InterviewSolution
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InterviewSolution
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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. `1.0mol dm^(-3)`C. `1.5mol dm^(-3)`D. `2.0mol^(-1) dm^(3)` |
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Answer» Correct Answer - A For first order reaction `k_(1)=(0.693)/(t_(1//2))=(0.693)/(40s)` For a zero order reaction: `k_(0)=([R]_(0))/(2t_(1//2))=(1.386)/(2xx20_(s))` Therefore `(k_(1))/(k_(0))=(0.693)/(40s)xx(40s)/(1.386)` `=(0.693(s^(-1))/(1.386(moldm^(-3)s^(-1)))` `=0.5mol^(-1)dm^(3)` |
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