Chapter 15: Kinetics 11/14/12

Transcription

1 Chapter 5: Kinetics /4/ Ch. 5: Chemical Kinetics: The Rates of Chemical Reactions Thermodynamics will a process occur? Considers only initial and final states; don t need to know about pathway between reactants and products. Kinetics how long will that process take? Focuses on pathway between reactants and products. Reaction rate is the change in the concentration of a reactant or a product with time (M/s). By convention, reaction rates are expressed as positive values. B rate Δ rate Δ change in concentration of over time period Because decreases with time, Δ is negative. change in concentration of B over time period rate Δ To determine the rate, measure disappearance of OR appearance of B time B Br (aq) + HCOOH (aq) Br - (aq) + H + (aq) + CO (g) Br (aq) + HCOOH (aq) Br - (aq) + H + (aq) + CO (g) Rate Expression and Rate Constant The rate of a reaction depends on concentration, since reactions usually occur as a result of collisions between reactant molecules. Rate expression relates reaction rate to concentration: Measure reaction rate by monitoring loss of Br color Δ[Br ] α Δbsorption time 393 nm Br (aq) average rate Δ[Br ] slope of tangent [Br ] final [Br ] initial t final t initial instantaneous rate rate at time t tangent to curve at time t (rate decreases as Br is used up) initial rate instantaneous rate at t0. Easy to use initial rate to determine instantaneous reaction rate before initial concentrations of reactants change significantly. rate k [Br ] k rate constant slope Each reaction has its own characteristic rate constant, k. The rate constant is independent of the concentrations of the reactants but depends on the temperature.

2 Chapter 5: Kinetics /4/ Reaction Rates and Stoichiometry B Two moles of disappear for each mole of B that is formed. rate rate a Δ Δ a + bb b cc + dd c Δ[C] d Δ[D] Write the rate expression for the following reaction: CH 4 (g) + O (g) rate Δ[CH 4 ] CO (g) + H O (g) Δ[O ] Δ[CO ] Δ[H O] N O 5 (g) NO (g) + ½ O (g) -Δ[N Rate O 5 ] Δ[NO ] Δ[O ] 0.5 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. a + bb cc + dd Rate k x [B] y reaction is xth order in reaction is yth order in B reaction is (x +y)th order overall F (g) + ClO (g) rate k [F ] x [ClO ] y Method of initial rates: change initial concentrations and determine initial rate for each run Double [F ], keep [ClO ] constant Rate doubles, x Quadruple [ClO ] with [F ] constant Rate quadruples, y rate k [F ][ClO ] FClO (g) Rate Data for reaction between F and ClO [F ] (M) [ClO ] (M) Initial Rate (M/s) Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F (g) + ClO (g) FClO (g) rate k [F ][ClO ]

3 Chapter 5: Kinetics /4/ Determine the rate law and calculate the rate constant for the following reaction from the following data: S O - 8 (aq) + 3I - (aq) SO - 4 (aq) + I - 3 (aq) Exp [S O 8 - ] [I - ] Initial Rate (M/s) x x x 0-4 Double [I - ], rate doubles (experiment & ) rate k [S O - 8 ] x [I - ] y y x Double [S O 8 - ], rate doubles (experiment & 3) k rate k [S O 8 - ][I - ] rate [S O - 8 ][I - ]. x 0-4 M/s (0.08 M)(0.034 M) 0.08/M s First Order Reaction plot of vs time The characteristic shape of the graph is an exponential decay. The larger the rate constant, the faster the decay from the same initial concentration. If we know k and the initial concentration,, how can we determine how much is left at a given time? Determining Reaction Order Mathematically (when inspection isn t enough) For the process products, measure the rates for two different starting concentrations: rate k m and Dividing the second rate by the first, m rate [ ] rate [ ] m [ ] [ ] & $ % m #! " rate k m plug in values, solve for m ClO + OH - ClO ClO - + H O a. Find the rate expression and solve for k. b. When [ClO ] 0.5M and [OH ] 0.036M, what is the rate of the reaction? Exp [ClO ] [OH - ] Initial Rate (M s - ) rate Δ First-Order Reactions product k Units for k: ΔM Δs k M so k is in s - The integrated rate law lets us predict the concentration of a reactant remaining after a given time!!! d k dt Integrate both sides: rearrange to d k dt ln kt + constant Δ rate t 0 e -kt First-Order Reactions product k ln ln t ln 0 kt rearranged: ln kt + ln y mx + b slope -k a. Rate.0 0 M - s - [ClO ] [OH - ] at t 0, let ln t kt + ln ln 0 + constant; constantln Integrated st order rate law b M s - ln k t You will be given this equation in this format: ln t ln -kt In a first-order reaction, decays exponentially with time. To verify that a reaction is first order, plot ln vs time and expect a straight line; the slope of the straight line is k, the y-intercept is ln. 3

