Question

For the reaction $$ 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) $$ The equilibrium constant, \(K\), has the value \(1.12 \times 10^{-54}\) at a particular temperature. a. What does the very small equilibrium constant indicate about the extent to which oxygen gas, \(\mathrm{O}_{2}(g),\) is converted to ozone gas, \(\mathrm{O}_{3}(g),\) at this temperature? b. If the equilibrium mixture is analyzed and \(\left[\mathrm{O}_{2}(g)\right]\) is found to be \(3.04 \times 10^{-2} \mathrm{M}\), what is the concentration of \(\mathrm{O}_{3}(g)\) in the mixture?

Short answer

a. The very small equilibrium constant (\(K = 1.12 \times 10^{-54}\)) indicates that only a very small amount of oxygen gas, \(\mathrm{O}_{2}(g)\), is converted to ozone gas, \(\mathrm{O}_{3}(g)\), at this temperature. b. The concentration of \(\mathrm{O}_{3}(g)\) in the mixture is approximately \(2.4 \times 10^{-19}\, \mathrm{M}\).

Step-by-step solution

Understand what a small equilibrium constant means

A small equilibrium constant, such as the value of \(K = 1.12 \times 10^{-54}\) given in this problem, indicates that the reaction does not proceed very far and, as a result, not much product is formed. Since the reaction we're considering here involves the conversion of oxygen gas to ozone gas, this means that only a small amount of ozone gas is formed at the given temperature.

Write the expression for the equilibrium constant

For the given reaction, we can write the expression for the equilibrium constant \(K\) in terms of the concentrations of the reactants and products. The equilibrium reaction in this case is: $$ 3 \,\mathrm{O}_{2}(g) \rightleftharpoons 2 \,\mathrm{O}_{3}(g) $$ So, the equilibrium constant can be expressed as: $$ K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3} $$

Calculate the concentration of O3

We are given the concentration of oxygen gas, \(\left[\mathrm{O}_{2}(g)\right] = 3.04 \times 10^{-2}\, \mathrm{M}\). Using the equilibrium constant equation, we can solve for the concentration of ozone gas: $$ \frac{[\mathrm{O}_3]^2}{(3.04 \times 10^{-2})^3} = 1.12 \times 10^{-54} $$ $$ [\mathrm{O}_3]^2 = 1.12 \times 10^{-54} \times (3.04 \times 10^{-2})^3 $$ Now, take the square root of both sides to find the concentration of \(\mathrm{O}_{3}(g)\): $$ [\mathrm{O}_3] = \sqrt{1.12 \times 10^{-54} \times (3.04 \times 10^{-2})^3} $$ Using a calculator, we get: $$ [\mathrm{O}_3] \approx 2.4 \times 10^{-19}\, \mathrm{M} $$ So the concentration of ozone gas in the mixture is approximately \(2.4 \times 10^{-19}\, \mathrm{M}\).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a vital concept in chemical equilibrium that helps to understand how far a reaction proceeds. It is calculated by taking the ratio of the concentrations of the products to the reactants, each raised to the power of their coefficients from the balanced chemical equation. For instance, in our case, the reaction is:

• \( 3 \,\mathrm{O}_{2}(g) \rightleftharpoons 2 \,\mathrm{O}_{3}(g) \)

Therefore, the equilibrium expression for this reaction is:\[ K = \frac{[\mathrm{O}_3]^2}{[\mathrm{O}_2]^3} \]A key point to note is the magnitude of \( K \). A very small \( K \) value, such as \( 1.12 \times 10^{-54} \), indicates that at equilibrium, the concentration of reactants (\( \mathrm{O}_2(g) \)) is significantly higher than that of products (\( \mathrm{O}_3(g) \)). This tells us that the reaction favors the formation of reactants. Thus, only a minute amount of \( \mathrm{O}_2 \) is converted into \( \mathrm{O}_3 \) at this specific temperature.

Concentration Calculation

Calculating the concentration of a chemical species at equilibrium involves using the equilibrium constant and the concentrations of other species involved in the reaction. In this context, knowing the concentration of oxygen gas, \([\mathrm{O}_{2}(g)]\), we aim to determine the concentration of ozone gas, \([\mathrm{O}_{3}(g)]\), within the reaction system.Starting with the given concentration of \([\mathrm{O}_{2}]\), which is \(3.04 \times 10^{-2} \mathrm{M}\), you can calculate \([\mathrm{O}_{3}(g)]\) using the equilibrium expression:\[ \frac{[\mathrm{O}_3]^2}{(3.04 \times 10^{-2})^3} = 1.12 \times 10^{-54} \]From this relation, rearrange to solve for \([\mathrm{O}_3]^2\):\[ [\mathrm{O}_3]^2 = 1.12 \times 10^{-54} \times (3.04 \times 10^{-2})^3 \]Taking the square root gives \([\mathrm{O}_3]\):\[ [\mathrm{O}_3] = \sqrt{1.12 \times 10^{-54} \times (3.04 \times 10^{-2})^3} \]Upon computing, you find \([\mathrm{O}_3] \approx 2.4 \times 10^{-19} \mathrm{M}\). This represents the very low concentration of ozone formed, consistent with the small value of \( K \).

Ozone Formation

Ozone (\( \mathrm{O}_3 \)) formation is an important atmospheric reaction, often involving the conversion of dioxygen (\( \mathrm{O}_2 \)) molecules. The equilibrium reaction of ozone formation is a classic reversible reaction where oxygen molecules combine to form ozone gas. The balanced equation for such a reaction is:

• \( 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) \)

This implies that three molecules of oxygen gas react to form two molecules of ozone. The reaction, however, is quite heavily dependent on conditions such as temperature and pressure, and most notably, on the equilibrium constant. In atmospheric chemistry, the formation of ozone is crucial for blocking harmful ultraviolet radiation from the Sun. However, under specific conditions, particularly at low temperatures or in the presence of certain catalysts, this equilibrium does not favor the production of large amounts of ozone, as indicated by a low equilibrium constant. Thus, understanding the factors that affect ozone formation is not only important for thermodynamics but also for its implications on environmental science and protection.