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At 193^(@)C , the ratelaw for the reaction2Cl_(2)O to 2Cl_(2) + O_(2) is rate = k[Cl_(2)O]^(2) . (a) How the ratechangesif [Cl_(2)O]is raised to therefold of the original ? (b) How should [Cl_(2)O] be changed in orderto order to double the rate ? |
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Answer» Solution :Original RATE is given as, rate `=k[Cl_(2)O]^(2)` (a) Rate with threefold concentration of `Cl_(2)O` is k `[3Cl_(2)O]^(2)`. This is NINE times to the original rate. (b) With `[Cl_(2)O]` the rate is `R=k[Cl_(2)O)^(2)`. Requirement is that .r. should be doubled at concentration of, `Cl_(2)O` as `x`. `(2r)/(r)=(kx^(2))/(k[Cl_(1)O]^(2))("or")x=sqrt(2)[Cl_(2)O]` The concentration of `Cl_(2)O` should be increased by `1.414` times. |
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