Study of Reaction Rates Grab your text book. Chemical Kinetics Study of Reaction Rates Grab your text book.

Reaction Rate Change in a concentration of product or reactant over a specific period of time [ ] = concentration in M Rate = [A]t2 – [A]t1 = D[A] t2 – t1 Dt Changes can be positive or negative

4PH3(g)  P4(g) + 6H2(g) Over time,. 0024 mol PH3 is consumed in a 2 4PH3(g)  P4(g) + 6H2(g) Over time, .0024 mol PH3 is consumed in a 2.0L container each second of the reaction. What are the rates of production of P4 and H2 in this experiment?

Reaction Rate We will always work with rate as a positive quantity. Concentrations of reactants always decrease with time, the rate expression will include a negative sign in order to keep the rate positive Instantaneous Rate – value of the rate at a particular time

Rate of a Chemical Reaction are affected by: Time Stoichiometry of the reaction ( which species is being studied)

Introduction to Rate Laws Chemical reactions are reversible. * After a period of time, enough products accumulate so that the reverse reaction becomes important. At this point the concentration of the reactants depends on the difference in the rates of the forward and reverse reactions. This tends to complicate matters, so we will focus on the reaction soon after the reactants are mixed, before the products have had time to build up to significant levels. If we choose conditions where the reverse reaction can be neglected, the reaction rate will depend only on the concentrations of the reactants. *reactants form products and products can form reactants

Rate Law Rate = k [reactant]n k = rate constant, determined experimentally n = order of the reactant, determined experimentally, can be an integer, including zero or a fraction [ ] = law only depends on concentration of reactants

Rate Law Key Points: [products] do not appear b/c conditions are set where reverse rxn is unimportant n, must be determined experimentally k, can be calculated once the form of the rate law has been determined by using data from one experiment We study rate laws in order to be able to infer the steps by which the reaction occurred.

Determining the form of the Rate Law Form of rate law Determine the power to which each reactant concentration must be raised in the rate law Order = how does changing concentration, change the rate

How to determine the rate law? Your data will tell you the method to use If you have initial concentrations and initial rates, as well as multiple experiments its METHOD of INITIAL RATES If you have concentration and time (and there is only 1 reactant) then its GRAPHICAL METHOD (or chart)

Method of Initial Rates Way to experimentally determine the form of the rate law Initial rate – instantaneous rate determined just after the reaction begins (before initial [reactants] have changed much) Look for the data table to list [initial] ([]0) and initial rates Several experiments are needed w/different [initial], each with a initial rate, then this info is compared to determine the form Rate = k [A]n [B]m [C]p* * Overall reaction order, n+m+p = overall reaction order

Sample Exercise 12.1 pg. 562 We will determine the order of each reactant by comparing initial concentrations and initial rates from several experiments. Compare experiments where the reactant your studying is changing but the others are constant. Go to white screen for example problem.

Second method for determining a rate law Graphical Method Data will be time and concentration only Only one reactant But you may also be told the order in the question and then you will be able to use all the info from the Kinetics Made Easy Chart

Integrated Rate Law Reactions involving a single reactant all have the same rate law form: Rate = k [A]n Integrated – concentration of a reactant as a function of time

First Order Rate Laws Doubling the reactant, doubles the product 2 N2O5  4 NO2 + O2 Rate = k [N2O5] When integrated with time the law becomes: ln[N2O5] = -kt + ln[N2O5]0 [N2O5] @t time [N2O5] @ t=0

First Order Rate Laws First Order Rate Law: Rate = k [A] Integrated First Order Rate Law: ln [A] = -kt + ln [A]0 If [A]0 and k are known, the [A] at any time can be calculated Uses the form y = mx + b, straight line graph y = ln [A] x = t m = -k b = ln[A]0 First order always gives a straight line graph, by plotting ln[A] vs. t Can also be expressed as a ratio of [A] and [A]0 ln [A]0 = kt [A] Sample exercise 12.2 and 12.3

The decomposition of N2O5 in the gas phase was studied at constant temperature. 2N2O5(g)  4NO2(g) + O2(g) [N2O5] (mol/L) Time (s) 0.1000 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 Using these data, verify that the rate law is first order in [N2O5] and calculate the rate constant. We will do this in class together, make sure you have your graphing calculator.

First Order Rate Law Half Life of a Reactant Time required for a reactant to reach half its original concentration (t1/2) First Order Rxn Half Life Law*: t1/2 = .693 k First order t1/2 doesn’t depend on concentration Example 12.4

A certain first-order reaction has a half-life of 20. 0 minutes A certain first-order reaction has a half-life of 20.0 minutes. Calculate the rate constant for this reaction. How much time is required for this reaction to be 75% complete?

Second-Order Rate Laws Rate Law  Rate = k[A]2 Doubling the conc of A, quadruples the reaction rate Tripling the conc of A, increases the rate 9 times Integrated Rate Law  1 = kt + 1 [A] [A]0

Second Order Rate Laws Plot of 1/[A] vs. t will produce a straight line m = k If k and [A]0 are known, then [A]t can be calculated. Second Order t1/2  t1/2 = 1 k[A]0 Each successive half-life is double the preceding one Sample exercise 12.5

Zero Order Rate Laws Rate is constant, does not change with concentration Rate Law  Rate = k Integrated Rate Law  [A] = -kt + [A]0

Zero Order Rate Laws Plot of [A] vs t produces a straight line slope m = -k Zero Order t1/2 t1/2 = [A]0 2k

Model of Chemical Kinetics Chemical rxns speed up when temperature is increased, all rate constants show an exponential increase w/ absolute temperature. Collision Model 1) Molecules must collide in order to react. 2) Molecules must possess a minimum amount of energy for collisions to be successful – Activation Energy 3) Molecular orientation – the reactants must be properly oriented to allow formation of new bonds of the product How does temperature speed up the reaction? What is the energy being used for?

How does a catalyst speed up the reaction? What is one more way to speed up the reaction?

Reaction Mechanisms The series of steps by which a chemical reaction occurs Not the balanced equation, which only tells reactants, products and stoichiometry

NO2(g) + CO(g)  NO(g) + CO2(g) This reaction occurs by elementary steps, a reaction whose rate law can be written from its molecularity. NO2(g) + NO2(g)  NO3(g) + NO(g) slow NO3(g) + CO(g)  NO2(g) + CO2(g) fast These are the two elementary steps by which the above reaction occurs. Intermediate – a species that is neither a reactant or product, but is formed and consumed during the reaction.

Molecularity The number of species that must collide to produce the reaction indicated by that step. Unimolecular – one molecule Bimolecular – two molecules Termolecular – three molecules (very rare, not likely to get three molecules to collide @once)

Reaction Mechanism – series of elementary steps that must: The rate law for an elementary step follows directly from the molecularity of that step. Reaction Mechanism – series of elementary steps that must: 1) Sum of the elementary steps must give the overall balance equation 2) The mechanism must agree w/ the experimentally determined rate law. Rate–Determining Step: slow step (highest Ea) A reaction is only as fast as its slowest step The slow step will be indicated to you. A mechanism can never be proven absolutely, it is possibly correct. Example 12.6

Material Not in Your Text Sometimes the rate expression obtained by the process involves a reactive intermediate ([intermediate] is to small to determine experimentally) The intermediate must be eliminated from the rate expression. See handwritten notes. Pg.5