Question

For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :

t/s	0	30	60
[CH3COOCH3] / mol L–1	0.60	0.30	0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.

(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(Given log 2 = 0.3010, log 4 = 0.6021)

Answer 1

(i) For the hydrolysis of methyl acetate to be a pseudo first-order reaction, the reaction should be first order with respect to ester when [H₂O] is constant. The rate constant k for a first order reaction is given by

`k=2.303/t`

where[R]_o = Initial concentration of reactant

[R] = Final concentration of reactant

At t₁ = 30 s,

`k_1=2.303/30`

At t₂ = 60 s,

`k_2=2.303/60`

It can be seen that the rate constant k for the reaction has a constant value under any given time interval. Hence, the given reaction follows the pseudo first-order kinetics.

 

(ii) Average rate of reaction between the time interval of 30–60 seconds is given by

`Rate=-(Delta[R])/(Deltat)`

         `=-((0.15-0.30)/(60-30))`

         `=0.15/30=0.005`