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Ozone, \(\mathrm{O}_{3}(g),\) is a form of elemental oxygen that plays an important role 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 \(\mathrm{O}_{3}(g) ?\) (b) Which has the higher enthalpy under these conditions, 2 \(\mathrm{O}_{3}(g)\) or 3 \(\mathrm{O}_{2}(g) ?\)

Short Answer

Expert verified
(a) The enthalpy change per mole of O₃(g) is \(-142.3 \, \text{kJ/mol}\). (b) 2 O₃(g) has a higher enthalpy than 3 O₂(g) under the given conditions.

Step by step solution

01

Calculate the enthalpy change per mole of O₃(g)

Divide the enthalpy change of the overall reaction by the number of moles of O₃ involved in the reaction. Reaction: \(2 O_{3}(g) \longrightarrow 3 O_{2}(g)\), \(\Delta H = -284.6 \, \text{kJ}\) The reaction involves 2 moles of Ozone (O₃). To get the enthalpy change per mole of O₃(g), we need to divide the value of \(\Delta H\) provided (the total change in enthalpy for the reaction) by 2: \(ΔH_{O_{3}}= \frac{-284.6 \, \text{kJ}}{2 }\)
02

Calculate the enthalpy change per mole of O₃(g)

Perform the division to find the enthalpy change per mole of O₃(g). \(ΔH_{O_{3}} = -142.3 \, \text{kJ/mol}\) The enthalpy change per mole of O₃(g) is -142.3 kJ/mol.
03

Determine which molecule has higher enthalpy

Since the reaction has a negative enthalpy change, the reactants (2 O₃(g)) have higher enthalpy than the products (3 O₂(g)) under these specific conditions. So, the answer is that 2 O₃(g) has higher enthalpy than 3 O₂(g) under the given conditions.

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Key Concepts

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

Thermochemistry
Thermochemistry is the branch of chemistry that deals with the energy changes associated with chemical reactions. In the context of our discussion on ozone decomposition, thermochemistry focuses on the enthalpy change, which is a measure of the heat content of a chemical system. This heat content changes as reactants transform into products during a reaction.

Enthalpy change (ΔH) is often reported in kilojoules per mole (kJ/mol) and can be either positive or negative. A negative ΔH indicates that the reaction is exothermic, meaning it releases heat to the surroundings; conversely, a positive ΔH signifies an endothermic reaction, where heat is absorbed from the surroundings. The magnitude of the enthalpy change provides key information about the energetics of the chemical process, allowing scientists and engineers to predict the behavior of reactions under various conditions.
Enthalpy per mole calculation
When calculating the enthalpy change per mole for a given reaction, it is crucial to understand stoichiometry and the idea of 'mole'. A mole is a unit that represents a specific number of particles, atoms, or molecules. In thermochemical equations, the coefficient before a chemical formula indicates the number of moles involved in the reaction.

To calculate the enthalpy per mole, we must first identify the total enthalpy change for the reaction (given as ΔH) and then divide this value by the number of moles of the substance in question. In the case of ozone decomposition, the reaction starts with 2 moles of O₃ and produces 3 moles of O₂, with a total enthalpy change of -284.6 kJ. Dividing -284.6 kJ by the 2 moles of ozone (O₃) gives us -142.3 kJ/mol, which is the enthalpy change per mole for O₃. This step-by-step approach ensures an accurate and clear understanding of the energy costs or gains per unit of reactant or product.
Stratospheric ozone
The role of ozone (O₃) in the stratosphere is a critical environmental concern as it acts like Earth's natural sunscreen, absorbing and scattering the potentially harmful ultraviolet (UV) radiation from the sun. Stratospheric ozone is found naturally in the upper layers of the atmosphere and performs a delicate balancing act, constantly being created and decomposed.

Ozone is formed through the reaction of oxygen molecules (O₂) with individual oxygen atoms (O), often generated by the interaction of UV light breaking down other ozone molecules. Conversely, ozone decomposes to O₂ in a reaction that can take place even at room temperature. This balance is what maintains the protective ozone layer. The enthalpy change for the decomposition of ozone (-142.3 kJ/mol) indicates that it's an exothermic reaction, releasing energy as it breaks into molecular oxygen. Understanding the energy changes during the formation and decomposition of ozone is critical for scientists monitoring the health of the ozone layer and assessing the impact of human activities such as the release of chlorofluorocarbons (CFCs).

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Most popular questions from this chapter

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is cooled by removing 0.655 \(\mathrm{kJ}\) of heat. It shrinks on cooling, and the atmosphere does 382 J of work on the balloon. (b) A 100.0 -g bar of gold is heated from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) during which it absorbs 322 \(\mathrm{J}\) of heat. Assume the volume of the gold bar remains constant.

Can you use an approach similar to Hess's law to calculate the change in internal energy, \(\Delta E,\) for an overall reaction by summing the \(\Delta E\) values of individual reactions that add up to give the desired overall reaction?

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g) :\) $$\mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{kJ}$$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of 1 mol of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l),\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

(a) What is meant by the term state function? (b) Give an example of a quantity that is a state function and one that is not. (c) Is the volume of a system a state function? Why or why not?

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

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