Chapter 5, Problem 5.100QE

(a)

Interpretation Introduction

Interpretation:

Balanced chemical reaction for the decomposition of ammonium nitrate should be written.

Explanation of Solution

In decomposition reaction, a single reactant undergoes reaction to form two or more products. Nitrogen gas, oxygen gas, and water vapor are the products obtained during the decomposition of ammonium nitrate.

Balanced chemical reaction for the decomposition of ammonium nitrate is given as:

  ${\text{2 NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(s\right)\underset{}{\overset{}{\to }}{\text{2 N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 4 H}}_{\text{2}}\text{O}\left(g\right)$

(b)

Interpretation Introduction

Interpretation:

Number of moles of gas are produced if $1.000kgN{H}_{4}N{O}_3$ is reacted should be calculated.

Answer to Problem 5.100QE

$\text{43}\text{.72 mol}$ of gas are produced if $1.000kgN{H}_{4}N{O}_3$ is reacted.

Explanation of Solution

Balanced chemical reaction for the decomposition of ammonium nitrate is given as:

  ${\text{2 NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(s\right)\underset{}{\overset{}{\to }}{\text{2 N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 4 H}}_{\text{2}}\text{O}\left(g\right)$

Number of moles in 1 kilogram of ammonium nitrate is determined as follows,

  $\begin{array}{c}n\text{ = }\frac{Mass}{Molar\text{}\text{ mass}}\\ \\ =\frac{{\text{10}}^3\text{ g}}{80.04 \text{g/mol}}\\ \\ =12.49\text{ mol}\end{array}$

Two moles ammonium nitrate reacts to give two moles of nitrogen gas, one mole of oxygen gas and four moles of water vapor. Hence, seven moles of gas is formed.

Number of moles of nitrogen, oxygen and water vapor formed on the decomposition of $1.000kgN{H}_{4}N{O}_3$ is determined as follows,

  $\begin{array}{c}{n}_{N_2}\text{= }12.49\text{ mol }N{H}_{4}N{O}_3\times \frac{2\text{ mol }{N}_2}{2\text{ mol }N{H}_{4}N{O}_3}\\ \\ =12.49\text{ mol }{N}_2\\ \\ n_{O_2}\text{= }12.49\text{ mol }N{H}_{4}N{O}_3\times \frac{1\text{ mol }{O}_2}{2\text{ mol }N{H}_{4}N{O}_3}\\ \\ =6.245\text{ mol }{O}_2\\ \\ n_{H_{2}O}\text{= }12.49\text{ mol }N{H}_{4}N{O}_3\times \frac{4\text{ mol }{H}_{2}O}{2\text{ mol }N{H}_{4}N{O}_3}\\ \\ =12.98\text{ mol }{H}_{2}O\end{array}$

Number of moles of gas are produced if $1.000kgN{H}_{4}N{O}_3$ is reacted is calculated as follows,

  $\begin{array}{c}{n}_{gas}\text{ = }{n}_{N_2}+n_{O_2}+n_{H_{2}O}\\ \\ =12.49\text{ mol}+6.245\text{ mol +}12.98\text{ mol}\\ \\ \text{= 43}\text{.72 mol}\end{array}$

Therefore, $\text{43}\text{.72 mol}$ of gas are produced if $1.000kgN{H}_{4}N{O}_3$ is reacted.

(c)

Interpretation Introduction

Interpretation:

Energy are released per pound of ammonium nitrate should be calculated.

Concept introduction:

Enthalpy$\left(H\right)$: It is the total amount of heat in a particular system.

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Where,

${\text{ΔH}}_{\text{f}}^{\text{ο}}$ is the standard enthalpies of formation.

n is the number of moles.

Answer to Problem 5.100QE

Energy are released per pound of ammonium nitrate is $-669\text{kJ}$.

Explanation of Solution

Balanced chemical reaction for the decomposition of ammonium nitrate is given as:

  ${\text{2 NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(s\right)\underset{}{\overset{}{\to }}{\text{2 N}}_{\text{2}}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right){\text{ + 4 H}}_{\text{2}}\text{O}\left(g\right)$

$\Delta H_f^o$ for ammonium nitrate is $-\text{87}\text{.37 kcal/mol}$.

Convert the unit of $\Delta H_f^o$ as follows,

  $\begin{array}{c}1\text{ kcal = 4}\text{.184 kJ}\\ \\ \Delta H_f^o=-\text{87}\text{.37 kcal/mol }\\ \\ \text{= }-\text{87}\text{.37 kcal/mol}\times \frac{\text{4}\text{.184 kJ/mol}}{1\text{ kcal/mol}}\\ \\ =-365.56\text{ kJ/mol}\end{array}$

Number of moles in 1 pound of ammonium nitrate is determined as follows,

  $\begin{array}{c}n\text{ = }\frac{Mass}{Molar\text{}\text{ mass}}\\ \\ =\frac{543.6\text{ g}}{80.04 \text{g/mol}}\\ \\ =\text{ 5}\text{.667 mol}\end{array}$

Standard enthalpy of formation values is given below,

  $\begin{array}{c}{\text{ΔH}}_f^o{\text{of O}}_{\text{2}}\left(g\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of H}}_{2}O\left(g\right)\text{=}-\text{241}\text{.82}\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o\text{of }{\text{N}}_{\text{2}}\left(g\right)\text{=}0\text{kJ}/\text{mol}\\ \\ {\text{ΔH}}_f^o{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(s\right)\text{=}-365.56\text{ kJ/mol}\end{array}$

Change in enthalpy can be calculated by the equation:

  ${\text{ΔH}}_{rxn}^{\circ }\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Substitute the values as follows,

  $\begin{array}{c}{\text{ΔH}}_{rxn}^o\text{=}\left[\left(\text{2 mol}\times 0\text{kJ}/\text{mol}\right)\text{+}\left(\text{1 mol×}0\text{kJ}/\text{mol}\right)\text{+}\left(\text{4 mol×}-\text{241}\text{.82}\text{kJ}/\text{mol}\right)\right]\\ -\left[\left(\text{2 mol}\times -365.56\text{ kJ/mol}\right)\right]\\ \\ \text{=}-236.16\text{kJ}\end{array}$

$\text{ΔH}°$ for the combustion of two moles of ammonium nitrate is $-236.16\text{kJ}$. Hence, $\text{ΔH}°$ during the combustion of $\text{5}\text{.667 mol}$ ammonium nitrate is determined as follows,

  $\begin{array}{c}\text{ΔH}°\text{ = 5}\text{.667 mol}\times \frac{-236.16\text{kJ}}{\text{2 mol}}\\ \\ =-669\text{kJ}\end{array}$

Therefore, the energy are released per pound of ammonium nitrate is $-669\text{kJ}$.