Reaction: Nitration of aniline using concentrated nitric acid and sulfuric acid.
\[ {C6H5NH2 + HNO3} \xrightarrow{{H2SO4}} {p-C6H4(NO2)NH2} \]
Explanation: Aniline reacts with nitrating mixture to give para-nitroaniline as the major product due to the activating nature of the –NH\(_2\) group.
Reaction: Chlorination of aniline in the presence of ferric chloride as a Lewis acid catalyst.
\[ {C6H5NH2 + Cl2} \xrightarrow{{FeCl3}} {C6H4ClNH2} \]
Explanation: Chlorine reacts with aniline to give substituted products. The reaction is usually controlled to obtain mono-chlorinated derivative such as chlorobenzene.
Reaction: Oxidation of aniline with potassium dichromate in acidic medium.
\[ {C6H5NH2 + K2Cr2O7 + H2SO4} \rightarrow {C6H5OH} \]
Explanation: The –NH\(_2\) group of aniline is oxidized to a hydroxyl group, yielding phenol.
Identify A in the following reaction. 
For the reaction, \(N_{2}O_{4} \rightleftharpoons 2NO_{2}\) graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is 5.40 kJ \(mol^{-1}\).
B. As \(\Delta G\) in graph is positive, \(N_{2}O_{4}\) will not dissociate into \(NO_{2}\) at all.
C. Reverse reaction will go to completion.
D. When 1 mole of \(N_{2}O_{4}\) changes into equilibrium mixture, value of \(\Delta G = -0.84 \text{ kJ mol}^{-1}\).
E. When 2 mole of \(NO_{2}\) changes into equilibrium mixture, \(\Delta G\) for equilibrium mixture is \(-6.24 \text{ kJ mol}^{-1}\).
Choose the correct answer from the following.
