The half life for a given reaction was doubled as the initial concentration of a reactant was doubled. What is the order of reaction.

Options

A. 0
B. 4
C. 1
D. 2

Answer

A

Explanation

For `n^(th)` order reaction
`t_(1//2)prop(1)/(a^(n-1))`
Since half life is doubled when the initial concentration `(a)` of the reaction is doubled, this implies that half-life is directly proportional to the initial concentration:
`t_(1//2)propa` or `t_(1//2)prop(1)/(a^(-1))`
Comparing the two expressions we get
`n-1=-1`
or `n=0`