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Question
The half-life period for the first order reaction is 1.7 hrs. How long will it take for 20% of the reactant to disappear?
Solution
Given: Half life (t1/2) = 1.7 hours, [A]0 = 100%, [A]t = 100 − 20 = 80%
To find: Time for 20% of reactant to react = t
Formulae:
- `"t"_(1/2) = 0.693/"k"`
- `"t" = 2.303/"K" log_10 ["A"]_0/["A"]_"t"`
Calculation: `"t"_(1/2) = 0.693/"k"`
k = `0.693/"t"_(1/2) = 0.693/(1.7 "h")` = 0.4076 h−1
t = `2.303/"k" log_10 ["A"]_0/["A"]_"t"`
= `2.303/(0.4076 "h"^-1) log 100/80`
t = `2.303/(0.4076 "h"^-1) xx 0.0969`
`= 0.5475 "h" xx (60 "min")/(1 "h")`
= 32.9 min
The time required for 20% of reaction to react is 0.5475 h or 32.9 min.
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