In a pseudo first order hydrolysis of ester in water the following result6s 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 `=(Delta[ester])/(Deltatime)`
`=(0.17-0.31)/(60-30)=-4.67xx10^(-3) mol L^(-1) s^(-1)`
(ii) `K=2.303/t "log" a/((a-x))`
Where `a` is initial conc. At time t.
`K_(1)=2.303/30"log"0.55/0.31=1.91xx10^(-2)`
`K_(2)=2.303/60"log"0.55/0.17=1.96xx10^(-2)`
`K_(3)=2.303/90"log"0.55/0.085=2.01xx10^(-2)`
Thus, `K=(K_(1)+K_(2)+K_(3))/(3)`
`=(1.91xx10^(-2)+1.96xx10^(-2)+2.01xx10^(-2))/(3)`
`=1.96xx10^(-2)`