Presentation on theme: "Chemical Kinetics Chapter 15"— Presentation transcript:
1Chemical Kinetics Chapter 15 H2O2 decomposition in an insectH2O2 decomposition catalyzed by MnO2
2Chemical KineticsWe can use thermodynamics to tell if a reaction is product- or reactant- favored.But this gives us no info on HOW FAST a reaction goes from reactants to products.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., the Reaction MECHANISM.
3Reaction MechanismsThe sequence of events at the molecular level that control the speed and outcome of a reaction.Br from biomass burning destroys stratospheric ozone.(See R.J. Cicerone, Science, volume 263, page 1243, 1994.)Step 1: Br + O3 ---> BrO + O2Step 2: Cl + O > ClO O2Step 3: BrO + ClO + light ---> Br + Cl + O2NET: 2 O > 3 O2
4Reaction RatesSection 15.1Reaction rate = change in concentration of a reactant or product with time.Three “types” of ratesinitial rateaverage rateinstantaneous rate
5Determining a Reaction Rate Blue dye is oxidized with bleach.Its concentration decreases with time.The rate — the change in dye concentration with time — can be determined from the plot.Dye ConcScreen 15.2Time
13Concentrations and Rates To postulate a reaction mechanism, we study• reaction rate and• its concentration dependence
14Concentrations and Rates Take reaction where Cl- in cisplatin [Pt(NH3)2Cl3] is replaced by H2O
15Concentrations & Rates Rate of reaction is proportional to [Pt(NH3)2Cl2]We express this as a RATE LAWRate of reaction = k [Pt(NH3)2Cl2]where k = rate constantk is independent of conc. but increases with T
16Concentrations, Rates, & Rate Laws In general, fora A + b B --> x X with a catalyst CRate = k [A]m[B]n[C]pThe exponents m, n, and p• are the reaction order• can be 0, 1, 2 or fractions• must be determined by experiment!
17Interpreting Rate Laws Rate = k [A]m[B]n[C]pIf m = 1, rxn. is 1st order in ARate = k [A]1If [A] doubles, then rate goes up by factor of 2If m = 2, rxn. is 2nd order in A.Rate = k [A]2Doubling [A] increases rate by 4If m = 0, rxn. is zero order.Rate = k [A]0If [A] doubles, rate is not changed
18Deriving Rate LawsDerive rate law and k forCH3CHO(g) --> CH4(g) + CO(g)from experimental data for rate of disappearance of CH3CHOExpt. [CH3CHO] Disappear of CH3CHO (mol/L) (mol/L•sec)
19Deriving Rate Laws Rate of rxn = k [CH3CHO]2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order.Now determine the value of k. Use expt. #3 data—0.182 mol/L•s = k (0.30 mol/L)2k = 2.0 (L / mol•s)Using k you can calc. rate at other values of [CH3CHO] at same T.
20Concentration/Time Relations What is concentration of reactant as function of time?Consider FIRST ORDER REACTIONSThe rate law isRate = - (∆ [A] / ∆ time) = k [A]
21Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get[A] / [A]0 =fraction remaining after time t has elapsed.Called the integrated first-order rate law.
22Concentration/Time Relations Sucrose decomposes to simpler sugarsRate of disappearance of sucrose = k [sucrose]If k = 0.21 hr-1and [sucrose] = MHow long to drop 90% (to M)?Glucose
23Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = M, how long to drop 90% or to M?Use the first order integrated rate lawln (0.100) = = - (0.21 hr-1) • timetime = 11 hours
24Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment.2 N2O5(g) ---> 4 NO2(g) + O2(g)Time (min) [N2O5]0 (M) ln [N2O5]0Rate = k [N2O5]
25Using the Integrated Rate Law 2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5]Plot of ln [N2O5] vs. time is a straight line!Data of conc. vs. time plot do not fit straight line.
26Using the Integrated Rate Law Plot of ln [N2O5] vs. time is a straight line!Eqn. for straight line:y = mx + bAll 1st order reactions have straight line plot for ln [A] vs. time.(2nd order gives straight line for plot of 1/[A] vs. time)
28Half-Life Section 15.4 & Screen 15.8 HALF-LIFE is the time it takes for 1/2 a sample is disappear.For 1st order reactions, the concept of HALF-LIFE is especially useful.Active Figure 15.9
29Half-Life Section 15.4 & Screen 15.8 Sugar is fermented in a 1st order process (using an enzyme as a catalyst).sugar + enzyme --> productsRate of disappear of sugar = k[sugar]k = 3.3 x 10-4 sec-1What is the half-life of this reaction?
