`1.0 mL` of ethyl acetate was added to `25 mL` of `N//2 HCl, 2 mL` of the mixture were withdraw form time to time during the progress of the hydrolyiss of the ester and titrated against standard `NaOH` solution. The amount of `NaOH` required for titration at various intervals is given below:

The value at `oo` time was obtained by completing the hydrolyiss on boiling. Show that it is a reaction of the first order and find the average value of the velocity constant.

Explanation

Amount of `NaOH` used at `t = 0`
`(V_(0))propHCl` present
Amount of `NaOH` used at any time `t=t`,
`(V_(t)) prop HCl "present" +CH_(3)COOH` formed.
`CH_(3)COOH` formed at any time, `t = prop`
Ethyl acetate reacted `= (V_(t)-V_(0)) prop x`.
Amount of `NaOH` used at time `t = oo`
`(V_(oo)) prop HCl "present + Maximum"CH_(3)COOH` formed
maximum `CH_(3)COOH` formed `prop` Initial concentration of ethyl acetate `(V_(oo)-V_(0)) prop a`
Hence, if the given reaction is of the first order, it must obey the equation
`k = (2.303)/(t)log.(a)/(a-x) = (2.303)/(t)log.(V_(oo)-V_(0))/(V_(oo)-V_(t))`
In the present case, `V_(0) = 20.24 mL, V_(oo) = 43.95 mL`
`:. V_(oo) - V_(0) = 43.95 - 20.24 = 23.71 mL`
The value of `k` at difference instant can be calculated as follows:

Since the value of `k` comes out to be nearly constant, it is a reaction of first order. The value of `k = 0.00312 min^(-1)`.