In a pseudo first order hydrolysis of an ester in water, the following results were obtained : {:("t/s",,,0,,,30,,,60,,,90),(["Ester"]//"mol L"^(-1),,,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.

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

1.	In a pseudo first order hydrolysis of an ester in water, the following results were obtained : {:("t/s",,,0,,,30,,,60,,,90),(["Ester"]//"mol L"^(-1),,,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.
Answer» Solution :(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).`