Question

The following reaction represents the removal of ozone in the stratosphere: $$ 2 \mathrm{O}_{3}(g) \rightleftarrows 3 \mathrm{O}_{2}(g) $$ Calculate the equilibrium constant \(\left(K_{P}\right)\) for this reaction. In view of the magnitude of the equilibrium constant, explain why this reaction is not considered a major cause of ozone depletion in the absence of humanmade pollutants such as the nitrogen oxides and CFCs. Assume the temperature of the stratosphere is \(-30^{\circ} \mathrm{C}\) and \(\Delta G_{\mathrm{i}}^{\circ}\) is temperature independent.

Short answer

Calculate \(K_P\) using \(\Delta G^\circ = -RT \ln K_P\). A small \(K_P\) implies the reaction isn't a major cause of ozone depletion without pollutants.

Step-by-step solution

Write the Equilibrium Constant Expression

The equilibrium constant expression for a gaseous reaction in terms of partial pressures is given by:\[ K_P = \frac{{(P_{O_2})^3}}{{(P_{O_3})^2}} \]where \(P_{O_3}\) and \(P_{O_2}\) are the partial pressures of ozone \(O_3\) and oxygen \(O_2\), respectively.

Determine the Temperature in Kelvin

The temperature of the stratosphere is given as \(-30^{\circ} \mathrm{C}\). We need to convert this to Kelvin using the formula:\[ T(K) = T(°C) + 273.15 \]Thus,\[ T = -30 + 273.15 = 243.15 \, K \]

Use Gibbs Free Energy to Calculate \(K_P\)

The relationship between the standard Gibbs free energy change \(\Delta G^\circ\) and the equilibrium constant \(K_P\) is described by the equation:\[ \Delta G^\circ = -RT \ln K_P \]where \(R\) is the ideal gas constant \(8.314 \, J\, mol^{-1}\, K^{-1}\). Since \(\Delta G^\circ\) is assumed to be temperature independent, use this to solve for \(K_P\):\[ \ln K_P = -\frac{\Delta G^\circ}{RT} \]

Calculate \(\Delta G^\circ\) for the Reaction

To calculate \(\Delta G^\circ\), use known standard Gibbs energies of formation for \(O_3(g)\) and \(O_2(g)\):\[ \Delta G^\circ = 3 \Delta G^\circ_{f}(O_2) - 2 \Delta G^\circ_{f}(O_3) \]Since \(\Delta G^\circ_{f}(O_2) = 0\) and \(\Delta G^\circ_{f}(O_3)\) is known from tables, substitute these values to find \(\Delta G^\circ\) for the reaction.

Solve for \(\ln K_P\) and Convert to \(K_P\)

Substitute \(\Delta G^\circ\), \(R\), and \(T\) into the equation from Step 3:\[ \ln K_P = -\frac{\Delta G^\circ}{(8.314 \, J\, mol^{-1}\, K^{-1})(243.15 \, K)} \]Calculate \(\ln K_P\) and then find \(K_P\) by exponentiating both sides:\[ K_P = e^{\ln K_P} \]

Interpret the Magnitude of \(K_P\)

A very small \(K_P\) would indicate that the reaction does not proceed significantly toward products (\(O_2\)), implying that under stratospheric conditions without pollutants, the regeneration of \(O_2\) from \(O_3\) is not a major pathway. Thus, in the absence of human-made pollutants, this reaction is not a significant cause of ozone depletion.

Key concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K_P \), is a vital concept in chemistry that helps us understand the extent of a chemical reaction. For reactions involving gases, it is expressed in terms of partial pressures. In the reaction \( 2 \text{O}_3(g) \rightleftarrows 3 \text{O}_2(g) \), the equilibrium constant is given by:

• \( K_P = \frac{(P_{O_2})^3}{(P_{O_3})^2} \)

The symbol \((P_{O_3})\) and \((P_{O_2})\) represent the partial pressures of ozone and oxygen, respectively. The equilibrium constant indicates the ratio of concentrations (or pressures in the case of gases) of products to reactants at equilibrium. A very small \( K_P \) value shows that the reactants are predominant and the reaction does not proceed significantly towards forming the products. This concept provides a clear picture of which direction the reaction favors under equilibrium conditions.

