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A graph between `log t_((1)/(2))` and log a (abscissa), a being the initial concentration of A in the reaction For reaction `Ato`Product, the rate law is : A. `(-d[A])/(dt)=K`B. `(-d[A])/(dt)=K[A]`C. `(-d[A])/(dt)=K[A]^2`D. `(-d[A])/(dt)=K[A]^3` |
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Answer» Correct Answer - C `t_(1//2)=Ka^(1-n)`,n being order K=Rate constant log `t_(1//2)=log K+(1-n) log a` 1-n=-1 n=2 |
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