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Collision Rate Model Three conditions must be met at the nano- scale level if a reaction is to occur: the molecules must collide; they must be positioned.

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Presentation on theme: "Collision Rate Model Three conditions must be met at the nano- scale level if a reaction is to occur: the molecules must collide; they must be positioned."— Presentation transcript:

1 Collision Rate Model Three 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.

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3 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

4 Activation Energy the minimum energy required to start a reaction

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6 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 (1 st condition). ……..molecules are often moving too slowly which prevents a close enough approach for rxn. to occur….

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8 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 SUCCESSFUL……….. Unsuccessful Collision

9 THREE conditions must be met for a reaction to occur : 1 st Condition: Two molecules must have sufficient kinetic energy (meaning they must be moving fast enough) 2 nd Condition: The sum of the molecular momentum vectors must be near zero (meaning molecules have to hit each other directly) 3 rd Condition: There must be overlap of the involved electronic orbitals of the molecules (meaning the molecules must be properly oriented) Product Successful Collision

10 What affect does TEMPERATURE have on the rate of a reaction?

11 Temperature & Reaction Rate

12 What is a Catalyst?

13 Catalyst substance which speeds up the rate of a reaction while not being consumed What are the two types of catalysts?

14 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

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16 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.

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20 ln k = ln A – (E a / 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.

21 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

22 Potential Energy Profile for an Exothermic Reaction How does the potential energy change in the course of a reaction ? …..consider a reaction where reactants A and B react to form products ……A and B collide, come into contact and distort, begin to exchange or discard atoms……………….. …….potential energy rises to maximum and cluster of atoms at this point is termed activated complex………….. ……potential energy then falls and reaches characteristic value for products Activated Complex

23 Arrhenius Parameters ln k = ln A – (E a / R T) or k = A e – Ea/RT Pre-exponential Factor ……the pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Activation Energy Fraction of collisions with a kinetic energy in excess of an Ea is given by the Boltzmann distribution as e – Ea/RT ….this is fraction of collisions with enough kinetic energy to lead to reaction

24 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 E a as the energy difference…………… (*Remember we discussed Boltzmann Distribution in Section 2) The ratio of the populations of molecules in any two energy levels (E 1 and E 2 ) is given by the Boltzmann Distribution Law: N 2 = e –(  E)/(kT) N 1 ………where  E = Ea k B = Boltzmann Constant = 1.381 x 10 -23 JK -1

25 N 2 = e –(Ea)/(k B T) N 1 ……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)

26 Arrhenius Parameters ln k = ln A – (E a / R T) or k = A e – Ea/RT Pre-exponential Factor ……the pre-exponential factor or frequency factor is a measure of the rate at which collisions occur irrespective of their energy. Activation Energy Fraction of collisions with a kinetic energy in excess of an Ea is given by the Boltzmann distribution as e –Ea/RT ….this is fraction of collisions with enough kinetic energy to lead to reaction

27 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

28 We can calculate the rate constant at different temperatures…….. ln k 1 = ln A – (E a / RT 1 ) ln k 2 = ln A – (E a / RT 2 ) Combine these two equations………… ln (k 2 /k 1 ) = - (Ea/R) (1/T 2 - 1/T 1 ) ……….if we know Ea and rate constant at one temperature we can calculate the rate constant at another temperature.


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