Page 181 - Essentials of physical chemistry

Basic Chemical Kinetics 143

already 1 M in HCl before the additional amount of acetic acid is produced by the hydrolysis. Thus,
the titration volumes in the table are mainly a titration of the 1 M HCl with a slowly increasing
amount of acetic acid. Note the final titration, which is usually done by using the same pipette for an
aliquot from the reaction solution several days after the start of the reaction, reaches a limiting value
at t ¼1. How can we get a quantitative rate constant, even if it is a pseudo-first-order rate constant,
from such ill-defined data? The key in this case and in many other reaction rates is to ignore a lot of
unnecessary information and focus on the ‘‘handle’’ of the data which connects the extent of the
reaction to time. In this case, the ‘‘handle’’ is the ‘‘amount yet to go’’ in the reaction. Reset your
clock to t ¼ 0 with the data at 339 s, then the ‘‘amount yet to react’’ is

(V HAc þ V HCl ) t¼1 (V HAc þ V HCl ) t¼339 s Š ¼ a 0 ¼ V 1 V 339 :
½
Then we can use the usual first-order equation a(t) ¼ a 0 e k t 0 and

a(t) 1 (V 1 V t 339 )
ln ¼ k ta(t) ¼ V 1 V t 339 so we write ln ¼ k and we can
0
0
a 0 t (V 1 V 339
make a table of k values:
0

1 39:81 27:80 1 (12:01) 4 1
ln ¼ ln ¼ 1:270491 10 s ,
(1242 339)s 39:81 26:34 903 s (13:47)

1 39:81 29:70 1 (10:11) 4 1
ln ¼ ln ¼ 1:192602 10 s ,
(2745 339)s 39:81 26:34 2406 s (13:47)

1 39:81 31:81 1 (8:00) 4 1
ln ¼ ln ¼ 1:238468 10 s :
(4546 339)s 39:81 26:34 4207 s (13:47)
____________________________
Average k ¼ 1.2325 10 4 s 1
0
We see that the calculated value of k fluctuates. The main reason for this is that the titrations
0
are subject to a large uncertainty due to the fact that even in the cold ice water of the titration flask
the reaction is still proceeding. Ideally, the end point of the titration should be the first visual
pink of the titration, but since the reaction is still proceeding at the lower temperature, the pink
will fade as more acetic acid is produced and then the titrator will add more base and then that will
fade, etc., so the end point of the titration is subject to considerable uncertainty. Even so this
experiment has been performed by hundreds of students and the results are remarkably reprodu-
cible, although with a large uncertainty.

Time, s 1242 2745 4546
(k =s) 10 4 1.270491 1.192602 1.238468
0

Example 4
Now we come to a more general treatment of a second-order reaction in which the coefficients in the
balanced reaction are not 1:1. This is a set of data which is reported in Ref. [7] (with permission).
This is the sort of quantitative data one needs for precise work and should be familiar from the use of
oxidation–reduction titrations in quantitative analysis. Although only the [(S 2 O 3 ) ] needs be
2
reported, we assume that the sodium salt is used for aqueous solubility.
H 2 O 2 þ 2Na 2 (S 2 O 3 ) þ 2HCl ! 2H 2 O þ Na 2 (S 4 O 6 ) þ 2NaCl: