Chapter 14 Chapter 14 by hedongchenchen

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									                      Chapter 14

                     Rates of Reaction



                                                        8–1




John A. Schreifels                       Chapter 14-1
Chemistry 212
                         Overview

• Reaction Rates
     – Definition of Reaction Rates
     – Experimental Determination of Rate
     – Dependence of Rate on Concentration
     – Change of Concentration with Time
     – Temperature and Rate; Collision and Transition-State
       Theories.
     – Arrhenius Equation
• Reaction Mechanisms
     – Elementary Reactions
     – Rate Law and the Mechanism
     – Catalysis                                                        8–2




John A. Schreifels                                       Chapter 14-2
Chemistry 212
                      Reaction Rates

• Deal with the speed of a reaction and controlled by:
     – Proportional to concentrations of reactants
     – Proportional to catalyst concentration; catalyst = a substance
       that increases the rate of reaction without being consumed in
       the reaction.
     – Larger surface area of catalyst means higher reaction rate
       (more sites for reaction to take place).
     – Temperature: Higher temperature of reaction means faster.




                                                                          8–3




John A. Schreifels                                         Chapter 14-3
Chemistry 212
                      Definition of Reaction Rate

• Reaction rate = increase in                                                    Concentration vs Reaction Time
  concentration of product of a                                                          A + 2B --> 3C
  reaction as a function of time or                                            0.090

  decrease in concentration of
  reaction as a function of time.




                                                            Concentration, M
                                                                                           Init               Inst.
                                                                                           Rate               Rate
• Thus the rate of a reaction is:                                              0.045
                                                                                                  Ave.
                                                                                                  Rate
                          conc change A [ A ]
           RateA                      
                           time change   t
                          [ A ]2  [ A ]1                                      0.000
                                                                                      0             250              500
                             t 2  t1                                                              Tim e, s


• Rates are expressed as positive numbers. For the reaction in
  the graph we have:
                     [ A ]                          [B]                                                [C]
        RA                                RB                                              RC  
                      t                              t                                                  t            8–4




John A. Schreifels                                                                                       Chapter 14-4
Chemistry 212
            Reaction Rates and Stoichiometry

• A + B  C; RC = RA = RB.
• A + 2B  3C; R  1 R  1 R
                      A       B       C
                          2       3
  E.g.Calculate the rate of decomposition of HI in the
  reaction: 2HI(g)  H2(g) + I2(g). Given: After a
  reaction time of 100 secs. the concentration of HI
  decreased by 0.500 M.
• For the general reaction: aA + bB  cC + dD
                          a    a     a
                     R A  RB  R C  RD
                          b    c     d
    E.g. For the reaction 2A + 3B  4C + 2D; determine
    the rates of B, C and D if the rate of consumption of A
    is 0.100 M/s.                                           8–5




John A. Schreifels                                  Chapter 14-5
Chemistry 212
             Rate Laws and Reaction Order

•   Rate Law – an equation that tells how the reaction rate depends on the
    concentration of each reaction.
•   Reaction order – the value of the exponents of concentration terms in
    the rate law.
•   For the reaction: aA + bB  cC + dD, the initial rate of reaction is
    related to the concentration of reactants.
•   Varying the initial concentration of one reactant at a time produces
    rates, which will lead to the order of each reactant.
•   The rate law describes this dependence: R = k[A]m[B]n where k = rate
    constant and m and n are the orders of A and B respectively.
     – m = 1 (A varied, B held constant) gives R = k’[A]. Rate is directly
       proportional to [A]. Doubling A doubles R
     – m = 2 (A varied, B held constant) gives R = k’[A]2. The rate is proportional
       to [A]2. Doubling A quadruples R.
    E.g. Determine order of each reactant:
    HCOOH(aq) + Br2(aq)  2H+(aq) + 2Br(aq) + CO2(g) R = k[Br2]
                                                                                        8–6
    E.g. The formation of HI gas has the following rate law: R = k[H2][I2].
    What is the order of each reactant?
John A. Schreifels                                                       Chapter 14-6
Chemistry 212
  Experimental Determination of a Rate Law: First
                     Order
• Varying initial concentration of reactants
                                                                         Concentration vs Reaction Time
  changes the initial rate (usually all but                                      A + 2B --> 3C
  one held constant) like one with two                                 0.100
  unknowns.
• Initial rate is the initial slope of the




