Chapter 14, Problem 14.114P

(a)

Interpretation Introduction

Interpretation:

The three steps of Ostwald process has to be given.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

Oxidation reaction: The loss of electrons or the gain of oxigen atoms. And also increase their oxidation number.

Disproportionation reaction: It is type of reaction where a substance can act as a reducing agent as well as an oxidizing agent since an atom present in the substance reacts to give atoms having and lower oxidation states.

Explanation of Solution

Oxidation takes place in the first two step of Ostwald process. Ammonium oxidizes to $NO$ which further oxidizes to give $N{O}_2.$  In the third step, nitric acid is formed by a disproportionation reaction.

The three steps of Ostwald process follows as,

  $\begin{array}{l}{\text{4NH}}_{\text{3}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{ 4NO}\left(\text{g}\right){\text{ + 6H}}_{\text{2}}\text{O}\left(\text{g}\right)\\ \\ \text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{3NO}}_{\text{2}}\left(\text{g}\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\overset{}{\to }}{\text{ 2HNO}}_{\text{3}}\left(l\right)\text{ + NO}\left(\text{g}\right)\end{array}$

(b)

Interpretation Introduction

Interpretation:

Moles of $N{H}_3$ are consumed per mole of $HN{O}_3$ produced has to be determined.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

Explanation of Solution

The three steps of Ostwald process follows as,

  $\begin{array}{l}{\text{4NH}}_{\text{3}}\left(\text{g}\right){\text{ + 5O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }\text{ 4NO}\left(\text{g}\right){\text{ + 6H}}_{\text{2}}\text{O}\left(\text{g}\right)\\ \\ \text{2NO}\left(\text{g}\right){\text{ + O}}_{\text{2}}\left(\text{g}\right)\underset{}{\to }{\text{ 2NO}}_{\text{2}}\left(\text{g}\right)\\ \\ {\text{3NO}}_{\text{2}}\left(\text{g}\right){\text{ + H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\overset{}{\to }}{\text{ 2HNO}}_{\text{3}}\left(l\right)\text{ + NO}\left(\text{g}\right)\end{array}$

Moles of $N{H}_3$ are consumed per mole of $HN{O}_3$ produced when $NO$ is not recycled is determined as given,

$\begin{array}{c}\frac{\text{No}{\text{.of moles of NH}}_3\text{ consumed}}{\text{No}{\text{.of moles of HNO}}_3\text{ produced}}\text{ = }\left(\frac{{\text{3 mol NO}}_2}{{\text{2 mol HNO}}_3}\right)\left(\frac{\text{2 mol NO}}{{\text{2 mol NO}}_2}\right)\left(\frac{{\text{4 mol NH}}_3}{\text{4 mol NO}}\right)\\ \\ =\text{ 1}{\text{.5 mol NH}}_{\text{3}}{\text{/mol HNO}}_3\end{array}$

Moles of $N{H}_3$ are consumed per mole of $HN{O}_3$ produced is $\text{1}{\text{.5 mol NH}}_{\text{3}}{\text{/mol HNO}}_{\text{3}}$ .

(c)

Interpretation Introduction

Interpretation:

Volume of concentrated aqueous nitric acid has to be determined using the given data.

Concept-Introduction:

Ostwald process: It is a chemical process where ammonium is converted to nitric acid.

According to Ideal gas equation,

$\text{Number}\text{of}\text{moles}\text{=}\frac{\text{Pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}\text{Yield in the first step = 96\%}\\ \\ \text{Yield in the second and third step = 100\%}\\ \\ \text{Moles of are consumed per mole of produced= 1}{\text{.5 mol NH}}_{\text{3}}{\text{/mol HNO}}_{\text{3}}\\ \\ \text{Pressure = 5}\text{.0 atm}\\ \\ {\text{Volume = 1 m}}^3\\ \\ Temperature={\text{ 850}}^{o}C\text{ = }\left(273+850\right)K=\text{ 1123K}\\ \\ Density\text{ = 1}\text{.37g/mol}\end{array}$

Number of moles of $N{H}_3$ is calculated as given,

$\begin{array}{c}No.{\text{ of moles of NH}}_3,\text{ n = }\frac{PV}{RT}\\ \\ =\frac{\left(5.0atm\right)\left(1{\text{ m}}^3\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}.K^{-1}\right)\left(1123\text{ K}\right)}\left(\frac{1\text{ L}}{{10}^{-3}{\text{ m}}^3}\right)\left(\frac{10\%{\text{ NH}}_3}{100\text{\% gas}}\right)\\ \\ =\text{ 5}{\text{.42309 mol NH}}_3\end{array}$

Number of moles of $N{H}_3$ is $\text{5}{\text{.42309 mol NH}}_3.$

Mass of $HN{O}_3$ is determined as follows,

Mass of $HN{O}_3$ can be obtained from the number of moles of $N{H}_3$ present where it is converted to moles $HN{O}_3$.

$\begin{array}{c}{\text{Mass (g) of HNO}}_3\text{ = }\left(\text{5}{\text{.42309 mol NH}}_3\right)\left(\frac{{\text{1 mol HNO}}_{\text{3}}}{\text{1}{\text{.5 mol NH}}_3}\right)\left(\frac{\text{63}{\text{.02 g HNO}}_3}{{\text{1 mol HNO}}_3}\right)\left(\frac{96\%}{100\%}\right)\\ \\ =\text{ 218}{\text{.728 g HNO}}_3\end{array}$

Volume of concentrated aqueous nitric acid is calculated as shown below,

$\begin{array}{c}Volume{\text{ of HNO}}_3\text{ = }\left(\text{218}{\text{.728 g HNO}}_3\right)\left(\frac{100\%}{60\%}\right)\left(\frac{\text{1 mL}}{\text{1}\text{.37 g}}\right)\\ \\ =\text{ 2}\text{.7}\times {\text{10}}^{\text{2}}{\text{ mL HNO}}_3\end{array}$

Volume of concentrated aqueous nitric acid is $2.7\times 10^{2}mL.$