Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g})=\) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container, so that the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the pressure present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) har; (b) \(1.00\) bar.

Understanding the equilibrium constant expression is a key part of mastering chemical equilibrium. In a general sense, the equilibrium constant, denoted as 'K', quantifies the ratio of product concentrations to reactant concentrations at equilibrium for a reversible chemical reaction. For the gas-phase reaction where phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas, the expression for K is formulated using partial pressures:
\[\begin{equation}K = \frac{P_{\text{PCl3}} \times P_{\text{Cl2}}}{P_{\text{PCl5}}}\end{equation}\]
In this case, we relate the degree of decomposition, represented by the symbol \(\alpha\), to the partial pressures of the reactants and products. By manipulating this equation, as shown in the step-by-step solution, we derive a formula solely in terms of \(\alpha\) and the total pressure, P. This approach simplifies the calculation and allows students to directly relate changes in pressure to changes in the equilibrium position.

Dalton's Law of partial pressures is crucial when dealing with gas-phase reactions. According to this principle, the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual gas component. A partial pressure, in turn, is the pressure a single gas would exert if it alone occupied the entire volume.
For the decomposition of phosphorus pentachloride (\(\mathrm{PCl}_5\)), Dalton's Law helps us link the mole fraction of each gas present at equilibrium to its respective partial pressure. For instance, if \(\alpha\) is the fraction of \(\mathrm{PCl}_5\) that decomposes, then the partial pressures of \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) are each \(P\alpha\), and the partial pressure of \(\mathrm{PCl}_5\) is \(P(1 - \alpha)\). By integrating Dalton's Law into our equilibrium constant expression, we create a clear relationship between mole fractions, partial pressures, and the equilibrium constant.

The reaction quotient, denoted as 'Q', serves as a predictor of which direction a reaction will proceed to reach equilibrium. It has the same form as the equilibrium constant expression but uses the current, non-equilibrium concentrations or pressures of the reactants and products.
In our textbook scenario, calculating Q would involve the same formula as the equilibrium constant but with the initial concentrations or partial pressures of \(\mathrm{PCl}_5\), \(\mathrm{PCl}_3\), and \(\mathrm{Cl}_2\). If Q is less than K, the reaction will proceed forwards, shifting towards products to reach equilibrium. If Q is greater than K, the reaction will shift towards reactants. When Q equals K, the system is at equilibrium. This concept is pivotal for students to recognize how a reaction will adjust under different conditions, such as changes in total pressure.

Chemical reaction stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. Understanding stoichiometry is essential for predicting the amounts of products formed or reactants required in a given reaction.
In our example, the decomposition of \(\mathrm{PCl}_5\) into \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) follows a straightforward stoichiometry where one mole of \(\mathrm{PCl}_5\) yields one mole of \(\mathrm{PCl}_3\) and one mole of \(\mathrm{Cl}_2\). Thus, the fraction \(\alpha\), representing the extent of the reaction, applies equally to \(\mathrm{PCl}_3\) and \(\mathrm{Cl}_2\) as they are formed in a 1:1 ratio. This concept drives the relationship between \(\alpha\) and the equilibrium constant expression we used in the solution, highlighting the interdependence of stoichiometry and equilibrium.