Chapter 14 Chemical Kinetics Dr. Peter Warburton peterw@mun.ca http://www.chem.mun.ca/zcourses/1051.php Rates Rates can be measured as the concentration change of a chemical (X) over a period (change) of time Rate = D[X] / Dt Rate Xt Xt 1 0 t1 t 0 Often use molL-1 (or M) as units of concentration so rate often has units molL-1s-1 (or Ms-1 ) All media copyright of their respective owners 2 Rate Rates can be expressed in two different ways. We can look at the formation of a product (increase in the concentration over time) or the disappearance of a reactant (decrease in the concentration over time). All media copyright of their respective owners 3 Rate of formation of a product 2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq) t0 = 0.0 s [Fe2+]t0 = 0.0000 M t1 = 38.5 s [Fe2+]t1 = 0.0010 M Δt = 38.5 s Δ[Fe2+] = (0.0010 – 0.0000) M rateof formationof Fe 2 Δ Fe 2 0.0010M 2.6 x 105 M s 1 Δt 38.5s All media copyright of their respective owners 4 Rate of disappearance of a reactant 2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq) t0 = 0.0 s [Sn2+]t0 = 0.0015 M t1 = 38.5 s [Sn2+]t1 = 0.0010 M Δt = 38.5 s Δ[Sn2+] = (0.0010 – 0.0015) M rateof disappearance of Sn 2 Δ Sn 2 (-0.0005M) 1.3 x 105 M s 1 Δt 38.5s The word “disappearance” implies the negative sign. Because of this, all rates are considered to be POSITIVE! All media copyright of their respective owners 5 Reaction rates aA+bBcC+dD 1 ΔA 1 ΔB 1 ΔC 1 ΔD reaction rate + + a Δt b Δt c Δt d Δt Reaction rates are always positive, so we must put negative signs in front of reactant concentration changes! To account for the stoichiometric relationships and their effect on rate, we must always divide by the stoichiometric coefficient for the chemical. All media copyright of their respective owners 6 Reaction rate 2 Fe3+(aq) + Sn2 +(aq) → 2 Fe2+(aq) + Sn4+(aq) 3 1 Δ Fe reactionrate Δt 2 1 ΔSn 2 1 Δt 1.3 x 10 5 M s 1 slide 5 2 4 1 Δ Fe 1 Δ Sn Δt 1 Δt 2 2.6 x 10 5 Ms 1 slide 4 reactionrate 1.3 x 105 M s 1 All media copyright of their respective owners 7 Measuring rates All media copyright of their respective owners 8 Measuring rates 1) Average rates 2) Slopes 3) Time = 0 All media copyright of their respective owners 9 Problem At some point in the reaction 2A+B3C [A] = 0.3629 M. At a time 8.25 minutes later [A] = 0.3187 M. What is the average rate of the disappearance of A in Ms-1, what is the average rate of the formation of C in Ms-1, and what is the average rate of reaction in Ms-1 over that time interval? All media copyright of their respective owners 10 Problem answers 5 averagerateof disappearance of A 8.93 x 10 M s 1 averagerateof formationof C 1.34 x 10 4 M s 1 averagereactionrate 4.46 x 105 M s 1 All media copyright of their respective owners 11 Instantaneous reaction rates What’s happening at “this instant in time”? We can use instantaneous reaction rates. The initial rate is the instantaneous reaction rate for a reaction at time zero. All media copyright of their respective owners 12 Rate laws The rate of a chemical reaction depends on the concentration of some or all of the reactants. A reactant might not affect the rate, regardless of its concentration. All media copyright of their respective owners 13 Rate laws rate law for a reaction is the equation showing the dependence of the reaction rate on The concentrations of the reactants. All media copyright of their respective owners 14 a A + b B + … products m n rate = k [A] [B] … k is a constant for the reaction AT A GIVEN TEMPERATURE, and is called the rate constant. The stoichiometry of the balanced reaction equation IS NOT ALWAYS the source of the rate equation exponents! m DOES NOT HAVE TO equal a n DOES NOT HAVE TO equal b All media copyright of their respective owners 15 Reaction order Reaction order with respect to a given reactant is the value of the exponent of the rate law equation for the specific reactant only. The overall reaction order is the sum of the reaction orders for all reactants. All media copyright of their respective owners 16 Reaction order example 2 rate = k [A] [B] The reaction order with respect to A is 2 or the reaction is second order in A The reaction order with respect to B is 1 or the reaction is first order in B The overall reaction order is 3 (2 + 1 = 3) or the reaction is third order overall All media copyright of their respective owners 17 “Sensitivity” to concentration change Rate change of the reaction if [A] is doubled depends on 2m rate doubles