Pseudo First Order Reactions

There are circumstances where a second order reaction might appear, in an experiment, to be first order. That is when one of the reactants in the rate equation is present in great excess over the other in the reaction mixture.

As an example, let's look again at the mixed second order rate law that we treated in the last section.

In the last section we had the rate law, , with a reaction whose stoichiometry is,

2A + B → P + etc (P = product).
Sometimes [A]o > > [B]o.
For example, say we prepare the system such that,

    [A]o = 2.0 M, and
    [B]o = 0.001 M.

Then, when the reaction has run to completion, (at time = infinity),

[A]inf = [A]o − 2 [B]o = 2.0 − 0.001 = 1.998 ≈ 2.0.
So [A] is approximately constant throughout the entire reaction.

Then the rate law becomes,

 ,                 (1)
which is effectively a first order rate equation. This rate equation has the solution,
 ,               (2)
where the "effective" first order rate constant is k[A]o.

You can actually solve this problem as a mixed second order rate problem, as we did in the last section, but it is a lot more work.

Pseudo first order reactions are sometimes used to find the rate constant of a second order reaction when one of your two components is very expensive and the other one is relatively cheap. You can use an excess of the inexpensive reagent and use a small amount of the expensive one. Notice that you can still obtain the second order rate constant by dividing the effective first order rate constant by the concentration of the excess component.

(3) Simple Third Order Reaction

A simple third order rate law, where component B is a reactant, has the form,

,               (3)
which can be rearranged for integration as,
,               (4)
and integrated to give,
,               (5)
or, in simplified form,
(6) .               (6)
As usual we define half-life as the time when [B] = ½ [B]o, which gives,
,               (7)

 ,               (8)

.               (9)

Note that in this case the half-life depends on [B]o2 as opposed to depending on [B]o for a simple second order reaction, and being independent of [B]o for a first order reaction.

Techniques for finding rate laws (NOT from the balanced equation.)

We emphasize that the form of the rate law for a chemical reaction cannot be deduced from the balanced reaction equation. It takes new experimental data (or new theories) to get the rate law for a reaction. We will mention three methods for determining the rate law for a reaction from experimental data.

1. Half-lives

Run several experiments at different initial concentrations to see how the half-life depends on [B]o

If the half-life does not depend on the initial concentration then the reaction is first order.
If the half-life of the reaction is proportional to 1/[B]o then the reaction is simple second order. If the half-life is proportinal to 1/[B]o2 then the reaction is simple third order, and so on. Mixed second and third order reactions are more complicated, but half-lives can be defined for these cases and the dependence of the half-life on [A]o and [B]o can be used to determine the rate law.

2. Initial Rate Method

The initial rate of a reaction is the reaction rate at t = 0. We will designate the initial rate by Ro. One can run several experiments and measure the rate as soon as possible after the reagents have been mixed to determine the initial rates.

For example, assume that

Ro = k[A]x[B]y[C]z .
Run several experiments at different starting concentrations and measure Ro in each case. We can place the data in a table as follows:
  [A]o   [B]o   [C]o   Ro
0.010  0.10   0.20  0.04   for experiment number 1,
0.020  0.10   0.20  0.08   for experiment number 2,
0.010  0.20   0.20  0.16   for experiment number 3, and
0.010  0.10   0.10  0.04   for experiment number 4.
Compare the first and second experiments. We double [A]o and the rate doubles. It looks like x = 1. Compare the first and third experiments. We double [B]o and the rate increases by a factor of 4. It looks like y = 2. Notice that there is no change in the rate when we cut [C]o in half so the rate must not depend on C. This says that z = 0.

Thus we determine that the rate = k[A]1[B]2[C]0.

We don't have to use simple whole number ratios for the concentrations, we can do it algebraically. Take the ratio of the rates from two successive experiments,

.               (10)
To find the exponent of [A] make all the concentrations in the second experiment the same as in the first experiment except [A]. This gives,
.               (11)
Take the logarithm of both sides to get,
,               (12)
and solve for x to get,
.               (13)
This process can be repeated for each of the components in turn.

3. Fit to Integrated Form (But Be Clever)

We have seen the integrated forms of the various rate equations. Simply plotting the concentrations as a function of time is not usually useful because it is sometimes hard to tell the difference between a first order decay and a second order decay for a limited set of real experimental data (which probably will contain some scatter). However, it is relatively easy to identify a straight line. So the best way to look at your experimental data is to plot things in such a way that you expect expect a straight line

For example, to test for first order plot ln[A] vs t, and see if you get a straight line. There may be some data scatter, but it should be random. If you see a definite curvature in your line the reaction is probably not first order.

To test for simple second order plot 1/[B] vs t and look for a straight line.

To test for simple third order plot 1/[B]2 vs t and see if you get a straight line. There are many possible varitions on this method, but this gives a general idea how the method works.

Arrhenius Theory - Temperature Dependence of the Rate Constant

(Named after its discoverer, Gustav Arrhenius)

It was known very early in the study of chemical kinetics that reaction rate constants were functions of temperature.  That is,

k = k(T).
It was found that an experimental plot of lnk vs 1/T  appeared to be approximately linear with a negative slope.  This implies that
.               (14)
Call c = lnA for simplicity, then,
.               (15)
We can solve Equatiion 15 for k by taking the antilog of this equation,
,               (16)
.               (17)
It is usual to set b = Ea/R to get
.               (18)
Ea is called the Arrhenius activation energy. The interpretation of this equation is that there is some intermediate configuration of the molecules involved in the reaction which is "half-way" between the reactants and products. This configuration of molecules, in the process of reacting, is called the "activated complex." The idea is that it takes energy to form this complex and the probability of the system attaining that energy at a temperature, T, is proportional to the exponential of (the negative of) the activation energy divided by RT.

This makes sense. Recall that kT and RT are thermal energies and we have seen exponentials of a mechanical energy divided by a thermal energy before. For example, the probability distribution function for molecular velocities in one dimension was given as

.               (19)
Actually, the preexponential factor, A = A(T), is also a function of temperature, but the temperature dependence is much weaker. Sometimes, for example,
But the major portion of the rate constant's temperature dependence is in the exponential term, 



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Last updated 4 Nov 04