Question

Consider the decomposition of barium carbonate: $$ \mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s)+\mathrm{CO}_{2}(g) $$ Using data from Appendix \(\mathrm{C}\), calculate the equilibrium pressure of \(\mathrm{CO}_{2}\) at (a) \(298 \mathrm{~K}\) and (b) \(1100 \mathrm{~K}\).

Short answer

(a) At 298 K, the equilibrium pressure of CO₂ (P₃₀₀) is approximately \(4.13 \times 10^{-5}\) atm. (b) At 1100 K, the equilibrium pressure of CO₂ (P₁₁₀₀) is approximately 12.94 atm.

Step-by-step solution

Find the standard Gibbs free energy change for the reaction

The standard Gibbs free energy change for the reaction can be calculated using the standard Gibbs free energies of formation of the products and reactants: $$ \Delta G^\circ_{rxn} = \sum \Delta G^\circ_{products} - \sum \Delta G^\circ_{reactants} $$ Using the values of standard Gibbs free energies of formation (ΔGf°) from Appendix C: ΔGf°(BaCO₃) = -1134 kJ/mol ΔGf°(BaO) = -548 kJ/mol ΔGf°(CO₂) = -394 kJ/mol Now calculate the standard Gibbs free energy change for the reaction at each temperature.

Calculate the equilibrium constant K at given temperatures

We can find the equilibrium constant (K) at each temperature using the standard Gibbs free energy change and the following equation: $$ K = e^{\left(-\frac{\Delta G^\circ_{rxn}}{RT}\right)} $$ Where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvins. Calculate the equilibrium constant K for both temperatures 298 K and 1100 K.

Calculate the equilibrium pressure of CO₂

Now that we have K for both temperatures, we can use the relationship between K and the partial pressure of CO₂ to find the equilibrium pressure: $$ K = \frac{P_{CO_{2}}}{1} $$ Solve for the partial pressure of CO₂ (P₃₀₀ and P₁₁₀₀) at both 298 K and 1100 K.

Present the results

After calculating the equilibrium pressure of CO₂ at 298 K and 1100 K using the above steps, present the results: (a) Equilibrium pressure of CO₂ at 298 K: P₃₀₀ = [insert value here] atm (b) Equilibrium pressure of CO₂ at 1100 K: P₁₁₀₀ = [insert value here] atm

Key concepts

Gibbs Free Energy

Understanding Gibbs free energy is crucial for students studying chemical reactions and their spontaneity. Gibbs free energy, represented as \( G \), combines enthalpy, entropy, and temperature to predict whether a process will occur spontaneously. It is a measure of the maximum amount of usable energy that can be extracted from a system at a constant temperature and pressure. The equation \( G = H - TS \) relates Gibbs free energy (\( G \)) to enthalpy (\( H \)), temperature (\( T \)), and entropy (\( S \)).

When you have a chemical reaction, you can calculate the change in Gibbs free energy, \( \Delta G \), using the standard Gibbs free energies of formation (\( \Delta G_f^\text{o} \)) from data tables like Appendix C in your textbook. A negative value for \( \Delta G \) implies that the reaction is spontaneous, while a positive value indicates that it is nonspontaneous. At equilibrium, \( \Delta G \) is zero, allowing us to connect this concept to the equilibrium constant and determine the conditions under which a reaction is at equilibrium.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a dimensionless value that quantifies the position of equilibrium in a chemical reaction. For a reaction where gases are involved, such as the decomposition of barium carbonate, this is often expressed in terms of partial pressures and is then specifically termed \( K_p \). The equilibrium constant is calculated using the concentrations or partial pressures of reactants and products at equilibrium.

The value of the equilibrium constant is related to the Gibbs free energy change by the formula: \[ K = e^{-\frac{\Delta G^\text{o}_{rxn}}{RT}} \] where \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvins. A large \( K \) value indicates that, at equilibrium, the products are favored, while a small \( K \) suggests that the reactants are favored. Understanding how to calculate and interpret \( K \) is essential for predicting the extent of a reaction and the concentrations of reactants and products at equilibrium.

Partial Pressure

Partial pressure is an important concept when dealing with gases, as seen in the decomposition of barium carbonate into barium oxide and carbon dioxide gas. The partial pressure refers to the pressure that a single gas component in a mixture of gases would exert if it occupied the entire volume alone at the same temperature. In the context of equilibrium, the partial pressure is directly related to the equilibrium constant \( K_p \) for gaseous reactions.

For a simple decomposition reaction producing a gas, the equilibrium constant can be expressed as the ratio of the partial pressure of the gas produced to that of a stable phase (as in the case of a solid product, which is typically considered to have a standard pressure). Therefore, \( K_p = \frac{P_{CO_2}}{1} \) for the given reaction simply shows that \( K_p \) is equal to the partial pressure of carbon dioxide at equilibrium. Calculating the partial pressure of gaseous products is crucial in predicting the direction and extent of a reaction under certain conditions.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. In the example of barium carbonate decomposing into barium oxide and carbon dioxide gas, this process is indicative of a decomposition reaction. These reactions can be influenced by factors such as temperature, pressure, and the presence of a catalyst.

From an educational perspective, it is vital to explore the conditions under which decomposition reactions occur and how they reach equilibrium. The equilibrium state of a decomposition reaction is reached when the rate of the forward reaction (decomposition) equals the rate of the reverse reaction (recombination). By learning about decomposition reactions, students can better understand the practical applications and implications of such changes, especially in industrial processes like the production of lime and cement from limestone or the breakdown of organic matter.