rate quadruples zeroth order in A first order in A second order in A rate no change halves NOTE: negative or non-integer orders All media copyright of their respective owners 18 Method of initial rates Reaction rate laws are often determined experimentally! We most commonly carry out a series of experiments in which the initial rate of the reaction is measured as a function of different initial concentrations of reactants All media copyright of their respective owners 19 Method of initial rates If you see a table like this with chemical concentrations or pressures and rate data, chances are good the question is a method of initial rates problem. All media copyright of their respective owners 20 Method of initial rates IGNORE THE REACTION with this type of problem. The chemicals in the TABLE are the interesting ones. You always require at least one more experimental reaction than your number of chemicals given in your table! Sometimes we are given a table with an extra experiment which we can use to check if we’ve done everything correctly. All media copyright of their respective owners 21 2 NO (g) + O2 (g) NO2 (g) Since rate laws are always expressed in terms of reactants (and sometimes catalysts – we’ll see these later), lets create a general form of the rate law for this reaction based on what chemicals the TABLE tells us are involved in the rate of the reaction. All media copyright of their respective owners 22 2 NO (g) + O2 (g) NO2 (g) m n rate = k [NO] [O2] All media copyright of their respective owners 23 Method of initial rates For our initial reactant order determination we need to choose a pair of reactions where only one reactant concentration changes. Experiments #1 and #2 fulfill this condition. All media copyright of their respective owners 24 Method of initial rates m n rate = k [NO] [O2] Since k is a constant then k for experiment 1 IS EQUAL TO k for experiment 2! m n k = rate / [NO] [O2] rate1 rate2 rate1 NO O so m n m n NO1 O2 1 NO2 O2 2 rate2 NO O m 1 m 2 All media copyright of their respective owners n 2 1 n 2 2 25 Reaction order w.r.t. NO rate 1 NO 1 O 2 1 rate 2 NO m2 O 2 n2 m m n 0.048M s -1 (0.015M) (0.015M) -1 0.192M s (0.030M)m (0.015M)n 0.25 (0.50)m log 0.25 log (0.50)m log 0.25 m log 0.50 All media copyright of their respective owners n log 0.25 log 0.50 0.602 m 0.301 m2 m 26 Reaction order w.r.t. O2 rate 1 NO 1 O 2 1 rate 3 NO 32 O 2 3n 2 2 n 0.048M s -1 (0.015M) (0.015M) -1 0.096M s (0.015M)2 (0.030M)n 0.50 (0.50)n log 0.50 log (0.50)n log 0.50 n log 0.50 All media copyright of their respective owners n log 0.50 log 0.50 0.301 n 0.301 n 1 n 27 ALTERNATE reaction order w.r.t. O2 rate 1 NO 1 O 2 1 rate 4 NO 24 O 2 n4 2 0.048M s -1 0.384M s (0.030M)2 (0.030M)n -1 (0.015M)2 (0.015M)n 0.125 (0.50)2 (0.50)n 0.125 [0.25](0.50)n All media copyright of their respective owners n 0.125 (0.50)n 0.25 0.50 (0.50)n n 1 28 Our rate law 2 1 rate = k [NO] [O2] All media copyright of their respective owners 29 Rate constant using experiment 1 2 1 k = rate / [NO] [O2] 0.048M s 1 k 2 2 NO O 2 0.015M 0.015M 3.38 x 106 M 3 rate 0.048M s 1 k 1.42 x 104 M -2 s 1 All media copyright of their respective owners 30 Rate constant using experiment 2 2 1 k = rate / [NO] [O2] rate 0.192M s 1 0.192M s 1 k 2 2 NO O 2 0.030M 0.015M 1.35 x 105 M 3 k 1.42 x 10 M s 4 -2 1 The rate constant is the same, as it should be! All media copyright of their respective owners 31 Check using extra experiment rate = (1.42 x 10 M s ) [NO] [O2] 4 rate 1.4 rate 1.4 -2 -1 2 1 0.030M 0.030M 2.7 x 10 M rate 1.42 x 104 M -2 s 1 NO O 2 2 1 2 x 10 M s 2 x 104 M -2 s 1 4 -2 2 -5 3 rate 3.83 x 10-1 M s 1 The rate is the same as the experimentally observed rate (within rounding errors). We MUST have done everything right! All media copyright of their respective owners 32 Units of rate constants Rate always has units in terms of concentration per time unit. Usually it is molL-1·s-1 (or M·s-1) To ensure we get the right units for rate means the rate constant must have different units depending on the overall reaction order. All media copyright of their respective owners 33 Problem H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l) The rate of formation of the red-coloured triiodide ion D[I3-]/Dt (and therefore the reaction rate – why?) can be determined by measuring the rate of appearance of the colour. All media copyright of their respective owners 34 Problem Expt Initial [H2O2] (M) Initial [I-] (M) Initial D[I3-]/Dt (Ms-1) 1 0.100 0.100 1.15 x 10-4 2 0.100 0.200 2.30 x 10-4 3 0.200 0.100 2.30 x 10-4 4 0.200 0.200 4.60 x 10-4 a) What is the rate law for the formation of I3-? b) What is the value for the rate constant? c) What is the initial rate of formation of triiodide when the concentrations are [H2O2] = 0.300 M and [I-] = 0.400 M? All media copyright of their respective owners 35 Problem answers a) rate = k [H2O2] [I-] b) k = 1.15 x 10-2 M-1s-1 c) rate = 1.38 x 10-3 Ms-1 All media copyright of their respective owners 36 Zero order reactions In a zero order reaction the rate of the reaction does NOT depend on the concentration of ANY of the reactants. 