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.

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.

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.