KINETICS The speed of a reaction

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

Reaction Rate Rate = Conc. of A at t2 -Conc. of A at t1 t2- t1 Rate = D[A] Dt Change in concentration per unit time For this reaction N2 + 3 H2  2 NH3

As the reaction progresses the concentration of H2 goes down Time

As the reaction progresses the concentration of N2 goes down 1/3 as fast. Check out the stoichiometry ! Concentration [N2] [H2] Time

As the reaction progresses the concentration of the product, NH3 goes up. Time

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. This is what is called in Calculus as a Derivative.

Average slope method Concentration D[H2] Dt Time

Instantaneous slope method. Concentration D[H2] D t Time

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. The stoichiometry is important. In our example: N2 + 3 H2  2 NH3 -D [N2] = - 3 D[H2] = 2 D[NH3] Dt Dt Dt

Rate Laws Reactions are reversible. As products accumulate they can begin to turn back into reactants. Early on the rate will depend on only the amount of reactants present. We want to measure the reactants as soon as they are mixed. This is called the Initial rate method and time = 0.

Two key points to remember. Rate Laws: Two key points to remember. The concentration of the products do not appear in the rate law because this is an initial rate so there are no products yet. The order must be determined experimentally because it can’t be obtained from the equation.

2 NO2  2 NO + O2 You will find that the rate will only depend on the concentration of the reactants. Rate = k [NO2]n This is called a rate law expression. k is called the rate constant. n is the order of the reactant - usually a positive integer.

2 NO2  2 NO + O2 The rate of appearance of O2 can be said to be. Rate' = D[O2] = k'[NO2]2 Dt Because there are 2 NO2 for each O2 Rate = 2 x Rate' So k[NO2]n = 2 x k'[NO2]n So k = 2 x k'

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

Determining Rate Laws The first step is to determine the form of the rate law (especially its order). Must be determined from experimental data which is usually given. For this reaction 2 N2O5 (aq)  4 NO2 (aq) + O2 (g) The reverse reaction won’t play a role

Now graph the data [N2O5] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000 Now graph the data

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

At .90 M the rate is (1.00 - .78) = 0.22 (0 - 400) - 400 = - 5.5x 10 -4

At .40 M the rate is (.52 - .31) = 0.22 (1000-1800) -800 = - 2.7 x 10 -4

Since the rate at twice the concentration is twice as fast, the rate law must be.. Rate = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt We say this reaction is first order in N2O5 First order means that the rate and the concentration are changing in a similar fashion. The only way to determine order is to run the experiment. So the Rate Law is Rate = k[N2O5]

The Method of Initial Rates This method requires that a reaction be run several times. The initial concentrations of the reactants are varied. The reaction rate is measured just after the reactants are mixed. Eliminates the effect of the reverse reaction. In our problems, this isn’t a concern.

An Example of a Rate Law For the reaction BrO3-1 + 5 Br-1 + 6 H+1  3 Br2 + 3 H2O The general form of the Rate Law is: Rate = k[BrO3-1]n[Br-1]m[H+1]p We use experimental data to determine the values of n, m, and p

Now we have to see how the rate changes with concentration Initial concentrations (M) Rate (M/s) Br-1 BrO3-1 H+1 0.10 0.10 0.10 8.0 x 10-4 0.20 0.10 0.10 1.6 x 10-3 0.20 0.20 0.10 3.2 x 10-3 0.10 0.10 0.20 3.2 x 10-3 Now we have to see how the rate changes with concentration Rate = k[BrO3-]1[Br-]1[H+]2

Remember our Formula ! Δ Concentration order = Δ Rate In words, the change in concentration raised to the order of the exponent = the change in the rate. Fortunately, most of the time, these can be solved on inspection. Eg. If the concentration is tripled and the rate is tripled, then 3x = 3 so x = 1. HOWEVER…

Sometimes more drastic measures are needed. Namely the law of logarithms. Δ Concentration order = Δ Rate becomes Order x (Log (conc.)) = Log (rate) It is simple and works every time. 23 = 8 3 log 2 = log 8 3 x .301 = .903

Using data to create linear relationships We can get more out of our data if we can create linear relationships (functions) from them. Data is manipulated to create these linear relationships. Obtaining a function gives us a way to determine many values in a straight-forward manner.

Integrated Rate Law Expresses the reaction concentration as a function of time. Form of the equation depends on the order of the rate law (differential). Changes Rate = D [A]n Dt We will mostly see n= 0, 1, and 2

First Order For the reaction 2 N2O5  4 NO2 + O2 We found the Rate = k[N2O5]1 If concentration doubles rate doubles. If we integrate this equation with respect to time we get the Integrated Rate Law ln[N2O5]t = - k t + ln[N2O5]0 ln is the natural log [N2O5]0 is the initial concentration. Using ln in 1st order reactions gives us a straight line.

