If initial concentration is doubled, the time for half-reaction is also doubled, the order of reaction is

Options

A. zero
B. first
C. second
D. third

Answer

A

Explanation

For nth order reaction, `t_(1//2)propa^(1-n)`
Here, `t_(1//2)propa`
Hence, 1 - n or n = 0 (i.e. zero order reaction)