Question

In a pseudo first order hydrolysis of ester in water, the following results are obtained:

t/sec	0	30	60	90
[ester] M	0.55	0.31	0.17	0.085

(i) Calculate the average rate of reaction between the first time intervals 30 to 60 second.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer 1

Hint:.i.) Rate is defined as the speed at which a chemical reaction occurs. Rate is generally expressed in the terms of concentration of reactant which is consumed during the reaction in a unit of time or the concentration of product which is produced during the reaction in a unit of time.
ii.) Pseudo first order reactions are the reactions which are not the first order reaction but approximated or appear to be of first order due to the higher concentration of the reactants than other reactants.

Complete step by step answer:
The data is given in the question
Steps to solve the first part: the average rate of reaction between the time interval of 30 to 60 second is calculated by subtracting the ratio of concentration and time between the interval 30 to 60 sec.
Average rate of reaction = $\dfrac{0.31-0.17}{60-30} = 4.67({{10}^{-3}})M/s$
Hence the average rate of reaction between the first time intervals 30 to 60 second = $4.67({{10}^{-3}})M/s$
Steps to solve the second part: rate constant is calculated using the given formulae:
\[k=\dfrac{2.303}{t}\log \dfrac{{{[ester]}_{0}}}{[ester]}\]
Now we have to calculate the value of rate constant at different time intervals at 30s and 60s and 90s

Rate constant when t = 30s
\[k=\dfrac{2.303}{t}\log \dfrac{[0.55]}{[0.31]}=1.91({{10}^{-2}})/s\]

Rate constant when t = 60s
\[k=\dfrac{2.303}{t}\log \dfrac{[0.55]}{[0.17]}=1.91({{10}^{-2}})/s\]

Rate constant when t = 90 s
\[k=\dfrac{2.303}{t}\log \dfrac{[0.55]}{[0.085]}=2.07({{10}^{-2}})/s\]

So the value of average rate constant will be = \[\dfrac{1.91({{10}^{-2}})/s+1.91({{10}^{-2}})/s+2.07({{10}^{-2}})/s}{3} = 1.98({{10}^{-2}})/s\]
So the value of average rate constant will be\[1.98({{10}^{-2}})/s\]

Note: If the reaction is a third order reaction, the unit for third order reaction is ${{M}^{-2}}h{{r}^{-1}}or\text{ mo}{{\text{l}}^{-2}}{{L}^{2}}h{{r}^{-1}}$. The negative and positive sign in the expression of the rate or reaction only means the change in concentration. A negative charge indicates that the concentration of the reactant is decreasing, similarly a positive charge means that the concentration of product is increasing.