In a pseudo first order hydrolysis of 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):}`
Calculate the pseudo first order rate constant for the hydrolysis of ester.

Explanation

For a pseudo first order reaction, rate constant,
`k'=(2.303)/(t)log.([A]_(0))/([A]), "where"[A]_(0)=0.55M` (given)
at `t=30s,k' = (2.303)/(30s)log.(0.55)/(0.31)=1.91xx10^(-2)s^(-1)`
at `t=60s,k' = (2.303)/(60s)log.(0.55)/(0.17)=1.96xx10^(-2)s^(-1)`
at `t=90s,k' = (2.303)/(90s)log.(0.55)/(0.085)=2.07xx10^(-2)s^(-1)`
`therefore"Average "k'=[(1.91+1.96+2.07)/(3)]xx10^(-2)`
`= 1.98xx10^(-2)s^(-1)`