Question

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 t_99% = 2t_90%.

Answer 1

For I^st order reaction k = `2.303/"t" log  (["A"_0])/(["A"_"t"])`

For 99% completion: If [A₀] = 100

[A_t] = 100 − 99

= 1

∴ t_99% = `2.303/"k" log  100/1`

t_99% = `2.303/"k" xx 2` .........(1)

∵ log 100 = 2

For 90% completion: If [A₀] = 100

[A_t] = 100 − 90

= 10

t_90% = `2.303/"k" log  100/10`

t_90% = `2.303/"k"` ...........(2)

∵ log 10 = 1

From (1) and (2),

t_99% = 2t_90%