Chapter 9, Problem 9.149QP

A new allotrope of oxygen, O₄, has been reported. The exact structure of O₄ is unknown, but the simplest possible structure would be a four-member ring consisting of oxygen-oxygen single bonds. The report speculated that the O₄ molecule might be useful as a fuel “because it packs a lot of oxygen in a small space, so it might be even more energy-dense than the liquefied ordinary oxygen used in rocket fuel.” (a) Draw a Lewis structure for O₄ and write a balanced chemical equation for the reaction between ethane, C₂H₆(g), and O₄(g) to give carbon dioxide and water vapor. (b) Estimate ΔH° for the reaction. (c) Write a chemical equation illustrating the standard enthalpy of formation of O₄(g) and estimate ΔH°_f. (d) Assuming the oxygen allotropes are in excess, which will release more energy when reacted with ethane (or any other fuel): O₂(g) or O₄(g)? Explain using your answers to parts (a)–(c).

Interpretation Introduction

Interpretation:

Lewis structure of $O_4$ has to be drawn.  Balanced chemical equation for the reaction of $O_4$ with ethane to give carbon dioxide and water vapour has to be written.

Concept Introduction:

Electron dot structure also known as Lewis dot structure represents the number of valence electrons of an atom or constituent atoms bonded in a molecule.  Each dot corresponds to one electron.

Balanced reaction is a chemical reaction in which number of atoms for each element in the reaction and the total charge are same on both reactant side and the product side.

Steps in balancing the information

• Step 1: Write the unbalanced equation
• Step 2: Find the coefficient to balance the equation.

According to the “Law of Conservation of Mass”, a chemical equation is balanced when the number of atoms involved in the reactant side is equal to that of the product side.

Explanation of Solution

Total valence electrons in $O_4$ is,

$\text{4}\left({\text{valence e}}^{\text{-}}\text{ on O}\right)\text{ = 4}\left(6\right)\text{ = 24}$

Accordingly Lewis structure of $O_4$ is drawn which shows $24$ electrons in the molecule including bonding and non-bonding electrons.

Balanced chemical equation for the reaction of $O_4$ with ethane to carbon dioxide and water vapor is written as,

$C_2{H}_{6(g)}+\frac{7}{4}{O}_{4(g)}\underset{}{\overset{}{\to }}2C{O}_{2(g)}+3H_2{O}_{(g)}$

Interpretation Introduction

Interpretation:

$\Delta H°$ for the reaction of $O_4$ with ethane to give carbon dioxide and water vapour has to be calculated.

Concept Introduction:

$\Delta H°$ refers to change in enthalpy.  Change in enthalpy in a reaction and bond enthalpy (BE) are related as,

$\text{ΔH°}\text{=}\text{Σ BE (reactants) - Σ BE (products)}$

Answer to Problem 9.149QP

$\Delta H°$ for the reaction of $O_4$ with ethane to give carbon dioxide and water vapor is calculated as $-2131kJ/mol.$

Explanation of Solution

$\begin{array}{l}\Delta H_{rxn}^{°}\text{=}\text{ΣBE(reactants)-ΣBE(products)}\\ \text{=}\left[\text{6BE}\left(\text{C-H}\right)+BE(C-C)+\frac{7}{4}\times 4BE\left(O-O\right)\right]-\left[2\times 2BE\left(\text{C=O}\right)+3\times 2BE\left(O-H\right)\right]\\ =\text{6}\left(414kJ/mol\right)+\left(347kJ/mol\right)+7\left(142kJ/mol\right)-4\left(799kJ/mol\right)-6\left(460kJ/mol\right)\\ =-2131kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

A chemical equation illustrating the reaction of formation of $O_4$ has to be written.  $\Delta H_f^{°}$ for $O_4$ has to be calculated.

Concept Introduction:

$\Delta H°$ refers to change in enthalpy.  Change in enthalpy in a reaction and bond enthalpy (BE) are related as,

$\text{ΔH°}\text{=}\text{Σ BE (reactants) - Σ BE (products)}$

$\Delta H°$ of the reaction also corresponds to the enthalpy of formation of the product.

Answer to Problem 9.149QP

A chemical equation illustrating the reaction of formation of $O_4$ is written as,

$2O_{2(g)}\stackrel{}{\to }{O}_{4(g)}$

$\Delta H_f^{°}$ for $O_4$ is calculated as $429kJ/mol.$

Explanation of Solution

A chemical equation illustrating the reaction of formation of $O_4$ is written as,

$2O_{2(g)}\stackrel{}{\to }{O}_{4(g)}$

Enthalpy of formation of $O_4$, $\Delta H_f^{°}$ is calculated as,

$\begin{array}{l}\Delta H_f^{°}\left[O_{4(g)}\right]\text{=}\text{ΣBE(reactants)-ΣBE(products)}\\ \text{=}\left[\text{2BE}\left(\text{O=O}\right)-4BE\left(O-O\right)\right]\\ =\text{2}\left(498.7kJ/mol\right)-4\left(142kJ/mol\right)\\ =429kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

The energy released by reaction of $O_{2(g)}$ with ethane and the energy released by reaction of $O_{4(g)}$ with ethane has to be compared.

Answer to Problem 9.149QP

The energy released by reaction of $O_{2(g)}$ with ethane will be lesser than the energy released by reaction of $O_{4(g)}$.

Explanation of Solution

$\Delta H_f^{°}$ for $O_4$ is calculated as $429kJ/mol.$  The positive value of $\Delta H_f^{°}$ indicates lesser amount of energy is required to break the four $O-O$ single bonds in $O_4$.  The energy required to break one $O=O$ double bond will be greater.  Thus the reaction that involves breaking four $O-O$ single bonds requires lesser energy and it releases more energy during product formation.

Hence the energy released by reaction of $O_{2(g)}$ with ethane will be lesser than the energy released by reaction of $O_{4(g)}$.