Chapter 5, Problem 5.119P

Interpretation Introduction

(a)

Interpretation:

The balanced chemical equation for decomposition of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ should be determined.

Concept Introduction:

In a balanced chemical equation, all the constituents present in the reaction have equal number of atoms on both side of the reaction arrow.

Answer to Problem 5.119P

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

Explanation of Solution

Decommission of solid Ammonium Nitrate generates gaseous dinitrogen oxide along with the water vapour.

The balanced chemical equation of the reaction taking place is as depicted below:

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

From the above balanced reaction, for each mole of NH₄ NO₃ produces one mole of di-nitrogen oxide and 2 moles of water.

Interpretation Introduction

(b)

Interpretation:

The partial pressure of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ produced should be determined.

Concept Introduction:

To calculate the partial pressure of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ and ${\text{H}}_{\text{2}}\text{O}\left(g\right)$ produced ideal gas equation is employed.

$P=\frac{nRT}{V}$

Where,

$\begin{array}{l}\text{R = Universal gas constant }\left(\text{0}{\text{.08206 L atm K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ \text{T = Temperature }\left(\text{K}\right)\\ \text{P = Pressure }\left(\text{atgm}\right)\\ \text{n = Number of the moles}\\ \text{V = Volume }\left(l\right)\end{array}$.

Answer to Problem 5.119P

Partial pressure of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ is 0.737 atm. Partial pressure of water is 1.47 atm.

Explanation of Solution

Calculate the number of moles of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ as follows:

$\text{0}{\text{.0312 mol NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{× }\frac{{\text{2 mol N}}_{\text{2}}\text{O}}{{\text{I mol NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{ = 0}{\text{.0312 mol N}}_{\text{2}}\text{O}$

Therefore, number of moles of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ is 0.0312 mol N₂ O.

Calculate the number of moles of water is as follows:

$\text{0}{\text{.0312 mol NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{× }\frac{{\text{2 mol H}}_{\text{2}}\text{O}}{{\text{I mol NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\text{ = 0}{\text{.0624 mol H}}_{\text{2}}\text{O}$

Therefore, number of moles of water is 0.0624 mol H₂ O.

From the ideal equation we have.

$\text{PV=}\text{nRT}$

$\begin{array}{l}\text{Volume of tank }\left(\text{V}\right)\text{ =1}\text{.75L}\\ \text{Number of moles of dinitrogen oxygen }\left(\text{n}\right)\text{ = 0}\text{.0312 moles}\\ \text{Temperature = 503 K}\text{.}\end{array}$

To calculate the partial pressure of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ substitute all the known value in the equation.

$\begin{array}{l}\text{P=}\frac{\text{nRT}}{\text{V}}\\ =\frac{\text{(0}\text{.0312}\text{mol)}\left(\text{0}\text{.08206}\frac{\text{L}\text{atm}}{\text{K}\text{mol}}\right)\text{(503}\text{K)}}{\text{(1}\text{.75}\text{L)}}\\ =0.737\text{a}\text{t}\text{m}\end{array}$

Therefore, the partial pressure of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ = 0.737 atm.

Volume of tank v = 1.75 L.

Number of moles of water n = 0.0624 moles.

The temperature t = 503 K.

To calculate the partial pressure of water, substitute all the known value in the equation.

$\text{P=}\frac{\text{nRT}}{\text{V}}$

$=\frac{(0.0624m\text{O}\text{l})\left(0.0821\frac{\text{l}\text{a}\text{t}\text{m}}{Km\text{O}\text{l}}\right)(503K)}{(1.75\text{l})}$

$\text{=1}\text{.47 atm}$

Therefore, partial pressure of water is 1.47 atm.

Interpretation Introduction

(c)

Interpretation:

The total gas pressure present in the flask at 230⁰ C should be determined.

Concept Introduction:

Dalton's law of partial pressure is state that the total pressure of a mixture gases is sum of the pressures that every gas would exert if it were present alone.

Answer to Problem 5.119P

Total pressure present in flask is 2.21 atm.

Explanation of Solution

In the provided reaction the gases molecules are ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ and water vapor. Thus, total pressure present in the flask is the sum of the partial pressures of ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ and water vapor.

$\begin{array}{c}{\text{P}}_{\text{total}}{\text{= P}}_{{\text{N}}_{\text{2}}\text{O}}{\text{+ P}}_{{\text{H}}_{\text{2}}\text{O}}\\ \begin{array}{l}\text{= 0}\text{.737 atm +1}\text{.47 atm}\hfill \\ \text{= 2}\text{.21 atm}\hfill \end{array}\end{array}$

Therefore, total pressure present in flask is 2.21 atm.

Interpretation Introduction

(d)

Interpretation:

The three equivalent resonance structure for ${\text{N}}_{\text{2}}\text{O }\left(g\right)$ should be drawn.

Concept Introduction:

The resonance structures show the arrangement of electrons and bonds in a molecule. The lone pair present on atom can show delocalization with pi electrons of double or triple bonds resulting formation of resonance structures. The position of atoms remains the same only position of bonds changes.

Answer to Problem 5.119P

Explanation of Solution

The resonance structures of N₂ O is shown in the following diagram.

Since, nitrogen has 3 valence electrons it can form three covalent bonds with other atoms. There are two valence electrons in oxygen thus, it can form one double or two single bonds with other atoms. Being more electronegative in nature, oxygen atom will be placed at the terminal position. Thus, there will be one double bond between two nitrogen atom and one double bond between nitrogen and oxygen atom resulting negative charge on one nitrogen atom and positive charge on other nitrogen atom.

The negative charge can delocalize with pi electrons of double bond resulting two resonance forms.