1.

Pseudo first order rate for the reaction, A+B toProduct, when studied in 0.1M of B is given by -(d[A])/(dt)=k[A], where , k=1.85xx10^(4)sec^(-1), calculate the value of second orderrate constant.

Answer»

Solution :`A+B to `Product
`-(d[A])/(DT)=K[A]`
`-(d[A])/(dt)=1.85xx10^(4)XX[A]`…………………..`(i)`
Assuming the reaction to be of second ORDER
`-(d[A])/(dt)=k.[A][B]`
`-(d[A])/(dt)=k.[A][0.1]` …………..`(ii)`
Dividing Eq. `(i)` by `(ii)`, we get
`1=(1.85xx10^(4))/(k.xx(0.1))`
`:.k.=1.85xx10^(5)Lmol^(-1)s^(-1)`


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