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

Answer: 0.01

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