(2) Simple Second Order Rate Equations
A typical simple second order reaction, where B is one of the reactants, would look like,
(1)This equation can be integrated to give,
which can be rearranged for integration as,
(4)We can make a quick check to see if this result fits what we know about the system. At t = 0 we get [B] = [B]o which is correct and at t = infinity we find that [B] = 0 which is also what we would expect.
We can define the half-life of a second order reaction in the same way as in first order reactions. That is, the half-life, t1/2, is the time required for [B] to fall from [B]o to [B]o/2. Thus we use Equation 3 to give
(6)Note that the half-life for a simple second order reactions depends on the initial concentration and that it is proportional to 1/[B]o. Contrast this to what we know about first order reactions where the half-life is independent of the initial concentration. This fact may come in useful later when we try to find experimental methods for determining rate laws.
(3) Mixed Second Order Rate Equations
A mixed second order rate law equation, where both components A and B are reactants, would look like,
(7)Equation 7 has two variables in it (besides time). In order to integrate this equation we need to know the stoichiometry so that we can tell how [A] depends on [B].
We will use the following stoichiometry as an example,
2A + B → P + etc (P = product). (8)We have a choice as to which component concentration to use for our variable, we can use [A], [B], or [P].
Let's try [P] as our variable. (The form of the answer will depend on the choice of variable. A different choice of variable will give an answer that looks different even though it is algebraically equivalent, see below.)
Let's assume that there is no product present at the beginning of the experiment, that is,
[P]o = 0. (9)We now need to express [A] and [B] in terms of [P]. We can see that,
[A] = [A]o − 2 [P]. (10)(We obtain Equation 10 from the stoichiometry, which says that every time we make one mole of P we use up two moles of A. That is, each P formed takes away two A's.) Also,
[B] = [B]o − [P], (11)because each P formed takes away only one B.
So our rate equation in terms of the single variable, [P] becomes,
(13)In order to make this equation easier to integrate we rearrange it as follows,
(15)The left-hand side can be integrated by partial fractions, or you can look it up in a book of integral tables. Since this is a common problem in chemical kinetics we will give a brief review of the method of partial fractions.
Review of partial fractions.
We would like to write the fraction appearing on the left side of Equation 15 as a sum of two simpler fractions. We can do this if we can find two quantities, X and Y, such that,
(16)Put the right-hand side over the common denominator,
(17)Notice that both sides have the same denominator. Then the numerators must be equal,
(19)Since in Equation 19 the left-hand side, 1, does not depend on [P] we conclude that the [P] on the right-hand side must cancel out. That is,
(21)We also conclude that
(22)or, using Equation 21,
-------------------- (End of review of partial fractions.)
which we can substitute back into Equation 15 to get
(26)Although this is messy, it is relatively easy to integrate to give
(27)Change sign inside the logarithms and combine them
(30)This result is probably good enough for our purposes, but you can, in fact, solve for [P],
(31)Note that [A]o = 2 [B]o is a special case. (Stoichiometrically equivalent amounts.) If you try to plug these initial concentrations into Equation 31 for [P] you get 0/0. Mathematically you can make this work by taking the limit as [A]o → 2 [B]o, but this is messy. It is easier to go back to the beginning and convert
(33)This is just a simple second order rate equation which we can set up to integrate as,
(34)which integrates to,
(36)Equation 36 can easily be solved for [P] and we can determine an expression for the half-life as we have done before.
We indicated above that we have a choice of variables in the integration of the mixed second order rate Equation 7. Different choices give results that look different on the surface, but which must be algebraically equivalent. Let's integrate Equation 7 again with the same stoichiometry, but this time using [B] as the variable. Our stoichiometry is determined by the balanced reaction,
2A + B → P + etc (P = product), (8)from which we conclude that the amount of B which is used up is given by,
B used = [B]o − [B]. (37)Since each B used takes 2A along with it, the amount of A used is
A used = 2 × (amount of B used) = 2([B]o − [B]). (38)Therefore, the concentration of A is,
[A] = [A]o − A used = [A]o − 2[B]o + 2[B]. (39)(Notice that [A] in Equation 39 can never be negative because a negative concentration is not physically possible. If the reactants are not present in their stoichiometrically equivalent amounts we would have to worry about which reagent was the "limiting reagent," that is, which one we will run out of first. For example, if 2[B]o is larger than [A]o the reaction would cease at the point when Equation 39 reached zero.)
Our mixed second order rate law was,
(7)Using equation 39 for [A] we get
(40)We set this up to integrate as,
(41)Using the method of partial fractions, which we have already reviewed above, we can rewrite this as
(43)which integrates to,
(46)Equation 46 is messy, but it can be solved for [B] if desired. The point is that this equation looks very different from our previous solution using [P] as a variable (Equation 30), even though according to the stoichiometry,
(47)This equation has the same problem we had before if the reactants are initially present in their stoichiometrically equivalent amounts.
It is an interesting exercise to show algebraically that Equations
30 and 46 are, indeed, equivalent.
From here you can:
Copyright 2004, W. R. Salzman
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