Chapter15 - Chemical Kinetics Chapter 15 We can use...

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Unformatted text preview: Chemical Kinetics Chapter 15 We can use thermodynamics to tell if a reaction is product - or reactant - favored. But this gives us no info on how fast reaction goes from reactants to products. Kinetics - the study of reaction rates and their relation to the mechanism. Reaction Mechanisms The sequence of events at the molecular level that control the speed and outcome of a reaction. Br from biomass burning destroys stratospheric ozone. Step 1: Step 2: Step 3: NET: Br + O3 ---> BrO + O2 Cl + O3 ---> ClO + O2 BrO + ClO + light ---> Br + Cl + O2 2 O3 ---> 3 O2 Reaction Rates Reaction rate = change in concentration of a reactant or product with time. We often determine: average rate, instantaneous rate, or initial rate Dye Conc Time Determining a Reaction Rate Factors Affecting Rates Concentrations Physical state of reactants and products Temperature Catalysts Concentrations & Rates Mg(s) + 2 HCl(aq) ---> MgCl2(aq) + H2(g) Physical State & Rate Catalysts & Rate CO2 in H2O Add dye Add NaOH 1. CO2(g) --> CO2 (aq) 2. CO2 (aq) + H2O(liq) --> H2CO3(aq) 3. H2CO3(aq) --> H+(aq) + HCO3(aq) Adding trace of NaOH uses up H+. Equilibrium shifts to produce more H2CO3. Enzyme speeds up reactions 1 and 2 Temperature Bleach at 54 C will decompose dyes at a much higher rate than Bleach at 22 C Iodine Clock Reaction 1. Iodide is oxidized to iodine H2O2 + 2 I- + 2 H+ -----> 2 H2O + I2 2. I2 reduced to I- with vitamin C I2 + C6H8O6 ----> C6H6O6 + 2 H+ + 2 I- When all vitamin C is depleted, the I2 interacts with starch to give a blue complex. Concentrations and Rates To postulate a reaction mechanism, we study reaction rate and its conc. dependence Rate of reaction is proportional to concentration We express this as a rate law Rate of reaction = k [A] where k is the rate constant k is independent of conc. but increases with T Concentrations, Rates, & Rate Laws In general, for a A + b B --> x X with a catalyst C Rate = k [A]m[B]n[C]p The exponents m, n, and p are the reaction order, can be 0, 1, 2 or fractions, can only be determined by experiment. Concentrations, Rates, & Rate Laws Rate = k [A]m[B]n[C]p If m = 1, rxn. is 1st order in A Rate = k [A]1 Doubles If [A] doubles, then rate ______ If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Quadruples Doubling [A] increases rate by ________ If m = 0, rxn. is zero order. Rate = k [A]0 Stays the Same If [A] doubles, rate _____________ Deriving Rate Laws Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of reduction of CH3CHO Expt. [CH3CHO] (mol/L) 0.10 0.20 0.30 0.40 Reduction of CH3CHO (mol/Lsec) 0.020 0.081 0.182 0.318 1 2 3 4 Deriving 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. 0.182 mol/Ls = k (0.30 mol/L)2 k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH3CHO] at same T. Integrated First-Order rate law [A] ln [A]o kt [A] / [A]0 =fraction remaining after time t. Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? ln (0.010M / 0.0010 M) = ln (0.100) = - 2.3 - 2.3 = - (0.21 hr-1) time time = 11 hours Using the Integrated Rate Law 2 N2O5(g) ---> 4 NO2(g) + O2(g) Rate = k [N2O5] Data of conc. vs. time plot do not fit straight line. Plot of ln [N2O5] vs. time is a straight line! Using the Integrated Rate Law Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b ln [N 2O5] = - kt + ln [N 2O5]o conc at time t rate const = slope conc at time = 0 All 1st order rxns have a line plot for ln [A] vs. t. 2nd order gives straight line for plot of 1/[A] vs. t Half-Life 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. Half-Life 1/2 remains 1/4 remains 1/8 remains 1/16 remains Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction? Half-Life 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 remaining, when t = t1/2 Therefore, ln (1/2) = - k t1/2 - 0.693 = - k t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 min Half-Life 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)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g Activation Energy Molecules need a minimum amount of energy to react. Visualized as an E barrier - activation energy, Ea. Reaction coordinate diagram MECHANISMS & Ea Mechanism: how reactants are converted to products at the molecular level. Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene] MECHANISMS Cis Transition state Trans Activation energy barrier Effect of Temperature Reactions generally occur slower at lower T. Room temperature In ice at 0 oC Iodine clock reaction H2O2 + 2 I- + 2 H+ --> 2 H2O + I2 Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. Differences in activation energy cause reactions to vary from fast to slow. Mechanisms 1.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. Arrhenius equation Temp (K) Rate constant k Ae -Ea /RT Frequency factor Activation 8.31 x 10-3 kJ/Kmol energy Frequency factor related to frequency of collisions with correct geometry. Ea 1 ln k = - ( )( ) + ln A R T Plot ln k vs. 1/T ---> straight line. slope = Ea/R Calculate the activation energy, Ea for N2O5(g) 2 NO2(g) + O2(g) k (at 25.0 C) = 3.46 10-5 M s-1 and k (at 45.0 C) = 5.79 10-4 M s-1. Need: lnk 2 ln k1 k2 ln k1 Ea 1 [ R T2 1 ] T1 5.79 10 M s ln 5 1 3.46 10 M s 42,051 J/mol 4 1 E a J /mol 1 [ 3.15J /mol K 318K 42 kJ/mol 1 ] 298K More on Mechanisms Isomerization of trans-butene --> cis-butene is unimolecular (only one reactant is involved) A bimolecular reaction bimolecular - two different molecules must collide to make products Exo- or endothermic? Collision Theory Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) ---> O2(g) + NO2(g) Activation energy Activation energy and geometry Mechanisms O3 + NO reaction occurs in a single elementary step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction. Mechanisms Most rxns involve a sequence of steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction. Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Proposed Mechanism Step 1 -- slow HOOH + I- --> HOI + OHStep 2 -- fast HOI + I- --> I2 + OHStep 3 -- fast 2 OH- + 2 H+ --> 2 H2O Rate of the reaction controlled by slow step Rate can be no faster than rate determining step Mechanisms 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Step 1 -- slow Step 2 -- fast Step 3 -- fast HOOH + I- --> HOI + OHHOI + I- --> I2 + OH2 OH- + 2 H+ --> 2 H2O Step 1 is bimolecular (I- and HOOH) This predicts the rate law should be dependant on both. The species HOI and OH- are rxn intermediates. Rate Laws and Mechanisms NO2 + CO reaction: Rate = k[NO2]2 Two possible mechanisms Two steps: step 1 Single step Two steps: step 2 Ozone Decomposition over Antarctica 2 O3 (g) ---> 3 O2 (g) Ozone Decomposition Mechanism 2 O3 (g) ---> 3 O2 (g) [O3 ]2 Rate = k [O2 ] Proposed mechanism Step 1: fast, equilibrium O3 (g) --> O2 (g) + O (g) Step 2: slow O3 (g) + O (g) ---> 2 O2 (g) CATALYSIS Catalysis and activation energy MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2 Uncatalyzed reaction Catalyzed reaction Isomerization of cis-2-Butene ...
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This note was uploaded on 04/07/2008 for the course CHEM 120 taught by Professor Tucker during the Spring '08 term at Siena College (Loudonville).

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