1.

A graph between log t_(1//2) and log a (abscissa), a being the initial concentration of A in the reaction A rarr product, is depicted in the figure. The rate law is

Answer»

`-(d[A])/(dt)=k`
`-(d[A])/(dt)=k[A]`
`-(d[A])/(dt)=k[A]^(2)`
`-(d[A])/(dt)=k[A]^(3)`

Solution : `t_(1//2)=ka^(1-n)`, n being ORDER, k=rate CONSTANT LOG `t_(1//2)`=log k+(1-n)log a
Slope, (1-n) = -2, n = 3


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