CHEMICAL KINETICS

Reaction rates

Consider the reaction A à B

Average rate = (change in moles)/(change in time) = D(B)/Dt = -D(A)/Dt

The amount of A or B can be expressed (and often is) in concentration units.

R = D[B]/Dt = -D[A]/Dt and this value represents the average rate over a particular time interval. The average rate decreases as the reaction proceeds.

The rate at a particular time (instantaneous rate) is the slope of the tangent at the point of interest. Often times scientists will select the instantaneous rate at time equals zero. This is called the initial rate and initial rates can be used to determine the rate law.

Reaction rates and stoichiometry

Define rate of disappearance of reactant or appearance of reactant such that a consistent value of the rate constant is obtained. For

2A + 3B à C + 2D

R = -1/2dA/dt = -1/3dB/dt = dC/dt = 1/2dD/dt

 Rate dependence on concentration R = k [A]n [B]m is called the rate law where k is the specific rate constant and n and m are the orders with respect to A and to B respectively

The orders are determined by experiment only - they are not determined by stoichiometry (the coefficients in the balanced equation)

 Units of k are determined by the rate law

For first order reaction the units are frequency, sec-1, min-1, etc.

If the reaction is overall second order then the units are M-1 sec-1

First order reactions

Consider the reaction A à B

R = d[B]/dt = -d[A]/dt = k[A] This equation can be solved using calculus - separate the variables and integrate to obtain   ln[A]t - ln[A]o = -kt (1) ..............................................(1) [A]t is the concentration of A at time t ln[A] = -ktt + ln[A]o represents the equation of a straight line   Plot ln[A]t against time to obtain a straight line with a slope equal to -k.   Can define a half-life, t1/2, the time it takes for half of the reactant to react.   when [A]t = 1/2[A]o then t = t1/2   ............................................(2)   can show that the half-life, t1/2, is independent of the initial concentration of reactant.   Plug equation (2) into equation (1) to obtain   t1/2 = ln2/k = 0.693/k ................................................(3) For second order reactions: -d[A]/dt = k[A]2 and using calculus 1/[A] = 1/[A]o + kt ............................................(4) Plot 1/[A] vs t to obtain a straight line with a slope of k.

Substitute [A] = 1/2[A]o at t = t1/2 and obtain the half life for a second order reaction: t1/2 = 1/k[A]o and as you can see the half life depends upon the initial concentration of A.

 Using initial rates to determine the rate law

For rate = k[A] when you double the concentration of [A] the rate will double; when you triple the concentration of [A] the rate will triple, etc.

For rate = k[A]2 when you double the concentration of [A] the rate will increase four times; when you triple the concentration of [A] the rate will increase nine times, etc.

Thus, often times you can just look at the data and determine the rate law.

Consider the data for the reaction A + B à products
 
 
Expt #  Initial rate  [A]    [B]
1  1 x 10-2     2    3
2 2 x 10-2     2    6 [A] constant, [B] is 2x
3 9 x 10-2     6    3 [B] constant, [A] is 3x

Notice that when the concentration of B is 2x, the rate is 2x hence first order in B
Notice that when the concentration of A is 3x, the rate is not 3x but (3)2  = 9x,  hence second order in A
                The rate law for this reaction would be rate = k[A]2 [B]

When the relative concentrations are not simple multiple one must use another approach.
Take the logs of both sides of the rate law.

For the first example above:

                R = k[A]n]B]m where n and m represent the orders

Take the log of both sides:

                log R = log k + n log[A] + m log[B]

hold the concentration of B constant

                log R3 - log R1 = n{log[A]3 - log[A]1} or
                log {R3 /R1} = n log {[A]3/ [A]1}
                log {9 x 10-2/1 x 10-2} = n log{6/2}

                n = log{9}/log{3} = 0.954/0.477 = 2

When rates and concentrations do not give simple whole number ratios you use the above approach.

Activation energy

Although the free energy change for a chemical reaction is negative (the reaction should occur spontaneously), the rate of the reaction can be so slow that there is no visible reaction taking place - the reaction must overcome an energy barrier. The magnitude of this energy barrier is called the activation energy.

The larger the activation energy, Ea, the slower the reaction and the smaller the rate constant.
The relationship between Ea and the rate constant, k, is given by the Arrhenius Equation:

k = A e-Ea/RT

Taking the log of both sides yields: ln k = - Ea/RT + ln A .................................(5)

This equation allows you to determine the activation energy if you know the rate constant at two
different temperatures. Or if you know the activation energy and the rate at some given temperature
you can calculate the rate constant at some other temperature.

As a rule of thumb it is often said that the rate doubles for every ten degree rise in temperature.
What would be Ea for a reaction around 300oK?

When k2 = 2k1 and T1 = 300, then T2 = 310

Using equation (5):         ln(k2/k1) = - Ea/R (1/T2 -1/ T1) ................................(6)

        substitute data above into equation (6) to get ln 2 = - Ea/R (1/310 -1/300)
solving one obtains 50 kJ. This "rule of thumb" is only a general comment.  Not all reactions
have an activation energy of 50 kJ and a different value would be obtained at a different temperature.

Reaction Mechanisms

The overall all balanced equation that describes a reaction provides no information about how the reaction occurs. Most reactions take place by a series of elementary reactions. This series of reactions is refered to as the mech anism of the overall reaction.

A possible mechanism for the reaction that describes the oxidation of HBr by oxygen,  4HBr + O2 à 2H2O + 2Br2,  is
         HBr + O2à HOOBr

        HOOBr + HBr à 2HOBr

        HOBr + HBr à H2O + Br2

Notice that each elementary reaction is bimolecular and the sum of the individual elementary reactions yields the overall stoichiometric reaction if the last reaction is counted twice. Those molecules that appear in the mechanism but are not in the overall reaction are called the intermediates. The intermediates in the above reaction are HOOBr and HOBr.

The reaction that has the smallest rate constant, that is, the slowest step in the mechanism determines the rate law. If the slowest step in the above reaction is the first reaction then the rate law is, R = k[HBr] [O2]. The reaction is overall second order, first order in HBr and first order in O2.

If one of the other two reactions is the slowest step, then getting the rate law is more difficult at this time.

Catalysts

A catalyst changes the path of a reaction such that the activation energy is lower and the reaction rate increases.

The catalyst can either be within the same phase as the reactants, homogeneous, or in a different phase, most likely the solid phase, heterogeneous.

Enzymes are catalysts - they speed up the rate of biological reactions.