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Under the same reaction conditions intit...

Under the same reaction conditions intital concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40 and 20 through first order and zero order kinetics respectively ratio `(k_(1)//k_(0))` of the rate constants for first order `k_(1)` and zero order `k_(0)` of the reaction is

A

`0.5 mol^(-1) dm^(3)`

B

`1.0 mol dm^(-3)`

C

`1.5 mol dm^(-3)`

D

`2.0 mol^(-1) dm^(3)`

Text Solution

Verified by Experts

Using half-ife period method, For first -order reaction, `k_(1) = 0.693//t_(1//2) = 0.693//40`
For zero-order reaction `k_(0) = [A]_(0)//2t_(1//2) = 1.386//(2 xx 20), (k_(1))/(k_(0)) = ((0.693//40))/((40//1.386)) = 0.5 mol^(-1)`
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