and represents the fraction of molecular collisions that have energy equal to or greater than the activation energy, Ea

• A, the frequency factor represents the frequency of collisions between reactant molecules

……….A is temperature dependent.

• plot of ln k versus 1/T gives straight line with…………

slope = - Ea/R

We can calculate the rate constant at different temperatures……..

ln k1 = ln A – (Ea / RT1)

ln k2 = ln A – (Ea / RT2)

Combine these two equations…………

ln (kln (k22/k/k11) = - (Ea/R) (1/T) = - (Ea/R) (1/T22 - 1/T - 1/T11))

……….if we know Ea and rate constant at one temperaturewe can calculate the rate constant at another temperature.