Page 179 - Essentials of physical chemistry

Basic Chemical Kinetics 141

At t ¼ 0, x ¼ 0, so we can solve for the value of C, the combined constant of integration.

(a x)
ln
ð ð
dx dx ln (a x) ln (b x) (b x)
¼ k 2 t þ C:
(a x)(b a) þ (b x)(a b) ¼ (b a) þ (a b) ¼ (a b)

(a) (a x) hi b(a x)
a
ln ln ln ln
(b) (b x) b a(b x)
¼ 0 þ C, so we have ¼ k 2 t þ or rearranged to ¼ k 2 t:
(a b) (a b) (a b) (a b)
This is not the most general form because the coefficients of A and B are both 1 but we are ready for
an example. Students could look up such integrals in tables or use a computer program to solve such
problems but here we offer a way to use a simple calculus trick which is general to many other cases
ð
d(cabin)
and only requires knowing ¼ ln (cabin) þ C followed by algebra.
cabin

Example 2
Data is given in a paper by W. J. Svirbely and J. F. Roth [5] for the kinetics of the reaction between
propionaldehyde and hydrocyanic acid in aqueous solution at 258C. This is excellent data for
teaching second-order kinetics because the concentrations of both reactants are given at different
times as follows:

Time, min 2.78 5.33 8.17 15.23 19.80 1
[HCN] 0.0990 0.0906 0.0830 0.0706 0.0653 0.0424
[C 3 H 6 O] 0.0566 0.0482 0.0406 0.0282 0.0229 0.0000

First, we notice that at t ¼1 there is still some HCN remaining (0.0424 mol=L) and it does no harm
to remind the reader that HCN is extremely poisonous. Perhaps you have encountered this type of
reaction before and recognize that the propionaldehyde is the ‘‘limiting reagent’’ but here we mainly
need to observe that in our model system [A] 6¼ [B]. Let us ‘‘reset our clock,’’ so that time zero is
measured from the first time when we know the concentrations of both reactants, namely at 2.78
min. Another good practice is to select data points which are far apart in time to get the best overall
picture of the rate process, namely at 19.80 min to reduce short term fluctuations in the data. The
next question is whether the process is really second order? The simple answer is that it is a good
guess that when two species react, the rate process is second order. Nevertheless, we will test this
assumption and if the rate constant (k 2 ) is not constant over the span of time in the data then our
assumption is incorrect. Here, we are only giving the essentials of the most common type of kinetics
but an excellent book on this subject is available in the text by Moore and Pearson [6]. Thus, we
assume second-order kinetics and see if the data fits that assumption. Let [A] ¼ a ¼ [HCN] and
[B] ¼ b ¼ [Pr] at t ¼ 2:78 min. Then, we can use the formula we derived above and insert the data
values relative to the concentrations at 19.80 min.

b(a x) 0:0566(0:0653)
ln ln
a(b x) 0:0990(0:0229)
¼ 0:677261 L=mol min:
(a b)t ¼ k 2 ¼ (0:0424 mol=L)(19:80 2:78 min)
Note that the concentrations [a x] and [b x] are the actual values in the table of data, while the
values for [a]and [b] are the values of the data at 2.78 min. In the denominator, the difference in the
two concentrations will always be 0.0424 mol=L, but the time is measured starting at 2.78 min,