2 NH3 (g) N 2 (g) 3 H 2 (g) 1130 K Pt catalyst 0 rate = k [NH3] = k (1) = k = constant All media copyright of their respective owners 37 Integrated rate law – 0th order rxn rat e k dA k dt dA k dt A t At A0 k t OR dA k dt A 0 A t k t A 0 t 0 y m x b A t A 0 k t - 0 All media copyright of their respective owners 38 Features of zero order reactions Concentration versus time graph gives a straight line. Integrated rate law A t k t A 0 m x y b Rate constant is the negative slope (UNITS!) All media copyright of their respective owners 39 First order reactions In a first order reaction the overall order of the reaction is 1. A common type of first order reaction is the decomposition of a chemical. 1 H 2 O 2 (aq) H 2 O (l) O 2 (g) 2 rate = k [H2O2] All media copyright of their respective owners 1 40 Integrated rate law – 1st order rxn rate kA dA kA dt dA k dt A A t dA k dt A 0 A 0 lnA t ln A 0 k t t A t ln k t - 0 A0 OR lnA t k t ln A 0 m x y All media copyright of their respective owners b 41 Integrated rate law – 1st order rxn Natural logarithm of concentration versus time graph gives a straight line. lnA t k t ln A 0 m x y b Rate constant is the negative slope (UNITS!) All media copyright of their respective owners 42 Problem The reaction A2B+C is first order. If the initial [A] = 2.80 M and k = 3.02 x 10-3 s-1, what is the value of [A] after 325 s? Answer: [A] = 1.05 M All media copyright of their respective owners 43 Half-life of 1st order reactions A t ln k t A0 1 A 0 ln 2 k t 1 2 A0 From the integrated rate law of a first order reaction we can look at what time [A] is onehalf of the initial concentration (½ [A]0). We call this time the half-life (t½). t1 2 ln 1 k t 1 2 2 ln 1 - 0.693 2 k k 0.693 t1 2 k All media copyright of their respective owners 44 Half-life of 1st order reactions A t ln k t A0 1 A 0 ln 2 k t 1 2 A0 The half-life (t½) of a first order reaction is constant! t1 2 ln 1 k t 1 2 2 ln 1 - 0.693 2 k k 0.693 t1 2 k All media copyright of their respective owners 45 Half-life of 1st order reactions [A]t = 1/2 [A]0 at t½ [A]t = 1/4 [A]0 at 2 x t½ [A]t = 1/8 [A]0 at 3 x t½ and so on All media copyright of their respective owners 46 Problem Consider the first order reaction A products with k = 2.95 x 10-3 s-1. What percent of A remains after 150 s? Answer: % A remaining = 64.2% All media copyright of their respective owners 47 Problem At what time (in minutes) after the start of the reaction is a sample of H2O2 (aq) twothirds decomposed if k = 7.30 x 10-4 s-1? Answer: time = 25.1 min All media copyright of their respective owners 48 1st order reactions of gases Amounts of gases are often measured by pressure instead of concentration molesA n A A volume V but P V nRT n P or V RT PA so A RT All media copyright of their respective owners 49 Problem Start with DTBP at a pressure of 800.0 mmHg at 147 C. What will the pressure be at t = 125 min if t½ = 8.0 x 10 min? ln A t A 0 k t PA RT t ln k t PA RT 0 P ln A t k t PA 0 Answer: k = 8.66 x 10-3 min-1 pressure = 271 mmHg All media copyright of their respective owners 50 Radioactive decay as 1st order rxns Many radioactive decay processes are first order… β - decay 14 14 half-life of 5730 yrs C N ν e 6 131 53 7 I Xe ν e half-life of 8.04 days 238 92 β - decay 131 54 U Th He α decay 234 90 4 2 2 half-life of 4.51 x 109 yrs All media copyright of their respective owners 51 Second order reactions In a second order reaction the overall order of the reaction is 2. If there is only a single reactant chemical, like in 2 NO2 (g) 2 NO (g) O2 (g) Then the rate is second order with respect to that chemical rate = k [NO2] All media copyright of their respective owners 2 52 Integrated rate law – 2nd order rxn rat e kA 2 dA 2 kA dt dA k dt 2 A A t dA 1 1 k t A t A 0 OR 1 1 k t A A mx t 0 1 1 k t - 0 A A y b A 2 A 0 t t k dt 0 0 All media copyright of their respective owners 53 Integrated rate law – 2nd order rxn Inverse of concentration versus time graph gives a straight line. 1 1 k t A t m x A 0 y b Rate constant is the slope (UNITS!) All media copyright of their respective owners 54 Half-life of 2nd order reactions From the integrated rate law of a second order reaction we can look at what time [A] is one-half of the initial concentration (½ [A]0). We also call this time the half-life (t½). 