Presentation on theme: "Kinetics The study of reaction rates."— Presentation transcript:
1 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.
2 Review- Collision Theory Particles have to collide to react.Have to hit hard enoughThings that increase this increase rateHigh temp – faster reactionHigh concentration – faster reactionSmall particles = greater surface area means faster reaction
3 Reaction Rate Rate = Conc. of A at t2 -Conc. of A at t1 t2- t1 Rate = D[A] DtChange in concentration per unit timeFor this reactionN2 + 3H NH3
4 As the reaction progresses the concentration H2 goes down N2 + 3H2 → 2NH3[H2]Time
5 As the reaction progresses the concentration N2 goes down 1/3 as fast N2 + 3H2 → 2NH3[N2][H2]Time
6 As the reaction progresses the concentration NH3 goes up 2/3 times N2 + 3H2 → 2NH3[N2][H2][NH3]Time
7 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 timeDerivative.
10 Defining RateWe can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.In our example N2 + 3H NH3D[H2] = 3D[N2] Dt DtD[NH3] = -2D[N2] Dt Dt
11 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.
12 Rate Laws Two key points The concentration of the products do not appear in the rate law because this is an initial rate.The order (exponent) must be determined experimentally,can’t be obtained from the equation
13 2 NO NO + O2You will find that the rate will only depend on the concentration of the reactants. (Initially)Rate = k[NO2]nThis is called a rate law expression.k is called the rate constant.n is the order of the reactant -usually a positive integer.
14 2 NO2 2 NO + O2 The rate of appearance of O2 can be said to be. Rate' = D[O2] = k'[NO2] DtBecause there are 2 NO2 for each O2Rate = 2 x Rate'So k[NO2]n = 2 x k'[NO2]nSo k = 2 x k'
15 Types of Rate LawsDifferential 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.
16 Determining Rate LawsThe first step is to determine the form of the rate law (especially its order).Must be determined from experimental data.For this reaction N2O5 (aq) NO2 (aq) + O2(g)The reverse reaction won’t play a role because the gas leaves
17 Now graph the data [N2O5] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400 Now graph the data
18 To find rate we have to find the slope at two points We will use the tangent method.
19 At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4 (200-600) -400
20 At. 40 M the rate is (. 52 -. 32) = 0. 20 =- 2. 5 x 10 -4
21 Since the rate at twice as fast when the concentration is twice as big the rate law must be.. First powerRate = -D[N2O5] = k[N2O5]1 = k[N2O5] DtWe say this reaction is first order in N2O5The only way to determine order is to run the experiment.
22 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.
23 An example For the reaction BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O The general form of the Rate Law is Rate = k[BrO3-]n[Br-]m[H+]pWe use experimental data to determine the values of n,m,and p
24 Now we have to see how the rate changes with concentration Initial concentrations (M)Rate (M/s)BrO3-Br-H+x 10-4x 10-3x 10-3x 10-3Now we have to see how the rate changes with concentration
25 Integrated Rate LawExpresses 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 DtWe will only work with n=0, 1, and 2
26 First Order For the reaction 2N2O5 4NO2 + O2 We found the Rate = k[N2O5]1If concentration doubles rate doubles.If we integrate this equation with respect to time we get the Integrated Rate Lawln[N2O5] = - kt + ln[N2O5]0ln is the natural ln[N2O5]0 is the initial concentration.
27 First Order General form Rate = D[A] / Dt = k[A] ln[A] = - kt + ln[A]0 In the form y = mx + by = ln[A] m = -kx = t b = ln[A]0A graph of ln[A] vs time is a straight line.
28 First OrderBy getting the straight line you can prove it is first orderOften expressed in a ratio
29 First OrderBy getting the straight line you can prove it is first orderOften expressed in a ratio
30 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
31 Half Life The time required to reach half the original concentration. If the reaction is first order[A] = [A]0/2 when t = t1/2ln(2) = kt1/2
32 Half Lifet1/2 = / kThe time to reach half the original concentration does not depend on the starting concentration.An easy way to find k
33 The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?
34 Second Order Rate = -D[A]/Dt = k[A]2 integrated rate law 1/[A] = kt + 1/[A]0y= 1/[A] m = kx= t b = 1/[A]0A straight line if 1/[A] vs t is graphedKnowing k and [A]0 you can calculate [A] at any time t
35 Second Order Half Life[A] = [A]0 /2 at t = t1/2
36 Zero Order Rate Law Rate = k[A]0 = k Rate does not change with concentration.Integrated [A] = -kt + [A]0When [A] = [A]0 /2 t = t1/2t1/2 = [A]0 /2k
37 Zero Order Rate LawMost often when reaction happens on a surface because the surface area stays constant.Also applies to enzyme chemistry.
