1.

t_(1//4) can be taken as the time taken for the concentration of a reactant of drop to (3)/(4) of its initial value. If the rate constant for a first order reaction is K, then t_(1//14) can be written as

Answer»

`0.10//K`
`0.29//K`
`0.69//K`
`0.75//K`

SOLUTION :`t_(1//4) = (2*303)/(k)LOG (a)/((3//4)a) = (2*303)/(K)log ""(4)/(3)`
`=(2*303xx 0*125)/(K) = (0*29)/(K)`


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