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Question
In a first-order reaction, 10% of the reactant is consumed in 25 minutes. Calculate:
(1) The half-life period of the reaction.
(2) The time required for completing 87.5% of the reaction.
Solution
(1)
t = 25 min
a = 100
a - x = 90
k = ?
For a first-order reaction,
k = `2.303/t log a/(a-x)`
= `2.303/25 log 100/90`
= `2.303/25 [log 10 - log9]`
k = `2.303/25 [1.000 - 0.9542]`
= `2.303/25 xx 0.0458`
k = 0.0042 min-1
`t_(1/2) = 0.693/k`
= `0.693/0.0042`
= 165 min
(2)
`t_(87.5%) = 2.303/k log 100/(100 - 87.5)`
= `2.303/0.0042 log 100/12.5`
= `2.303/0.0042 xx log 8`
`t_(87.5%)` = 495 min
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