Page 180 - Essentials of physical chemistry

142 Essentials of Physical Chemistry

so we have to subtract that from the 19.80 min. Now let us check to see if the data supports the idea
of a second-order process using the data at 8.17 min.

(0:0566)(0:0830)
ln
(0:0990)(0:0406)
¼ 0:682438 L=mol min:
(0:0424 mol=L)(8:17 2:78 min)
That is close enough to the previous value to declare this process is second order and we would
ordinarily expect that the earlier data would be less precise than the later data since the reaction is
occurring faster in the beginning leading to more uncertainty in the data. Thus we can round the two
values (or even compute an average of all four values possible from the given data) and report that
the reaction is indeed second order with a rate constant of k 2 ¼ 0:68 L=mol min.

Example 3
In some cases, a reaction may seem to be first order even though we know two species are involved.
These rates are called pseudo-first-order reactions and can occur when the concentration of one of
the reactants is in great excess. Consider the following data from Ref. [7] (with permission).
A reaction flask is set up containing methyl acetate and 1 M HCl at 258C. The almost unspoken
second reactant is the water in the HCl solution which is of the order of 55 M in H 2 O. Actually the
HCl is a catalyst and is not consumed during the reaction but aids the hydrolysis.

CH 3 COOCH 3 þ H 2 O þ HCl ! CH 3 OH þ CH 3 COOH þ HCl
(a x) (b x) x x
It should be obvious that a ‘‘hydrolysis’’ reaction involves water and in fact some water is used up
during the reaction. However, if the concentration of the methyl acetate is less than 1M the loss of
1 mol of water out of a concentration of 55 M will hardly be missed and further 1 mol of water is
only about 18 mL. Thus, on the basis of moles and volume, the amount of water in the 1 M HCl
solution seems almost constant. The rate constant really should be

dx
¼ k 2 [H 2 O] [CH 3 COOCH 3 ] ¼ k 2 [55][CH 3 COOCH 3 ] ¼ k [CH 3 COOCH 3 ]
0
þ
dt
but since the water concentration changes very little during the reaction we can just absorb the water
concentration into an effective pseudo-first-order constant k ¼ (k 1 [55]). Consider the data carefully.
0

Time, s 339 1242 2745 4546 1
Volume, mL 26.34 27.80 29.70 31.81 39.81

You need to understand the way the experiment is set up. Suppose a 2 L flask is set up about half full
with a solution of 1 M HCl and a clock is started when an unknown amount of methyl acetate is
added to the solution. Then at times which are noted by the clock reading, a pipette of unknown size
(but the same quantitative volume for each aliquot) is used to draw out an aliquot which is drained
into a 250 Erlenmeyer flask containing about 50 mL of ice and water to ‘‘quench’’ (i.e., slow down)
the hydrolysis reaction at a lower temperature. Then the acid in the flask is titrated with carefully
standardized 0.100 M NaOH to a phenolphthalein pink end point and the volume of the titration is
recorded in the table above. There is uncertainty in the titrations caused by the ‘‘creeping end point’’
which makes it important to obtain a lot of titration points in the hope that the error will average out.
Usually it will be helpful to have many more points than given here. First, we are getting a ‘‘handle’’
on this reaction rate using a titration of the acid in this solution as a function of time but it was