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Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively. The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction isA. `0.5 "mol" ^(-1)dm^(3)`B. `1"mol dm"^(-3)`C. `1.5 "mol dm"^(-3)`D. `2 "mol"^(-1) dm^(3)` |
Answer» Correct Answer - A `t_(1//2)=(0.693)/(k_(1))" " t_(1//2)=(a_(0))/(2k_(0))` `40=(0.693)/(k_(1))," " 20=(1.386)/(2k_(0))=(0.693)/(k_(0))` `(20)/(40)=(0.693//k_(0))/(0.693//k_(1))=(k_(1))/(k_(0))` `(k_(1))/(k_(0))=0.5 (sec^(-1))/("mol dm"^(-3)sec^(-1))=0.5 "mol"^(-1) "dm"^(3)` |
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