30Half-Life Section 15.4 & Screen 15.8 Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half- life of this reaction?Solution[A] / [A]0 = fraction remainingwhen t = t1/2 then fraction remaining = _________Therefore, ln (1/2) = - k • t1/2= - k • t1/2t1/2 = / kSo, for sugar,t1/2 = / k = sec = 35 min
31Half-Life Section 15.4 & Screen 15.8 Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)?Solution2 hr and 20 min = 4 half-livesHalf-life Time Elapsed Mass Left1st 35 min g2nd g3rd g4th g
32Half-Life Section 15.4 & Screen 15.8 Radioactive decay is a first order process.Tritium ---> electron helium3H e 3Het1/2 = yearsIf you have 1.50 mg of tritium, how much is left after 49.2 years?
33Half-Life Section 15.4 & Screen 15.8 Start with 1.50 mg of tritium, how much is left after years? t1/2 = yearsSolutionln [A] / [A]0 = -kt[A] = ? [A]0 = mg t = yNeed k, so we calc k from: k = / t1/2Obtain k = y-1Now ln [A] / [A]0 = -kt = - ( y-1) • (49.2 y)=Take antilog: [A] / [A]0 = e-2.77 == fraction remaining
34Half-Life Section 15.4 & Screen 15.8 Start with 1.50 mg of tritium, how much is left after years? t1/2 = yearsSolution[A] / [A]0 =is the fraction remaining!Because [A]0 = 1.50 mg, [A] = mgBut notice that 49.2 y = 4.00 half-lives1.50 mg ---> mg after 1 half-life---> mg after 2---> mg after 3---> mg after 4
35MECHANISMS A Microscopic View of Reactions Sections 15.5 and 15.6 Mechanism: how reactants are converted to products at the molecular level.RATE LAW > MECHANISMexperiment > theory
36Reaction coordinate diagram Activation EnergyMolecules need a minimum amount of energy to react.Visualized as an energy barrier - activation energy, Ea.Reaction coordinate diagram
37MECHANISMS & Activation Energy Conversion of cis to trans-2-butene requires twisting around the C=C bond.Rate = k [trans-2-butene]
38ACTIVATION ENERGY, Ea = energy req’d to form activated complex. MechanismsReaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product.ACTIVATION ENERGY, Ea = energy req’d to form activated complex.Here Ea = 262 kJ/mol
39MECHANISMSAlso note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol.Therefore, cis ---> trans is EXOTHERMICThis is the connection between thermo- dynamics and kinetics.
40Effect of Temperature Reactions generally occur slower at lower T. In ice at 0 oCReactions generally occur slower at lower T.Room temperatureIodine clock reaction, Screen 15.11, and book page 705.H2O2 + 2 I- + 2 H+ --> 2 H2O + I2
41Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules.In general, differences in activation energy cause reactions to vary from fast to slow.
42Mechanisms1. Why is trans-butene <--> cis-butene reaction observed to be 1st order?As [trans] doubles, number of molecules with enough E also doubles.2. Why is the trans <--> cis reaction faster at higher temperature?Fraction of molecules with sufficient activation energy increases with T.
43More on Mechanisms Exo- or endothermic? A bimolecular reaction Reaction oftrans-butene --> cis-butene is UNIMOLECULAR - only one reactant is involved.BIMOLECULAR — two different molecules must collide --> productsExo- or endothermic?
44Collision Theory 1. Activation energy 2. Correct geometry Reactions require(a) activation energy and(b) correct geometry.O3(g) + NO(g) ---> O2(g) + NO2(g)1. Activation energy2. Correct geometry
45MechanismsO3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps.Adding elementary steps gives NET reaction.
46Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O H+ ---> I H2ORate = k [I-] [H2O2]NOTE1. Rate law comes from experiment2. Order and stoichiometric coefficients not necessarily the same!3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction.
47Most rxns. involve a sequence of elementary steps. MechanismsMost rxns. involve a sequence of elementary steps.2 I- + H2O H+ ---> I H2ORate = k [I-] [H2O2]Proposed MechanismStep 1 — slow HOOH + I- --> HOI + OH-Step 2 — fast HOI + I- --> I2 + OH-Step 3 — fast 2 OH H+ --> 2 H2ORate of the reaction controlled by slow step —RATE DETERMINING STEP, rds.Rate can be no faster than rds!
48Mechanisms2 I- + H2O H+ ---> I H2ORate = k [I-] [H2O2]Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH-Step 3 — fast 2 OH H+ --> 2 H2OElementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should beRate = k [I-] [H2O2] — as observed!!The species HOI and OH- are reaction intermediates.