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.

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 this equation?
Rate `=k'[CH_(3) COOCH_(3) ][H_(2)O]`

Explanation

For pseudo first order raction, the reaction should be first order with respect to ester when `[H_(2)O]` is constant. The rate constant k for pseudo first order reaction is
`k=(2.303)/(t)log ""(C_(0))/(C)` where` k=k' [H_(2)O]`
From the above data we note

It can be seen that k' `[H_(2)O]` is constant and equal to `2.004 xx10^(-3) min ^(-1)` and hence, it is pseudo first order reaction. We can now determine k from
`k'[H_(2)O] =2.004 xx10^(-3) min ^(-1)`
`k'[55 mol L ^(-1)]=2.004xx10^(-3)min ^(-1)`
`k'=3.64 xx10^(-5)mol ^(-1)L min ^(-1)`