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Question
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution
99% completion means that x = 99% of [R]0
or, [R] = [R]0 − 0.99[R]0 = 0.01[R]0
For 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%
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