Engineering Chemistry S Datta

124 ENGINEERING CHEMISTRY
Example 7. The following is the data for the hydrolysis of ethyl acetate by NaOH solution.
An aliquot (25 cm) of the reaction mixture was titrated against a standard acid at regular time
intervals.
Time (in minute) 0 5 15 25 35
Vol. of acid (a – x) 16 10.24 6.13 4.32 3.41
Show that the reaction is of the 2nd order.
Sol. The 2nd order rate equation is
k = 1 x
at ( a– x)
Here a = 16 c.c.
T a – xx k
5 10.24 5.76 k =
15 6.13 9.87 k =
1 576 .
× = 0.007 conc
16 × 5 10.
24
–1 sec –1
1 987
× . = 0.0067 conc –1 sec –1
16 × 15 613 .
25 4.32 11.68 k =
1 1168 .
×
16 × 25 432 .
= 0.0069 conc –1 sec –1
1 12.
59
35 3.41 12.59 k = × = 0.0066 conc
16 × 35 341 .
–1 sec –1
Since the k values are fairly constant by putting in the 2nd order rate equation i.e., the
hydrolysis of ethylacetate by NaOH is of the 2nd order.
Example 8. Deduce the rate constant and t 1/2 for nth order reaction.
Sol. Let the reaction be
nA ⎯⎯→ Products.
The rate equation is
– d[ A ] = k . [A]
n
dt
Let a be the initial concentration and x be the concentration which has reacted in the
time ‘t’. So the concentration of A at time t is (a – x).
∴ – d[ A ] – da ( – x ) dx
= = = k(a – x)
n
dt dt dt
dx
∴
( a – x)
n = k dt
z z
x dx
t
= k
0
n
0
dt
( a – x)
Q t = 0 , x = 0
L
N
M
1 1 1
−
n–
n
( n – 1) ( a – x)
a
1 −1
O
Q
P = k.
t