41 More Complicated Reactions BrO Br- + 6H Br2 + 3 H2OFor this reaction we found the rate law to beRate = k[BrO3-][Br-][H+]2To investigate this reaction rate we need to control the conditions
42 Rate = k[BrO3-][Br-][H+]2 We set up the experiment so that two of the reactants are in large excess.[BrO3-]0= 1.0 x 10-3 M[Br-]0 = 1.0 M[H+]0 = 1.0 MAs the reaction proceeds [BrO3-] changes noticeably[Br-] and [H+] don’t
43 Rate = k[BrO3-][Br-][H+]2 This rate law can be rewrittenRate = k[BrO3-][Br-]0[H+]02Rate = k[Br-]0[H+]02[BrO3-]Rate = k’[BrO3-]This is called a pseudo first order rate law.k = k’ [Br-]0[H+]02
44 Reaction MechanismsThe series of steps that actually occur in a chemical reaction.Kinetics can tell us something about the mechanismA balanced equation does not tell us how the reactants become products.
45 Reaction Mechanisms 2NO2 + F2 2NO2F Rate = k[NO2][F2] The proposed mechanism isNO2 + F NO2F + F (slow)F + NO NO2F (fast)F is called an intermediate It is formed then consumed in the reaction
46 Reaction MechanismsEach 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.Elementary steps add up to the balanced equation
47 Unimolecular step involves one molecule - Rate is first order. Bimolecular step - requires two molecules - Rate is second orderTermolecular step- requires three molecules - Rate is third orderTermolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
49 How to get rid of intermediates They can’t appear in the rate law.Slow step determines the rate and the rate lawUse the reactions that form themIf 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.
50 Formed in reversible reactions 2 NO + O NO2Mechanism2 NO N2O2 (fast)N2O2 + O NO2 (slow)rate = k2[N2O2][O2]k1[NO]2 = k-1[N2O2] (equilibrium)Rate = k2 (k1/ k-1)[NO]2[O2]Rate =k [NO]2[O2]
51 Formed in fast reactions 2 IBr I2+ Br2MechanismIBr I + Br (fast)IBr + Br I + Br2 (slow)I + I I2 (fast)Rate = k[IBr][Br] but [Br]= [IBr] because the first step is fastRate = k[IBr][IBr] = k[IBr]2
53 Collision theory Molecules must collide to react. Concentration affects rates because collisions are more likely.Must collide hard enough.Temperature and rate are related.Only a small number of collisions produce reactions.
58 Reaction Coordinate Br---NO Potential Br---NO Transition State Energy 2NO + Br2Reaction Coordinate
59 TermsActivation 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.
60 Arrhenius Said the at reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier.The number of collisions with the necessary energy increases exponentially.
61 Arrhenius Number of collisions with the required energy = ze-Ea/RT z = total collisionse is Euler’s number (inverse of ln)Ea = activation energyR = ideal gas constant (in J/K mol)T is temperature in Kelvin
62 Problem with this Observed rate is too small Due to molecular orientation- the have to be facing the right waywritten into equation as p the steric factor.
63 O O O O O O O O O O No Reaction Br N N N Br Br Br N Br N N Br Br N Br
64 k = zpe-Ea/RT = Ae-Ea/RT Arrhenius Equationk = zpe-Ea/RT = Ae-Ea/RTA is called the frequency factor = zpk is the rate constantln k = -(Ea/R)(1/T) + ln AAnother line !!!!ln k vs 1/T is a straight lineWith slope Ea/R so we can find EaAnd intercept ln A
65 A reaction is found to have a rate constant of 8 A reaction is found to have a rate constant of 8.60x10-1sec-1 at 523 K and an activation energy of kJ/mol. What is the value of the rate constant at 270 K?
66 Which statement is true concerning the plot of rate constants at various temperatures for a particular reaction?A) A steep slope of the ln k versus 1/T plot is indicative of small changes in the rate constant for a given increase in temperature. B) Different sections of the ln k versus 1/T plot show different Ea values. C) The plot of k versus T shows a linear increase in k as the temperature increases. D) A steep slope of the ln k versus 1/T plot is indicative of a large Ea. E) The y-intercept of the ln k versus 1/T plot is the Ea value for that reaction
68 Mechanisms and ratesThere is an activation energy for each elementary step.Activation energy determines k.k = Ae- (Ea/RT)k determines rateSlowest step (rate determining) must have the highest activation energy.
76 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.
77 How Catalysts WorkCatalysts 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.Does not change EShow up as a reactant in one step and a product in a later step
78 Heterogenous Catalysts Hydrogen bonds to surface of metal.Break H-H bondsHHHHHPt surface