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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) - Chemistry

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प्रश्न

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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उत्तर

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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