Chapter 6, Problem 6.176AP

(a)

Interpretation Introduction

Interpretation: Ammonium nitrate given to decompose on heating and the product is stated to being dependent on the reaction temperature. A sample of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ decomposes at an unspecified temperature, and the resulting gases are collected over water at $20°\text{C}$. The given questions on the basis of above statement are to be answered.

Concept introduction: Charles law states that “the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant”.

To determine: The explanation about the statement that the volume of gases collected can be used to distinguish between the two reactions pathways.

Answer to Problem 6.176AP

Solution

The given compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ decomposes to give a mixture of ${\text{N}}_{\text{2}}$ and ${\text{O}}_2$ gas, then the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create ${\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

The given equation for ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ decomposing to give ${\text{N}}_{\text{2}}\text{O}$ is,

${\text{NH}}_4{\text{NO}}_3(s)\to {\text{N}}_2\text{O}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In this path, $1\text{mol}$ of ${\text{N}}_2\text{O}$ gas is formed and $1\text{mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ gas is formed.

This reaction proceeds to form ${\text{N}}_{\text{2}}$ and ${\text{O}}_2$ gas,

${\text{NH}}_4{\text{NO}}_3(s)\to {\text{N}}_2\left(g\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(\text{g}\right)+2{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In this path, $1\text{mol}$ of ${\text{N}}_2$ gas is formed and $\frac{1}{2}$ mol of ${\text{O}}_2$ gas is formed.

At constant temperature and pressure, the volume of gas is proportional to the numbers of moles.

Therefore, if ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ decomposes to give us a mixture of ${\text{N}}_{\text{2}}$ and ${\text{O}}_2$ gas, than the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create ${\text{N}}_{\text{2}}\text{O}$.

(b)

Interpretation Introduction

To determine: The gas produced during the thermal decomposition of $0.256\text{g}$ ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ displaces $79\text{mL}$ of water at $20°\text{C}$ and $760\text{mmHg}$ of atmospheric pressure.

Answer to Problem 6.176AP

Solution

The volume of gas is $79\text{mL}$, it must be ${\text{N}}_{\text{2}}\text{O}$.

Explanation of Solution

The number of moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{in}\text{grams}}{\text{Molar}\text{mass}\text{in}\text{grams}}$

Substitute the given values of mass and molar mass in the above expression.

$\begin{array}{c}n=\frac{0.256\text{g}}{80\text{g/mol}}\\ =0.0032\text{mol}\end{array}$

The stated reaction proceeds to $1\text{mol}$ of ${\text{N}}_2\text{O}$ gas.

The number of moles of ${\text{N}}_{\text{2}}\text{O}$ is calculated as,

$1\text{mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ produces $1$ mole of ${\text{N}}_{\text{2}}\text{O}$.

Therefore, $0.0032\text{mol}$ ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ produces moles of ${\text{N}}_{\text{2}}\text{O}$ $\begin{array}{l}=\left(\frac{0.0032}{1}\times 1\right)\text{mol}\\ =\text{0}\text{.0032}\text{mol}\end{array}$

One mole of any gas occupies $22.4\text{L}$ at STP.

So, the volume occupied by $0.0032\text{mol}\text{of}{\text{N}}_{\text{2}}\text{O}$ at STP $\begin{array}{c}=\frac{22.4}{1}\times 0.0032\\ =0.07168\text{L}\end{array}$

The conversion of temperature from $°\text{C}$ to $\text{K}$ is done as,

$K=°C+273.15$

For $T_2$,

$\begin{array}{c}{T}_2=20°\text{C}+\text{273}\text{.15}\\ =\text{290}\text{.15}\text{K}\\ \approx 290\text{K}\end{array}$

The volume is calculated by the formula,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where,

• $T_1$ is the initial temperature $273\text{K}$.
• $T_2$ is the new final temperature $290\text{K}$.
• $V_{\text{1}}$ is the initial volume $0.07168\text{L}$.
• $V_2$ is the new final volume.

Substitute these values of $T_1$, $T_2$ and $V_{\text{1}}$ in above expression.

$\begin{array}{c}\frac{V_1}{T_1}=\frac{V_2}{T_2}\\ V_2=\frac{0.07168\text{L}\times 290\text{K}}{273\text{K}}\\ =0.077\text{L}\\ =\text{77}\text{mL}\end{array}$

If the reaction proceeds to create a mixture of ${\text{N}}_{\text{2}}$ and ${\text{O}}_2$, then $1\text{mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ will produce a mixture of ${\text{O}}_{\text{2}}$ and ${\text{N}}_{\text{2}}$.

Therefore, the moles of the mixture produced from $0.0032\text{mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is calculated as,

$\begin{array}{c}\text{Moles}\text{of}\text{mixture}= 0.0032\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\times \frac{\frac{3}{2}\text{moles}\text{of}\text{mixture}}{1\text{mol}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\\ =0.0048\text{moles}\text{of}\text{mixture}\end{array}$

One mole of any gas occupies $22.4\text{L}$ at STP.

So, the volume occupied by $0.0048\text{mol}\text{of}\text{mixture}$ at STP $\begin{array}{c}=\frac{22.4}{1}\times 0.0048\\ =0.1075\text{L}\end{array}$

The volume occupied by a mixture is calculated by the formula,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where,

• $T_1$ is the initial temperature $273\text{K}$.
• $T_2$ is the new final temperature $290\text{K}$.
• $V_{\text{1}}$ is the initial volume $0.1075\text{L}$.
• $V_2$ is the new final volume.

Substitute these values of $T_1$, $T_2$ and $V_{\text{1}}$ in above expression,

$\begin{array}{c}\frac{V_1}{T_1}=\frac{V_2}{T_2}\\ V_2=\frac{0.1075\text{L}\times 290\text{K}}{273\text{K}}\\ =0.01153\text{L}\\ =115.3\text{mL}\end{array}$

As the given volume of gas is $79\text{mL}$, it must be ${\text{N}}_{\text{2}}\text{O}$.

Conclusion:

1. a) The given compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ decomposes to give a mixture of ${\text{N}}_{\text{2}}$ and ${\text{O}}_2$ gas, than the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create ${\text{N}}_{\text{2}}\text{O}$.
2. b) The gas present is ${\text{N}}_{\text{2}}\text{O}$