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Question
Solve
The half-life of a first-order reaction is 1.7 hours. How long will it take for 20% of the reactant to react?
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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:
i. t1/2 = `0.693/"k"`
ii. `"t" = 2.303/"K" "log"_10 ["A"]_0/["A"]_"t"`
Calculation:
t1/2 = `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 32.9 min.
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