Engineering Chemistry S Datta

122 ENGINEERING CHEMISTRY
Since the ‘k’ values are fairly constant by putting the data in the Ist order-rate equation
the reaction i.e., decomposition of H 2
O 2
is of the Ist order.
Example 2. A 20% sugar solution is taken. 50 cc of this solution and 50 cc of (N) HCl is
mixed at constant temperature. The mixture is poured in a polarimeter tube. The optical rotation
of this solution is measured at regular intervals and the final reading is taken after 24 hrs. The
data is given below:
The optical rotation of sucrose in 1(N) HCl at various times is given in the following
table:
Time (in minute) 0 7.18 18 27.05 α
Rotation (degrees) + 24.09 + 21.4 + 17.7 + 15 – 10.74
Show that the reaction is of the Ist order.
Sol. Here, a = r 0
– r α
(a – x) = r t
– r α
where r 0
, r t
and r α
are initial rotation, rotation after time ‘t’ and rotation after infinite time
where the reaction is assumed to be completed.
r 0
– r α
= 24.09 + 10.79 = 34.83
Time r t
– r α
, k = 1 r0
– rα
ln
t rt
– rα
1 34.83
7.18 32.14 k = log = 0.00486 min–1
7.18 32.14
18 28.14 k = 1 34.83
log = 0.00519 min–1
18 28.14
1 34.
83
27.05 25.74 k = log = 0.00485 min –1 .
27.
05 25.
79
Since, the k-values are fairly constant by putting the data in the Ist order rate equation
i.e., the hydrolysis of methyl acetate is of the Ist order.
Example 3. 1 cc methyl acetate was added to a flask containing 20 ml 0.5(N) HCl kept
at a temperature of 25°C. 2 ml of aliquot (reaction mixture) was withdrawn at different intervals
and titrated against a standard alkali. The data are as follows:
Time ( in minutes): 0 75 119 183 α
Vol. of alkali used: 19.24 24.20 26.60 29.32 42.03
From the above data show that the hydrolysis is of the Ist order.
Sol. Here a = V α
– V o
a – x = V α
– V t
where V o
, V t
and V α
are the volumes of alkali initially, after a time t and after an infinite time,
respectively.
V α
– V o
= 42.03 – 19.24 = 22.79
Time V α
– V t
k
75 17.83 k = 2 . 303 22.
79
log = 0.00395 min –1
75 17.
83
119 15.43 k = 2 . 303 22.
79
log = 0.00321 min –1
119 15.43
183 12.71 k = 2 . 303 22.
79
log = 0.00316 min –1
183 12.
71