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Under the same reaction 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 is |
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Answer» `0.5" mol"^(-1)" dm"^(3)` `t_(1//2) = (0*693)/(K) or k_(1) = (0.0693)/(t_(1//2)) = (0*693)/(40S)` For zero order reaction, `t_(1//2) = ([A]_(0))/(2K) or k_(0) = ([A]_(0))/(2t_(1//2)) = (1*386 mol dm^(-3))/(2xx 20s)` `therefore (k_(1))/(k_(0)) = (0*693)/(40s) xx (2xx 20s)/(1*386 mol dm^(-3)) = 0*5 mol^(-1) dm^(3)` |
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