SUMMARY OF THE KINETICS FOR aA------>Prods
 

                                               0th Order                1st Order            2nd Order

Rate Law:                               rate = k[A]^o            rate = k[A]¹         rate = k[A]²

Integrated Rate Law:
                                 [A] = [A]_o - akt         ln[A] - ln[A]_o = -akt     (1/[A])-(1/[A]_o)=akt
                                 y    =   b   +  mx            y    =     b   +   mx          y     =    b    +   mx

Plot needed to give            [A] vs. t                          ln[A] vs. t                     1/[A] vs. t
     straight line:

What does slope give?    slope = -ak                     slope = -ak                  slope = +ak

What does intercept give?     [A]_o                             ln[A]_o                             1/[A]_o

Half-life:                         t_1/2 = [A]_o/2ak             t_1/2 = 0.693/ak             t_1/2 = 1/ak[A]_o

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14.4 Activation Energy and Temperature Dependence
        of Rate Constants

COLLISION THEORY OF CHEMICAL KINETICS:

Temperature and Rate:
 As temperature increases, the rate increases.
 Since the rate law has no temperature term in it, the rate
        constant must depend on temperature.

 Consider the first order reaction CH₃NC -----> CH₃CN.
    As temperature increases from 190°C to 250°C, the
        rate constant increases from 2.52 x10^-5 s^-1 to
        3.16 x 10^-3 s^-1.
 The temperature effect is quite dramatic, about a factor
        of 10².  Why?

The Collision Model:
 Observations: rates of reactions are affected by concentration
    and temperature.
 Goal: develop a model which explains why rates of reactions
    increase as concentration and temperature increases.

 The collision model: in order for molecules to react, they
        must collide.
 The greater the number of collisions per time, the faster
        the rate.
 Also, the more molecules present, the greater the probability
        of collision and the faster the rate.
 The higher the temperature, the more energy available to the
        molecules and the faster the rate.

Complication: not all collisions lead to products; they are not
    all effective collisions.  In fact, only a small fraction of
    collisions lead to products.

In order for reaction to occur, the reactant molecules must
    collide in the correct orientation and with enough energy
    to form products; this still does not guarantee that a
    reaction will occur.

EXAMPLE in text using NO + N₂O ---> NO₂ + N₂ showing
    the various molecule orientations:
            a)  favorable:  N-N-O + N-O
            b) ineffective: O-N-N + N-O
            c) ineffective: O-N-N + O-N

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TRANSITION STATE THEORY:

We have already seen plots of potential energy versus the
    progress of the reaction.  If the reaction is exothermic,
    then the total energy of the products will be at a lower
    value than the total energy of the reactants.

Vice-versa, if the reaction is endothermic, the total energy
    of the products will be higher in energy than the total
    energy of the reactants.

The products pass through a short-lived, high-energy
    intermediate state called the TRANSITION STATE, before
    the products are formed.

ACTIVATION ENERGY, E_a, is the kinetic energy that reactant
    molecules must have to allow them to reach the transition
    state.

The activation energy, E_a, is the difference in energy between
    reactants and the transition state.

Ex.:  H₂(g) + I₂(g) ---->  2 HI(g)

  E_a (forward reaction) - E_a (reverse reaction) = DE (reaction)

When DE = - , then the reaction is exothermic; and vice-versa.

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THE ARRHENIUS EQUATION
        EFFECT OF TEMPERATURE ON RATE:

Activation Energy:

Arrhenius: molecules must possess a minimum amount of
    energy to react.
        Why?
 In order to form products, bonds must be broken in the
        reactants.
 Bond breakage requires energy.

BOLTZMANN DISTRIBUTION OF ENERGIES:  (show graph) -
    see Fig. 5.14(a)  on p. 147.
 

 Activation energy, E_a, is the minimum energy required to
    initiate a chemical reaction.

The activation energy, E_a, is the difference in energy between
    reactants and the transition state.

The rate depends on E_a.

Notice that if a forward reaction is exothermic
                        CH₃NC -------> CH₃CN,
    then the reverse reaction is endothermic
                        CH₃NC <------- CH₃CN.

    Frequency of collisions affects the rate as well as the
    orientation of molecules during collisions (remember
    different types of car crashes:  head-on, side swipe,
    rear-end, etc.)
 

