InterviewSolution
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InterviewSolution
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Under the same reactions conditions , initial concentration of 1.386 mol `dm^(-3)` of a substance becomes half in 40 seconds and 20 seconds 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 reactions isA. `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)` |
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Answer» Correct Answer - a For first order `(t_(1//2))_(1) = (0.693)/(k_(1)) implies K_(1) = (0.693)/((t_(1//2))_(1))` for zero `(t_(1//2))_(0) = (A_(0))/(2k_(0)) implies k_(0) = (A_(0))/(2(t_(1//2))_(0))` `(K_(1))/(K_(0)) = (0.693 xx 2(t_(1//2))_(0))/((t_(1//2))_(1)A_(0)) implies (K_(1))/(K_(0)) = (0.693)/(1.386) = 0.5 mol^(-1) dm^(3)`. |
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