Kinetics l The study of reaction rates. l Spontaneous reactions are reactions that will happen - but we can’t tell how fast. l Diamond will spontaneously turn to graphite – eventually. l Reaction mechanism- the steps by which a reaction takes place.

Reaction Rate l Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1 Rate =  [A] Dt l Change in concentration per unit time l For this reaction l N 2 + 3H 2 2NH 3

l As the reaction progresses the concentration H 2 goes down ConcentrationConcentration Time [H 2 ]

l As the reaction progresses the concentration N 2 goes down 1/3 as fast ConcentrationConcentration Time [H 2 ] [N 2 ]

l As the reaction progresses the concentration NH 3 goes up. ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ]

Calculating Rates l Average rates are taken over long intervals l 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 l Derivative.

l Average slope method ConcentrationConcentration Time D[H 2 ] DtDtDtDt

l Instantaneous slope method. ConcentrationConcentration Time  [H 2 ] D t D t

Defining Rate l We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. l In our example N 2 + 3H 2 2NH 3 -  [N 2 ] = -3  [H 2 ] = 2  [NH 3 ]  t  t  t

Rate Laws l Reactions are reversible. l As products accumulate they can begin to turn back into reactants. l Early on the rate will depend on only the amount of reactants present. l We want to measure the reactants as soon as they are mixed. l This is called the Initial rate method.

l Two key points l The concentration of the products do not appear in the rate law because this is an initial rate. l The order must be determined experimentally, l can’t be obtained from the equation Rate Laws

l You will find that the rate will only depend on the concentration of the reactants. l Rate = k[NO 2 ] n l This is called a rate law expression. l k is called the rate constant. l n is the order of the reactant -usually a positive integer. 2 NO 2 2 NO + O 2

l The rate of appearance of O 2 can be said to be. Rate' = D[O 2 ] = k'[NO 2 ] Dt l Because there are 2 NO 2 for each O 2 l Rate = 2 x Rate' l So k[NO 2 ] n = 2 x k'[NO 2 ] n l So k = 2 x k' 2 NO 2 2 NO + O 2

Types of Rate Laws l Differential Rate law - describes how rate depends on concentration. l Integrated Rate Law - Describes how concentration depends on time. l For each type of differential rate law there is an integrated rate law and vice versa. l Rate laws can help us better understand reaction mechanisms.

Determining Rate Laws l The first step is to determine the form of the rate law (especially its order). l Must be determined from experimental data. l For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g) The reverse reaction won’t play a role

[N 2 O 5 ] (mol/L) Time (s) Now graph the data

l To find rate we have to find the slope at two points l We will use the tangent method.

At.90 M the rate is ( ) = 0.22 =- 5.5x (0-400) -400

At.40 M the rate is ( ) = 0.22 =- 2.7 x ( ) -800

l Since the rate at twice the concentration is twice as fast the rate law must be.. Rate = -D[N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] Dt l We say this reaction is first order in N 2 O 5 l The only way to determine order is to run the experiment.

The method of Initial Rates l This method requires that a reaction be run several times. l The initial concentrations of the reactants are varied. l The reaction rate is measured bust after the reactants are mixed. l Eliminates the effect of the reverse reaction.

An example l For the reaction BrO Br - + 6H + 3Br H 2 O l The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p l We use experimental data to determine the values of n,m,and p

Initial concentrations (M) Rate (M/s) BrO 3 - Br - H+H+H+H x x x x x x x x Now we have to see how the rate changes with concentration

Integrated Rate Law l Expresses the reaction concentration as a function of time. l Form of the equation depends on the order of the rate law (differential). Changes Rate =  [A] n  t l We will only work with n=0, 1, and 2

First Order l For the reaction 2N 2 O 5 4NO 2 + O 2 l We found the Rate = k[ N 2 O 5 ] 1 l If concentration doubles rate doubles. l If we integrate this equation with respect to time we get the Integrated Rate Law l ln[N 2 O 5 ] = - kt + ln[N 2 O 5 ] 0 l ln is the natural log l [N 2 O 5 ] 0 is the initial concentration.