1 1 k t A t A 0 1 k t 1 1 A 0 A 0 2 2 1 1 k t 1 A 0 1 1 2 2 2 1 k t 1 A 0 1 2 1 t1 2 k A 0 All media copyright of their respective owners 55 Half-life of 2nd order reactions The half-life (t½) of a second order reaction is NOT constant! 1 1 k t A t A0 1 k t 1 1 A 0 A 0 2 2 1 1 k t 1 A 0 1 1 2 2 2 1 k t 1 A 0 1 2 1 t1 2 k A 0 All media copyright of their respective owners 56 Half-life of 2nd order reactions The greater the initial conc. of A ([A]0), the smaller the half-life! 1 1 k t A t A 0 1 k t 1 1 A 0 A 0 2 2 1 1 k t 1 A 0 1 2 1 2 2 1 k t 1 A 0 1 2 1 t1 2 k A 0 All media copyright of their respective owners 57 Half-life of 2nd order reactions 1 t1 2 k A 0 Since reactant concentration is halved over a half-life, the next half-life is twice as long compared to the previous half-life. All media copyright of their respective owners 58 Pseudo first-order reactions If we carry out some second (or higher) order reactions under conditions where some reactant concentrations do not change significantly, then the reaction may APPEAR to act like a lower order reaction (pseudo-first order, …) CH3COOC2H5 H2O CH3COOH C2H5OH All media copyright of their respective owners 59 Kinetics summary - rates •You can calculate rate with known rate law •If you don’t know a rate law, the rate can be determined from the tangent of a [A] versus time graph or by –D[A]/Dt over a small DT All media copyright of their respective owners 60 Kinetics summary – rxn orders •Method of initial rates •Plot graphs to find a straight line •First order reactions have constant half-life •Use data in integrated rate laws – k MUST NOT change All media copyright of their respective owners 61 Kinetics summary – rate constants •Rate constants are related to slope of appropriate straight line graphs •Use data in integrated rate laws to get k. Once you know k you can use the integrated rate law to solve for concentrations or times. •Use half-life of a first order reaction to determine k All media copyright of their respective owners 62 Problem Without graphing, find the order and the rate constant for the following reaction using the given data B products Time (s) [B] (M) Time (s) [B] (M) 0 0.88 100 0.44 25 0.74 150 0.31 50 0.62 200 0.22 75 0.52 250 0.16 All media copyright of their respective owners 63 Problem answer Reaction is first order. -3 -1 k = 6.93 x 10 s All media copyright of their respective owners 64 Bumper cars A gasp of surprise could be a “reaction” when riding in a bumper car “Reactions” generally occur when the bumps are very hard and when they come from behind All media copyright of their respective owners 65 Collision theory A + BC AB + C At some point in time, the B-C bond starts to break, while the A-B bond starts to form. At this point, all three nuclei are weakly linked together. All media copyright of their respective owners 66 Collision theory Molecules tend to repel each other when they get close. It takes energy to force the molecules close together! This is like forcing together the north poles of two magnets. This energy is the kinetic energy of the molecules. It converts to potential energy as the molecules get closer. All media copyright of their respective owners 67 Collision theory A---B---C has a higher potential energy than either A + B-C or A-B + C A---B---C is the transition state or the activated complex All media copyright of their respective owners 68 All media copyright of their respective owners 69 The difference in energy between products and reactants is DE The difference in energy between the transition state and the reactants is Ea – the activation energy All media copyright of their respective owners 70 Temperature is the average kinetic energy of molecules. Collisions between molecules at higher temperatures are more likely to have energy GREATER THAN the activation energy. Higher temperatures mean higher rates of reaction! All media copyright of their respective owners 71 Collisions 1 billion collisions per molecule per second