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

Writing the integrated rate law for first order reaction, we have
`k' = (2.303)/(t) log (([R]_(0))/([R]_(t)))`
Where `k' = K[H_(2)O]`. Now we calculate k' using the above data. At the end of 30 minutes
`k' = (2.303)/(30 min)log((0.8500)/(0.8004))`
`= 2.004xx10^(-3)min^(-1)`
At the end of 60 minutes
`k' = (2.303)/60 min)log(0.8500)/(0.7538)`
`= 2.002xx10^(-3)min^(-1)`
At the end of 90 minutes
`k' = (2.303)/(90min)log((0.8500)/(0.7096))`
`= 2.005xx10^(-3)min^(-1)`
Since k' is more or less constant, it follows that it is a pseudo first order reaction. Its average values is equal to `2.004xx10^(-3)min^(-1)`. To determine k (second order rate constant) we use the following relationship :
`k'= K[H_(2)O]`
or `k = k'//[H_(2)O]`
`= (2.004xx10^(-3)min^(-1))/(55.5 mol L^(-1))`
`= 3.61xx10^(-3)mol^(-1) L min^(-1)`