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Question
A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed? (Given: log 5 = 0.6990, log 8 = 0.9030, log 2 = 0.3010)
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Solution
Half-life `t_(1/2) = 0.693/k`
k = `0.693/69.3 = 1/100` = 0.01 min–1
For first-order reaction
k = `2.303/t log ([Ro])/([R])`
t = `2.303/0.01 log 100/20`
t = 230.3 log 5 (log 5 = 0.6990)
t = 160.9 min
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