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 a pseudo first order reaction, the reaction should be first order w.r.t. ester when `[H_(2)O]` is constant `k` for pseudo first order reaction is:
For first order reaction: rate `= k[CH_(3)COOCH_(3)]` …(i)
For pseudo first order reaction:
Rate `= k'[CH_(3)COOCH_(3)][H_(2)O]`
Compare Eqs. (i) and (ii),
`k[CH_(3)COOCH_(3)] = k'[CH_(3)COOCH_(3)][H_(2)O]`
`:. k = k' [H_(2)O]`
Thus rate constant `k` for pseudo first order reaction is:
`k = (2.303)/(t)log.(c_(0))/(c_(t))`, where `c_(0)` is the initial concentration of ester and `c_(1)` is the concentration of ester left after time `t`.
Thus, when `t = 30 min`
`k = (2.3)/(30) log.((0.85)/(0.8004)) = 2.004 xx 10^(-3) min^(-1)`
When `t = 60 min`
`k = (2.3)/(60) log.((0.85)/(0.7538)) = 2.002 xx 10^(-3) min^(-1)`
When `t = 90 min`
`k = (2.3)/(90) log.((0.85)/(0.7096)) = 2.005 xx 10^(-3) min^(-1)`
It can be seen that `k'[H_(2)O]` is constant and equal to `2.004 xx 10^(-3) min^(-1)` and hence it is pseudo first order reaction. Thus, `k'` can be determined as follows:
`k'[H_(2)O] = 2.004 xx 10^(-3) min^(-1)`
`k'(55 min L^(-1)) = 2.004 xx 10^(-3) min^(-1)`
`:. k' = (2.004 xx 10^(-3) min^(-1))/(55 mol L^(-1)) = 3.64 xx 10^(-5) mol^(-1) L min^(-1)`
Note: Units of k' represent second order reaction.