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The rate of a certain reaction depends on concentration according to equation -(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A]) What will be the order of reaction when (i) concentration is very high (ii) very low ? |
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Answer» Solution :Given, `-(d[A])/(dt)=(k_(1)[A])/(1+k_(2)[A])` `IMPLIES(-d[A])/(dt)=(k_(1))/((1)/([A])+k_(2))` `(i)` When `[A]` is very high `(1)/([A])` is very SMAL , and thus negligible `:.-(d[A])/(dt)=(k_(1))/(k_(2))=` constant Thus, ORDER of reaction is ZERO. `(ii)` When `[A]` is very low. `[A+k_(2)[A]=k.` `-(d[A])/(dt)=(k_(1)[A])/(k.)=k.[A]` Thus order of reaction is one. |
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