1.

The ration of the time required for 3/4 of the reaction and half ofthe reaction is :

Answer»

`4:3 `
`3:2`
`2:1`
`1:2`

Solution :(C ) TIME required for 3/4 of reaction
`t_(3//4) = (2.303)/(k) "log" (a)/(1//4a) = (2.303)/(k) log 4 `
Time required for half of reaction
`t_(1//2) = (2.303)/(k ) "log" (a)/((1)/(2)a)=(2.303)/(k)"log"2`
`(t_(3//4))/(t_(1//2))=("log"4)/("log"2)=(2"log"2)/("LOG2)=2:1`


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