1.

For a given reaction, Ararr"Product, rate is" 1xx10^(-4)"M s^(-1) when [A] = 0.01 M and rate is 1.41xx10^(-4)"M s"^(-1) when[A] = 0.02 M. Hence, rate law is :

Answer»

`-(d[A])/(DT)=K[A]^(2)`
`-(d[A])/(dt)=k[A]`
`-(d[A])/(dt)=(k)/(4)[A]`
`-(d[A])/(dt)=k[A]^(1//2)`

ANSWER :D


Discussion

No Comment Found

Related InterviewSolutions