91Ó°ÊÓ

The equilibrium constant, often denoted as \( K_p \) for gaseous reactions, is a crucial factor in understanding chemical reactions at equilibrium. For any reaction occurring in a closed system, the equilibrium constant provides a ratio of the concentration of products to reactants, each raised to the power of their coefficients in the balanced equation. In the context of our exercise, we're dealing with partial pressures because the reaction involves gases. In a gaseous equilibrium such as \[ 2 \mathrm{H}_2 \mathrm{S}(g) \leftrightharpoons 2 \mathrm{H}_2(g) + \mathrm{S}_2(g) \]the equilibrium constant expression \( K_p \) can be determined using the partial pressures of the gaseous components:\[ K_p = \frac{(P_{\mathrm{H}_2})^2 \cdot (P_{\mathrm{S}_2})}{(P_{\mathrm{H}_2 \mathrm{S}})^2} \]This expression reflects the underlying principle of mass action, where pressure terms replace concentration terms typical in solutions. Understanding \( K_p \) helps predict the direction of the reaction under specific conditions, making it an indispensable tool in chemical thermodynamics.

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more simpler substances. This can be due to various factors such as heat, light, or electricity. In our exercise, the decomposition of hydrogen sulfide (\( \mathrm{H}_2 \mathrm{S} \)) occurs at a high temperature of 1400 K, resulting in simpler molecules: dihydrogen (\( \mathrm{H}_2 \)) and disulfur (\( \mathrm{S}_2 \)).During decomposition, the reactant \( \mathrm{H}_2 \mathrm{S} \) is initially under a set pressure. As it decomposes, its pressure decreases, while the pressures of the products, \( \mathrm{H}_2 \) and \( \mathrm{S}_2 \), increase. This change in pressure signifies that the reaction proceeds and reaches a state where forward and reverse rates are equal - the equilibrium state.Understanding decomposition reactions can help explain various industrial processes, including the breakdown of compounds under high temperatures or in catalytic converters.

Partial pressure is the pressure contributed by each gas in a mixture to the total pressure exerted by the mixture. It is based on the idea that each gas behaves independently of the others.In our equilibrium scenario for the decomposition of \( \mathrm{H}_2 \mathrm{S} \), we track changes in partial pressures to understand how the system evolves. Initially, the system contains only \( \mathrm{H}_2 \mathrm{S} \) with a total pressure of 0.956 bar. As the reaction progresses and it decomposes, new gases \( \mathrm{H}_2 \) and \( \mathrm{S}_2 \) form, subsequently increasing the total pressure to 1.26 bar at equilibrium.We use the change in partial pressures to calculate the specific pressures of each gas at equilibrium, necessary for finding \( K_p \). This calculation accounts for the stoichiometry of the reaction and ensures accurate representation of each gas's contribution to the system's overall pressure.

Equilibrium calculations are essential to finding out how a reaction behaves when it reaches a state of balance at a given temperature. These calculations involve setting up expressions using the initial conditions and stoichiometry of the reaction.Let's break it down step-by-step:- Initially, you record the pressures of the reactants and products. For \( \mathrm{H}_2 \mathrm{S} \), the initial pressure is 0.956 bar, while \( \mathrm{H}_2 \) and \( \mathrm{S}_2 \) start at 0 bar.- The total pressure at equilibrium tells us how the pressure distributes among the gases. We formed an equation based on the initial pressures and the pressure changes due to decomposition.- Solving for \( x \), the change in pressure, lets us determine each component's equilibrium pressure. Partial pressures are adjusted accordingly: \( P_{\mathrm{H}_2 \mathrm{S}} = 0.956 - 2x \), \( P_{\mathrm{H}_2} = 2x \), and \( P_{\mathrm{S}_2} = x \).- By substituting these values into the \( K_p \) expression, we find our equilibrium constant, providing insight into the reaction's behavior at given conditions.These calculations not only determine the state of equilibrium but also guide predictions for shifts in reaction conditions.