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The equilibrium constant, denoted as \( K_p \), is a fundamental aspect of understanding chemical equilibrium in reactions involving gases. This constant is calculated using the partial pressures of gases in a reaction, serving as a measure of the extent to which a reaction proceeds before reaching equilibrium. In the decomposition of copper(II) oxide \((\text{CuO})\), the chemical reaction is:\[4 \text{CuO} (s) \rightleftarrows 2 \text{Cu}_2 \text{O} (s) + \text{O}_2 (g)\]For gaseous reactions, we express the equilibrium constant \( K_p \) as the ratio of the product's partial pressure to that of the reactants. Here, only \( \text{O}_2 \) is a gas, so \( K_p = P_{\text{O}_2} \). For this reaction, since the pressure of \( \text{O}_2 \) is \( 0.49 \) atm at equilibrium, \( K_p \) equals \( 0.49 \) atm. Understanding and calculating \( K_p \) helps in predicting how the reaction will behave under different conditions.

A decomposition reaction is a chemical process where a single compound breaks down into two or more simpler substances. In the context of \( \text{CuO} \), the decomposition is represented by its transformation into copper(I) oxide and oxygen gas:\[4 \text{CuO} (s) \rightarrow 2 \text{Cu}_2 \text{O} (s) + \text{O}_2 (g)\]This type of reaction is crucial in understanding how compounds decompose when subjected to certain conditions, such as high temperatures. In the given exercise, the decomposition of \( \text{CuO} \) is significant at \( 1024^{\circ} \text{C} \), resulting in the formation of \( \text{O}_2 \). Knowing how decomposition reactions work allows chemists to predict the quantities of products formed, which is essential for carrying out further calculations, like determining equilibrium constants.

The mole fraction is an important concept in chemistry that represents the ratio of the number of moles of a component to the total number of moles in the mixture. In gaseous systems, it is used to calculate partial pressures required for chemical equilibrium calculations. For the decomposition reaction of \( \text{CuO} \), initially, there are pure reactants and no products. As \( \text{O}_2 \) forms, its mole fraction becomes relevant in determining how much of \( \text{CuO} \) decomposes. The approach involves determining how moles of \( \text{O}_2 \) relate to the reaction stoichiometry and how they influence the equilibrium state. Understanding the mole fraction helps in calculating the partial pressures and subsequently the equilibrium constant \( K_p \). It is an essential component of evaluating the extent of reaction progress.

Partial pressure is the pressure exerted by a single species in a gas mixture and plays a pivotal role in gas-phase equilibrium reactions. Each gas in a mixture contributes to the total pressure in proportion to its mole fraction. In the example of \( \text{CuO} \) decomposition, knowing the partial pressure of \( \text{O}_2 \) is crucial as it is the only gaseous product. The partial pressure of \( \text{O}_2 \) is given as \( 0.49 \) atm, which is used directly to determine the equilibrium constant \( K_p \). Calculating partial pressures allows for understanding and predicting how a system will reach equilibrium and how changing conditions might shift this balance. By monitoring partial pressures, chemists can apply Le Chatelier's principle to predict shifts in equilibrium.