Every reaction should be almost instantaneous. This is not the case. Not every collision breaks the activation energy barrier! All media copyright of their respective owners 72 Collisions The fraction of collisions that have enough energy to overcome the activation energy barrier is f = e-Ea/RT e is approximately 2.7183, Ea is the activation energy, T is the temperature in Kelvin, R is the gas law constant (8.3145 JK-1mol-1) All media copyright of their respective owners 73 All media copyright of their respective owners 74 Bumper cars and energy bumper car - a more energetic collision is more likely to make us gasp (our “reaction”) molecular collisions higher energy collisions are more likely to lead to reaction (by overcoming the activation energy) All media copyright of their respective owners 75 Bumper cars and orientation You are also more likely to gasp if you are hit from behind by another bumper car. The orientation of how the collision occurs is also important to get a “reaction.” The same is true for molecules where the fraction of collisions that have the right orientation is p. We call this fraction p the steric factor. All media copyright of their respective owners 76 General reaction A + BC AB + C Atom A MUST collide with the B side of BC to form the transition state A---B---C. If atom A hits the C side, we get a different transition state A---C---B. All media copyright of their respective owners 77 General reaction A + BC AB + C About half of our collisions won’t give the right reaction EVEN IF THEY BREAK THE ACTIVATION ENERGY BARRIER! The steric factor p will be about 0.5. All media copyright of their respective owners 78 General reaction A + BC AB + C Collision rate = Z [A] [BC] Z is a constant related to the collision frequency. Recall only a fraction (f) of the collisions have a collision energy greater than or equal to the activation energy. Of those collisions, only a fraction (p) have the correct orientation to proceed through the transition state to the products. All media copyright of their respective owners 79 General reaction A + BC AB + C Reaction rate = p x f x Collision rate Reaction rate = pfZ [A] [BC] If for our general reaction rate = k [A] [BC] k = pfZ = pZ e-Ea/RT = A e-Ea/RT (where A = pZ) All media copyright of their respective owners 80 Arrhenius Equation pZ = A As T increases k increases All media copyright of their respective owners 81 Problem AB + CD AC + BD What is the value of the activation energy for this reaction? Is the reaction endothermic or exothermic? Suggest a plausible structure for the transition state. All media copyright of their respective owners 82 Using the Arrhenius Equation If we know the rate constants for a reaction at two different temperatures, we can then calculate the activation energy. k = A e-Ea/RT ln k = ln (A e-Ea/RT) ln k = ln (A) + ln (e-Ea/RT) All media copyright of their respective owners 83 ln k = ln (A) – (Ea/RT) This is the equation for a straight line! All media copyright of their respective owners 84 slope = -Ea/R so Ea = - slope x R A graph of the natural logarithm of the rate constant versus inverse temperature will give a straight line All media copyright of their respective owners 85 If we know the rate constants (k1 and k2) at two temperatures (in Kelvin!) T1 and T2 we will see the rate constant has changed because the temperature has changed. All media copyright of their respective owners 86 One of the things that HAS NOT CHANGED with the temperature is the ACTIVATION ENERGY -Ea/R (at T1) = -Ea/R (at T2) It is constant just like slope is constant for a straight line! All media copyright of their respective owners 87 ln k2 = ln (A) – (Ea/RT2) ln k1 = ln (A) – (Ea/RT1) Subtract the bottom from the top on both sides ln k2 – ln k1 = [-(Ea/RT2)] – [-(Ea/RT1)] Pull out –Ea/R and put it in front ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1) All media copyright of their respective owners 88 Your textbook says ln (k2/k1) = (Ea/R) (1/T1 – 1/T2) This is absolutely correct as well! Use whichever form of the relation that you feel more comfortable with mathematically. There are more forms given in the text. All media copyright of their respective owners 89 Problem Rate constants for the decomposition of gaseous dinitrogen pentaoxide are 3.7 x 10-5 s-1 at 25 °C and 1.7 x 10-3 s-1 at 55 °C 2 N2O5 (g) 4 NO2 (g) + O2 (g) What is the activation energy of this reaction in kJmol-1? What is the rate constant at 35 C? All media copyright of their respective owners 90 Problem answer -1 kJmol Ea = 104 -4 -1 k35C = 1.4 x 10 s All media copyright of their respective owners 91 Reaction Mechanisms A reaction mechanism is the sequence of molecular events (elementary steps or elementary processes) that defines the pathway from the reactants to the products in the overall reaction. The elementary processes describe the behaviour of individual molecules while the overall reaction tells us stoichiometry. All media copyright of their respective owners 92 NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction) The reaction actually takes place in two elementary processes! 