Ncert Exercises & Solutions

4.8 In a pseudo first order reaction 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.

NEETprep Answer:

(i) Average rate of reaction between the time interval, 30 to 60 seconds,

= $\frac{0.31 - 0.17}{60 - 30}$
$\frac{0.14}{30}$
$= 4.67 \times 10^{- 3} m o l L^{- 1} s^{- 1}$

(ii) For a pseudo first order reaction,

$K = \frac{2.3.3}{t} \log \frac{[R]}{[R]}$

$f o r K_{1} = \frac{2.30 .}{30} \log \frac{0.55}{0.31 t = 30 s,}$
$= 1.311 \times 10^{- 2} s^{- 1}$
$K_{2} = \frac{2.303}{60} \log \frac{0.55}{0.17}$
$= 1.957 \times 10^{- 2} s^{- 1}$
$K_{3} = \frac{2.303}{90} \log \frac{0.55}{0.085}$
$F o r t = 60 s,$
$= 1.957 \times 10^{- 2} s^{- 1}$
$K = \frac{K_{1} + K_{2} K_{3}}{3}$
$= 1.98 \times 10^{- 2} s^{- 1}$

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