The hydrolyiss of methyl acetate in aqueous solution is has been studied by titrating the liberated acetic acid against `NaOH`. The concentration of 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 remains nearly constant `(55 mol L^(-1))` during the course of the reaction . What is the value of `k'` in the equation ?
Rate `= k'[CH_(3)COOCH_(3)][H_(2)O]`

Explanation

For the given reaction, rate `=k'["CH"_(3)"COOCH"_(3)]["H"_(2)O]`
But as `["H"_(2)O]` remains constant, we put `k'["H"_(2)O]=k` so that rate `=k["CH"_(3)"COOCH"_(3)]`
Now, it is a pseudo first order reaction. Hence, for the given data, `k=(2.303)/(t)log""(C_(0))/(C_(t))`
We are given `C_(0)=0.8500" mol L"^(-1)`
`{:("t/min",,,C_(t)//"mol L"^(-1),,,k=(2.303)/(t)log""(C_(0))/(C_(t))),(30,,,0.8004,,,k=(2.303)/(30min)log""(0.8500)/(0.8004)=2.004xx10^(-3)min^(-1)),(60,,,0.7538,,,k=(2.303)/(60min)log""(0.8500)/(0.7538)=2.002xx10^(-3)min^(-1)),(90,,,0.7096,,,k=(2.303)/(90min)log""(0.8500)/(0.7096)=2.005xx10^(-3)min^(-1)):}`
Average value of `k=2.004xx10^(-3)min^(-1)`
Thus, `k'["H"_(2)"O"]=k" or "k'[55" mol L"^(-1)]=2.0074xx10^(-3)min^(-1)`
`:.k'=(2.004xx10^(-3)min^(-1))/(55" mol L"^(-1))=3.64xx10^(-5)" L mol"^(-1)" min"^(-1).`