Question

Given following series of reactions:
(I) $N{H}_3+O_2\to NO+H_{2}O$
(II) $NO+O_2\to N{O}_2$
(III) $N{O}_2+H_{2}O\to HN{O}_3+HN{O}_2$
(IV) $NH{O}_2\to HN{O}_3+NO+H_{2}O$
Select the correct option(s).

• A. Moles of $HN{O}_3$ obtained is half of moles of Ammonia used if $HN{O}_2$ is not used to produce $HN{O}_3$ by reaction (IV)
• B. $\frac{100}{6}$% more $HN{O}_3$ will be produced if $HN{O}_2$ is used to produce $HN{O}_3$ by reaction (IV) than if $HN{O}_2$ is not used to produce $HN{O}_3$ by reaction (IV)
• C. If $HN{O}_2$ is used to produce $HN{O}_3$ then $\frac{1}{4}th$ of total $HN{O}_3$ is produced by reaction (IV)
• D. Moles of $NO$ produced in reaction (IV) is $50$% of moles of total $HN{O}_3$ produced

Answer 1

The correct option is C If $HN{O}_2$ is used to produce $HN{O}_3$ then $\frac{1}{4}th$ of total $HN{O}_3$ is produced by reaction (IV)

$\begin{array}{c}\text{(A)}{NH}_3⟶{HNO}_3+{HNO}_2\text{(till reaction III)}\\ \text{by nitrogen balance,}\\ \begin{array}{c}{n}_{HN{O}_2}=\frac{1}{2}{n}_{{NH}_3}\\ n_{{HNO}_3}=\frac{1}{2}{n}_{{NH}_3}\end{array}\end{array}$

$\begin{array}{c}\left(B\right)3HN{O}_2⟶HN{O}_3+2NO+H_{2}O\\ {\text{Let's 1 mole of NH}}_3\text{is initialy taken}\end{array}$

$\begin{array}{c}\text{If makes}(\frac{1}{2};\frac{1}{2})\text{mole of}HN{0}_2\text{and}{HNO}_3\text{till reaction III}\\ \frac{1}{2}\text{mole}{HNO}_2\text{make}\frac{1}{6}\text{mole of}{HNO}_3\text{in reaction}-IV\text{so}{HNO}_3\\ \text{made.}\end{array}$

$\begin{array}{cc}\Rightarrow \text{i.e.}\frac{1}{2}+\frac{1}{6}=& \frac{2}{3}\text{mole}\\ \text{\% increase}& =\frac{\frac{1}{6}}{\frac{1}{2}}\times 100=\frac{100}{3}\%\end{array}$

$\begin{array}{c}\text{(C) By above data it is correct.}\\ \text{(D) Moles of NO produced}=\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}\text{mole.}\\ \%=\frac{\frac{1}{3}}{\frac{2}{3}}\times 100=50\%\end{array}$