General form Rate =  [A] /  t = k[A] l ln[A] = - kt + ln[A] 0 l In the form y = mx + b l y = ln[A]m = -k l x = tb = ln[A] 0 l A graph of ln[A] vs time is a straight line. First Order

l By getting the straight line you can prove it is first order l Often expressed in a ratio First Order

l By getting the straight line you can prove it is first order l Often expressed in a ratio First Order

Half Life l The time required to reach half the original concentration. l If the reaction is first order l [A] = [A] 0 /2 when t = t 1/2

Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A] 0 /2 when t = t 1/2 l ln(2) = kt 1/2

Half Life l t 1/2 = 0.693/k l The time to reach half the original concentration does not depend on the starting concentration. l An easy way to find k

Second Order Rate = -  [A] /  t = k[A] 2 l integrated rate law l 1/[A] = kt + 1/[A] 0 l y= 1/[A] m = k l x= tb = 1/[A] 0 l A straight line if 1/[A] vs t is graphed l Knowing k and [A] 0 you can calculate [A] at any time t

Second Order Half Life l [A] = [A] 0 /2 at t = t 1/2

Zero Order Rate Law l Rate = k[A] 0 = k l Rate does not change with concentration. l Integrated [A] = -kt + [A] 0 l When [A] = [A] 0 /2 t = t 1/2 l t 1/2 = [A] 0 /2k

l Most often when reaction happens on a surface because the surface area stays constant. l Also applies to enzyme chemistry. Zero Order Rate Law

Time ConcentrationConcentration

ConcentrationConcentration  A]/  t tt k =  A]

More Complicated Reactions l BrO Br - + 6H + 3Br H 2 O l For this reaction we found the rate law to be l Rate = k[BrO 3 - ][Br - ][H + ] 2 l To investigate this reaction rate we need to control the conditions

Rate = k[BrO 3 - ][Br - ][H + ] 2 l We set up the experiment so that two of the reactants are in large excess. l [BrO 3 - ] 0 = 1.0 x M l [Br - ] 0 = 1.0 M l [H + ] 0 = 1.0 M l As the reaction proceeds [BrO 3 - ] changes noticably l [Br - ] and [H + ] don’t

l This rate law can be rewritten l Rate = k[BrO 3 - ][Br - ] 0 [H + ] 0 2 l Rate = k[Br - ] 0 [H + ] 0 2 [BrO 3 - ] l Rate = k’[BrO 3 - ] l This is called a pseudo first order rate law. l k = k’ [Br - ] 0 [H + ] 0 2 Rate = k[BrO 3 - ][Br - ][H + ] 2

Reaction Mechanisms l The series of steps that actually occur in a chemical reaction. l Kinetics can tell us something about the mechanism l A balanced equation does not tell us how the reactants become products.

l 2NO 2 + F 2 2NO 2 F l Rate = k[NO 2 ][F 2 ] l The proposed mechanism is l NO 2 + F 2 NO 2 F + F (slow) l F + NO 2 NO 2 F(fast) l F is called an intermediate It is formed then consumed in the reaction Reaction Mechanisms

l Each of the two reactions is called an elementary step. l The rate for a reaction can be written from its molecularity. l Molecularity is the number of pieces that must come together. Reaction Mechanisms

l Unimolecular step involves one molecule - Rate is rirst order. l Bimolecular step - requires two molecules - Rate is second order l Termolecular step- requires three molecules - Rate is third order l Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

l A products Rate = k[A] l A+A productsRate= k[A] 2 l 2A productsRate= k[A] 2 l A+B productsRate= k[A][B] l A+A+B Products Rate= k[A] 2 [B] l 2A+B Products Rate= k[A] 2 [B] l A+B+C Products Rate= k[A][B][C]

How to get rid of intermediates l Use the reactions that form them l If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry. l If it is formed by a reversible reaction set the rates equal to each other.

