Question

For the hydrolysis of methyl acetate in aqueous solution, the above tabulated results were obtained:
(i)- Show that if follows a pseudo first order reaction as the concentration of water remains constant.
(ii)- Calculate the average rate of reaction between the time intervals of 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)

t/s	0	30	60
$C{{H}_{3}}COOC{{H}_{3}}/mol{{L}^{-1}}$	0.60	0.30	0.15

Answer 1

Hint: To prove the reaction is a pseudo first order reaction, we have to calculate the rate constant of the reaction by using the formula given below:
$K=\dfrac{2.303}{t}\log \dfrac{{{[A]}_{o}}}{[A]}$
Where t is the time, ${{[A]}_{o}}$ is the initial concentration of the reactant, and [A] is the final concentration of the reactant, if the rate constant comes out to be the same then it is pseudo first order reaction.
The average rate of the reaction can be calculated by:
$K=\dfrac{-\Delta [C{{H}_{3}}COOC{{H}_{3}}]}{\Delta t}$

Complete answer:
(a)- The given reaction is the hydrolysis of methyl acetate, and we have to prove that this reaction is a pseudo first order reaction, in which the concentration of water is constant.
To prove the reaction is a pseudo first order reaction, we have to calculate the rate constant of the reaction by using the formula given below:
$K=\dfrac{2.303}{t}\log \dfrac{{{[A]}_{o}}}{[A]}$
Where t is the time, ${{[A]}_{o}}$ is the initial concentration of the reactant, and [A] is the final concentration of the reactant.
In first case,
t = 30 – 0 = 30 sec
${{[A]}_{o}}$ = 0.60
[A] = 0.30
Putting the values, we get:
$K=\dfrac{2.303}{30}\log \dfrac{0.60}{0.30}$
$K=0.07677\text{ x }\log 2$
$K=0.0231{{s}^{-1}}$
Now, for the second case,
t = 60 – 30 = 30 sec
${{[A]}_{o}}$ = 0.30
[A] = 0.15
Putting the values, we get:
$K=\dfrac{2.303}{30}\log \dfrac{0.30}{0.15}$
$K=0.07677\text{ x }\log 2$
$K=0.0231{{s}^{-1}}$
So, in both cases rate constant (K) is the same, hence it is a pseudo first order reaction.
(b)- In this we have to find the average rate of the reaction between time interval 30 to 60 sec, we have to use the formula:
$K=\dfrac{-\Delta [C{{H}_{3}}COOC{{H}_{3}}]}{\Delta t}$
At time 30 sec the concentration is 0.30 and at time 60 sec the concentration is 0.15, therefore,
$\Delta [C{{H}_{3}}COOC{{H}_{3}}]$ will be (0.15 – 0.30 )
$\Delta t$ will be 60 – 30 sec
Putting the values, we get:
$K=-\dfrac{(0.15-0.30)}{(60-30)}$
$K=\dfrac{0.15}{30}=0.005\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}$
Therefore, the average rate of reaction is $0.005\text{ mol }{{\text{L}}^{-1}}\text{ }{{\text{s}}^{-1}}$.

Note:
Pseudo first order reaction means there are two reactants in the reaction but the rate depends only on one reactant but not on the second one because the concentration of the second reactant is very large due to which any changes will not affect the rate of the reaction.