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प्रश्न
Show that in case of a first order reaction, the time taken for completion of 99% reaction is twice the time required for 90% completion of the reaction. (log 10 = 1)
संख्यात्मक
उत्तर
For a first-order reaction, we have
`t = 2.303/k log [A]_0/[[A])`
When the reaction is 99% complete,
`[A] = a - (99a)/100 = 0.01a`
∴ `t_99 = 2.303/k log a/(0.001a)`
= `2.303/k xx 2` ...(i)
When the reaction is 90% complete,
`[A] = a - (90a)/100 = 0.1a`
`t_90 = 2.303/k log a/(0.1a)`
`t_90 = 2.303/k xx 1` ...(ii)
By dividing equations (i) by (ii), we get
`t_99/t_90 = 2 ⇒ t_99 = 2t_90`
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