Problem 81 - Chemistry The Central Science

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 \((\mathbf{b}) 1100 \mathrm{K} .\)

Step-by-Step Solution

Answer

$$ \Delta G^{\circ}_{reaction} = 177.7 \: \text{kJ/mol} $$ #Step 2: Calculate Kp at 298 K and 1100 K# Using the formula: $$ K_p = e^{-\frac{\Delta G^\circ_{reaction}}{RT}} $$ we can find the Kp at 298 K and 1100 K: At 298 K: $$ K_p(298 \: \text{K}) = e^{-\frac{177.7 \times 10^3 \: \text{J/mol}}{8.314 \: \text{J/mol} \cdot \text{K} \times 298 \: \text{K}}} $$ $$ K_p(298 \: \text{K}) = 4.42 \times 10^{-12} $$ At 1100 K: $$ K_p(1100 \: \text{K}) = e^{-\frac{177.7 \times 10^3 \: \text{J/mol}}{8.314 \: \text{J/mol} \cdot \text{K} \times 1100 \: \text{K}}} $$ $$ K_p(1100 \: \text{K}) = 4.40 \times 10^{-2} $$ #Step 3: Find the Equilibrium Pressure of CO2# From the balanced equation, we know that the stoichiometric coefficients are all equal to 1. Thus, for every mole of BaCO3 that decomposes, one mole of CO2 is produced. We can then equate the Kp values with the equilibrium partial pressure of CO2 (P_CO2): At 298 K: $$ K_p(298 \: \text{K}) = P_{CO2} $$ $$ 4.42 \times 10^{-12} = P_{CO2}(298 \: \text{K}) $$ At 1100 K: $$ K_p(1100 \: \text{K}) = P_{CO2} $$ $$ 4.40 \times 10^{-2} = P_{CO2}(1100 \: \text{K}) $$ So, the equilibrium pressure of CO2 at 298 K is \(4.42 \times 10^{-12}\: \text{atm}\) and at 1100 K is \(4.40 \times 10^{-2}\: \text{atm}\).

Step 1: Find the Gibbs Free Energy Change of Reaction

We will use the standard Gibbs free energies of formation for each compound at 298 K to find the change in standard Gibbs free energy for the reaction: $$ \Delta G^{\circ}_{reaction} = G^{\circ}_{products} - G^{\circ}_{reactants} $$ From Appendix C, we have the standard Gibbs free energies of formation for the compounds: - \(\Delta G^{\circ}_{BaCO3}\) = -1135.6 kJ/mol - \(\Delta G^{\circ}_{BaO}\) = -563.5 kJ/mol - \(\Delta G^{\circ}_{CO2}\) = -394.4 kJ/mol Now, we can find the \(\Delta G^{\circ}_{reaction}\): $$ \Delta G^{\circ}_{reaction} = \left( -563.5 \: \text{kJ/mol} + (-394.4 \: \text{kJ/mol}) \right) - \left( -1135.6 \: \text{kJ/mol} \right) $$

Key Concepts

Gibbs Free Energy

Gibbs free energy is a crucial concept in chemistry used to predict the direction of chemical reactions. It combines enthalpy, temperature, and entropy into a single value, represented by \( \Delta G \). This value helps determine whether a reaction is spontaneous or non-spontaneous.

• If \( \Delta G < 0 \), the reaction is spontaneous, occurring without external influence.
• If \( \Delta G > 0 \), the reaction is non-spontaneous, meaning it requires external energy input.
• If \( \Delta G = 0 \), the system is at equilibrium.

In a standard reaction, \( \Delta G \) can be calculated using the equation:\[ \Delta G^{\circ}_{reaction} = G^{\circ}_{products} - G^{\circ}_{reactants} \]This expression demands the standard Gibbs free energies of formation for each product and reactant, which are typically found in tables included in many chemical appendices.
The relevancy of Gibbs free energy is evident in the calculation of equilibrium constants and the prediction of equilibrium pressures, like in the decomposition reactions of various compounds.

Decomposition Reaction

A decomposition reaction involves a single compound breaking down into two or more simpler substances. In these reactions, the original compound usually transforms under external conditions, such as heat, light, or electricity.
The general form of a decomposition reaction is:\[ \text{AB} \rightarrow \text{A} + \text{B} \]In this type of reaction:

• Energy (heat, light, or electricity) typically drives the reaction.
• Products are simpler substances compared to the reactant.

These reactions are common in chemistry, particularly in the laboratory setting and industrial processes. They are essential for understanding how compounds transform under various conditions and influences.
Decomposition reactions can result in the release of gases. This aspect is significant when discussing equilibrium pressures and gases involved in reactions, as seen in the decomposition of barium carbonate into barium oxide and carbon dioxide.

Barium Carbonate Decomposition

Barium carbonate decomposition is a particular example of a chemical decomposition reaction. The reaction proceeds as follows:\[ \mathrm{BaCO}_{3}(s) \rightleftharpoons \mathrm{BaO}(s) + \mathrm{CO}_{2}(g) \]
In this process, solid barium carbonate breaks down into solid barium oxide and gaseous carbon dioxide. Several factors influence the extent of this reaction and the conditions under which it occurs.

• The temperature plays a significant role; as temperature increases, the decomposition becomes more favorable.
• Equilibrium pressure of carbon dioxide affects the reaction, shifting the position of equilibrium by Le Chatelier's Principle.

This reaction provides an excellent context for understanding equilibrium processes and the practical applications of Gibbs free energy calculation. By calculating \( \Delta G^{\circ}_{reaction} \) at various temperatures, one can predict the spontaneity of the decomposition and the equilibrium pressures of the gaseous products formed. Understanding these principles is key to mastering chemical equilibrium in both academic and practical scenarios.