                                                            [A] 0, M
                                                                       0.050

  graph shown.
• As the initial concentration of that                                 0.000
                                                                               0             250                 500
  compound increases so does the rate.                                                      Tim e, s


                                                                                   Initial Rate vs. [A]o
     – Initial rate vs. [A]o plotted.
     – If straight line then reaction is first order             0.0005
       and slope is rate constant.                               0.0004

• Second order rate law determined in                            0.0003




                                                       Ro
                                                                 0.0002
  like manner.                                                   0.0001
                                                                                                              8–7
                                                                 0.0000
                                                                       0.00        0.03    0.05        0.08   0.10
                                                                                            [A]o

John A. Schreifels                                                                        Chapter 14-7
Chemistry 212
                     Rate Law for All Reactants

• Order for all components done same way.
E.g. Determine the reaction order for each reactant from the table.
       (aq)+5Br(aq)+6H+(aq)3Br2(aq)+3H2O(l)
 BrO    3                       
                           [BrO3 ]o   [Br]o    [H+]o     Ro
                           0.10       0.10      0.10      1.2
                           0.20       0.10      0.10      2.4
                           0.10       0.30      0.10      3.5
                           0.20       0.10      0.15      5.4
Eg. 2: Determine the reaction orders for the reaction indicated from
  the data provided.
                       A + 2B + C  Products.
                           [A]o   [B]o   [C]o     Ro
                           2.06   3.05   4.00     3.7
                           0.87   3.05   4.00     0.66
                           0.50   0.50   0.50     0.013                        8–8
                           1.00   0.50   1.00     0.072

John A. Schreifels                                              Chapter 14-8
Chemistry 212
    Integrated Rate Law: First–Order Reaction
•    For a first order reaction, Rate = [A]/t = k[A] or RA = d[A]/dt = k[A].

                                      [A]                 [A]       k
•    Use of calculus leads to: ln           kt or log               t
                                    [ A ]o              [ A ]o    2.303
•    Allows one to calculate the [A] at any time after the start of the reaction.
     E.g. Calculate the concentration of N2O remaining after its decomposition
     according to 2N2O(g)  2N2(g) + O2(g) if it’s rate is first order and [N2O]o = 0.20M,
     k = 3.4 s1 and T = 780°C. Find its concentration after 100 ms.

     Linearized forms: ln[ A ]  kt  ln[ A ]o or log[ A ]  
                                                                  k
•                                                                     t  log[ A ]o
•    Plot ln[A] vs t.                                           2.303
•    Slope of straight line leads to rate constant, k.
     E.g. When cyclohexane(let's call it C) is heated to 500 oC, it changes into
     propene. Using the following data from one experiment, determine the first order
     rate constant.:
                       t,min      0.00 5.00 10.00 15.00
                       [C],mM 1.50 1.24 1.00               0.83                            8–9




John A. Schreifels                                                          Chapter 14-9
Chemistry 212
                Half-Life: First Order Reaction

• Half-life of First order reaction,
                                                      1 / 2[ A]o 
  t1/2 = 0.693/k. the time required for            ln                 k  t1/ 2
  the concentration of the reactant to                 [ A]o 
  change to ½ of its initial value.                            1 
                                                            ln     k  t1/ 2
  i.e. at t1/2 , [A] = ½ [A]o                                  2
  E.g. For the decomposition of N2O5                            t1/ 2  0.693 / k
  at 65 °C, the half-life was found to
  be 130 s. Determine the rate
  constant for this reaction.
                                                          1 
• For n half-lives t = n*t1/2 [A] =   2n   [A]o   n  ln   k  n  t1 / 2
                                                          2
                                                        1
                                                    ln   k  tn
                                                         2n                         8–10
                                                         1  [ A ]t
                                                         n   [A]
                                                        2        o
John A. Schreifels                                                    Chapter 14-10
Chemistry 212
    Second–Order Reactions: Integrated Rate
                    Law
• Rate law: R = k[A]2 and the integrated rate equation is:
                            1              1
                                  kt 
                          [ A ]t        [ A ]o
            1
• Plot of [ A ] vs. t gives a straight line with a slope of k.
               t
                             1
• Half-life is: t1/ 2  k  [ A ]
                                  o

