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प्रश्न
Show that for a first-order reaction the time required to complete 75% of reaction is about 2 times more than that required to complete 50% of the reaction.
उत्तर
For a first-order reaction
t = `2.303/"k" "log" "a"/("a" - "x")`
`"t"_(75%) = 2.303/"k" "log" 100/(100 - 75)`
`"t"_(75%) = 2.303/"k" "log" 100/25`
`"t"_(75%) = 2.303/"k" "log" 4`
`"t"_(50%) = 2.303/"k" "log" 100/(100 - 50)`
`= 2.303/"k" "log" 100/50`
`"t"_(50%) = 2.303/"k" "log" 2`
Dividing equation (1) by equation (2), we get
`"t"_(75%)/"t"_(50%) = (2.303/"k" "log" 4)/(2.303/"k" "log" 2)`
`= "log 4"/"log 2" = "log2"^2/"log 2" = "2log 2"/"log 2"`
`therefore "t"_(75%)/"t"_(50%) = 2/1`
`therefore "t"_(75%)= 2 xx "t"_(50%)`
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