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.

1.0 mL of ethyl acetate was added to 25 mL of N//2 HCl, 2 mL of the mi

1.	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.
Answer» Solution :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 isa reaction of first order. The value of `k = 0.00312 min^(-1)`.