E.g. At 330°C, the rate constant for the decomposition of
  NO2 is 0.775 L/(mol*s). If the reaction is second-order,
  what is the concentration of NO2 after 2.5x102 s if the
  starting of concentration was 0.050 M?                  8–11




John A. Schreifels                                   Chapter 14-11
Chemistry 212
                     Reaction Mechanisms
•   Give insight into sequence of reaction events leading to product
    (reaction mechanism).
•   Each of the steps leading to product is called an elementary
    reaction or elementary step.
•   Consider the reaction of nitrogen dioxide with carbon dioxide which
    is second order on NO2:
           NO2(g) + CO(g)  NO(g) + CO2(g) Rate = k[NO2]2.
•   Rate law suggests at least two steps.
•   A proposed mechanism for this reaction involves two steps.
               Step 1          2NO2(g)  NO3(g) + NO(g)
               Step 2    NO3(g) +CO(g) NO2(g) + CO2(g)
               Overall       NO2 + CO  NO + CO2
     – NO3 is a reaction intermediate = a substance that is produced and
       consumed in the reaction so that none is detected when the reaction is
       finished.
•   The elementary reactions are often described in terms of their
    molecularity.
     – Unimolecular One particle in elementary.                                   8–12
     – Bimolecular = 2 particles and
     – Termolecular = 3 particles
John A. Schreifels                                                    Chapter 14-12
Chemistry 212
      Rate Laws and Reaction Mechanisms
•   Overall reaction order is often determined by the rate determining step.
•   Use rate law of limiting step; No intermediates!
                 2NO2(g)  NO3(g) + NO(g),            R1 = k1[NO2]2             Slow
         NO3(g) +CO(g) NO2(g) + CO2(g)               R2 = k2[NO3][CO]          Fast
               NO2 + CO  NO + CO2                    Robs = k[NO2]2

E.g. Determine the rate law for the following mechanism:
                                        k1                               Fast
                        2*[N2O5(g)  NO2 (g)  NO3 (g) ]
                                        k 1
                                 k2
                 NO3(g) +NO2(g)   NO 2 (g)  NO(g)  O2 (g)           Slow
                                  
                                   k3
                         NO3 + NO   2NO 2 (g)                         Fast
                                    
                                    k obs
                          2N2O5(g)   4NO 2 (g)  O2 (g)
                                       
 Use steady state approximation. at “equilibrium” rates of forward and
  reverse reactions are same. Use to eliminate intermediates from rate law
  equations.
                     R1  R1                                  k 1 [N2O5 ]                    8–13
                                                    [NO3 ] 
           k 1 [N2O5 ]  k 1[NO 2 ][NO3 ]     or              k 1 [NO 2 ]


John A. Schreifels                                                                Chapter 14-13
Chemistry 212
    Reaction Rates and Temperature: The Arrhenius
                      Equation
•    Rate (rate constant) increases exponentially with temperature.
•    Collision theory indicates collisions every 109s – 1010s at 25°C and 1
     atm.
     i.e. only a small fraction of the colliding molecules actually react.
•    Collision theory assumes:
      – Reaction can only occur if collision takes place.
      – Colliding molecules must have correct orientation and energy.
      – Collision rate is directing proportional to the concentration of colliding
         particles.
      A + B  Products; Rc = Z[A][B]
      2A + B  Products; Rc = Z[A]2[B], etc.
•    Only a fraction of the molecules, p (“steric factor”), have correct
     orientation; multiply collision rate by p.
•    Particle must have enough energy. Fraction of those with correct
     energy follows Boltzmann equation f  e E a / RT where Ea = activation
     energy, R = gas constant and T = temp. (Kelvin scale only please).      8–14
•    This gives: k = Zpf

John A. Schreifels                                                          Chapter 14-14
Chemistry 212
                     Transition State Theory

• Explains the reaction resulting from the collision of molecules to
  form an activated complex.
• Activated complex is unstable and can break to form product.




