The initial concentration of the reactant in a first order reaction is a . It takes time t for the completion of noth fraction of the reaction. Will it take time 2t for the completion of the same fraction if the initial concentration of the reactant is made twice ?

Explanation

Let , time taken to complete `n^(th)` fraction of the reaction is t.
`thereforet=(2.303)/(k)log.(a)/(a-(a)/(n))=(2.303)/(k)log.(n)/(n-1)`
Again, time taken to complete `n^(th)` fraction of the reaction is `t_(1)` if concentration is 2a.
`thereforet_(1)=(2.303)/(k)log.(2a)/(2a-(2a)/(n))=(2.303)/(k)log.(n)/(n-1)=t`
Thus, the time required for a first order reaction to complete a certain fraction of the initial concentration, does not depend on the substrate concentration . In both the cases where the initial concentrations are a and 2a respectively , time required is same .