2 NO2 NO and NO3 NO3 + CO NO2 and CO2 All media copyright of their respective owners 93 NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction) Elementary processes must add together to give the overall equation! All media copyright of their respective owners 94 NO2 (g) + CO (g) NO (g) + CO2 (g) (Overall Reaction) Some of the “crossed-out” chemicals are neither reactants nor products in the overall reaction. For example, in the above reaction NO3 is formed in one elementary step and consumed in a later elementary step. All media copyright of their respective owners 95 Reaction intermediate A reaction intermediate is a species that is formed in an elementary process, that is consumed in a later elementary process. We never see reaction intermediates in the overall reaction! All media copyright of their respective owners 96 Molecularity The molecularity of an elementary process is the number of molecules on the reactant side of the elementary process reaction. A one molecule elementary reaction is unimolecular. A two molecule elementary reaction is bimolecular. A three molecule elementary reaction is termolecular. All media copyright of their respective owners 97 Molecularity All media copyright of their respective owners 98 Chances for molecularity The chances of a unimolecular reaction occurring only depends on the one molecule, and are good. A bimolecular reaction requires that two molecules collide with each other. This isn’t difficult and happens quite often. A termolecular reaction requires that three molecules collide with each other at the same time. The chances of this happening are not very good. All media copyright of their respective owners 99 Bumper cars Consider bumper cars. Very often, you will hit one other bumper car. It is a very rare occurrence to have a “bumper car” pile-up where three or more cars hit at exactly the same time. All media copyright of their respective owners 100 Problem A suggested mechanism for the reaction of nitrogen dioxide and molecular fluorine is shown a) Give the chemical equation for the overall reaction, and identify any reaction intermediates. b) What is the molecularity of each of the elementary processes? All media copyright of their respective owners 101 Problem answer 2 NO2 (g) + F2 (g) 2 NO2F (g) F is a reaction intermediate (formed in step 1 then consumed in step 2) Step 1 is bimolecular. Step 2 is bimolecular. All media copyright of their respective owners 102 Rate Laws and Reaction Mechanisms UNLIKE an overall reaction the rate law for an elementary process follows directly from the molecularity of the elementary reaction! For a general elementary process a A + b B products rate = k [A]a [B]b All media copyright of their respective owners 103 Ozone Unimolecular decomposition of ozone. O3 (g) O2 (g) + O (g) The rate law will respect to ozone be first order with rate = k [O3] All media copyright of their respective owners 104 Bimolecular reaction A + B products Reaction depends on collisions between molecules A and B Increase A, you increase # collisions Increase B, you increase # collisions rate = k [A] [B] All media copyright of their respective owners 105 All media copyright of their respective owners 106 Bimolecular reactions for one reactant A + A products Increase A, you increase # collisions for EACH A rate = k [A] [A] = k [A]2 All media copyright of their respective owners 107 Mechanisms and overall rate law The mechanism of the overall reaction is predicted through the elementary processes therefore the elementary processes will determine the rate law of the overall reaction! All media copyright of their respective owners 108 One step mechanisms If the overall reaction occurs in one elementary process, then the two reactions are the same. The rate law for the overall reaction is the same