When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is

Options

A. zero
B. first
C. second
D. more than zero but less than first

Answer

B

Explanation

for a zero order reaction `t_(1//2)` is directly proportional to the initial concentration of the reactant ,`[R]_(0)`
`t_(1//2)prop[R]_(0)`
For a first order reaction ,
`k=(2.303)/(t)"log"([R]_(0))/([R])`
`"at" " t"_(1//2),[R]=[R]_(0)/(2)`
So, the above question becomes
`k=(2.303)/(t_(1//2))"log"[R]_(0)/([R]_(0)//2)`
`t_(1//2=(2.303)/(k)"log"2=(2.303)/(k)xx.3010=(0.693)/(k))`
i.e. half-life period is independent of initial concentration of a reactant.