Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below.
`{:("t(min)",0,30,60,90),(C("mol.L"^(-1)),0.8500,0.8004,0.7538,0.7096):}`
Show that it follows a pseudo first order reaction as, the concentration of water remain nearly constant `(55"mol.L"^(-1))` , during the course of the reaction what is the value of k' in this equation ? Rate `=k'[CH_(3)COOCH_(3)][H_(2)O]`

Explanation

As the concentration of `H_(2)O` remains constant during the course of the reaction , the reaction follows pseudo first order kinetics with respect to `CH_(3)CO_(2)CH_(3)` .
Hence , rate `k'[CH_(3)CO_(2)CH_(3)][H_(2)O]=k[CH_(3)CO_(2)CH_(3)]`
[ Where, k = k' `[H_(2)O]` = constant]
The integrated rate equation for this reaction is
`thereforek=(2.303)/(t)log.([A]_(0))/([A])`
The following are the values of k obtained from the given data

Thus, we see that `k'[H_(2)O]` is constant and equal to `2.004xx10^(-3) "min"^(-1)` . Hence , it is a pseudo first order reaction .
`thereforek=k'[H_(2)O]`
or, `k'=(k)/([H_(2)O])=(2.004xx10^(-3))/(55)=3.64xx10^(-5)"mol"^(-1)."L.min"^(-1)`