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Question
Show that the time required for 99.9% completion of a first-order reaction is three times the time required for 90% completion.
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Solution
For a first-order reaction,
`"t" = 2.303/"k" "log"_10 ["A"]_0/["A"]_"t"`
i. Time is taken for 99.9% completion:
Let the time taken for 99.9% completion of the reaction be t99.9%.
Let initial concentration, [A]0 = a
The final concentration, [A]t = a - 99.9% of a
= `"a" - (99.9/100 xx "a") = 0.001 "a"`
t99.9% = `2.303/"k" "log"_10 ["A"]_0/["A"]_"t"`
= `2.303/"k" "log"_10 "a"/(0.001 "a")`
= `2.303/"k" "log"_10 1000` ...(1)
ii. Time is taken for 90% completion:
Let the time taken for 90% completion of the reaction be t90%.
Let initial concentration, [A]0 = a
Then, final concentration, [A]t = a - 90% of a
= a - `(90/100 xx "a") = 0.1 "a"`
t90% = `2.303/"k" "log"_10 ["A"]_0/["A"]_"t" = 2.303/"k" "log"_10 "a"/(0.1 "a")`
= `2.303/"k" "log"_10 10` ...(2)
Dividing (1) by (2), we get
`("t"_99.9%)/("t"_90%) = (2.303/"k" "log"_10 1000)/(2.303/"k" "log"_10 10) = ("log"_10 1000)/("log"_10 10) = 3/1`
∴ `("t"_99.9%)/("t"_90%) = 3`
∴ t99.9% = 3 t90%
Therefore, for a first-order reaction, the time required for 99.9% completion is 3 times that required for 90% completion.
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