Question

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

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

• A. $3.0\times 10^{-2}{s}^{-1}$
• B. $1.98\times 10^{-2}{s}^{-1}$
• C. $4\times 10^{-3}{s}^{-1}$
• D. $5.20\times 10^3{s}^{-1}$

Answer 1

Answer: (B)

$k=\frac{2.303}{t}log\frac{\left[R_0\right]}{\left[R\right]}where\left[R_0\right]=0.55M$
$At30sec{k}_1=\frac{2.303}{t}log\frac{0.55}{0.31}=1.91\times 10^{-2}{s}^{-1}$
$At60sec{k}_2=\frac{2.303}{60}log\frac{0.55}{0.17}=1.96\times 10^{-20}{s}^{-1}$
$At90sec{k}_3=\frac{2.303}{90}log\frac{0.55}{0.085}=2.07\times 10^{-2}{s}^{-1}$
$\frac{\left(1.91\times 10^{-2}\right)+\left(1.96\times 10^{-2}\right)+\left(2.07\times 10^{-2}\right)}{3}$
$=1.98\times 10^{-2}{s}^{-1}$