As further examples of the determination of rate laws from proposed mechanisms for chemical reactions we will consider two examples of a class of reactions called "chain reactions."

Chain Reactions

Chain reactions usually involve free radicals. We will show how to deal with chain reactions by working out several examples. Our first example is the gas phase reaction of hydrogen with bromine to give HBr. (The temperature must be high enough that bromine is a gas and not a liquid.)

H2 + Br2 → 2 HBr                (1)
The experimental rate law is
(2)
Our task is to show that the postulated mechanism will give this rate law. (We will not try to invent mechanisms for particular reactions in this course.) The postulated mechanism has five elementary reaction steps:
(3)

(4)

(5)

(6)

(7)

Our goal is to find the rate law implied by this mechanism and see how it comparaes to the experimental rate law, Equation 2. Our first task will be to define the reaction rate. We will define the reaction rate as rate of production of the product, HBr.
(8)
Looking at the mechanism and using only the steps which yield or consume HBr we find that
(9)
We see that the rate contains two transient species, Br• and H• , so we set up the equations which show the rate of change of the concentrations of these two species and apply the steady state approximation. The two equations are,
(10)
and
(11)
In Equation 11 we have set three of the terms apart by braces { } in order to make it easier to see that the part inside the braces is just the negative of the middle of Equation 10. Since the part inside the braces is zero we conclude that
(12)
or
(13)
We can solve Equation 10 for [H• ] to get
(14)
which, on substituting the result of Equation13, becomes,
(15)
We now go back to the equation for the Rate, Equation 9, and note that the right hand side of Equation 9 is the same as Equation 10 except for the sign of the kc term. Rewrite Equation 10 as,
(10b)
or
(10c)
Combining Equations 9 and 10c we find that,
(16)
Since Equation 15 gives us an expression for [H• ] we can combine Equations 15 and 16 to eliminate all the transient concentrations,
(17)
which rearranges to
(18)
Equation 18 has the same form as Equation 2 with,
(19)
and
(20)

The two phenomenological rate constants, k' and k", can be written in terms of the rate constants for the elementary reaction steps in the mechanism.

Another Example of a Chain Reaction

The reaction (gas phase)

C2H6 → C2H4 + H2     (21)
is a very important commercial reaction because ethylene (ethene) is a precursor to many important products (of which polyethylene is just one). Because of its commercial importance the reaction has been much studied. In the temperature range 700 K to 900 K the early stages of the reaction appear to be governed by the mechanism,
(22)

(23)

(24)

(25)

(26)

Define the rate as the rate of production of C2H4 from Equation 24,
(27)
There are three transient species, CH3•, C2H5•, and H• . We will use the steady state approximation on each of these:
(28)            (29)

(30)

From Equation 28 we conclude that
(31)
(This equation, Equation 31, says that the concentration of the methyl radical remains constant throughout the course of the reaction.  This, of course, is impossible because the concentration of the methyl must be zero at the beginning and end of the reaction.  However, the steady state approximation is still useful if the concentration builds up to this constant concentration fairly quickly, remains essentially constant throughout the portion of the reaction for which this mechanism is valid.)

If we add Equations 29 and 30 and substitute in the result of Equation 31 we get,

(32)
which simplifies to
(33)
We can get another expression for [C2H5•] from Equation 29
(34)
When we equate Equations 33 and 34 and simplify we get
(35)
Equation 35 is clearly a quadratic in [H• ], but the coefficients (in other words, everything else in the equation) are all constants so [H•] must equal a constant, call it k',
.
Then
(36)
and the rate becomes, combining Equations 27 and 36,
(37)
which is first order in ethane. That is, in the early stages of the cracking of ethane to ethylene (ethene) and hydrogen the reaction is first order in ethane.

(If you want to get picky, the above quadratic can be solved to give,

(38)
where we will have to pick the plus sign because concentrations cannot be negative.)

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Copyright 2004, W. R. Salzman
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Last updated 11 Nov 04
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