In a pesudo first order hydrolysis of ester in water, the following results are obtained

Calculate the averagte rate of reaction between the time inverval 30 to 60 s.

Explanation

i) Average rate during the interval `30-60sec . =-(c_(2)-c_(1))/(t_(2)-t_(1))`
`=((0.17-0.3)mol L^(-1))/((60-30)s)`
`=(0.14)/(30)mol L^(-1) s^(-1)`
Average rate `=4.67xx10^(-3) mol L^(-1) s^(-1)`
ii) `k'=(2.303)/(t)log ""([A_(0)])/([A])` in which `[A_(0)]=0.55M`
At `=30s,k'=(2.303)/((30s))log ""(0.55)/(0.17)=1.*91xx10^(-2) s^(-1)`
At `t=60s, k'=(2.303)/((60s))log ""(0.55)/(0.17)=1.96xx10^(-2)s^(-1)`
At `t=90 s,k'=(2.303)/((90s))log ""(0.55)/(0.085)=2.07xx10^(-2)s^(-1)`
Average `k' =((1.91+1.96+2.07)xx10^(-2)s^(-1))/(3)`
Pseudo first order rate constant `=1.98 xx10^(-2) s^(-1)`