Page 182 - Essentials of physical chemistry

144 Essentials of Physical Chemistry

As usual, the oxidizing agent [H 2 O 2 ] is reduced and the reducing agent [(S 2 O 3 ) ] is oxidized. The
2
important part of this reaction for the kinetic study is that the coefficients on the peroxide and the
thiosulfate are in 1:2 ratio. Thus, the reaction is of the type A þ 2B ! C. The data is given as

Time, min 16 36 43 52
[(S 2 O 3 ) ] 0.01030 0.00518 0.00416 0.00313
2

At t ¼ 0, [H 2 O 2 ] ¼ 0.03680 and [(S 2 O 3 ) ] ¼ 0.02040 at pH 5.0.
2
This time, we are given the initial concentrations and the intermediate concentrations of only
one reactant. We do not know the order of the reactions but we will guess it is second order
because there are two reacting species and if the rate constant checks out to be a constant that will
confirm the process is second order. Let [H 2 O 2 ] ¼ a and [(S 2 O 3 ) ] ¼ b, then we can write
2
þdx
¼ k 2 (a x)(b 2x) and that leads to an integrated expression as
dt
ð ð ð ð
dx dx dx
using the theorem above.

þ
¼ k 2 dt ¼
(a x)(b 2x) (a x)(b 2a) b
(b 2x) a
2

(a x)
ln
ln (a x) ln (b 2x) (b 2x)
Then we find þ ¼ ¼ k 2 t þ C and if x ¼ 0 when t ¼ 0we
(b 2a) b (2a b)
2 a
2

(a) (a x) hi b(a x)
a
ln ln ln ln
(b) (b 2x) b a(b 2x)
have ¼ 0 þ C so we find ¼ k 2 t þ and ¼ k 2 .
(2a b) (2a b) (2a b) (2a b)t
This can be generalized for any reaction of the type A þ nB ! C with the value of ‘‘n’’ as

b(a x)
ln
a(b nx)
¼ k 2 but here n ¼ 2 so we can use the data for a second-order rate process.
(na b)t
This problem is made more difficult than the previous example by the fact that we need to use
quantitative reasoning to obtain the concentrations. We have concentrations for what we are
calling ‘‘b’’ here and according to the coefficients in the balanced equation ‘‘a’’ will disappear only
half as fast as ‘‘b’’ is being used up. Once again we choose the longest time period at t ¼ 52 min to get
the best overall view of the process. At t ¼ 52 min, the concentration of the [H 2 O 2 ] will be given by
[H 2 O 2 ] ¼ 0:03680 {0:02040 [(S 2 O 3 ) ]}=2
2
That is the key to the whole problem. We can now fill in the table of values for the [H 2 O 2 ].

Time, min 16 36 43 52
[H 2 O 2 ] 0.03175 0.02910 0.02868 0.028165

b(a x) (0:02040)(0:028165)
ln ln
a(b 2x) (0:03680)(0:00313)
¼ 0:580929 L=mol min:
(2a b)t [2(0:03680) 0:02040 mol=L](52 min )
¼ k 2 ¼