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InterviewSolution
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For a chemical reaction `Ararr B`,the rate of the reaction is `2.0xx10^(-3) sec^(-1)`, when the initial concentration is `0.05 mol dm^(-3)`. The rate of the same reaction is `1.6xx10^(-2) mol dm^(-3) sec^(-1)`. When the initial concentration is `0.1` mol `dm^(3)`, find the order of reaction. |
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Answer» Correct Answer - 3 Let the rate equation for the reaction be rate `=K[A]^(n)` `r_(1)=2.0xx10^(-3) mol dm^(3) mol dm^(-3) sec^(-1)`, `[A_(0)]=0.05 mol dm^(-3)` `:. 2.0xx10^(-3)=K[0.05]^(n) …(1)` and `r_(2)=1.6xx10^(-2) mol dm^(-3) sec^(-1)`, `[A_(0)]=0.1 mol dm^(-3)` `:. 1.6xx10^(-2)=K[0.1]^(n)` eq. (1) divide by. `(2)`, `(2.0xx10^(-3))/(1.6xx10^(-2))=([0.05]^(n))/([0.1]^(n))implies 1/8=(1/2)^(n)` `implies n=3` `:.` Order of reaction `=3` |
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