1.

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

Answer»

zero
one
Fraction
none

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 n`t_(1//2) = 2 t_(1//2)`

`2t_(1//2)PROP(1)/([2A_0]^(n-1))""...(2)`
Dividing Eq. (2) by Eq.(1)
Dividing Eq. 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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