In a pseudo first order hydrolysis of ester in water the following results were obtained:
`{:(t//s,0,30,60,90),(["Ester"],0.55,0.31,0.17,0.085):}`
(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.

Explanation

(i) Average rate of reaction between the time interval , 30 to 60 seconds , = `(d ["Ester"])/("dt")`
= `(0.31 - 0.17)/(60 - 30)`
`= (0.14)/(30)`
`= 4. 67 xx 10^(-3) "mol" L^(-1) s^(-1)`
(ii) For a pseudo first order reaction ,
`k = (2.303)/(t) "log" ([R]_(0))/([R])`
For t = 30 s , `k_(2) = (2.303)/(30) "log" (0.55)/(0.31)`
`= 1.911 xx 10^(-2) s^(-1)`
For t = 60 s `k_(2) = (2.303)/(60) "log" (0.55)/(0.17)`
`= 1.957 xx 10^(-2) s^(-1)`
For t = 90 s , `k_(3) = (2.303)/(90) "log" (0.55)/(0.085)`
=`2.075 xx 10^(-2) s^(-1)`
Then , average rate constant , `k = (k_(1) + k_(2) + k_(3))/(3)`
= `((1.911 xx 10^(-2)) + (1.957 xx 10^(-2)) + (2.075 xx 10^(-2)))/(3)`
`= 1.98 xx 10^(-2) s^(-1)`