Question

The following data were obtained for the saponification of ethyl acetate using equal concentration of ester alkali.

Time $(\min)$	$0$	$4$	$10$	$20$
Vol. of acid used $(ml)$	$8.04$	$5.3$	$3.5$	$2.22$

Show that the reaction is of second order.

Answer 1

applying second order kinetics
$k=\frac{1}{t} \frac{x}{a(a-x)} a=8.04$
for $t =4\, \min ;( a - x )=8.04-5.30=2.74$
$k =\frac{1}{4} \times \frac{2.74}{8.04 \times 5.30}=0.016\, L\, mol^{-1} \min^ { - 1 }$
for $t =10\, \min ; x =8.04-3.50=4.54$
$k =\frac{1}{10} \times \frac{4.54}{8.04 \times 3.5}=0.016\, , mol ^{-1} \min ^{-1}$.
similarly for $t =20\, \min$.
$k =0.016\, L\, mol ^{-1} \min ^{-1} .$
Since $k$ is throughout constant. Hence reaction is of second order.