The decomposition of a compound is found to follow the first order rate law. If it takes 15 minutes for 20 per cent of the original material to react, calculate
i) the specific rate constant
ii) the time in which 10 percent of the original material remains unreacted.
iii) The time it takes for the next 20 percent of the reactant left to react.

Answer

(i) 0.01488 `min^(-1)``" "` (ii) 154.77 `min^(-1)``" "` (iii) 15 `min^(-1)`

Explanation

(i) `k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(15" min")log""(a)/(a-0.20a)=0.01488" min"^(-1)`
(ii) `t=(2.303)/(k)log""(a)/(0.10a)=(2.303)/(0.01488)log10=154.77" min".`
(iii) In first 15 minutes, `20%` has reacted. `:.` Amount left unreacted `=80%` of `a=0.80` a
This is now initial concentration. For next `20%` of this amount to react means `x=20%` of `0.80a=0.16a`
`:." "t=(2.303)/(k)log""(0.80a)/(0.80a-0.16a)=(2.303)/(0.01488" min"^(-1))log""(0.80)/(0.64)`
`=154.77" min "xxlog1.25=154.77xx0.0969" min "=15 "minutes"`