In an aqueous solution containing sodium bicarbonate, aniline reacts quickly with bromine to give 2,4,6-tribromoaniline. nitrati
on of aniline requires very strong conditions, however, and the yields (mostly m-nitroaniline) are poor.
1 answer:
Answer:-
In bromination of aniline, due to the strongly activating NH2 group 2,4,6-tribromoaniline is formed.
In nitration of aniline the reagent used is concentrated nitric acid. Nitric acid reacts with aniline Nitrogen and protonates it. Due to NH3+ group formation, the benzene ring becomes strongly deactive in the ortho and para positions.
Hence the yields towards nitro nucleophile is poor.
The meta position being more electron rich than ortho or para due to the strongly deactivating group hence yields m-nitroaniline.
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<u>Answer:</u> The of the reaction at given temperature is -12.964 kJ/mol.
<u>Explanation:</u>
For the given chemical reaction:
The expression of for the given reaction:
We are given:
Putting values in above equation, we get:
To calculate the Gibbs free energy of the reaction, we use the equation:
where,
= Gibbs' free energy of the reaction = ?
= Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)
R = Gas constant =
T = Temperature =
= equilibrium constant in terms of partial pressure =
Putting values in above equation, we get:
Hence, the of the reaction at given temperature is -12.964 kJ/mol.