A \(0.10-\mathrm{mol}\) sample of pure czone, \(\mathrm{O}_{3}\), is placed in a sealed \(1.0\) - L container and the reaction \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{~g})\) is allowed to reach equilihrium. A \(0.50-\mathrm{mol}\) sample of pure ozone is placed in a second 1.0-L container at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same? Explain each of your answers: (a) amount of \(\mathrm{O}_{2}\); (b) concentration of \(\mathrm{O}_{2}\); (c) the ratio \(\left[\mathrm{O}_{2}\right] /\left[\mathrm{O}_{3}\right]\); (d) the ratio \(\left.\left[\mathrm{O}_{2}\right]^{3} / \mathrm{O}_{3}\right]^{2} ;\) (e) the ratio \(\left|\mathrm{O}_{3}\right|^{2} /\left[\mathrm{O}_{2}\right]^{3}\).

Short Answer

Expert verified
The amount of O2 (a) and its concentration (b) will be different in the two containers, as they depend on the initial quantity of O3. The ratios O2/O3 (c), O2^3/O3^2 (d), and O3^2/O2^3 (e), which are related to the equilibrium constant, will remain the same in both containers.

Step by step solution

01

- Understanding the Reaction

The chemical reaction given is \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})\). It shows that two molecules of ozone (O3) decompose into three molecules of oxygen (O2). Note that since the reaction takes place in a sealed container with a fixed volume, the concentration of each species is solely determined by its mole number in the given volume.
02

- Analyzing Part (a) Amount of O2

The amount of O2 produced depends on the initial amount of O3 that's available for the reaction. As there is more O3 in the second container, it will produce more O2 upon complete reaction. Therefore, the amount of O2 will be different in the two containers.
03

- Analyzing Part (b) Concentration of O2

Concentration is the mole number divided by the volume. Since both reactions occur in 1.0-L containers but with different initial amounts of O3, the final concentrations of O2 will be different because more O2 will be produced in the second container which had a higher initial amount of O3.
04

- Analyzing Part (c) Ratio of O2/O3

The ratio O2/O3 at equilibrium is determined by the stoichiometry of the reaction and the equilibrium constant, which is the same for both reactions at the same temperature. Since both reactions proceed with the same stoichiometry in the same volume, the O2/O3 ratio at equilibrium will be the same.
05

- Analyzing Part (d) Ratio of O2^3/O3^2

Since this ratio is simply the equilibrium constant expression raised to the reaction coefficients of the respective substances, the ratio O2^3/O3^2 must be the same for both containers as it is equivalent to the equilibrium constant for the reaction, which is a temperature-dependent, not concentration-dependent property.
06

- Analyzing Part (e) Ratio of O3^2/O2^3

This ratio is the inverse of the equilibrium constant of the reaction raised to the reaction coefficients. Similarly, to the argument presented in the previous step, this ratio will be the same for both containers.

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Key Concepts

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

Le Châtelier's Principle
Le Châtelier's principle is a fundamental concept in chemistry that predicts how a system at equilibrium reacts to changes in concentration, temperature, volume, or pressure. It states that if an external stress is applied to a system in dynamic equilibrium, the system adjusts in a way that opposes the change.

When a system undergoes a change, such as the addition of more reactants or removal of products, the equilibrium shifts to either produce more products or consume more reactants until a new equilibrium is established. This shift helps to 'counteract' the stress imposed on the system. For example, in a gaseous reactions, reducing the volume increases pressure and shifts the equilibrium to the side with fewer moles of gas.
Equilibrium Constant
The equilibrium constant, represented as K, is a number that relates the concentrations of reactants and products of a reversible chemical reaction at equilibrium at a given temperature. It is a measure of how far a reaction proceeds before reaching equilibrium.

For the given gas-phase reaction, the equilibrium constant expression would be written as: \(K = \frac{{[O_2]^3}}{{[O_3]^2}}}\) where [O2] and [O3] are the molar concentrations of oxygen and ozone, respectively. The value of K does not change with the initial concentrations of reactants or products—it solely depends on temperature. This is why the ratios involving the equilibrium constant will be the same for both containers irrespective of the different starting amounts of ozone.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the principle that matter is neither created nor destroyed in chemical reactions. Stoichiometry uses the mole concept to help scientists determine how much of each reactant is needed and how much of each product will be made in a reaction.

The reaction provided demonstrates stoichiometry, as two moles of ozone decompose to form three moles of oxygen. This stoichiometric relationship is constant and dictates the ratios of reactants to products at equilibrium, such as the O2/O3 ratio. Calculations involving stoichiometry require a careful balance of these ratios to adequately predict the outcomes of a reaction.
Gas-phase Reactions
Gas-phase reactions are chemical reactions that occur between substances in the gas state. These reactions are particularly interesting because they can be affected by changes in pressure and volume, in addition to concentration and temperature. The behavior of gases is described by the ideal gas law, which relates the pressure (P), volume (V), temperature (T), and amount of gas (n) in moles: \(PV = nRT\), where R is the ideal gas constant.

For gas-phase reactions, changing the volume or pressure can shift the equilibrium. In the provided problem, ozone decomposes into oxygen in a sealed container, so no gases can enter or escape, meaning the total number of gas particles is conserved. Keeping the temperature constant allows us to focus on the change in the number of moles of gas to predict shifts in equilibrium based on stoichiometry.

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Most popular questions from this chapter

At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\) An analysis of a reaction vessel at \(2500 \mathrm{~K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{Cl}_{2}, 0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward cquilibrium?

Write the equilibrium expressions \(K\) for the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \Rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Explain why the following cquilibria are heterogeneous and write the reaction quotient \(Q\) for each one. (a) \(2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{FcCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{KNO}_{3}\) (s) \(\rightleftharpoons 2 \mathrm{KNO}_{2}\) (s) \(+\mathrm{O}_{2}\) (g)

A 0.500-L reaction vessel at \(700 \mathrm{~K}\) contains \(1.20 \times 10^{-3} \mathrm{~mol} \mathrm{SO}_{2}(\mathrm{~g}), 5.0 \times 10^{-4} \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\), and \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{SO}_{3}(\mathrm{~g}) .\) At \(700 \mathrm{~K}, K_{c}=1.7 \times 10^{6}\) for the cquilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}) .\) (a) Calculate the reaction quotient \(Q_{e}\) (b) Will more \(\mathrm{SO}_{3}(\mathrm{~g})\) tend to form?

Calculate the standard reaction frec cnergy for each of the following reactions: (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \neq 2 \mathrm{HI}(\mathrm{g}), K=160\) at \(500 \mathrm{~K}\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), K=47.9\) at \(400 \mathrm{~K}\)

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