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Questions
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Show that for a first order reaction, time required for 99% completion is twice the time required for 90% completion of the reaction.
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Solution 1
99% completion means that x = 99% of [R]0
or, [R] = [R]0 − 0.99 [R]0 = 0.01[R]0
For a first-order reaction, t = `2.303/k log [R]_0/[[R]]`
∴ t99% = `2.303/k log [R]_0/(0.01[R]_0)`
= `2.303/k log 10^2`
= `2 xx 2.303/k`
90% completion means that [R] = [R]0 − 0.99[R]0
= 0.1[R]0
∴ t90% = `2.303/k log [R]_0/(0.1[R]_0)`
= `2.303/k log 10`
= `2.303/k`
∴ `t_(99%)/t_(90%) = ((2 xx 2.303/k)/((2.303/k)))`
= 2
or, t99% = 2 × t90%
Solution 2
If the initial concentration is [A]0, for 99% completion,
[A] = `[A]_0 - [A]_0 xx 99/100`
= `[A]_0 xx 1/100`
Time required for 99% completion is given by
t99% = `2.303/k log_10 ([A]_0)/([A]_0 xx 1/100)`
= `2.303/k log_10 100/1` ...(i)
For 90% completion,
[A] = `[A]_0 - [A]_0 xx 90/100`
= `[A]_0 xx 10/100`
Time required for 90% completion is given by
t90% = `2.303/k log_10 ([A]_0)/([A]_0 xx 10/100)`
= `2.303/k log_10 100/10` ...(ii)
Dividing eq. (i) by eq. (ii), we get
`t_(99%)/t_(90%) = (log_10 100/1)/(log_10 100/10)`
`t_(99%)/t_(90%)` = 2
∴ t99% = 2 × t90%
Solution 3
For a first order reaction,
t = `2.303/k log_10 [A]_0/([A])`
For 99% completion, [A]0 = 100,
[A] = 100 − 99 = 1
∴ t99 = `2.303/k log_10 100/1`
∴ t99 = = `(2 xx 2.303)/k` ...(i)
For 90% completion, [A]0 = 100,
[A] = 100 − 90 = 10
∴ t90 = `2.303/k log_10 100/10`
∴ t99 = = `(1 xx 2.303)/k` ...(ii)
From equation (i) and (ii),
t99 = 2 × t90.
Thus, the time required for 99% completion for a first order reaction is twice the time required for 90% completion.
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