InterviewSolution
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InterviewSolution
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The half-lives for two samples are 0.1 and 0.8 s whose concentrations are 400 and 50 mol `L^(-1)` respectively. The order of the reaction is |
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Answer» Correct Answer - C The ratio of two half-lives and their concentrations are related as `((t_(1//2))_(1))/((t_(1//2))_(2))=[(a_(2))/(a_(1))]^(n-1)` [Here, n = order] Therefore, `(0.1)/(0.8)=[(50]/(400)]^(n-1)` On taking log of both sides `log .(0.1)/(0.8)=(n-1)log. (50)/(400)` `1 = n-1implies n=2` |
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