At \(T \mathrm{~K}\), a compound \(\mathrm{AB}_{2}(\mathrm{~g})\) dissociates according to the reaction: \(2 \mathrm{AB}_{2}(\mathrm{~g})\) \(\rightleftharpoons 2 \mathrm{AB}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})\), with a degree of dissociation ' \(\mathrm{x}\) ' which is small compared with unity. The expression for 'x' in terms of the equilibrium constant, \(K_{\mathrm{p}}\) and the total pressure, \(P\), is (a) \(\frac{K_{P}}{P}\) (b) \(\left(K_{P}\right)^{1 / 3}\) (c) \(\left(\frac{2 K_{P}}{P}\right)^{1 / 3}\) (d) \(\left(\frac{K_{P}}{P}\right)^{1 / 3}\)

The equilibrium constant, represented as either K_c (in terms of molar concentrations) or K_p (in terms of partial pressures), is a fundamental aspect of chemical equilibrium. It quantifies the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

For a reaction where the reactants A and B transform into products C and D, the equilibrium constant in terms of partial pressure is given by: \[\begin{equation}K_{p} = \frac{P_{C}^{c} \cdot P_{D}^{d}}{P_{A}^{a} \cdot P_{B}^{b}}\end{equation}\]where P represents the partial pressure of each species and a, b, c, and d are their respective stoichiometric coefficients in the balanced chemical equation.

In the case of the dissociation of compound AB2, the equilibrium constant would reflect the ratio of the products' partial pressures (squared for AB and once for B2) over the reactant's partial pressure squared. Understanding how to utilize this expression is key to solving problems that involve calculating the equilibrium composition of a reaction mixture.

The degree of dissociation, typically denoted by 'x', is a measure of the extent to which a compound separates into its constituent elements or simpler compounds. It is a dimensionless quantity, representing the fraction of a mole of reactant that has dissociated at equilibrium.

For a simple dissociation reaction, \[\begin{equation}A \rightleftharpoons nB\end{equation}\]if we start with one mole of A and 'x' is the degree of dissociation, at equilibrium we will have (1-x) moles of A and nx moles of B.

The step-by-step solution given above illustrates how 'x' is used to express the concentrations of reactants and products in terms of an initial concentration 'a' and the relation to the equilibrium constant. It's important for students to understand that while 'x' provides a useful way to express concentrations at equilibrium, it is also tied to how the equilibrium constant is defined and how pressures or concentrations are calculated.

Partial pressure is the pressure that would be exerted by one of the gases in a mixture if it occupied the entire volume on its own. In a chemical reaction involving gases, partial pressures play a crucial role, as they relate to the concentrations of the gases when dealing with gaseous equilibrium.

For a gas mixture, the total pressure P is the sum of the partial pressures of each individual gas. According to Dalton's Law, \[\begin{equation}P_{total} = P_{1} + P_{2} + P_{3} + \ldots\end{equation}\]where P₁, P₂, P₃, etc., are the partial pressures of the gases.

In context with the solved exercise, the degree of dissociation 'x' influenced the partial pressures of the reactants and products. Recognizing how to connect the partial pressures with the initial pressure and the degree of dissociation is essential when using the equilibrium constant to find the composition of the reaction at equilibrium.