Question

Formation of ${\text{HN}}{{\text{O}}_{\text{3}}}$ when ${\text{N}}{{\text{O}}_2}$ is dissolved in ${{\text{H}}_{\text{2}}}{\text{O}}$ takes place through various reactions. Select the reaction not observed in this step.
A. ${\text{N}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{HN}}{{\text{O}}_{\text{3}}}{\text{ + HN}}{{\text{O}}_{\text{3}}}$
B. ${\text{HN}}{{\text{O}}_{\text{2}}} \to {{\text{H}}_{\text{2}}}{\text{O + NO + HN}}{{\text{O}}_{\text{3}}}$
C. Both (A) and (B)
D. None of these

Answer 1

Write down the entire step from initiation, in the formation of ${\text{HN}}{{\text{O}}_{\text{3}}}$and examine the above options. There are four steps involved in this process and one of the products get recycled in the process to yield ${\text{HN}}{{\text{O}}_{\text{3}}}$. The reactions initiate with oxidation of ammonia.

Complete step by step answer:
No doubt that ${\text{N}}{{\text{O}}_2}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ combine to form ${\text{HN}}{{\text{O}}_{\text{3}}}$. But this step is not the first step of the reaction. Let us look stepwise how this reaction occurs.
Step 1: Ammonia and oxygen both react to form nitric oxide (NO). The reaction is as follows.
${\text{4N}}{{\text{H}}_3}{\text{ + 5}}{{\text{O}}_{\text{2}}} \to 4{\text{NO + 6}}{{\text{H}}_2}{\text{O}}$
Step 2: Now, the product of the above reaction that is NO further reacts with oxygen to yield nitrogen dioxide. The reaction is as follows.
$2{\text{NO + }}{{\text{O}}_{\text{2}}} \to 2{\text{N}}{{\text{O}}_2}$
This is the actual reactant given in the equation.
Step 3: Reaction between ${\text{N}}{{\text{O}}_2}$ and ${{\text{H}}_{\text{2}}}{\text{O}}$ occurs as follows.
${\text{3N}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{HN}}{{\text{O}}_{\text{3}}}{\text{ + HN}}{{\text{O}}_{\text{3}}}{\text{ + NO}}$
The same reaction is given to us in option A. So, option A is correct because the reaction is involved in the formation of ${\text{HN}}{{\text{O}}_{\text{3}}}$.
Step 4: As you can see in step 3 that NO is regenerated, it again combines with oxygen to form nitrogen dioxide which then combines with oxygen and water to form higher concentration of nitric acid.
${\text{4N}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{O}}_{\text{2}}} + 2{{\text{H}}_2}{\text{O }} \to {\text{ 4HN}}{{\text{O}}_{\text{3}}}$
Clearly, option B is not in any step in the formation of nitric acid from nitrogen dioxide and water. Moreover, option B is a disproportionation reaction of ${\text{HN}}{{\text{O}}_2}$, but is not a part of the above process. It is a different reaction.

So, the correct answer is Option B .

Note:
Nitric acid (${\text{HN}}{{\text{O}}_{\text{3}}}$) is also known as aqua fortis and spirit of niter. It is a mineral acid and highly corrosive. It is colourless in pure state, but turns yellowish if placed for too long because of the formation of oxides. The process shown above is known as the Ostwald process.