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Question
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
(Given : log = 2 = 0·3010, log 3 = 0·4771, log 4 = 0·6021)
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Solution
For a first order reaction
`t = 2.303/k log [R]_0/[R]`
`k = 2.303/"20 min" log 100/(100 - 25)`
`= 2.303/"20 min" log 4/3`
`= 2.303/"20 min" (log 4 - log 3)`
`= 2.303/"20 min" (0.6021 - 0.4771)`
= 1.44 x 10-2 min-1
The time when 75% of the reaction completed can be calculated as
`t = 2.303/k log 100/(100 - 75)`
`= 2.303/(1.44 xx 10^(-2)) log 4`
`= 2.303/(1.44 xx 10^(-2)) (0.6021)`
= 96.3 min (approximately)
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