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. three
B. zero
C. one
D. two

Answer

B

Explanation

For `nth` order reaction
`t_(50) prop ((1)/(a))^(n-1)`
`(t_(2))/(t_(1)) = ((a_(1))/(a_(2)))^(n-1)`
`(t_(2))/(t_(1)) = 2, ((a_(1))/(a_(2))) = (1)/(2)`
`(2) = ((1)/(2))^(n-1)`
`((1)/(2))^(-1) = ((1)/(2))^(n-1)`
`:. n-1 = -1`
`n = 0`