Problem 54 - Chemistry and Chemical Reactivity

Question

Lanthanum oxalate decomposes when heated to lanthanum(III) oxide, \(\mathrm{CO},\) and \(\mathrm{CO}_{2}\) \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}(\mathrm{s}) \rightleftharpoons \mathrm{La}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{g})\) (a) If, at equilibrium, the total pressure in a \(10.0-\mathrm{L}\) flask is 0.200 atm, what is the value of \(K_{p} ?\) (b) Suppose 0.100 mol of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) was originally placed in the 10.0-L. flask. What quantity of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) remains unreacted at equilibrium at \(373 \mathrm{K} ?\)

Step-by-Step Solution

Answer

(a) \(K_p = 1.55 \times 10^{-6}\) atm. (b) 0.0667 mol of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) remains unreacted.

Step 1: Identify the Reaction

The given chemical reaction of lanthanum oxalate decomposing is \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}(\mathrm{s}) \rightleftharpoons \mathrm{La}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{g})\). This is a decomposition reaction that converts solid lanthanum oxalate into solid lanthanum(III) oxide and gaseous products \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\).

Step 2: Derive the Expression for \(K_p\)

Since \(K_p\) only includes gases, \[ K_p = P_{\mathrm{CO}}^3 \cdot P_{\mathrm{CO}_2}^3 \]Given that the stoichiometry is 1:3:3, the partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_2\) are equal. Let \(x\) be the partial pressure of each, so \[ K_p = x^3 \times x^3 = x^6 \].

Step 3: Calculate Partial Gas Pressures at Equilibrium

The total pressure given is 0.200 atm, which is the sum of the partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_2\) since they produce no other gases. From our set up:\[ 6x = 0.200 \text{ atm} \]\[ x = \frac{0.200}{6} = 0.0333 \text{ atm} \].

Step 4: Calculate \( K_p \) Value

Substitute \(x\) into the expression for \(K_p\):\[ K_p = (0.0333)^6 \]Calculating this value gives \(K_p = 1.55 \times 10^{-6}\) atm.

Step 5: Determine Initial Amounts and Reaction Quotient

Initially, 0.100 mol of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) is placed in a 10.0 L flask, equating to an initial concentration of 0.0100 M (since \(\frac{0.100}{10.0} = 0.0100\)). At equilibrium, a quantity \(x\) will decompose. We find the equilibrium concentration of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) to be \(0.0100 - x/10\) mol/L.

Step 6: Solve for \(x\) using Equilibrium Conditions

At equilibrium, the mole fractions for gases relate by stoichiometry (equal amounts of \(\mathrm{CO}\) and \(\mathrm{CO}_2\) total 0.2 atm). The amount of solid decomposed equals gas pressure divided by stoichiometry coefficient relation:\[ x = \text{mole fraction of CO + CO}_2 = 0.0333 \text{ mol}\].This means at equilibrium \(0.1 - 0.0333 = 0.0667\) mol of \(\mathrm{La}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\) remains.

Key Concepts

Decomposition Reaction

In a decomposition reaction, a single compound breaks down into two or more simpler substances. These reactions often involve heat, light, or electricity to break the chemical bonds. In the example of lanthanum oxalate decomposition, the solid compound splits into lanthanum(III) oxide and gases (carbon monoxide and carbon dioxide). This transformation highlights the breaking of oxalate ions into simpler carbon-based gases and solid lanthanum oxide. Decomposition plays a key role in chemistry by providing a pathway for chemical transformation and energy release depending on the type of decomposing material. Understanding decomposition reactions is crucial for predicting products and balancing chemical equations, especially in complex reactions involving multiple reactants and products.

Partial Pressure Calculation

Partial pressure is a concept that refers to the pressure exerted by a single type of gas in a mixture of gases. It is crucial in analyzing gaseous reactions, where the equilibrium constant ( \( K_p \)) is involved. In this scenario, we calculate partial pressures to find the equilibrium constant of the reaction. The total pressure in the system is the sum of the individual partial pressures of carbon monoxide and carbon dioxide, both gases resulting from the decomposition. Given their stoichiometric ratio, they exert equal partial pressures when at equilibrium. Thus, calculating partial pressures lets us better understand how gases behave under set conditions, ensuring that we can accurately predict and utilize the values in equilibrium conduction.

Gaseous Equilibrium

Gaseous equilibrium refers to the state of a chemical reaction in which the concentrations of reactants and products remain constant over time, despite the ongoing interconversion of these species. It plays a critical role in reactions, especially where gases are involved, such as the decomposition of lanthanum oxalate. At equilibrium, the amount of each gas has reached a steady state where their rates of formation are equal to their rates of decomposition. The equilibrium position can be quantified using the equilibrium constant ( \( K_p \)) for gaseous substances given their partial pressures. Understanding gaseous equilibrium is essential for controlling and predicting the behavior of reactions in industrial processes, as well as ensuring optimal conditions for the reaction to proceed with the desired efficiency.

Stoichiometry

Stoichiometry involves using balanced chemical equations to determine the relative amounts of reactants and products involved in a chemical reaction. It is like a recipe that guides how much of each element or compound is needed to ensure a reaction occurs efficiently. In the case of lanthanum oxalate decomposition, stoichiometry helps us deduce the proportional release of gases relative to the initial reactants: three moles each of CO and CO\(_2\) from one mole of lanthanum oxalate. By employing stoichiometry, we can accurately predict the amounts of gases produced and remaining reactants, making it an invaluable tool in calculating yields and designing chemical processes. This understanding also enables adjustments in reactant quantities to achieve different reaction outcomes.