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Under the same reaction condition, initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40s and 20s through first order and zero order kinetics respectively. Find out the `k_(1)/k_(0)` ratio for first order `(k_(1))` and zero order `(k_(0))` of the reaction.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)` |
Answer» Correct Answer - A a) 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 dm^(-3))/(2 xx20)` `k_(1)/k_(0) = 0.693/(40) xx 40/(1.386 "mol"dm^(-3))` `0.5 mol^(-1)dm^(3)` |
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