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

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

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

Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A]0/2 when t = t1/2

Half Life The time required to reach half the original concentration. If the reaction is first order [A] = [A]0/2 when t = t1/2 ln(2) = kt1/2

Half Life t1/2 = 0.693/k The time to reach half the original concentration does not depend on the starting concentration for first order reactions. An easy way to find k.

Second Order Rate = -D[A] / Dt = k[A]2 Integrated rate law 1/[A] = kt + 1/[A]0 y= 1/[A] m = k x= t b = 1/[A]0 A straight line if 1/[A] vs. t is graphed Knowing k and [A]0 you can calculate [A] at any time.

Second Order Half Life [A] = [A]0 /2 at t = t1/2

Zero Order Rate Law Rate = k[A]0 = k Rate does not change with concentration. Integrated [A] = -kt + [A]0 When [A] = [A]0 /2 t = t1/2 t1/2 = [A]0 / 2k

Zero Order Rate Law Most often when reaction happens on a surface because the surface area stays constant. Also applies to enzyme chemistry. The material is said to be at the zeroeth order.

Concentration Time

Concentration k = D[A]/Dt D[A] Dt Time

More Complicated Reactions BrO3-1 + 5 Br -1 + 6H+1  3 Br2 + 3 H2O For this reaction we found the rate law to be: Rate = k[BrO3-][Br-][H+]2 To investigate this reaction rate we need to control the conditions.

Rate = k[BrO3-][Br-][H+]2 We set up the experiment so that two of the reactants are in large excess. [BrO3-1]0= 1.0 x 10-3 M [Br-1]0 = 1.0 M [H+1]0 = 1.0 M As the reaction proceeds [BrO3-1] changes noticably [Br-1] and [H+1] don’t

Rate = k[BrO3-][Br-][H+]2 This rate law can be rewritten Rate = k[BrO3-1][Br-1]0[H+1]02 Rate = k[Br-]0[H+1]02[BrO3-1] Rate = k’[BrO3-] This is called a pseudo first order rate law. k = k’ [Br-]0[H+]02

Summary of Rate Laws Know your integrated rate laws. Know how to find the order for the reaction Know how to find k after you determine the orders. As always, read the question carefully to make sure you are using all the given information.

Reaction Mechanisms The series of steps that actually occur in a chemical reaction. Kinetics can tell us something about the mechanism. A balanced equation does not tell us how the reactants become products. We will always want to find the rate determining step, which will usually be the slowest step in the mechanism.

Reaction Mechanisms 2 NO2 + F2  2 NO2F Rate = k[NO2][F2] The proposed mechanism is NO2 + F2  NO2F + F (slow) F + NO2  NO2F (fast) F is called an intermediate It is formed then consumed in the reaction, so it isn’t present in the given equation.

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

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

A  Products Rate = k[A] A + A  Products Rate= k[A]2 2A  Products Rate= k[A]2 A + B  Products Rate= k[A][B] A + A + B  Products Rate= k[A]2[B] 2A + B  Products Rate= k[A]2[B] A + B + C  Products Rate= k[A][B][C]

How to get rid of intermediates Use the reactions that form them. If the reactions are fast and irreversible - the concentration of the intermediate is based on stoichiometry. If it is formed by a reversible reaction set the rates equal to each other. A picture is worth 1,000 words here, so let’s check out the next slide.

Formed in reversible reactions 2 NO + O2  2 NO2 Mechanism 2 NO  N2O2 (fast) Step 1 N2O2 + O2  2 NO2 (slow) Step 2 rate = k2[N2O2][O2] k1[NO]2 = k-1[N2O2] rate = k2 (k1/ k-1)[NO]2[O2] =k[NO]2[O2] NOTICE: Step 1 is in equilibrium so the rate of formation of the N2O2 is equal to the rate of formation of the 2 NO.

Formed in fast reactions 2 IBr  I2+ Br2 Mechanism IBr  I + Br (fast) IBr + Br  I + Br2 (slow) I + I  I2 (fast) What is the rate determining step ? Rate = k[IBr][Br] but [Br]= [IBr] Rate = k[IBr][IBr] = k[IBr]2

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

Potential Energy Reactants Products Reaction Coordinate

Activation Energy Ea Reaction Coordinate Potential Energy Reactants Products Reaction Coordinate

Reaction Coordinate Activated complex Potential Energy Reactants Products Reaction Coordinate

Potential Energy } Reactants DE Products Reaction Coordinate

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

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

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

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

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

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

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

Activation Energy and Rates The final saga

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

This reaction takes place in three steps

Ea First step is fast Low activation energy

High activation energy Ea Second step is slow High activation energy

Ea Third step is fast Low activation energy

Second step is rate determining

Intermediates are present

Activated Complexes or Transition States

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

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

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

Heterogenous Catalysts Pt surface

Heterogenous Catalysts The double bond breaks and bonds to the catalyst. H H H C C H H H H H Pt surface

Heterogenous Catalysts The hydrogen atoms bond with the carbon H H H C C H H H H H Pt surface

Heterogenous Catalysts Pt surface

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

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. Zero Order

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