1.

If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is

Answer»

zero
one
Fraction
none of these

Solution :For a first ORDER reaction `t_(1//2)` is independent of initial CONCENTRATION.
i.e., `N NE 1` , for such cases
`t_(1//2)prop1/([A_0]^(n-1))""....(1)`
If `[A_0]=2[A_0]`,then `t_(1//2)=2t_(1//2)`
`2t_(1//2)prop1/([2A_0]^(n-1))""...(2)`
Dividing Eq. (2) by Eq. (1)
`2=1/([2A_0]^(n-1))xx([A_0]^(n-1))/1=([A_0]^(n-1))/([2A_0]^(n-1))`
`2 = (1/2)^(n-1)=(2^(-1))^(n-1)`
`2^1=(2^(-n+1))`
n = 0


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