Saponification of ethyl acetate by NaOH is a second order reaction with rate constant K $ = \,6.36\,litre\,mo{l^{ - 1}}\,{\min ^{ - 1}}$at ${25^\circ }C$.
The initial rate of the reaction when the base and the ester have concentration $ = \,0.02\,mol\,litr{e^{ - 1}}?$
What will be the rate of reaction $10$minutes after the commencement of the reaction$?$
A) $2.544 \times {10^{ - 3}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}$$\left( b \right)\,4.925 \times {10^{ - 4\,}}mol\,litr{e^{ - 1}}\,{\min ^{ - 1\,}}$
B)$\,2.544 \times {10^{ - 4}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}\,\left( b \right)\,49.25 \times {10^{ - 4}}\,mol\,litr{e^{ - 1\,}}\,{\min ^{ - 1}}$
C) $\,25.44 \times {10^{ - 3}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}\,\left( b \right)\,4.925 \times {10^{ - 5}}\,mol\,litr{e^{ - 1}}\,{\min ^{ - 1}}$
D) None of these

Hint:‌The reaction is a second order reaction thus we can use differential rate law for second order reactions to calculate initial rate and rate after time t and also, we can use integrated rate law expression for second order reactions to calculate concentration of the reactants after time t.As clearly provided in the question, second order rate kinetics is involved in this question.


Complete step by step solution:
So, the Initial rate using differential rate law expression is
 ${r_0} = k{\left[ A \right]_0}{\left[ B \right]_0}$$6.36\,litre\,mo{l^{ - 1}}{\min ^{ - 1}} \times 0.02\,mol\,litr{e^{ - 1}} \times 0.02\,mol\,litr{e^{ - 1\,}} = 2.544 \times {10^{ - 3}}mol\,litr{e^{ - 1\,}}{\min ^{ - 1}}$
Now, Concentration of reactants after time t=10 minutes is calculated using integrated rate law expression is $\dfrac{1}{{{A_t}}} = \dfrac{1}{{{A_0}}} + kt$
$\dfrac{1}{{{A_t}}} = \dfrac{1}{{0.02}} + \left( {6.36 \times 10} \right)\,litre\,mo{l^{ - 1}}$
$\dfrac{1}{{{A_t}}} = 50 + 63.6$
${A_t} = \dfrac{1}{{113.6}}\,mol\,litr{e^{ - 1}}$
Concentration of reactants after time t=10 minutes =$\dfrac{1}{{113.6}}mol\,litr{e^{ - 1}}$$ = 0.0088\,mol\,litr{e^{ - 1}}$
Also, Rate after time t=10 minutes
$r = k\left[ A \right]\left[ B \right]$
$6.36 \times 0.0088 \times 0.0088 = 4.925 \times {10^{ - 4\,}}\,mol\,litr{e^{ - 1}}{\min ^{ - 1}}$

The correct option is Option A.

Additional Information:For a second order reaction,
$A\, + B\xrightarrow{{}}\,P$
The differential rate equation is given as follows:
$Rate = \dfrac{{ - d\left[ A \right]}}{{dt}} = \dfrac{{ - d\left[ B \right]}}{{dt}} = \dfrac{{d\left[ P \right]}}{{dt}} = k\left[ A \right]\left[ B \right]$
On, integrating the above equation, we get
$\frac{1}{{{A_t}}} = \dfrac{1}{{{A_0}}} + kt$

Note:We use the differential rate law expression and integrated rate law expression to calculate rate of the reaction and concentration of the reactants after time t.The overall order of the reaction is 2, first order with respect to A(NaOH) and first order with respect to B(ester) so the total order of the reaction is 2.