Question

Saponification of ethyl acetate by $NaOH$ is a second order reaction with rate constant $K=6.36litremo{l}^{-1}mi{n}^{-1}$ at ${25}^{o}C$.
The initial rate or the reaction when the base and the ester have concentration is $=0.02mollitr{e}^{-1}$?
What will be the rate of reaction, 10 minute after the commencement of the reaction?

• A. (a) $2.544\times {10}^{-3}mollitr{e}^{-1}mi{n}^{-1},$ (b) $4.925\times {10}^{-4}mollitr{e}^{-1}mi{n}^{-1}$
• B. (a) $2.544\times {10}^{-4}mollitr{e}^{-1}mi{n}^{-1},$ (b) $49.25\times {10}^{-4}mollitr{e}^{-1}mi{n}^{-1}$
• C. (a) $25.44\times {10}^{-3}mollitr{e}^{-1}mi{n}^{-1},$ (b) $4.925\times {10}^{-5}mollitr{e}^{-1}mi{n}^{-1}$
• D. None of these

Answer 1

Answer: (A)

(a) $2.544\times {10}^{-3}mollitr{e}^{-1}mi{n}^{-1},$ (b) $4.925\times {10}^{-4}mollitr{e}^{-1}mi{n}^{-1}$

(a) The initial rate or the reaction when the base and the ester have concentration
$rate=k\left[A{]}_0\right[B{]}_0=6.36litremo{l}^{-1}mi{n}^{-1}\times 0.02mollitr{e}^{-1}\times 0.02mollitr{e}^{-1}=0.002544mollitr{e}^{-1}mi{n}^{-1}$
(b) Since the concentrations of two reactants are same, the reaction $A+B\to products$ can be approximated to $2A\to products$
Hence, the following expression can be applied.
$\frac{1}{\left[A\right]}=\frac{1}{[A{]}_0}+kt$
$\frac{1}{\left[A\right]}=\frac{1}{0.02}+6.36\times 10=56.36$
$\left[A\right]=0.0177M$
The rate of reaction, 10 minute after the commencement of the reaction,
$rate=k\left[A\right]\left[B\right]=6.36litremo{l}^{-1}mi{n}^{-1}\times 0.0177mollitr{e}^{-1}\times 0.0177mollitr{e}^{-1}=0.002002mollitr{e}^{-1}mi{n}^{-1}$