Question

For the hydrolysis of methyl acetate in aqueous solu- tion, the following results were obtained.

(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 10 to 20 s.

(Given: log2 = 0.3010, log4 = 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 = \(\frac{2.303}{10} \ log\ \frac{[R]_0}{[R]}\)

where, [R]₀ = initial concentration of the reactant 

[R] = final concentration of the reactant

At t₁ = 10s

k₁ = \(\frac{2.303}{10} \ log\ \frac{0.10}{0.05}\)

= 6.93 x 10^-2 s^-1

t₂ = 20s

k₂ = \(\frac{2.303}{10} \ log\ \frac{0.10}{0.025}\)

= 6.93 x 10^-2 s^-1

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 10-20s is given by

Average rate = \(\frac{-\triangle[Easter]}{\triangle t}\)

= \(-(\frac{0.025 - 0.05}{20 - 10})\)

= 0.0025 mol L^-1 s^-1