Formed in reversible reactions l 2 NO + O 2 2 NO 2 l Mechanism l 2 NO N 2 O 2 (fast) l N 2 O 2 + O 2 2 NO 2 (slow) l rate = k 2 [N 2 O 2 ][O 2 ] l k 1 [NO] 2 = k -1 [N 2 O 2 ] l rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]=k[NO] 2 [O 2 ]

Formed in fast reactions l 2 IBr I 2 + Br 2 l Mechanism l IBrI + Br (fast) l IBr + Br I + Br 2 (slow) l I + II 2 (fast) l Rate = k[IBr][Br] but [Br]= [IBr] l Rate = k[IBr][IBr] = k[IBr] 2

Collision theory l Molecules must collide to react. l Concentration affects rates because collisions are more likely. l Must collide hard enough. l Temperature and rate are related. l Only a small number of collisions produce reactions.

Potential EnergyPotential Energy Reaction Coordinate Reactants Products

Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activation Energy E a

Potential EnergyPotential Energy Reaction Coordinate Reactants Products Activated complex

Potential EnergyPotential Energy Reaction Coordinate Reactants Products EE }

Potential EnergyPotential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO 2 Transition State

Terms l Activation energy - the minimum energy needed to make a reaction happen. l Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.

Arrhenius l Said the at reaction rate should increase with temperature. l At high temperature more molecules have the energy required to get over the barrier. l The number of collisions with the necessary energy increases exponentially.

Arrhenius l Number of collisions with the required energy = ze -E a /RT l z = total collisions l e is Euler’s number (opposite of ln) l E a = activation energy l R = ideal gas constant l T is temperature in Kelvin

Problems l Observed rate is less than the number of collisions that have the minimum energy. l Due to Molecular orientation l written into equation as p the steric factor.

O N Br O N O N O N O N O N O N O N O N O N No Reaction

Arrhenius Equation l k = zpe -E a /RT = Ae -E a /RT l A is called the frequency factor = zp l ln k = -(E a /R)(1/T) + ln A l Another line !!!! l ln k vs t is a straight line

Activation Energy and Rates The final saga

Mechanisms and rates l There is an activation energy for each elementary step. l Activation energy determines k. l k = Ae - (E a /RT) l k determines rate l Slowest step (rate determining) must have the highest activation energy.

+ This reaction takes place in three steps

+ EaEa First step is fast Low activation energy

Second step is slow High activation energy + EaEa

+ EaEa Third step is fast Low activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

Catalysts l Speed up a reaction without being used up in the reaction. l Enzymes are biological catalysts. l Homogenous Catalysts are in the same phase as the reactants. l Heterogeneous Catalysts are in a different phase as the reactants.

How Catalysts Work l Catalysts allow reactions to proceed by a different mechanism - a new pathway. l New pathway has a lower activation energy. l More molecules will have this activation energy. Do not change  E

Pt surface HHHH HHHH l Hydrogen bonds to surface of metal. l Break H-H bonds Heterogenous Catalysts

Pt surface HHHH Heterogenous Catalysts C HH C HH

Pt surface HHHH Heterogenous Catalysts C HH C HH l The double bond breaks and bonds to the catalyst.

Pt surface HHHH Heterogenous Catalysts C HH C HH l The hydrogen atoms bond with the carbon

Pt surface H Heterogenous Catalysts C HH C HH HHH

Homogenous Catalysts l Chlorofluorocarbons catalyze the decomposition of ozone. l Enzymes regulating the body processes. (Protein catalysts)

Catalysts and rate l Catalysts will speed up a reaction but only to a certain point. l Past a certain point adding more reactants won’t change the rate. l Zero Order

Catalysts and rate. Concentration of reactants RateRate l Rate increases until the active sites of catalyst are filled. l Then rate is independent of concentration