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) Calcualte the pseudo first order rate constant for the hydrolysis of ester.

In a pseudo first order hydrolysis of ester in water, the following re

1.	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) Calcualte the pseudo first order rate constant for the hydrolysis of ester.
Answer» SOLUTION :Average rate of reaction = `("Change in molar concentration")/("Change in time INTERVAL")` `=(0.17 - 0.31)/(60-30) = (-0.14M)/(30s) = -4.67 xx 10^(-3)MS^(-1)` the rate constant, `K=(2.303)/t log[A_(0)]/[A], [A_(0)]=0.55 M` For t=30s , `k_(1)=2.303/(30s) log[0.55]/[0.31]= 2.303/(30s) xx 0.249 = 1.91 xx 10^(2)s^(-1)`. For t=60s, `k_(2) = 2.303/(60s) log [0.55]/[0.31]=2.303/(60s)xx0.5099 = 1.96 xx 10^(-2)s^(-1)` For t=60s , `k_(2)=2.303/(60s), k_(2) = 2.303/(60s)log[0.55]/[0.17]=2.303/(60s) xx 0.5099 = 1.96 xx 10^(-2)s^(-1)`. For t=90 s, `k_(3)=2.303/(90S)log[0.55]/[0.085]=2.303/(90s) xx 0.811 = 2.07 xx 10^(-2)s^(-1)`. Average rate constant (k) = `(k_(1)+k_(2)+k_(3))/(3) = (1.91 + 1.96+2.07)/(3) xx 10^(-2)s^(-1)` `=1.98 xx 10^(-2)s^(-1)`.