Gibbs Free Energy

Gibbs free energy, symbolized as \( \,\Delta G^{\circ} \,\), is a thermodynamic property that tells us whether a chemical reaction will occur spontaneously under constant pressure and temperature. It relates directly to the equilibrium constant with the equation:

• \( \,\Delta G^{\circ} = -RT \ln K_P \, \)

where \( R \) is the ideal gas constant and \( T \) is the temperature in Kelvin. For the reaction \( 2 \text{O}_3(g) \rightleftarrows 3 \text{O}_2(g) \), it's crucial to calculate \(\Delta G^{\circ}\) to understand the spontaneity and equilibrium condition. If \( \,\Delta G^{\circ} \,\) is negative, the reaction tends to proceed in the forward direction, favoring product formation. The relationship also demonstrates how energy and chemical reactions are interlinked, playing a critical role in determining how favorable a reaction is at a given temperature.

Stratosphere Chemistry

The stratosphere is a layer of the Earth's atmosphere that plays a crucial role in protecting life from the sun's harmful ultraviolet radiation. Here, ozone, \( \text{O}_3 \), absorbs this radiation and decomposes into oxygen \( \text{O}_2 \), which in turn regenerates ozone, thus maintaining a balance. This delicate balance is part of what we call Stratosphere Chemistry, where various reactions occur, predominantly involving ozone.Stratosphere Chemistry is unique compared to the troposphere due to its lower density and higher UV radiation exposure. Most ozone depletion reactions occur here when pollutants such as Nitrogen Oxides or CFCs neutralize ozone, leading to significant environmental concerns. Understanding these reactions is vital because they alter the equilibrium of ozone, potentially increasing the amount of UV radiation reaching Earth.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward reaction and the reverse reaction are equal, resulting in constant concentrations of products and reactants. In the reaction \( 2 \text{O}_3(g) \rightleftarrows 3 \text{O}_2(g) \), reaching equilibrium means the conversion between ozone and oxygen occurs at the same rate in both directions.This concept is central to many areas of chemistry and environmental science. It means the system is stable, and no net change in concentrations of substances will occur unless disrupted by external factors like temperature, pressure, or catalysts. In the context of ozone depletion, equilibrium informs us about how quickly existing ozone can regenerate, especially when disrupted by human-made pollutants.

Ozone (O3)

Ozone \( \text{O}_3 \) is a molecule composed of three oxygen atoms and is a crucial component of the stratosphere. It's responsible for absorbing the majority of the sun's harmful ultraviolet radiation, thereby safeguarding life on Earth. However, ozone is also susceptible to various natural and anthropogenic factors that can lead to its depletion.Naturally, ozone forms when \( \text{O}_2 \) molecules are split by UV light, and these atomic oxygens recombine with \( \text{O}_2 \) to form \( \text{O}_3 \). Despite being a minor component of the atmosphere, its presence in the stratosphere is critical for filtering UV radiation. Understanding the chemistry of ozone, including its formation and breakdown, is essential for predicting changes in the ozone layer due to human activities.

Oxygen (O2)

Oxygen, symbolized as \( \text{O}_2 \), is a diatomic molecule and a fundamental part of Earth's atmosphere, comprising about 21% of its volume. It's a key reactant in the conversion process involving ozone in the stratosphere. In the ozone cycle, \( \text{O}_2 \) is both a starting point and a product, acting as an essential player in balancing the ozone layer. As UV radiation breaks \( \text{O}_3 \) into \( \text{O}_2 \) and atomic oxygen, these species can readily recombine to form ozone again, demonstrating the dynamic nature of the stratosphere. This constant interchange helps maintain an equilibrium that shields Earth from excessive solar UV radiation. Understanding the role of \( \text{O}_2 \) and its interactions with \( \text{O}_3 \) is vital to grasp the full picture of atmospheric chemistry.