We can relate the temperature (T) and Activation Energy
    (E_a) of a reaction  to the rate constant (k) using the
        Arrhenius Equation:        k = Ae^-Ea^/RT

 R is the gas constant (8.314 J/K-mol) and T is the
    temperature in K.
A is called the frequency factor and is a measure of the
    probability of a favorable collision.
 Both A and E_a are specific to a given reaction.

If we have a lot of data, we can determine E_a and A
    graphically by rearranging the Arrhenius equation:

                  ln k = (-E_a/R)(1/T) + ln A
                    y   =     (m)     (x)   +   b

By plotting ln k versus 1/T, the slope is:    -E_a/R
       (see Fig. 14-12, p. 451).  Therefore, can get E_a from
        rate data.

 Alternatively,  ln (k₁/k₂) = (E_a/R)(1/T₂ - 1/T₁)

    Or:   ln k₁ - ln k₂ = (E_a/R)[(T₁ - T₂)/T₁T₂]

Same kind of equation as Clausius-Clapeyron equation,
        which is:  ln P₁ - ln P₂ = (DH_vap/R)[(T₁ - T₂)/T₁T₂]

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14.5 REACTION MECHANISMS AND RATE LAWS

The reaction mechanism gives a detailed pathway for a
    chemical reaction.
Mechanisms consist of one or more elementary steps.

ELEMENTARY STEP:  any process occurring in a single step.

MOLECULARITY:   # of molecules present in an elementary
    step;  e.g., unimolecular (one molecule) , bimolecular
    (two molecules), termolecular (three molecules, very
    rare!).

The SUM of the elementary steps must give the balanced
    chemicalequation.

INTERMEDIATE:something that appears in an elementary
    stepbut not in the final balanced chemical equation.

Ex.:  NOBr₂  shown below:
 

RELATIONSHIP BETWEEN ELEMENTARY STEPS AND THE RATE
LAW:

Rate laws of elementary steps determine the overall rate law
    for a reaction, AND the molecularity of a step determines
    the rate law for a step!

Thus, unimolecular gives FIRST ORDER, bimolecular gives
    SECOND ORDER, etc.

If a reaction has one unimolecular elementary step, the rate
    law for the reaction is first order:

   e.g., CH₃NC -------> CH₃CN is one step, is first order in
        [CH₃NC] and there is no intermediate.
 

Likewise, a bimolecular reaction with one elementary step is
    second order.

However, for multistep reactions, the overall rate law is
    determined by the slowest elementary step; this step is
            called the     RATE-DETERMINING STEP (RDS).

When the RDS is the first step, determining the rate law is
    easy.

             Ex.:    NO₂ + CO ------> NO + CO₂

        This has two steps, for which k₂ >>>>> k₁

                             k₁
        NO₂ + NO₂------> NO + NO₃    (slow step)

                             k₂
        NO₃ + CO ------> NO₂ + CO₂   (fast step)

     The Rate Law is:         rate = k₁[NO₂]²

            because the first step is BImolecular.
 
 

When a later step is the RDS, it is a little more complicated.

    Ex.:    2NO    +    Br₂ -------> 2NOBr    (all gases)

Experimentally, we find:       Rate = k[NO]²[Br₂]

What is the mechanism?  Termolecular is unlikely!

1st step:

                        k₁
   NO + Br₂ ------> NOBr₂ (fast)
                  <------
                       k_-1

    rate of forward reaction = k₁[NO][Br₂]
    rate of backward reaction = k_-1[NOBr₂]
 

2nd step:

                              k₂
      NOBr₂ + NO ------> 2NOBr (slow)

      rate = k₂[NOBr₂][NO]

Since the 2nd step is the SLOW step in the reaction, this
    latter step's rate law should lead to the overall rate
    law....but how do we get [NOBr₂]?   This is an
    intermediate, whose concentration is very difficult to
    get.

Solution:  express [NOBr₂] in terms of [NO] or [Br₂], and
    then use that to solve the problem.

