Problem 80 - Chemistry: The Central Science

Question

Consider the following reaction: $$ \mathrm{PbCO}_{3}(s) \rightleftharpoons \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) $$ Using data in Appendix \(C\), calculate the equilibrium pressure of \(\mathrm{CO}_{2}\) in the system at (a) \(400^{\circ} \mathrm{C}\) and (b) \(180^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The equilibrium pressure of CO₂ in the system at \(400^{\circ}\mathrm{C}\) is approximately 1.48 × 10⁻⁴ bar, and at \(180^{\circ}\mathrm{C}\), it is approximately 1.53 × 10⁻² bar.

Step 1: Find the standard Gibbs free energy change of the reaction

To find the standard Gibbs free energy change, ∆G°, of the reaction, we can use the equation: ∆G° = ∆Gf°(products) - ∆Gf°(reactants) Using the data in Appendix C, we find the following values for the standard Gibbs free energy of formation, ∆Gf°, at 298 K, ∆Gf°(PbCO₃) = -699.5 kJ/mol ∆Gf°(PbO) = -218.4 kJ/mol ∆Gf°(CO₂) = -394.4 kJ/mol Now, substitute the values into the equation, ∆G° = (-218.4 - (-394.4)) - (-699.5) = 175.5 kJ/mol

Step 2: Use the van't Hoff equation to find the equilibrium constant, K

The van't Hoff equation relates the equilibrium constant, K, to the standard Gibbs free energy change, ∆G°, and temperature, T: K = \(\exp\left(\frac{-\Delta G^{\circ}}{RT}\right)\) Where R is the universal gas constant (8.314 J/mol⋅K). We need to find the equilibrium constant, K, at the two given temperatures, (a) 400°C and (b) 180°C. Since the given ∆Gf° values are at 298 K, we will use the same value of ∆G° for both cases as an approximation. (a) For 400°C: T = 400 + 273.15 = 673.15 K K = \(\exp\left(\frac{-175500}{8.314\times673.15}\right)\) K ≈ 1.48 × 10⁻⁴ (b) For 180°C: T = 180 + 273.15 = 453.15 K K = \(\exp\left(\frac{-175500}{8.314\times453.15}\right)\) K ≈ 1.53 × 10⁻²

Step 3: Find the equilibrium pressure of CO₂ using the expression for K

The reaction involves one mole of a gaseous product (CO₂) and no gaseous reactants. Therefore, the expression for the equilibrium constant, K, is: K = (P_CO₂)/P° Where P_CO₂ is the equilibrium pressure of CO₂, and P° is the standard pressure (1 bar). Now, we can solve for the equilibrium pressure of CO₂ at the given temperatures. (a) For 400°C: 1.48 × 10⁻⁴ = (P_CO₂)/1 bar P_CO₂ = 1.48 × 10⁻⁴ bar (b) For 180°C: 1.53 × 10⁻² = (P_CO₂)/1 bar P_CO₂ = 1.53 × 10⁻² bar Thus, the equilibrium pressure of CO₂ in the system at 400°C is about 1.48 × 10⁻⁴ bar, and at 180°C, it is about 1.53 × 10⁻² bar.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, indicated by \( \Delta G \), is a key concept in determining the feasibility of a chemical reaction and its tendency to reach equilibrium. In simple terms, it embodies the balance between the energy released by forming products and the energy needed to break down reactants. It's represented by the equation:\[ \Delta G^{\circ} = \Delta G^{\circ}_f(\text{products}) - \Delta G^{\circ}_f(\text{reactants}) \]Here's why Gibbs Free Energy is crucial:

• **A negative \( \Delta G \):** Suggests that a reaction is spontaneous and is likely to proceed in the forward direction towards products.
• **A positive \( \Delta G \):** Indicates a non-spontaneous reaction under standard conditions, tending not to occur without external input.
• **Zero \( \Delta G \):** Implies that the system is at equilibrium, the point where neither the reactants nor products are favored.

In practice, for the reaction involving lead carbonate \((\text{PbCO}_3)\), lead oxide \((\text{PbO})\), and carbon dioxide \((\text{CO}_2)\), solving for \( \Delta G^{\circ} \) gives insight into how much energy is involved in the process and whether the reaction can proceed without outside influence. Calculations using standard Gibbs Free Energy formations help predict the behavior of the reaction at equilibrium.

Equilibrium Constant

The Equilibrium Constant, denoted \( K \), is a vital parameter to describe the status of a chemical system at equilibrium. It connects the concentrations or pressures of reactants and products to provide a snapshot of where equilibrium lies. In simpler terms, it's a mathematical expression that quantifies the balance of materials.For reactions involving gases, the equilibrium constant can often be expressed using partial pressures, specifically like this:\[ K = \frac{(P_{\text{CO}_2})}{P^{\circ}} \]Here’s how the value of \( K \) works:

• **\( K > 1 \):** Indicates that the products are favored at equilibrium, meaning the reaction tends to go forward.
• **\( K < 1 \):** Signals that the reactants are favored, so the reaction does not proceed much past its initial state.
• **\( K = 1 \):** Suggests that neither reactants nor products are favored significantly, achieving a perfect balance.

In our reaction \( \text{PbCO}_3(s) \rightleftharpoons \text{PbO}(s)+\text{CO}_2(g) \), calculating \( K \) allows us to find the equilibrium pressure of \( \text{CO}_2 \), which indicates how much \( \text{CO}_2 \) will be present in equilibrium conditions. This concept guides our understanding of how far a reaction will proceed before it stabilizes.

Van't Hoff Equation

The Van't Hoff Equation is a powerful tool in understanding how temperature affects the equilibrium constant \( K \). By relating \( K \) to the standard Gibbs free energy change \( \Delta G^{\circ} \), it allows chemists to predict how reactions respond to changes in temperature. The equation is:\[ K = \exp\left(\frac{-\Delta G^{\circ}}{RT}\right) \]Where:

• \( R \) is the universal gas constant, valued at 8.314 J/mol⋅K.
• \( T \) stands for temperature in Kelvin.

Here's how the Van't Hoff Equation helps:

• **Predicts behavior under different temperatures:** A rise in temperature can shift \( K \), indicating alternate equilibrium positions.
• **Temperature-dependent equilibrium constant:** By computing the effect of temperature on \( \Delta G^{\circ} \), it lets us calculate new equilibrium constants for conditions other than standard temperature.

In the context of our reaction of \( \text{PbCO}_3 \) decomposing into \( \text{PbO} \) and \( \text{CO}_2 \), the equation assists in adjusting calculations to different temperatures, like 400°C and 180°C, thereby helping forecast the equilibrium state at these temperatures. This ensures a deeper insight into how temperature promotes or hinders the progression of chemical reactions.