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The rate of decomposition of ammonia is found to depend upon the concentration of NH_(3) according to the equation -(d[NH_(3)])/(dt)=(k_(1)[NH_(3)])/(1+k_(2)[NH_(3)]) What will be the order of reaction when (i) concentration of NH_(3) is very high ? (ii) concentration of NH_(3) is very low ? |
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Answer» Solution :The given rate law equation can be written as `-(d[NH_(3)])/(DT)=(k_(1))/(1//[NH_(3)]+k_(2))` (i) If `[NH_(3)]` is very high, `1//[NH_(3)]` becomes negligible `:.-(d[NH_(3)])/(dt)=(k_(1))/(k_(2))=k` i.e., rate becomes independent of concentration. Hence,it is of zero ORDER. (ii) If `[NH_(3)]` is very SMALL, `1//[NH_(3)]` will be very large `(gtgtk_(2))`, so that `k_(2)` can be neglected in comparison to `1//[NH_(3)]`. Hence, `-(d[NH_(3)])/(dt)=(k_(1))/(1//[NH_(3)])=k_(1)[NH_(3)]` Thus, REACTION is of 1st order. |
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