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							CHE 112 - MODULE 3




     CHAPTER 15
   LECTURE NOTES
             Chemical Kinetics
   Chemical kinetics - study of the rates of
    chemical reactions and is dependent on the
    characteristics of the reactants

   Reaction mechanisms - detailed pathway that
    atoms and molecules take as a chemical
    reaction proceeds

   CD-ROM Screen 6.2 & 15.2
     Effects on Chemical Kinetics
   Concentration of reactants – usually the rate
    of a reaction increases with increased
    concentration of reactants
   Concentration of a catalyst - catalyst speeds
    up the reaction
   Temperature – rate will increase with an
    increase in temperature (increased KE)
   Surface area – as the surface area increases,
    the reaction will proceed at a faster rate
     Rates of Chemical Reactions
   Rates - the change in concentration of a
    chemical per unit of time ex. (M/sec)
    – much like the speed of a car ex. (miles/hr)
    – much like interest rates ex. (5.4%/year)
    – much like sales taxes ex. (7cents/$)


   Rates are just a ratio or fraction of one
    thing changing with respect to another
       Decomposition of N2O5
           2 N2O5  4 NO2 + O2
   We look at the disappearance of N2O5

   Rate of Reaction = change in [N2O5]
                        change in time

   Rate of Reaction =  [N2O5]
                          t
          Stoichiometry

         2 N2O5  4 NO2 + O2
         2moles : 4moles : 1mole
Therefore, for every 2 moles of N2O5
decomposed, you have 1 mole of O2
formed. The rate of formation of O2 is
equal to ½ the rate of decomposition of
N 2O 5
       [O2]/t = -½ [N2O5]/ t
         Decomposition of N2O5

   Therefore we can determine the rate of
    decomposition by the rate of formation of
    either NO2 or the O2 with consideration to
    stoichiometry.
               2 N2O5  4 NO2 + O2

Rf =  [NO2] = -2Rd      Rf =  [O2] = -½ Rd
        t                     t
                Calculations
                2 N2O5  4 NO2 + O2
   Where ti = 600s and tf = 1200s and the
    concentration of N2O5 decomposed from 1.24
    x 10-2 M to 0.93 x 10-2 M
   Rate of decomposition = [N2O5 ] /t
        = (0.93 x 10-2 M -1.24 x 10-2 M)
                (1200s – 600s)
        = -5.2 x 10-5 M/s
   Rate of formation O2 = -½ [-5.2 x 10-5 M/s]
        = 2.6 x 10-5 M/s
    Plot of Concentration vs. Time
   Figure 15.2 shows the graph of the
    disappearance of N2O5.

   Average rate = change in concentration
    over an interval in time (c/t)

   Instantaneous rate = concentration at
    an instant in time; tangent to the curve
    at a particular point in time (dc/dt)
                  Rate Law

   Rate Law – equation that relates the
    rate of reaction to the concentration of
    reactants raised to various powers

            Rate = k [N2O5]x

    – where k is the rate constant and x is the
      order of the reaction
       Determining Rate Law
 We can perform an experiment to decompose
 N2O5. If we doubled the concentration of the
 reactant we can observe the change in rate
 as follows:

      Initial Conc.   Rate of decomposition
Exp.1 1x10-2 mol/l     4.8x10-6 mol/l•sec
Exp.2 2x10-2 mol/l     9.6x10-6 mol/l•sec

Observation: The rate of reaction doubled.
            Determining Order
   When a reaction has this observable result, it
    is said to be first order.

   As we double the concentration, the rate is 2x,
    where x is the classification of the order of
    that particular reaction.

   Or R2/R1 = 2x, where x=1, then we can see
    that it is clearly first order.
   Overall Rate Law for N2O5

We previously stated that:


    Rate = k [N2O5]x

Where x = 1 determined experimentally,
therefore this reaction is a first order reaction
and follows the rules of first order kinetics.
          Order of Reactions

  Considering other reactions where we double
  the initial reactant concentration, we can
  observe the rate and order as follows:
Quadruple rate = 4 (2x=4, x=2) second order
Double rate       = 2 (2x=2, x=1) first order
No change         = 1 (2x=1, x=0) zero order
Half rate       = 1/2 (2x=1/2, x=-1) -1 order
1.4 the rate = 1.4 (2x=1.4, x=1/2) half order
          Review of Logarithms
   Logs are the inverse function of an
    exponential function (y = 3x)
   Algebraically it is written as x = 3y or y =
    log3x where y is the exponent on 3 that
    results in a value of x
    – ex. 52 = 25 or log5 25 = 2
   Written as a function logs are expressed as
    f(x) = logbx
   If b is not designated, b=10 is assumed.
   If b = e it is called the natural log (ln).
                First Order Kinetics

                2 N2O5  4 NO2 + O2
   Rate of decomposition = - [N2O5]
                                t
   Rate = kt

   Therefore - [N2O5] = -kt (by substitution)
                   t
   ln [N2O5]t = -kt + ln [N2O5]0 (derive by integration)
      Integrated Rate Equations
   For a zero order reaction:
       [A]t = -kt + [A]0

   For a first order reaction:
       ln [A]t = -kt + ln [A]0

   For a second order reaction:
       1/[A]t = kt + 1/[A]0
   Iodine Clock Reaction Prep

  Each solution is prepared in a 100ml
  volumetric flask and qs with distilled
  water. Accurately weigh the following:
2.998g of NaI = 0.2M
1.169g of NaCl = 0.2M
0.2482g of sodium thiosulfate = 0.01M
3.485g potassium sulfate = 0.2M
5.404g potassium persulfate = 0.2M
           Half-life of Reaction

   Half-life (t½) = the time it takes for the
    initial concentration of a reaction to
    disintegrate to half its original
    concentration

   First order reaction:   t½ = 0.693/k
             Collision Theory

   Three conditions must be met for a
    reaction to take place:
    – The reacting molecules must collide with
      one another
    – Molecules must collide with sufficient
      energy
    – Molecules must collide with the proper
      orientation
           Activation Energy

   Ea = the minimum amount of energy
    that must be absorbed by a system to
    cause it to react where k = Ae -Ea/RT
   Activation energy can be determined
    from the Arrhenius Equation:
           ln k = ln A - Ea/R (1/T)


     ln (k 2/k1) = - Ea/R [1/T2 - 1/T1]
      Conditions Increasing Rates
   Temperature - increases the KE and
    enough energy to overcome Ea

   Presence of a catalyst - provides
    different pathways with lower Ea

   Surface area - increases the probability
    of a collision with proper orientation
         Reaction Mechanisms

   Reaction mechanisms = sequence of
    steps that show the intermediates
    formed ( bonds broken and bonds
    formed) between the reactant side and
    product side of any reaction
   See CD-ROM Screens 15.12 – 15.13
Rate Equations for Elementary Steps

    Elementary Step - a singular molecular
     event classified by the # of reactant
     molecules involved (molecularity)
     – unimolecular: AP      where R = k[A]
     – bimolecular: A + BP where R = k[A][B]
                 or A + AP where R = k[A]2
     – termolecular: 2A +B P where R = k[A]2[B]


    See Table at the bottom of page 634
          Rate-determining Step

Step 1:        A + B  X + M; k1 is slow
Step 2:        M + A  Y;     k2 is fast

Overall Rxn.   2A + B  X + Y

 Because step 2 is fast, it does not
 contribute to the overall rate, therefore:
               R = k1[A][B]