1.

The rate constant for forward and backward reactions of hydrolysis of ester are 1.1xx10^(-2) and 1.5xx10^(-3) per minute respectively. Equilibrium constant for the reaction is CH_(3)OOC_(2)H_(5)+H_(2)OhArrCH_(3)COOH+C_(2)H_(5)OH

Answer»

`4.33`
`5.33`
`6.33`
`7.33`

Solution :`K_(f)=1.1xx10^(-2),K_(B)=1.5xx10^(-3)`
`K_(C)=(K_(f))/(K_(b))=(1.1xx10^(-2))/(1.5xx10^(-3))=7.33.`


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