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

Options

A. zero
B. one
C. Fraction
D. none of these

Answer

A

Explanation

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