as the elementary reaction rate law. All media copyright of their respective owners 109 Single step mechanism rate = k [CH3Br] All media copyright of their respective owners [OH ] 110 Rate-determining step For reaction mechanisms of more than one elementary step, one of the elementary step reactions MAY HAVE a much slower rate than any of the other steps. All media copyright of their respective owners 111 Rate-determining step The rate-determining step of an overall reaction is an elementary step reaction which has a much slower rate than any other step. The overall rate law will match the rate law of the rate-determining step in this case. All media copyright of their respective owners 112 Proposing mechanisms A proposed mechanism must satisfy two criteria to be considered plausible: 1) the elementary processes must add up to give the appropriate overall reaction 2) the rate law that arises from the proposed mechanism must be consistent with the observed rate law All media copyright of their respective owners 113 H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g) Observed rate law rate = k [H2][ICl] SLOW : H 2 ICl HI HCl FAST : HI ICl I 2 HCl H 2 2 ICl I 2 2 HCl All media copyright of their respective owners 114 H2 (g) + 2 ICl (g) I2 (g) + 2 HCl (g) The elementary SLOW : H 2 ICl HI HCl reactions DO add up FAST : HI ICl I 2 HCl to the overall reaction. H 2 2 ICl I 2 2 HCl The rate law of the rate-determining step is rate = k [H2][ICl] It turns out this is the same as the experimentally observed rate law, so this mechanism is plausible. All media copyright of their respective owners 115 2 NO (g) + O2 (g) 2 NO2 (g) Observed rate = k [NO]2 [O2] The rate law is termolecular. The chances of a single slow termolecular step reaction occurring are very poor! We need a plausible mechanism where the slow step cannot be the first step. All media copyright of their respective owners 116 Fast reversible step – slow step A reversible reaction is one where reactants can become products and the products can become reactants AT THE SAME TIME. Eventually both reactions have the same rate (the system is at equilibrium) All media copyright of their respective owners 117 Fast reversible step – slow step FAST : 2 NO k1 k2 N 2O 2 k3 SLOW : N 2 O 2 O 2 2 NO2 k 2 NO O 2 2 NO2 Rate law of slow step rate3 = k3 [N2O2] [O2] BUT N2O2 is a reaction intermediate (concentration is effectively constant!) We can’t have it in the rate law! All media copyright of their respective owners 118 Fast reversible step – slow step k Rate law of fast FAST : 2 NO N 2O 2 forward step k k rate1 = k1 [NO]2 SLOW : N 2O 2 O 2 2 NO2 k Rate law of fast 2 NO O 2 2 NO2 reverse step rate2 = k2 [N2O2] 1 2 3 At equilibrium rate1 = rate2 All media copyright of their respective owners 119 Fast reversible step – slow step FAST : 2 NO k1 k2 N 2O2 k3 SLOW : N 2 O 2 O 2 2 NO 2 k3 2 NO O 2 2 NO 2 k1 [NO]2 = k2 [N2O2] so [N2O2] = k1/k2 [NO]2 Let’s say that k1/k2 = K All media copyright of their respective owners 120 Fast reversible step – slow step FAST : 2 NO k1 k2 N 2O 2 k3 SLOW : N 2 O 2 O 2 2 NO2 2 NO O 2 2 NO2 k Rate law of slow step rate3 = k3 [N2O2] [O2] We can substitute in [N2O2] from fast rxn! rate3 = k3 K [NO]2 [O2] If we let k = k3 K then we see the rate law of the slow reaction does match the observed rate law! All media copyright of their respective owners 121 Just because a mechanism is plausible doesn’t mean it is right! All media copyright of their respective owners 122 Problem In a proposed two-step mechanism for CO + NO2 CO2 + NO, the second, fast step is NO3 + CO NO2 + CO2. What must be the slow step? What would you expect the rate law of the rxn to be? All media copyright of their respective owners 123 Problem answer Slow step NO2 + NO2 NO3 + NO. Rate law: rate = k [NO2]2 All media copyright of their respective owners 124 Problem Show that the proposed mechanism is plausible for the reaction 2 NO2 (g) + F2 (g) 2 NO2F (g) if the rate law is rate = k [NO2][F2] FAST : NO2 (g) F2 (g) NO2 F2 (g) SLOW : FAST : NO2 F2 (g) NO2 F (g) F (g) F (g) NO2 NO2 F (g) All media copyright of their respective owners 125 Steady-state approximation More than one elementary process may determine the rate of an overall reaction. No rate-determining step! Can use the steady state approximation in this case. All media copyright of their respective owners 126 2 NO (g) + O2 (g) 2 NO2 (g) Observed rate = k [NO]2 [O2] We’ve already seen this as the fast reversible