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In pseudo unimolecule reactions:
(A) Both the reactants are present in low concentration
(B) Both the reactants are present in same concentration
(C) One of the reactant is present in excess
(D) One of the reactant is non-reactive

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Last updated date: 27th Apr 2024
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Answer
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Hint:
In pseudo unimolecular reactions, the rate of the reaction depends only upon the concentration of one reactant. Actually a second reactant is present in the pseudo unimolecular type of reactions but somehow it does not alter the rate of the reaction.

Complete Step-by-Step Solution:
Let’s see what is meant by the pseudo unimolecular reaction.
Pseudo unimolecular reactions are those reactions in which the rate of the reaction depends only upon the concentration of one reactant even though the second reactant is present in the reaction. Here the other reactant is present in excess of amount and hence its value can be taken as constant, so it has a rate equation exactly like a unimolecular reaction.
Here is an example of a Pseudo unimolecular reaction.
\[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\to C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\]
Here in this reaction, hydrolysis of the ester is done in presence of an excess of water. The C-O bond is cleaved and we get a carboxylic acid and alcohol as a product. water is present in excess amount and hence its concentration at the starting and completion of the reaction is almost the same, so we can take its value as a constant and so, the rate of the reaction will only depend upon the concentration of ester in the given reaction.
- Thus we can say that one of the reactants is present in excess amounts in pseudo unimolecular reactions.
So, the correct answer is (C) One of the reactant is present in excess

Note:
Always remember that in any given reaction, if the concentration of any species is not altering much with time, then we can consider them as constant values and proceed with the calculations. Be careful that this type of reactions look like second-order reactions at first sight but knowing the concentration values changes will help you predict the correct answer.
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