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The rate law of a chemical reaction : 2NO+O_(2) to 2NO_(2), is given as rate =k[NO]^(2)[O_(2)]. How will the rate of reaction change if the volume of reaction vessel is reduced to 1//4th of its original value ? |
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Answer» Solution :For, `2NO+O_(2)to2NO_(2)` Rate `=k[NO]^(2)[O_(2)]` Let a MOLES of `NO` and `B` moles of `O_(2)` be taken to start a reaction at any time in a vessel of V litre `r_(1)=k[(a)/(V)]^(2)[(b)/(V)]`…………`(i)` If VOLUME of vessel is reduced to `V//4`, then for same moles of `NO` and `O_(2)` `r_(2)=k[(a)/(V//4)]^(2)[(b)/(V//4)]=64k[(a)/(V)]^(2)[(b)/(V)]`................`(ii)` By Eqs. `(i)` and `(ii)` , `(r_(2))/(r_(1))=64` `r_(2)` is `64` times of `r_(1)` |
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