Question

At \(1024^{\circ} \mathrm{C},\) the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}:\) $$4 \mathrm{CuO}(s) \rightleftarrows 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g)$$ (a) What is \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) that will decompose if \(0.16 \mathrm{~mol}\) of it is placed in a 2.0 -L flask at \(1024^{\circ} \mathrm{C}\). (c) What would the fraction be if a 1.0 -mol sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

Short answer

(a) \( K_P = 0.49 \) atm; (b) 61.3% decomposed; (c) 9.8% decomposed; (d) 0.098 mol minimum.

Step-by-step solution

Write the equilibrium expression for the reaction

For the reaction given, the equilibrium expression in terms of partial pressures is \( K_P = \frac{P_{\text{O}_2}}{1} \). This is because the solid reactants and products do not appear in the expression. Since \( P_{\text{O}_2} = 0.49 \) atm, we conclude that \( K_P = 0.49 \) atm.

Calculate moles of \( \mathrm{CuO} \) at equilibrium (Part b)

Initially, we have 0.16 mol of \( \mathrm{CuO} \) in a 2.0 L flask, so the initial concentration is \( \frac{0.16}{2.0} = 0.08 \) M. The moles of \( \mathrm{CuO} \) that decompose is equal to \( x \), producing \( \frac{x}{4} \) mol of \( \mathrm{O}_2 \), since 4 mol \( \mathrm{CuO} \) produce 1 mol \( \mathrm{O}_2 \). At equilibrium, \( P_{\text{O}_2} = \frac{x}{4} \times \frac{RT}{2.0} = 0.49 \) atm. Solving gives \( x = 0.098 \) mol.

Calculate the fraction of \( \mathrm{CuO} \) decomposed (Part b)

The fraction decomposed is \( \frac{\text{moles decomposed}}{\text{initial moles}} = \frac{0.098}{0.16} \approx 0.613 \). Thus, 61.3% of \( \mathrm{CuO} \) decomposes.

Calculate moles of \( \mathrm{CuO} \) at equilibrium with 1.0 mol initial (Part c)

Repeat the process with 1.0 mol initial: the initial concentration is \( \frac{1.0}{2.0} = 0.50 \) M. The moles of \( \mathrm{CuO} \) decomposed \( x \), producing \( \frac{x}{4} \) mol \( \mathrm{O}_2 \), and solve \( \frac{x}{4} \times \frac{RT}{2.0} = 0.49 \) atm to get \( x = 0.098 \) mol.

Calculate the fraction with 1.0 mol \( \mathrm{CuO} \) (Part c)

The fraction decomposed is \( \frac{0.098}{1.0} = 0.098 \). So 9.8% of the \( \mathrm{CuO} \) decomposes.

Determine minimum \( \mathrm{CuO} \) to establish equilibrium (Part d)

For equilibrium to establish, enough \( \mathrm{CuO} \) must decompose to reach \( P_{\text{O}_2} = 0.49 \) atm. Using \( \frac{x}{4} \times \frac{RT}{2.0} = 0.49 \) to solve for smallest \( x \), the smallest amount of \( \mathrm{CuO} \) required is \( 0.098 \) mol.

Key concepts

equilibrium constant (Kp)

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.

decomposition reaction

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.

mole fraction

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

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.