step – slow step mechanism example. Let’s propose a mechanism WITHOUT making any initial assumptions about the step rates! All media copyright of their respective owners 127 2 NO (g) + O2 (g) 2 NO2 (g) P rocess1 : P rocess2 : k1 NO NO N 2O2 N 2 O 2 NO NO k2 P rocess3 : N 2 O 2 O 2 2 NO 2 k3 This is essentially the same as the fast reversible step – slow step mechanism, but we treat the reversible reaction as two separate forward reactions. Each step has its own rate law. All media copyright of their respective owners 128 2 NO (g) + O2 (g) 2 NO2 (g) P rocess1 : P rocess2 : NO NO N 2 O 2 k1 N 2 O 2 NO NO k2 P rocess3 : N 2 O 2 O 2 2 NO 2 k3 We know a reaction intermediate ALWAYS has a near-zero concentration that does not change much with time (otherwise we would see it build up!) This means D[N2O2]/Dt 0 (steady state) All media copyright of their respective owners 129 2 NO (g) + O2 (g) 2 NO2 (g) P rocess1 : P rocess2 : NO NO N 2 O 2 k1 N 2 O 2 NO NO k2 P rocess3 : N 2 O 2 O 2 2 NO 2 k3 Choose our starting point as the step which has a rate law closest to the observed one, while including the reaction intermediate. rate = k3 [N2O2] [O2] All media copyright of their respective owners 130 2 NO (g) + O2 (g) 2 NO2 (g) D[N2O2]/Dt 0 rateof formationN 2 O 2 rateof disappearance N 2 O 2 0 rateof formationN 2 O 2 - rateof disappearance N 2 O 2 We form the intermediate in process 1 and consume it in processes 2 and 3 rateof formationN 2 O 2 rateprocess1 k1 NO 2 rateof disappearance N 2 O 2 - rateprocess2 rateprocess3 BE CAREFUL! These are two different negative signs! negative because we used positive reactionrates for disappearance! k 2 N 2 O 2 k 3 N 2O 2 O 2 All media copyright of their respective owners 131 2 NO (g) + O2 (g) 2 NO2 (g) rateof formationN 2 O 2 - rateof disappearance N 2 O 2 k1 NO - k 2 N 2 O 2 k 3 N 2 O 2 O 2 2 k1 NO k 2 N 2 O 2 k 3 N 2 O 2 O 2 2 Solve for [N2O2] k 1 NO N 2 O 2 k 2 k 3 O 2 2 k 1 NO N 2 O 2 k 2 k 3 O 2 2 All media copyright of their respective owners 132 2 NO (g) + O2 (g) 2 NO2 (g) Substitute into starting point rate k 3 N 2 O 2 O 2 k 1 NO O 2 rate k 3 k k O 3 2 2 k 1k 3 2 NO O 2 rate k 2 k 3 O 2 2 All media copyright of their respective owners 133 2 NO (g) + O2 (g) 2 NO2 (g) k1k 3 2 NO O 2 rate k 2 k 3 O 2 Very close to the observed rate law! How do we make the first part constant? Now we assume something about step rates! All media copyright of their respective owners 134 2 NO (g) + O2 (g) 2 NO2 (g) Assume rate2 >> rate3 so k2 [N2O2] >> k3 [N2O2] [O2] k2 >> k3 [O2] SO k2 + k3 [O2] k2 k1k 3 2 NO O2 rate k2 k All media copyright of their respective owners 135 Catalysis Reaction rates are not just affected by reactant concentrations and temperatures. A catalyst is a substance that increases the rate of a reaction WITHOUT undergoing permanent change in the reaction. All media copyright of their respective owners 136 How does a catalyst work? A catalyst makes available a different reaction pathway that is more efficient than the uncatalyzed mechanism because this pathway has a lower activation energy. All media copyright of their respective owners 137 All media copyright of their respective owners 138 Homogeneous catalysts A homogeneous catalyst exists in the same phase as the reactants. I- is a homogeneous catalyst for the decomposition of hydrogen peroxide All media copyright of their respective owners 139 Decomposition of H2O2 All media copyright of their respective owners 140 Decomposition of H2O2 Catalysts don’t appear in the overall balanced equation! All media copyright of their respective owners 141 Heterogeneous catalysts A heterogeneous catalyst exists in a different phase (usually solid) than the reactants. All media copyright of their respective owners 142 Heterogeneous catalysts All media copyright of their respective owners 143 Heterogeneous catalysts Most catalysts used in industry are heterogeneous It is much easier to separate a solid from a gas or liquid (for example) than two liquids or gases). All media copyright of their respective owners 144 Enzymes are catalysts In living beings catalysts are usually called enzymes Carbonic anhydrase catalyzes the reaction of carbon dioxide with water CO2 (g) + H2O (l) H+ (aq) + HCO3- (aq) The enzyme increases the rate of this reaction by a factor of 106. Equivalent to about a 200 K increase … All media copyright of their respective owners 145 Enzymes Lockand-key model All media copyright of their respective owners 146