Question

For a given reaction `A rarr ` Product, rate is `1xx10^(-4)M s^(-1)` when `[A]=0.01M` and rate is `1.41xx10^(-4)M s^(-1)` when `[A]=0.02 M`. Hence, rate law is `:`
A. `-(d[A])/(dt)=k[A]^(2)`
B. `-(d[A])/(dt)=k[A]`
C. `-(d[A])/(dt)=(k)/(4)[A]`
D. `-(d[A])/(dt)=k[A]^(1//2)`

Answer 1

Correct Answer - 4
`A rarr `Product
We know, Rate `=K[` conc.`]^(n)`
`1xx10^(-2)=K[.01]^(n)" ".....(i)`
`1.41xx10^(-4)=K[.02]^(n)" "....(ii)`
`(i)//(ii)" "(1)/(1.41)=((1)/(2))^(n)`
`n=(1)/(2)`
Then `(-d(A))/(dt)=K[A]^(1//2)`