`t_(1//2)` of a first order reaction is 15 min . Calculate the time for 80% completion of the reaction . If the initial concentration of the reactant is doubled , calculate the time taken for 80% completion of the reaction Give reasons.

Explanation

For a first reaction , `t_(1//2)=(0.693)/(k)" " ...[1]`
`thereforek=(0.693)/(t_(1//2))=(0.693)/(15)=0.0462"min"^(-1)`
Also, `t=(2.303)/(k)log.(a)/(a-x)" " ...[2]`
When the reaction is 80% complete, the remaining amount of the reactant (a - x) = (100-80) = 20%
That is, `(a)/(a-x)=(100)/(20)=5`
Substituting for k and `(a)/(a-x)` in equation [2] we get,
`t=(2.303)/(0.0462)xxlog5=(2.303)/(0.0462)xx0.699"min"=34.84"min"`
So , time required for 80% completion of reaction is 34.84min.
Last part: At a given temperature , the value of rate constant (k) of a reaction is fixed. For any initial concentration of the reactant , 80% completion of a first order reaction means,
`(a)/(a-x)=(100)/(100-80)=(100)/(20)=5`
Thus, it can be easily shown from equation [2] that if the initial concentration of the reactant is doubled , the time for 80% completion will be 34.84min , as the values of k and `(a)/(a-x)` remain unaltered.