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Newton's Laws of Motion and the corresponding concepts
of kinetic and potential energy can be used to calculate
on a computer the motion of molecules and the interactions
of molecules with other molecules. Click on the above
to see a movie of atoms on a molecule. You can find this
and much more at the
Wilson group's Internet site.
This method was originally developed by Bodenstein in 1913 - as a result of having investigated the H_2(g) + Br_2(g) --» 2HBr(g) reaction (among others). The section on Derivatives and chemistry discusses the empirical rate law determined by applying the isolation method to initial rate data. The steady-state approximation (SSA) is particularly useful when any intermediates that are present in the reaction mechanism are present in "small" amounts. By "small" we mean concentrations much less than the major species in the mechanism. When this condition is met, the rate of change of the concentration of the intermediate(s) can be considered negligible. In mathematical language:
d[intermediate] ~
--------------- = 0 of [intermediate] << [major species]
dt
Example #2 - Applying the Steady-State Approximation (SSA)
Solution:
Note that comparison of the steady-state solution
[Z(t)]_{SS} above) with the previously stated
exact solution for Z(t), will closely correspond only
if k_2 >> k_1, i.e. if the intermediate ("Y") is so
reactive that it has little time to accumulate. Note
that if k_2 >> k_1, then the expression for [Y(t)]_{SS}
above is very small - as required.
Let's do another example to show how the SSA can be used
to derive the experimental rate law for a reaction that
we have seen before.
Example #3: Consider the gas phase reaction of
H_2(g) with Br_2(g), to form HBr(g). We saw this
reaction in
Derivatives and chemistry. The overall reaction
is:
The observed (experimental) rate law during the initial
stages of the reaction (initial rate) is:
i.e., first order with respect to H_2(g), one-half
order with respect to Br_2(g) and 3/2 order overall.
At later times, the rate is found to be inhibited by
HBr(g) product. The rate law at later times is:
Using the mechanism below and the SSA as needed, attempt
to derive the above rate laws.
The proposed mechanism - termed a free radical chain
mechanism - is a fairly common one for gas-phase
reactions which give rise to rate laws with fractional
order. Free radicals are atoms that are frequently
proposed as intermediates in gas-phase reactions with
rate laws of fractional order. These free radicals
have an odd number of electrons and hence are extremely
reactive intermediates. Thus, the SSA can usually be
successfully applied to such intermediates. The
steps in a free radical chain mechanism are termed
initiation, propagation, inhibition, and
termination. In the mechanism below, the
free radical intermediates are Br* and H*. The single
dot (.) depicts the odd electron free radical. Note
that when two free radicals unite - such as 2 H* or
2Br* - they form an electron-pair or covalent bond
(H..H or Br..Br). Chemists usually depict such
electron-pair bonds with a dash (-). Thus, H..H
and Br..Br would be depicted H-H and Br-Br, respectively.
Of course, if we are just interested in the molecular
formulas, we usually write H-H and Br-Br as H_2 and
Br_2, respectively. The proposed mechanism is:
Let's attempt to derive the observed rate law from the
above mechanism and see if the mechanism is a plausible
one. In other words, we want to find +d[HBr]/dt in
terms of only [H_2], [Br_2], and [HBr].
Solution:
Paying attention to the stoichiometry of each step and
remembering the rate law for an E.R. going left to right
(reactants to products, i.e. "forward") is a function
of the reactant species only. Also, the rate law for
an E.R. going right to left (products to reactants,
i.e. "backward") is a function of the product species
only. We will set up our differential equations such
that terms contributing to the increase of a species
are explicitly positive and terms contributing to the
decrease of a species are explicitly negative. Be sure
you can arrive at the following two expressions.
Now we have to solve for the [H] and [Br] in terms of [H_2],
[Br_2] and [HBr]. We could, of course, use Maple to solve it,
but if you look carefully at the above two expressions it is
easy to see that the second equation is just the negative of
the middle three terms in the first equation. Since these
three terms (collectively) add to zero, by design, we can
immediately rewrite the first equation as:
and so
Show for yourself that the steady-state [H] is:
From the mechanism, we have to write down the expressions for
the net rate of production of HBr, i.e. +d[HBr]/dt. Note
that it is produced by each of the two propagation steps and
consumed by the inhibition step. Therefore, prove for
yourself that the following expression results.
Although we could just proceed by brute force (or "Maple force")
and just substitute our expressions for [Br]_{SS} and
[H]_{SS} into the above expression for the net rate of
production of HBr, it is possibly easier to just note the
following. From our original equation for d[H]/dt:
it is perhaps easy to see that:
So, a simple substition gives:
Plugging in for [H]_{SS} gives:
Thus, dividing numerator and denominator by "k_3[Br_2]" gives
the result:
Comparing the expression derived above to the empirical
expressions given at the beginning of the problem:
we see that the correspondence is pretty good - although not
a perfect match! Notice also that the E.R.'s of the
mechanism do not add up to the given overall reaction!
Finally, what do you suppose is the purpose of the "Wall of
vessel" in the termination reaction? Hint: for this
stay tuned for week #4 of physics! Stay tuned for more fun
with kinetics and mechanisms!
In order to demonstrate the use of the SSA, let's return
to our two-step unimolecular mechanism that was proposed
to describe X --» Z. Recall that the mechanism is
given by:
The system of differential equations are:
Let's use the SSA to find an approximate expresssion for
the rate of formation of the product in the overall reaction
("Z") in terms of the reactant in the overall reaction
("X"). Finally, let's see how the [Y(t)] and [Z(t)] are
altered from their exact values as a result of the application
of the SSA.
In this example, recall that "Y" is the intermediate and
hence the focus of our application of the SSA. If the
SSA is valid for this system - recall the conditions above -
we can write:
where the subscript "SS" refers to "steady-state" conditions.
If this is valid, we can then write:
Our differential equations for [X(t)] and [Z(t)] then
become (prove for yourself):
We also assume the same initial conditions that we
stated previously: [X]_0 = "A", [Y]_0 = 0, and
[Z]_0 = 0. You will prove in a homework problem
that the steady-state solutions are:
We begin by applying the SSA to the intermediates H
(i.e. H*) and Br (i.e. Br*). Since we are to use
the SSA on the intermediates H and Br, it is necessary
to write down the differential equation for the net
rate of production of H and for the net rate of
production of Br - directly from the above mechanism.
Here we go!
Using the SSA, we set the above two rates equal to zero.
An example of how Maple can be used to solve the
differential equations.
Do problem 3-13.