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Questions
Show that the time required for 99% completion is double of the time required for the completion of 90% reaction.
For a first-order reaction, derive the relationship t99% = 2t90%.
Solution
For Ist order reaction k = `2.303/"t" log (["A"_0])/(["A"_"t"])`
For 99% completion: If [A0] = 100
[At] = 100 − 99
= 1
∴ t99% = `2.303/"k" log 100/1`
t99% = `2.303/"k" xx 2` .........(1)
∵ log 100 = 2
For 90% completion: If [A0] = 100
[At] = 100 − 90
= 10
t90% = `2.303/"k" log 100/10`
t90% = `2.303/"k"` ...........(2)
∵ log 10 = 1
From (1) and (2),
t99% = 2t90%
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