We can assume that NOBr₂ is unstable and returns to NO
    and Br₂ in step 1; thus, the rates of the forward and
    backward reactions in step one are equal, and
            k₁[NO][Br₂] = k_-1[NOBr₂]

 Rearranging and solving for [NOBr₂] = (k₁/k_-1)[NO][Br₂]

    Thus, the second step rate law becomes:

   rate = (k₂.k₁/k_-1)[NO]²[Br₂] = k[NO]²[Br₂], as observed!
       since k₂.k₁/k_-1  is another constant, k

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14.6 CATALYSIS
    Catalysts are substances that are added to reactions to
    increase the rate of reaction (usually); they allow reactions
    to occur via alternative pathways that increase the reaction
    rates by lowering the activation energies; see Fig. 14.16,
    p. 458.  They may be consumed in the early steps in a
    reaction, but are then regenerated in an equal amount in
    later steps, so that they seem not to be used up at all.

 Generally, catalysts operate by lowering the activation
    energy for a reaction.
 However, catalysts can operate by increasing the number
    of effective collisions (enzymes, etc.)

Catalysts lower the activation energy, E_a, of a reaction. They
    do not affect the equilibrium constant or the concentrations
    of the reacting species.
    A catalyst shortens the time it takes to reach equilibrium
    by increasing the rate constants for both the forward and
    reverse reactions.

Heterogeneous Catalysis:  catalyst is in a different phase
    than the reactants and products.

   Typical example: solid catalysts with gaseous reactants
        and products (ex.: catalytic converters in cars).  Most
        industrial catalysts are heterogeneous.

      Ex.:  Pt catalyst for the hydrogenation of ethylene:
                    C₂H₄(g) + H₂(g) ----> C₂H₆(g)

    It is thought that both the ethylene and the hydrogen gas
    molecules diffuse to the surface of the catalyst where they
    undergo "chemisorption".  The pi electrons of ethylene
    form bonds to the metal, and hydrogen molecules break
    into H atoms that bond to the metal.  Then the H atoms
    migrate to the ethylene molecule, and react to form ethane. 
    The product, ethane, since it no longer contains double
    bonds, does not adhere to the metal, and diffuses away.

    Vegetable oils containing carbon-carbon double bonds
    (unsaturated) are changed to solid fats (shortening =
    saturated) when the bonds are catalytically hydrogenated.

Homogeneous Catalysis:  catalyst exists in the same phase
    as the reactants; i.e., the reactants AND the catalyst are
    all soluble in the solvent.

  Ex.:  Oxidation of thallium(I) to thallium(III) by
           cerium(IV) in aqueous solution:

      2 Ce⁴⁺(aq) + Tl⁺(aq) --->  2 Ce³⁺(aq) + Tl³⁺(aq)
      This reaction is very slow; presumably, it involves the
    collision of three (3) positive ions.  The reaction can be
    catalyzed by manganese(II) ion.  The mechanism is
    thought to be:

           Ce⁴⁺(aq) + Mn²⁺(aq) --->  Ce³⁺(aq) + Mn³⁺(aq)
           Ce⁴⁺(aq) + Mn³⁺(aq) --->  Ce³⁺(aq) + Mn⁴⁺(aq)
           Mn⁴⁺(aq) + Tl⁺(aq)   --->   Mn²⁺(aq) + Tl³⁺(aq)

    where each step is bimolecular, and relatively fast.
 

A catalyst may add intermediates to the reaction.
   Ex.: in the absence of a catalyst, H₂O₂ decomposes
           directly  to water and oxygen.
 BUT.... In the presence of Br^-, Br₂(aq) is generated as an
        intermediate, which reacts with peroxide rapidly and
        therefore accelerates the rate.   Here, E_a is reduced.
                2H₂O₂ ----> 2H₂O + O₂

             2Br^-+ H₂O₂ + 2H⁺ ----> Br₂ + 2H₂O

                Br₂+ H₂O₂ ----> 2Br^- + 2H⁺ + O₂

Bromide is not consumed and can be recovered after the
    reaction.
 

Enzyme Catalysis
Biological catalysts are called enzymes.

    Most enzymes are protein molecules with large molecular
    masses; they have very specific shapes and often
    catalyze very specific reactions.  Substrates react at
    the active site of an enzyme; this increases A in the
    Arrhenius equation by increasing the probability of
    effective collisions.

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