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प्रश्न
40% of a first-order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction?
उत्तर
For a first-order reaction
t = `2.303/"k" log "a"/("a"-"x")`
At t40% a = 100
x = 40
t40% = 50 minutes
At t80% a = 100
a = 80
t80% = ?
t80% = `2.303/"k" log 100/(100-80)`
t80% = `2.303/"k" log 100/20`
t80% = `2.303/"k" log 5`
t40% = `2.303/"k" log 100/(100-40)`
t40% = `2.303/"K" log 100/60`
`= 2.303/"k" log 5/3`
`"t"_(80%)/"t"_(40%) = (2.303/"k" log 5)/(2.303/"k" log 5/3`
`"t"_(80%)/50 log 5/(log5-log3)`
`"t"_(80%) = 50 [0.6990/(0.6990-0.4771)]`
`= (50xx0.6990)/0.2219`
t80% = 157.5 minutes
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