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Ozone, \(\mathrm{O}_{3}(g)\), is a form of elemental oxygen that is important in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{~kJ} $$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?(\mathbf{b})\) Which has the higher enthalpy under these condi- $$ \text { tions, } 2 \mathrm{O}_{3}(g) \text { or } 3 \mathrm{O}_{2}(g) ? $$

Short Answer

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(a) The enthalpy change per mole of O₃(g) for this reaction is -142.3 kJ/mol. (b) 2 O₃(g) has a higher enthalpy than 3 O₂(g) under these conditions.

Step by step solution

01

Calculation of Enthalpy Change per Mole of O₃(g)

We are given the enthalpy change for the reaction (∆H) as -284.6 kJ. This value corresponds to the enthalpy change for 2 moles of O₃(g), as shown in the balanced chemical equation. To find the enthalpy change per mole of O₃(g), we need to divide this value by the number of moles, which is 2 moles of O₃(g). $$ \frac{-284.6 \mathrm{kj}}{2 \,\mathrm{moles \, of \, O_3(g)}} = -142.3 \mathrm{kJ/mol \, of \, O_3(g)} $$ The enthalpy change per mole of O₃(g) for this reaction is -142.3 kJ/mol.
02

Comparison of Enthalpies of 2 O₃(g) and 3 O₂(g)

Since the given enthalpy change (∆H) for the reaction is negative (-284.6 kJ), this means that the reaction is exothermic. A negative enthalpy change indicates that energy is released as the reaction proceeds, which implies that the reactants have higher enthalpy compared to the products. In this case, that means 2 O₃(g) has a higher enthalpy than 3 O₂(g) under these conditions. In conclusion, (a) The enthalpy change per mole of O₃(g) for this reaction is -142.3 kJ/mol. (b) 2 O₃(g) has a higher enthalpy than 3 O₂(g) under these conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ozone Decomposition
The decomposition of ozone into oxygen gas is a fascinating chemical reaction that takes place in our environment. Ozone, which is represented chemically as \(\mathrm{O}_3(g)\), is a form of oxygen that plays a crucial role in protecting life on Earth by absorbing most of the sun's harmful ultraviolet radiation. However, under certain conditions, such as at room temperature and standard atmospheric pressure, ozone is not very stable and can decompose into oxygen gas \(\mathrm{O}_2(g)\). This reaction can be represented as \(2 \mathrm{O}_3(g) \longrightarrow 3 \mathrm{O}_2(g)\).
Understanding this decomposition process is important because it helps explain how ozone's protective layer is maintained and balanced in the atmosphere. As ozone decomposes, it releases energy, indicating the transformation from a less stable form (ozone) to a more stable form (oxygen gas). This natural process is part of a complex cycle involving the creation and destruction of ozone, which is crucial for the atmospheric equilibrium.
Exothermic Reaction
An exothermic reaction is one in which energy is released, usually in the form of heat, into the surroundings. In this context, the decomposition of ozone to form oxygen gas is an exothermic process. The reaction's enthalpy change \(\Delta H\) is given as \(-284.6 \mathrm{~kJ}\). This negative value tells us that the reaction releases energy. For every 2 moles of ozone that decompose, 284.6 kJ of energy is liberated.
To determine the energy released per mole of ozone, simply divide the total energy change by 2, which gives \(-142.3 \mathrm{kJ/mol}\) of \(\mathrm{O}_3(g)\). This energy release is a sign that the reactants, in this case, ozone molecules, possess higher energy compared to the products, which are oxygen molecules. An exothermic reaction like this suggests that after the reaction, the system and surroundings are in a lower energy state, contributing to the stability of \(\mathrm{O}_2(g)\) over \(\mathrm{O}_3(g)\).
Ultraviolet Radiation Absorption
Ozone's critical environmental function is its ability to absorb ultraviolet (UV) radiation from the sun. In the stratosphere, a region of the Earth's atmosphere, ozone plays a key protective role by blocking harmful UV rays that could otherwise reach the Earth’s surface, causing damage to living organisms. This process is essential for life on Earth, as UV radiation can lead to health issues like skin cancer and cataracts, as well as adversely affecting ecosystems.
The UV absorption process involves the energy from UV radiation splitting ozone molecules into oxygen molecules and a singlet oxygen atom, which can quickly recombine to form ozone again. This cyclical process of ozone formation and decomposition maintains ozone levels and continues to provide a shield against UV radiation. Understanding the absorption properties of ozone is crucial for environmental science, as it helps us comprehend the effects of pollutants that can diminish the ozone layer, emphasizing the importance of ozone in our atmosphere.

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Most popular questions from this chapter

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

What is the connection between Hess's law and the fact that \(H\) is a state function?

The automobile fuel called E85 consists of \(85 \%\) ethanol and \(15 \%\) gasoline. \(\mathrm{E} 85\) can be used in so-called "flex-fuel" vehicles (FFVs), which can use gasoline, ethanol, or a mix as fuels. Assume that gasoline consists of a mixture of octanes (different isomers of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ), that the average heat of combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) is \(5400 \mathrm{~kJ} / \mathrm{mol}\), and that gasoline has an average density of \(0.70 \mathrm{~g} / \mathrm{mL}\). The density of ethanol is \(0.79 \mathrm{~g} / \mathrm{mL}\). (a) By using the information given as well as data in Appendix C, compare the energy produced by combustion of \(1.0 \mathrm{~L}\) of gasoline and of \(1.0 \mathrm{~L}\) of ethanol. (b) Assume that the density and heat of combustion of \(\mathrm{E} 85\) can be obtained by using \(85 \%\) of the values for ethanol and \(15 \%\) of the values for gasoline. How much energy could be released by the combustion of \(1.0 \mathrm{~L}\) of E85? (c) How many gallons of E85 would be needed to provide the same energy as 10 gal of gasoline? (d) If gasoline costs \(\$ 3.10\) per gallon in the United States, what is the break-even price per gallon of \(\mathrm{E} 85\) if the same amount of energy is to be delivered?

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are +185.4 \(+20.4,\) and \(-103.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) Calculate the heat evolved on combustion of \(1 \mathrm{~kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

At \(20^{\circ} \mathrm{C}\) (approximately room temperature) the average velocity of \(\mathrm{N}_{2}\) molecules in air is \(1050 \mathrm{mph}\). (a) What is the average speed in \(\mathrm{m} / \mathrm{s}\) ? (b) What is the kinetic energy (in J) of an \(\mathrm{N}_{2}\) molecule moving at this speed? (c) What is the total kinetic energy of \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) molecules moving at this speed?

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