         Exothermic Reaction              Endothermic Reaction         8–15




John A. Schreifels                                        Chapter 14-15
Chemistry 212
                         The Arrhenius Equation
•     Summary:                     E        where A = frequency factor.
                      k  A exp   a 
                                   RT 
•     Linear form: .                E                                        Arrhenius Plot
                       ln k  ln A  a                                    -5
                                    RT
•     Plot ln k vs. 1/T; the slope gives Ea/R.
                                                                          -6
      E.g. determine the activation energy for the decomposition of N2O5 from the




                                                                       ln k
      temperature dependence of the rate constant.
         k, s1
                                                                          -7
                                     Temp., °C               Temp., K
         4.8x104                 45.0                      318.15            -8
                                                                                   0.00300   0.00305     0.00310   0.00315

         8.8x104                 50.0                      323.15                              1/T, K
                                                                                                       




         1.6x103                 55.0                      328.15
         2.8x103                 60.0                      333.15
                                                  k    E 1        1 
•  Two point equation sometimes used also: ln 2  a                 
                                                  k1    R  T1 T2 
                                                                     
E.g.2: Determine the rate constant at 35°C for the hydrolysis of sucrose, given that
   at 37°C it is 0.91mL/(mol*sec). The activation energy of this reaction is 108
   kJ/mol.
• Rate constant increases when T2>T1                                                                          8–16




    John A. Schreifels                                                                 Chapter 14-16
    Chemistry 212
                              Catalysis
• Catalysts a substance that increases the rate of a reaction
  without being consumed in the reaction.
• Catalyst provides an alternative pathway from reactant to
  product which has a rate determining step that has a lower
  activation energy than that of the original pathway.
• E.g. Hydrogen peroxide and bromine:
                  2H2O2(aq)  2H2O(l)+ O2(g).
• Mechanism is believed to be :
  1. Br2 red                 Br2(aq) + H2O2(aq)  2Br(aq) +2H+(aq)+O2(g)
  2. Br oxid        2H+(aq)+2Br(aq)+H2O2(aq)  Br2(aq) + 2H2O(l).
          Overall                   2H2O2(aq)  2H2O(l)+O2(g)

• Notice that bromine is not consumed, even though it has
  participated in the reaction.
                                                                              8–17




John A. Schreifels                                                Chapter 14-17
Chemistry 212
     Homogeneous and Heterogeneous Catalysts

• Homogeneous catalyst: catalyst existing in the same
  phase as the reactants.
• Heterogeneous catalysis: catalyst existing in a
  different phase than the reactants.
     – The previous section gave an example of a homogeneous
       catalyst since the catalyst Br2 was in the same phase as
       the hydrogen peroxide.
• The catalytic hydrogenation of ethylene is an example
  of a heterogeneous catalysis reaction:
                                       Pt
                                         
                 H2C  CH2 (g)  H2 (g)  H3C  CH3(g)
• ENZYMES (biological catalysts)
     – They are proteins (large organic molecules that are
       composed of amino acids).
     – Slotlike active sites. The molecule fits into this slot and          8–18
       reaction proceeds. Poisons can block active site or reduce
       activity by distorting the active site.
John A. Schreifels                                              Chapter 14-18
Chemistry 212
                           “Steric Factor”

• Molecules must have the correct orientation before
  a reaction can take place.




               Figure 14.12 Importance of Molecular Orientation               8–19




John A. Schreifels             Return to p. 14-14                 Chapter 14-19
Chemistry 212

								
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