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 during the interval 30-60 sec.
`=(C_(2)-C_(1))/(t_(2)-t_(1))=(0.31-0.17)/(60-30)=(0.14)/(30)" mol L"^(-1)s^(-1)=4.67xx10^(-3)" mol L"^(-1)s^(-1)`
(ii) `k'=(2.303)/(t)log""([A_(0)])/([A])" in which "[A_(0)]=0.55" M"`
`t=30" sec",k'=(2.303)/(30s)log""(0.55)/(0.31)=1.91xx10^(-2)s^(-1)`
`t=60" sec",k'=(2.303)/(60s)log""(0.55)/(0.17)=1.96xx10^(-2)s^(-1)`
`t=90" sec",k'=(2.303)/(90s)log""(0.55)/(0.085)=2.07xx10^(-2)s^(-1)`
Average `k'=(1.91+1.96+2.07)/(3)xx10^(-2)=1.98xx10^(-2)s^(-1).`