1.

In a pesudo first order hydrolysis of ester in water, the following results were obtained. {:("1",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.

Answer»

Solution :For a PSEUDO FIRST order reaction,
`k=(2.303)/(t)"log"([R]_(0))/([R])`
For, `t = 303`
`k_(1)=(2.303)/(t)"log"(0.55)/(0.31)`
`=1.911xx10^(-2)s^(-1)`
For, `t = 603`
`k_(2)=(2.303)/(60)"log"(0.55)/(0.17)`
`=1.957xx10^(-2)s^(-1)`,
For, `t = 90 s`
`k_(3)=(2.303)/(90)"log"(0.55)/(0.085)`
`=2.075xx10^(-2)s^(-1)`.
Then, average rate constant,
`k=(k_(1)+k_(2)+k_(3))/(3)`
`=((1.911xx10^(-2))+(1.957xx10^(-2))(2.075xx10^(-2)))/(3)`
`= 1.98xx10^(-2)s^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions