The equilibrium constant \(K=3.5 \times 10^{4}\) for the reaction \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})=\mathrm{PCl}_{5}(\mathrm{~g})\) at \(760^{\circ} \mathrm{C}\). At equilibrium, the partial pressure of \(\mathrm{PCl}_{5}\) was \(2.2 \times\) \(10^{4}\) bar and that of \(\mathrm{PCl}_{3}\) was \(1.33\) bar. What was the equilibrium partial pressure of \(\mathrm{Cl}_{2}\) ?

The equilibrium constant, denoted as K, is a crucial number in chemistry that quantifies the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. In our context, it is calculated using the partial pressures of the gaseous components, since we are dealing with a gas phase reaction.

For the reaction \( \mathrm{PCl}_3(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) = \mathrm{PCl}_5(\mathrm{g}) \), an equilibrium constant with a high value, such as \(3.5 \times 10^{4}\), suggests that at equilibrium, the reaction will favor the formation of products, in this case \(\mathrm{PCl}_5\).

It's also worth noting that the equilibrium constant is temperature-dependent and will vary if the temperature changes. However, at a fixed temperature, it remains constant, regardless of the initial concentrations or partial pressures of the reactants and products.

Partial pressure is a term that describes the pressure contributed by a single gas in a mixture of gases. It's proportionate to the mole fraction of that gas in the entire mixture and is used when dealing with gases at equilibrium, much like concentrations are used for solutions.

In our example, we are given the partial pressures of \(\mathrm{PCl}_5\) and \(\mathrm{PCl}_3\), and we are required to find the partial pressure of \(\mathrm{Cl}_2\). Knowing the individual partial pressures is key to calculating the equilibrium constant expression for gaseous reactions.

Understanding how to manipulate these pressures to isolate a single gas's pressure—like we did in our example by rearranging the equilibrium expression—is an essential skill in solving equilibrium problems.

The equilibrium expression is a mathematical representation of the equilibrium constant. For gas-phase reactions, it often involves partial pressures, as seen in the example with \( K = \frac{P_{\mathrm{PCl}_5}}{P_{\mathrm{PCl}_3} \times P_{\mathrm{Cl}_2}} \).

This formula illustrates how each gas's partial pressure at equilibrium contributes to a balanced system. When dealing with the equilibrium expression, it is important to remember that it only includes gases and aqueous solutions in the equation, omitting pure solids and liquids.

By following a step-by-step approach, you can rearrange the equilibrium expression to solve for a missing partial pressure, as demonstrated with \( P_{\mathrm{Cl}_2} = \frac{P_{\mathrm{PCl}_5}}{K \times P_{\mathrm{PCl}_3}} \). This level of understanding can help students unlock complex equilibrium challenges.