4 Chapter 5: Kinetics /4/ The reaction B is first order in with a rate constant of.8 x 0 - s - at 80 0 C. How long will it take for to decrease from 0.88 M to 0.4 M? k.8 x 0 - s - ln t ln 0 kt kt ln 0 ln t t ln 0 ln k ln 0 k M 0.4 M solve for t ln 0.88 M 0.4 M 66 s.8 x 0 - s - Half Life for First-Order Reactions The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ t when 0 / ln ln 0 - kt t ½ ln 0 0 / ln k k What is the half-life of N O 5 if it decomposes with a rate constant of 5.7 x 0-4 s -? ln t ½ s 0 minutes k 5.7 x 0-4 s - How do you know decomposition is first order? units of k (s - ) k / /4 First-order reaction product # of half-lives 0 /n 3 4 / /4 /8 /6 time Second-Order Reactions + product rate - Δ k rate M/s Units for k : k /M s L mol - s - M d Rearrange rate expression: -k dt Integrate both sides: - /(-+) -kt + constant or / kt - constant Since the original concentration was present at t 0, constant -/ kt + Integrated second-order rate law t ½ t when 0 / t ½ k 0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order Zero First Second Rate law: rate k rate k rate k Integrated rate law (conctime): -kt + ln -kt + ln / kt + / Linear plot: vs. t ln vs. t / vs. t Slope of straight line: Half life: t / /k Slope -k Slope -k Slope k t / ln/k 0.693/k t / /k Summary of plots that give straight lines for first- and second-order reactions 4

5 Chapter 5: Kinetics /4/ Collision Theory of Chemical Kinetics Molecules must collide to react threshold energy, the activation energy E a, must be overcome for a collision to produce a reaction E a is the energy needed to rearrange the bonds to form products Both the rate of a reaction and the effect of temperature on the rate depend on the size of E a Molecular orientation is important; not every collision will have molecules effectively aligned rrhenius equation describes factors affecting k: k e -E a /RT frequency factor prob. collision frequency and orientation Exp(-E a /RT) fraction of collisions with energy for rxt Temperature Dependence of the Rate Constant ll rate constants show an exponential increase with absolute temperature k e -Ea/RT E a is the activation energy (J/mol) R is the gas constant (8.34 J/K mol) T is the absolute temperature is the frequency factor lnk - (rrhenius equation) E a R T + ln + B Exothermic Reaction C + D Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. t higher temperature, a larger fraction of molecules have a kinetic energy > E a rrhenius plot: plot of lnk vs. /T is a straight line; E a can be determined from slope E a k e -Ea/RT take ln of lnk + ln both sides R T y m x + b ln Variation of rate constant, k, with temperature for different values of E a lnk E a R slope -E a /R T + ln slope -E a /R n rrhenius plot of ln k vs. /T is used to determine E a. large activation energy signifies a high sensitivity of k to changes in temperature. 5

6 Chapter 5: Kinetics /4/ Orientation of reacting molecules is important ICH 3 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. NO (g) + O (g) NO (g) N O is detected during the reaction -- predict mechanism Elementary step: NO + NO N O + Elementary step: N O + O NO Overall reaction: NO + O NO Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. n intermediate is always formed in an early elementary step and consumed in a later elementary step. Elementary step: NO + NO N O + Elementary step: N O + O NO Overall reaction: NO + O NO The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction elementary step with molecule Bimolecular reaction elementary step with molecules Termolecular reaction elementary step with 3 molecules Rate Laws and Elementary Steps Unimolecular reaction products rate k Bimolecular reaction + B products Rate k [B] Bimolecular reaction + products rate k Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation. Reaction rate is controlled by the rate-determining step (rds) reaction intermediate Path of reaction Whenever a fast step precedes a slow step, solve for the concentration of an intermediate by assuming an equilibrium is established in the fast step. Steps after the rds do not affect the rate or the rate law. The experimental rate law for the reaction between NO and CO to produce NO and CO is rate k[no ]. The reaction is believed to occur via two steps: Step : NO + NO NO + NO 3 Step : NO 3 + CO NO + CO What is the equation for the overall reaction? NO + CO NO + CO What is the intermediate? NO 3 What can you say about the relative rates of steps and? rate k[no ] is the rate law for step so step must be slower than step 6

7 Chapter 5: Kinetics /4/ Elimination of Intermediates in Rate Expression Reaction between nitric oxide and chlorine proceeds through a reactive intermediate, NOCl. Find the rate expression for the reaction, which includes only those species in the balanced equation. k Step : NO + Cl k - NOCl (fast) k Step : NOCl + NO NOCl (slow) NO + Cl NOCl Overall rate rate of Step k [NOCl ][NO] To eliminate [NOCl ], assume Step is fast equilibrium k [NO][Cl ] k - [NOCl ] Solve for [NOCl ] and substitute in rate eq: [NOCl ] Overall rate k [NOCl ][NO] k k [NO] [Cl ] k - Reaction is second-order in NO and first-order in Cl, as predicted by mechanism. k [NO][Cl ] k - catalyst is a substance that increases the rate of a chemical reaction without itself being consumed k e -Ea/RT When E a is lowered, the reaction rate increases. Note that the catalyst lowers E a for both forward and reverse reactions. 7