Problem 88 - Chemistry

Question

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

Step-by-Step Solution

Answer

(a) \(K_P = 0.49\)\n(b) 0.5 or 50%\n(c) 0.5 or 50%\n(d) No smallest limit. Even with a very small amount of CuO, equilibrium would be achieved as per the stoichiometry of the reaction.

Step 1: Calculation of KP

KP is calculated using the formula \[KP = \frac{(P_{O_2})^{\Delta n}}{1}\] Since there is only oxygen gas involved in the equation, the equation simplifies to KP = P_{O_2}. Given that the pressure is 0.49 atm, KP is therefore KP = 0.49.

Step 2: Calculation of the stoichiometric factor

Undertake a stoichiometric analysis to find out how much CuO decomposes to produce the observed amount of O2. Given that 2 moles of O2 are formed from 4 moles of CuO, using the stoichiometry of the reaction, the factor \(f\) is calculated as \(f = moles\, of\, CuO/total\, moles = \frac{2}{4} = 0.5\)

Step 3: Analyzing the fraction that decomposes for 0.16 moles in 2.0L flask

While doubling the amount of CuO, the amount of O2 produced is also halved to maintain the equilibrium. The moles of CuO that decompose is therefore \(0.5 \times 0.16 = 0.08\, moles\). Thus, the fraction that decomposes is \(0.08/0.16 = 0.5 (or 50%)\).

Step 4: Analyzing the fraction that decomposes for 1.0 mole

For a 1.0 mole sample of CuO, the same stoichiometry applies since the reaction is independent of the initial amount of CuO. So the fraction that decomposes is still 50%.

Step 5: Determining the smallest amount of CuO (in moles) that would establish the equilibrium

The equilibrium would be established even with very small amounts of CuO. This is because the reaction quotient Q would equal KP at any point where the stoichiometric requirements of the reaction are met. So theoretically, even with a vanishingly small amount of CuO, equilibrium would still be achieved as per the stoichiometry of the reaction. Thus, there is no smallest limit.

Key Concepts

Equilibrium Constant (Kp)

Understanding the equilibrium constant, represented as Kp when dealing with gases, is fundamental in the study of chemical equilibrium. Kp is a measure of the extent to which a chemical reaction has achieved equilibrium at a particular temperature, expressed in terms of the partial pressures of the gaseous reactants and products. The value of Kp remains constant at a given temperature and is a key indicator of how far a reaction will proceed before reaching equilibrium.

For the decomposition of copper(II) oxide, the equilibrium constant Kp is calculated by using the partial pressure of oxygen gas (\( P_{O_2} \) and the change in the number of moles of gas (\( \)Delta n) in the reaction. Since the reaction involves the decomposition of a solid to form another solid and oxygen gas, Kp simplifies to the partial pressure of oxygen at equilibrium. The value obtained for Kp is vital to predict how changes in conditions—like pressure and temperature—will affect the system's equilibrium.

Stoichiometry

Stoichiometry is the section of chemistry that involves quantifying the relationships between reactants and products in a chemical reaction. It is like a recipe for a chemical reaction, detailing the amounts of each substance involved. By understanding the molar ratios provided by the balanced chemical equation, one can predict the amount of reactants needed to produce a certain amount of products, or vice versa.

In the given problem, stoichiometry is used to determine the fraction of copper(II) oxide (CuO) that decomposes in the reaction. By following the stoichiometric coefficients from the balanced equation—4 moles of CuO producing 2 moles of \(Cu_2O\) and 1 mole of \(O_2\)—it is possible to calculate the changes in mole quantities of the reactants and the fraction of the compound that reacts to reach equilibrium.

Decomposition Reaction

A decomposition reaction occurs when a single compound breaks down into two or more elements or new compounds. This type of reaction is integral to understanding how complex substances are broken into simpler ones—crucial in processes ranging from geological transformations to industrial applications.

In the exercise, copper(II) oxide decomposes to form copper(I) oxide and oxygen gas when heated. The reaction exemplifies a thermal decomposition, where heat energy facilitates the breaking of chemical bonds. Knowing the nature of the decomposition reaction helps determine how energy changes and reactant-to-product ratios affect the reaction progress and the conditions under which equilibrium is attained.

Partial Pressure

Partial pressure is the pressure that a single gas in a mixture of gases would exert if it occupied the whole volume on its own. It is a fundamental concept in gas laws and is directly related to the concentration of the gas in a mixture, which in turn affects reaction rates and chemical equilibria in gaseous systems.

In the context of the reaction involving CuO, the partial pressure of oxygen (\(P_{O_2}\)) is significant because it represents the driving force of the reaction towards equilibrium. Calculating the partial pressure allows us to understand how variations in pressure can shift the equilibrium position, according to Le Chatelier's principle, and influence the reaction's extent or the fraction of reactants that will decompose.

Le Chatelier's Principle

Le Chatelier's principle is a vital concept in chemical equilibrium, stating that if an external change is applied to a system at equilibrium, the system adjusts in a way that opposes the change. This is true for changes in concentration, pressure, volume, or temperature of the system. It's like a balancing act where the reaction shifts to relieve the imposed stress and restore equilibrium.

When considering the decomposition of CuO, if external conditions such as the pressure of oxygen gas change, the position of equilibrium will shift to counteract this change. For example, increasing the pressure of \(O_2\) would shift the equilibrium to the left, reducing the amount of oxygen gas by forming more CuO. This principle helps us predict how the reaction will respond to changes in the system's conditions and consequently how much of CuO will decompose under varying scenarios.