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Diazonium salt decomposes as C6H5N2+Cl → C6H5Cl + N2 . At 0ºC, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it is
  • a)
    a first order reaction
  • b)
    a second order reaction
  • c)
    independent of the initial concentration
  • d)
    a zero order reaction
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At ...
For a first order reaction
r = k[A]
Hence when conc. of (C6H5N2+Cl)is doubled the conc. of nitrogen gets doubled.
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Most Upvoted Answer
Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At ...
The decomposition of diazonium salts occurs through a process called diazonium coupling. In the case of C6H5N2Cl, it decomposes into a phenyl radical (C6H5•) and molecular nitrogen gas (N2).

The reaction can be represented as follows:

C6H5N2Cl → C6H5• + N2

In this reaction, the diazonium salt loses the N2 group and forms a phenyl radical. This reaction is often used in organic synthesis to introduce phenyl groups into various compounds. The resulting phenyl radical can then react with other organic molecules to form new compounds.
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There are many reactions which obey a first order rate equation although it reality they are bi- or ter-molecular. As an example of these may be taken the decomposition of Carbonyl sulfide in water, namely, COS + H20 002 + H2SAccording to the law of mass action this reaction should be second order with the rate dependent on the concentration of both the carbonyl sulfide and the water. Actually however, the rate is found to be first order with respect to the carbonyl sulfide and independent of the water Reactions exhibiting such behaviour are said to be pseudo-molecular.The pseudo-unimoecuar nature of this reaction is explainable by the fact that water is present in such excess that its concentration remains practically constant during the course of the reaction. Under these condition b x = b, and the rate equation becomesOn integration this leads towhich is the equation for a first order reaction. It is evident, however, that the now constant k is not independent of the concentration, as is the case with true first order constants, but may vary with b if the latter is changed appreciably, When such is the case, the true constant k2 can be obtained from k by dividing the latter by b. pseudo-molecular reactions are encountered whenever one or more of the reactants remain constants during the course of an experiment. This is the case with reactions conducted in solvents which are themselves one of the reactants, as in the decomposition of carbonyl sulfide in water, or in the esterification of acetic anhydride in alcohol(CH3C0)20 + 2C2H5OH 2CH3C00C2H5 + H20Again, this is also true of reactions subject to catalysis, in which case the concentration of the catalyst does not change. The decomposition of diacetone alcohol to acetone in aqueous solution is catalysed by hydroxyl ions, with the rate proportional to the concentration of the alcohol and that of the base. Since the concentration of the base does not change within any one experiment, however, the rate equation reduces to one of first order with respect to the alcohol. But the rateconstant k obtained for various concentrations of base are not identical, as may be seen from table. To obtain from these the true second order velocity constant, the ks must be divided by the hydroxyl ion concentration. When this is done excellent k2 values result, as column 3 indicatesTable : Decomposition of diacetone alcohol in water at 25C (Catalyst : NaOH)Q.By what factor does the rate of reaction of diacetone alcohol in water solution change if p0H is increased by 2 units other things remaining same ?

Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At 0ºC, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isa)a first order reactionb)a second order reactionc)independent of the initial concentrationd)a zero order reactionCorrect answer is option 'A'. Can you explain this answer?
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Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At 0ºC, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isa)a first order reactionb)a second order reactionc)independent of the initial concentrationd)a zero order reactionCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At 0ºC, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isa)a first order reactionb)a second order reactionc)independent of the initial concentrationd)a zero order reactionCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Diazonium salt decomposes as C6H5N2+Cl− → C6H5Cl + N2 . At 0ºC, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isa)a first order reactionb)a second order reactionc)independent of the initial concentrationd)a zero order reactionCorrect answer is option 'A'. Can you explain this answer?.
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