1.

In a reaction the initial concentration of the reactants increase four fold and the rate becomes eight times its initial value . The order of the reaction is

Answer»

`2.0`
`3.5`
`2.5`
`1.5`

Solution :Rate = `KC^(n)`
Or r = `KC^(n) ""… (i)`
`8R = K(4C)^(n) "" … (ii)`
Dividing EQ. (ii) by (i) , we GET
`2^(3) = 2^(2N)`
or 2 n =3
n = 1.5.


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