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The rate constant for a first order reaction is "60 s"^(-1). How much time will it take to reduce the initial concentration of the reactant to its (1)/(16) th value ? |
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Answer» SOLUTION :If we carry out HALF - change FOUR times, the CONCENTRATION will reduce to `(1)/(16)`. `t_(1//2)=(0.693)/(K)=(0.693)/("60 s"^(-1))=0.01155s` Total time taken `=4xxt_(1//2)=4xx0.01155s=0.0462s` |
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