Question

Consider the gas-phase reaction \(3 \mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{O}_{3}(g) .\) Suppose \(K_{\mathrm{eq}}\) for this reaction is \(\sim 1\) (it is not, but assume it is for this problem). Suppose you want pure ozone \(\left(\mathrm{O}_{3}\right)\) that is uncontaminated with oxygen \(\left(\mathrm{O}_{2}\right)\). (a) Why can't you simply remove the oxygen from the reaction vessel once the reaction has come to equilibrium to obtain pure ozone? (b) In fact, \(K_{\text {eq }}\) for this reaction at room temperature is \(2.5 \times 10^{-29}\). Knowing this, how important would you say Le Châtelier's principle is for this reaction when it comes to influencing the amount of ozone present at equilibrium? Explain.

Short answer

In conclusion, (a) we cannot obtain pure ozone by just removing \(O_2\) after the reaction comes to equilibrium because according to Le Chatelier's principle, the system will respond to oppose the change, and \(O_3\) will break down to form more \(O_2\) to re-establish equilibrium. (b) Le Chatelier's principle is not very important for this reaction when it comes to influencing the amount of ozone present at equilibrium because the equilibrium constant, \(K_{eq}\), is extremely small, indicating a limited extent of ozone formation.

Step-by-step solution

Part (a): Reasons for impurity at equilibrium state

In any chemical reaction at equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of the reactants and products remain constant. In this case, it means that there is always a presence of both \(O_2\) and \(O_3\) in the reaction vessel when equilibrium is established. If we try to remove \(O_2\) from the reaction vessel at equilibrium, Le Chatelier's principle states that the system will respond to oppose the change, and in this case, it means that \(O_3\) will break down to form more \(O_2\) to re-establish equilibrium. Thus, we cannot obtain pure ozone by just removing \(O_2\) after the reaction comes to equilibrium.

Part (b): Importance of Le Chatelier's principle when K_eq is provided

Given \(K_{eq} = 2.5 \times 10^{-29}\) at room temperature, we can analyze the importance of Le Chatelier's principle when it comes to influencing the amount of ozone present at equilibrium. Firstly, we can write the expression for \(K_{eq}\) for this reaction as: \[K_{eq} = \frac{[O_3]^2}{[O_2]^3}\] With the provided value of \(K_{eq}\), the concentration of \(O_3\) is very small compared to the concentration of \(O_2\), as the value of \(K_{eq}\) is very close to 0. This indicates that ozone is not favored at room temperature, and the extent of the reaction towards ozone formation is very limited. When applying Le Chatelier's principle to this reaction at room temperature, we can try to influence the amount of ozone present at equilibrium by changing the temperature, pressure, or concentration of \(O_2\). However, given the extremely small value of \(K_{eq}\), it would be difficult to cause a significant shift in the reaction towards ozone formation, and thus pure ozone will not be easily obtainable. In conclusion, Le Chatelier's principle isn't very important for this reaction when it comes to influencing the amount of ozone present at equilibrium, because the equilibrium constant, \(K_{eq}\), is extremely small, indicating a limited extent of ozone formation.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a central concept in understanding how chemical reactions respond to changes in their environment, especially when they are at equilibrium. Whenever a system at equilibrium experiences a change, such as in concentration, temperature, or pressure, the system will adjust itself to counteract this change and re-establish equilibrium. This means the system "wants" to maintain a balance.

In the case of the reaction between oxygen (\( \mathrm{O}_{2}\)) and ozone (\( \mathrm{O}_{3}\)), Le Chatelier's principle helps us to understand why we can't simply remove oxygen to produce pure ozone. If oxygen is removed from the system, the equilibrium will shift to the left, meaning more ozone will decompose back into oxygen. The system does this to replace the missing oxygen, thus maintaining the balance described by the equilibrium conditions. As a result, attempting to "purify" the ozone by removing oxygen actually leads to more ozone breaking down, leaving us with a mix rather than pure ozone.

Equilibrium Constant (K_eq)

The equilibrium constant, \( K_{eq}\), quantifies the ratio of the concentration of products to reactants at equilibrium for a given chemical reaction. It provides insights into the position of equilibrium and the extent of a reaction.

For the gas-phase reaction \( 3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g) \), the equilibrium constant expression is:

• \( K_{eq} = \frac{[O_3]^2}{[O_2]^3} \)

A very small \( K_{eq}\), like \( 2.5 \times 10^{-29}\), indicates that the concentration of ozone will be significantly lower than that of oxygen at equilibrium. Thus, the formation of ozone is not favored at room temperature. This tells us that under typical conditions, very little ozone is present compared to oxygen, reflecting a reaction that lies significantly to the left.

Large or small \( K_{eq}\) values can tell us a lot about the feasibility and direction of a reaction. In cases where \( K_{eq}\) is close to zero, forward reactions are limited, and product formation is not favored.

Gas Phase Reactions

Gas phase reactions involve reactants and products that are all in the gas phase. These reactions often occur at specific pressures and temperatures that significantly influence their behavior and the position of equilibrium.

For example, changes in the concentration of gases can be easily achieved by adjusting the pressure. Increasing the pressure usually favors the side of the reaction with fewer moles of gas, while decreasing it favors the side with more moles. In the reaction \( 3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{O}_3(g) \), increasing pressure would favor the formation of ozone due to fewer gas molecules on the product side.

Gas phase reactions are also sensitive to temperature changes. Raising the temperature for an endothermic reaction shifts the equilibrium towards the products, whereas lowering it favors the reactants. Therefore, temperature and pressure are key factors in manipulating gas phase reactions to achieve desired outcomes. Understanding these principles allows chemists to control and optimize reactions, ensuring the efficient production of desired products.