Question

For the hydrolysis of methyl acetate in aqueous solution the above tabulated results were obtained:
(a) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(b) Calculate the average rate of reaction between the time interval ${\text{10}}$ to ${\text{20}}$ seconds.
(Given: $\log 2 = 0.3010,\,\log 4 = 0.6021$)

${\text{t/s}}$	0	10	20
$\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{H}}_{\text{3}}}} \right]{\text{/mol }}{{\text{L}}^{ - 1}}$	0.10	0.05	0.025

Answer 1

Hint: A bimolecular reaction behaves like a first order reaction is known as a pseudo-first order reaction. A reaction is said to be a pseudo first order reaction when one reactant is present in excess.

Formula Used: $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
${\text{Average rate}} = - \dfrac{{{\text{Change in concentration}}}}{{{\text{Change in time}}}}$

Complete step by step answer:
(a) For the hydrolysis of methyl acetate, the reaction will be a pseudo first order reaction if it is first order with respect to methyl acetate when the concentration of water is constant.
The equation for the rate constant of a first order reaction is,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ a \right]}^0}}}{{\left[ a \right]}}$
Where k is the rate constant of a first order reaction,
t is time,
${\left[ a \right]^0}$ is the initial concentration of the reactant,
$\left[ a \right]$ is the final concentration of the reactant.
At $t = 10{\text{ s}}$:
${k_1} = \dfrac{{2.303}}{{10{\text{ s}}}}\log \dfrac{{0.10}}{{0.05}}$
${k_1} = \dfrac{{2.303}}{{10{\text{ s}}}}\log 2$
${k_1} = \dfrac{{2.303}}{{10{\text{ s}}}} \times 0.3010$
${k_1} = 0.0693{\text{ }}{{\text{s}}^{ - 1}}$
Thus, at $t = 10{\text{ s}}$, the rate constant of first order reaction is $0.0693{\text{ }}{{\text{s}}^{ - 1}}$.
At $t = 20{\text{ s}}$:
${k_2} = \dfrac{{2.303}}{{20{\text{ s}}}}\log \dfrac{{0.10}}{{0.025}}$
${k_2} = \dfrac{{2.303}}{{20{\text{ s}}}}\log 4$
${k_2} = \dfrac{{2.303}}{{20{\text{ s}}}} \times 0.6021$
${k_2} = 0.0693{\text{ }}{{\text{s}}^{ - 1}}$
Thus, at $t = 20{\text{ s}}$, the rate constant of first order reaction is $0.0693{\text{ }}{{\text{s}}^{ - 1}}$.
We can say that at any time interval, the value of rate constant for the reaction is constant. Thus, the reaction is a pseudo first order reaction.
(b) Calculate the average rate of reaction between the time interval ${\text{10}}$ to ${\text{20}}$ seconds as follows:
${\text{Average rate}} = - \dfrac{{{\text{Change in concentration}}}}{{{\text{Change in time}}}}$
The change in time $ = \left( {20 - 10} \right){\text{s}} = 10{\text{ s}}$
Change in concentration $ = \left( {0.025 - 0.05} \right){\text{mol }}{{\text{L}}^{ - 1}} = - 0.025{\text{ mol }}{{\text{L}}^{ - 1}}$
Substitute $10{\text{ s}}$ for the change in time, $ - 0.025{\text{ mol }}{{\text{L}}^{ - 1}}$ for the change in concentration. Thus,
${\text{Average rate}} = - \dfrac{{ - 0.025{\text{ mol }}{{\text{L}}^{ - 1}}}}{{10{\text{ s}}}}$
${\text{Average rate}} = 2.5 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$
Thus, the average rate of reaction between the time interval ${\text{10}}$ to ${\text{20}}$ seconds is $2.5 \times {10^{ - 3}}{\text{ mol }}{{\text{L}}^{ - 1}}{\text{ }}{{\text{s}}^{ - 1}}$.

Note: The unit of rate constant for first order reaction is ${{\text{s}}^{ - 1}}$. The units do not contain concentration terms. Thus, we can say that the rate constant of a first order reaction is independent of the concentration of the reactant.