For the following rate law determine the unit of rate constant.
Rate `k[A]^((1)/(2))[B]^(2)=[R]^((5)/(2))`

Explanation

The total order of reaction `n = 1/2 + 2 = 5/2 = 2.5 `
Rate `K[A]^(1/2) [B]^(2)= [R]^(5/2)`
`therefore k=(Rate)/([5]^((5)/(2)))`
`therefore` unit of k=`("unit of rate")/(("unit of concentration")^((5)/(2)))`
`=((mol L^(-1))^(1)s^(-1))/((mol L^(-1))^((5)/(2)))`
`(mol L^(-1))^(-(3)/(2))s^(-1)`
`=(mol)^(-(3)/(2))(L^(-1))^(-(3)/(2))S^(-1)`
`=(mol)^(+(3)/(2))(L)^(-(3)/(2))S^(-1)`
If the order of reaction =`(5)/(2)` then unit of rate constant k is `L^((+3)/(2)) mol^((-3)/(2))s^(-1)`