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Reaction Rates: Chemical Kinetics Chemical Kinetics Reaction Rates: Reaction Rate: The change in the concentration of a reactant or a product with time (M/s). Reactant → Products A → B change in number of moles of B Average rate = change in time ∆()moles of B ∆[B] = = ∆t ∆t ∆[A] Since reactants go away with time: Rate=− ∆t 1 Consider the decomposition of N2O5 to give NO2 and O2: 2N2O5(g)→ 4NO2(g) + O2(g) reactants products decrease with increase with time time 2 From the graph looking at t = 300 to 400 s 0.0009M −61− Rate O2 ==× 9 10 Ms Why do they differ? 100s 0.0037M Rate NO ==× 3.7 10−51 Ms− Recall: 2 100s 0.0019M −51− 2N O (g)→ 4NO (g) + O (g) Rate N O ==× 1.9 10 Ms 2 5 2 2 25 100s To compare the rates one must account for the stoichiometry. 1 Rate O =×× 9 10−−61 Ms =× 9 10 −− 61 Ms 2 1 1 −51−−− 61 Rate NO2 =×× 3.7 10 Ms =× 9.2 10 Ms Now they 4 1 agree! Rate N O =×× 1.9 10−51 Ms−−− = 9.5 × 10 61Ms 25 2 Reaction Rate and Stoichiometry In general for the reaction: aA + bB → cC + dD 11∆∆∆∆[AB] [ ] 11[CD] [ ] Rate ====− − ab∆∆∆ttcdtt∆ 3 Rate Law & Reaction Order The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants. Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms−1). (It also has deeper significance, which will be discussed later) For the general reaction: aA+ bB → cC+ dD x and y are the reactant orders determined from experiment. Rate = k [A]xy [B] x and y are NOT the stoichiometric coefficients. 4 The Overall Order of a reaction is the sum of the individual orders: Rate (Ms−1) = k[A][B]1/2[C]2 Overall order: 1 + ½ + 2 = 3.5 = 7/2 or seven−halves order note: when the order of a reaction is 1 (first order) no exponent is written. Units for the rate constant: The units of a rate constant will change depending upon the overall order. The units of rate are always M/s or Ms–1 To find the units of a rate constant for a particular rate law, simply divide the units of rate by the units of molarity in the concentration term of the rate law. Rate (Ms–1) = k[A] 1st order Ms−1 k(units)= = s−1 M 5 Reaction Rates Instantaneous rate (tangent line) ∆Y∆ [concentration] Average Rate = ∆∆Xtime concentration (M) time (s) Rules of logarithms log (1) = 0 ln (1) = 0 log (10) = 1 ln (e) = 1 log (100) = 2 ln (ex) = x Log (10x) = x log Ax = xlog A ln Ax = xln A x AAx A logx == log x log BB B A log(AB) = log A + log B log=− log A log B B 6 Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280°C is: 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) From the following data, determine the rate law and rate constant. Initial Rate Run [NO] (M) [H ] (M) o 2 o (M/min) 1 0.0100 0.0100 0.00600 2 0.0200 0.0300 0.144 3 0.0100 0.0200 0.0120 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) The rate law for the reaction is given by: x y Rate(M/min) = k [NO] [H2] Taking the ratio of the rates of runs 3 and 1 one finds: xy Rate(3) k [NO](3) [H 2 ] (3) = xy Rate(1) k [NO](1) [H 2 ] (1) 0.0120M / min k [0.0100]xy [0.0200] [0.0200]y = 2 = = 0.00600M / min k [0.0100]xy [0.0100] [0.0100]y y [0.0200]y 0.0200 y = [0.0100] 0.0100 7 xy y Rate(1) k [NO](1) [H 2 ] (1) 0.0200 = log = 2 Rate(2) k [NO]xy [H ] 0.0100 (2) 2 (2) x y 0.00600 k [0.0100] [0.0100] 0.0200 = x log= log 2 0.144 k [0.0200] [0.0300] 0.0100 x 1 11 = × ylog( 2) = log 2 24 23 x 11 y = 1 = 28 Now that “y” is known, we 11 xlog= log may solve for x in a similar 28 manner: x = 3 The Rate Law is: 3 Rate(M/min) = k [NO] [H2] To find the rate constant, choose one set of data and solve: M 3 0.0120= k()() 0.0100M 0.0200M min M M 0.0120 0.0120 min min k = = 3 ()()0.0100M3 0.0200M ()()0.0100 0.0200 M4 M−3 k6.0010=×5 min 8 Integrated Rate Laws: time dependence of concentration For a first order process, the A → products rate law can be written: ∆[A] Rate(Ms−1 )=− = k[A] This is the “average rate” ∆t If one considers the infinitesimal changes in concentration and time the rate law equation becomes: d[A] Rate(Ms−1 )=− = k[A] This is the “instantaneous rate” dt Integrating in terms of d[A] and dt: [A]d[A] t where [A] = [A] at time t = 0 =−kdt o ∫∫[A]o [A] 0 and [A] = [A] at time t = t [A]d[A] t [A] =−kdt ln= − kt ∫∫[A]o [A] 0 [A]o Taking the exponent to each side of the equation: [A] = e−kt or −kt [A]= [A]o e [A]o Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time. 9 Recognizing a first order process: AÆproducts Whenever the conc. of a reactant falls off exponentially, the kinetics follow first order. −kt [A]= [A]o e Determining the Rate constant for a first order process Taking the log of the integrated rate law for a first order process we find: −kt ln() [A]= [A]o e ln[A]= ln[A]o −× k t A plot of ln[A] versus time (t) is a straight line with slope -k and intercept ln[A]o 10 Example: The rate of decomposition of azomethane (C2H6N2) was studied by monitoring the partial pressure of the reactant as a function of time. Determine if the data below support a first order reaction. Calculate the rate constant for the reaction. Time P ln (P) (s) (mmHg Plot of lnP vs. time 0 284 5.65 5.70 100 220 5.39 5.60 y = -0.0026x + 5.6485 150 193 5.26 5.50 R2 = 1 200 170 5.14 5.40 250 150 5.01 5.30 ln P 5.20 5.10 5.00 -3 -1 k = 2.6x10 s 4.90 0 50 100 150 200 250 300 time (s) Reaction Half-Life 11 Reaction Half-life: Half-life is the time taken for the concentration of a reactant to drop to half its original value. [A] [A] = o 2 For a first order process the half life (t½ ) is found mathematically from: (1) ln A=− kt + ln A [ ] [ ]0 []A (5) ln0 = kt (2) ln[ A] −=− ln[ A] kt [A] 1 0 0 2 2 [A] (3) ln=− kt [A] 0 ln 2 0.693 t == [A] 1 kk (4) ln0 = kt 2 [A] A certain reaction proceeds through t first order kinetics. The half-life of the reaction is 180 s. What percent of the initial concentration remains after 900s? Step 1: Determine the magnitude of the rate constant, k. ln 2 ln 2 ln 2 0.693 k== = 0.00385s−1 t 1 == t1 180s 2 kk 2 Using the integrated rate law, substituting in the value of k and 900s we find: [A] −kt [A] −×0.00385 s−1 900 s = e = e = 0.0312 [A]o [A]o Since the ratio of [A] to [A]0 represents the fraction of [A]o that remains, the % is given by: 100 × 0.0312 = 3.12% 12 For a Second Order Process: A → Products Rate = k[A]2 d[A] Rate(Ms−12 )=− = k[A] dt Integrating as before we find: 11 =+kt []AA[]0 A plot of 1/[A] versus t is a straight line with slope k and intercept 1/[A]0 For a second order reaction, a plot of ln[A] vs. t is not linear. 11 =+kt []AA[]0 Non−linearity indicates that the reaction is not first order. Slope = k (rate constant) 13 Half-life for a second-order reaction Unlike a first order reaction, the rate constant for a second order process depends on and the initial concentration of a reactant. 11 =+kt AA []t0[] 1 at the half–life, []AA=×[] t02 1 Substituting and solving, t1/2 = kA []0 EXAMPLE: The reaction 2 NOBr (g) → 2 NO (g) + Br2 (g) is a second order reaction with respect to NOBr. k = 0.810 M-1⋅s-1 at 10o C. -3 If [NOBr]o = 7.5 × 10 M, how much NOBr will be left after a reaction time of 10 minutes? Determine the half-life of this reaction. 14 SOLUTION: One can solve for the amount of NOBr after 10 minutes by substituting the given data into the integrated rate law for a second-order reaction. 11 =+kt NOBr NOBr (Second Order) []t0[] 1 1 = (0.810 M-1⋅ s -1 )× (600 s) + NOBr −3 []t 7.5× 10 M 1 =×6.19 102-1 M NOBr []t NOBr= 1.6× 10−3 M [ ]t To determine the half-life for this reaction, we substitute the initial concentration of NOBr and the rate constant for the reaction into the equation for the half-life of a second-order reaction.
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