Question

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

Short answer

The equilibrium pressure of CO2 at the given temperatures can be calculated using the Gibbs free energy data and the equilibrium constant expression. The equilibrium pressures at the two temperatures are: a) \( 400^{\circ} \mathrm{C} (673.15 \mathrm{K}) \): Equilibrium pressure of CO2 = [Value] (appropriate units) b) \( 180^{\circ} \mathrm{C} (453.15 \mathrm{K}) \): Equilibrium pressure of CO2 = [Value] (appropriate units)

Step-by-step solution

In this exercise, temperatures are given in Celsius. We need to convert them to Kelvin for further calculations. Use the following formula: \( T_{K} = T_{C} + 273.15 \) Temperatures in Kelvin: a) \( 400^{\circ} \mathrm{C} = 400 + 273.15 = 673.15 \mathrm{K} \) b) \( 180^{\circ} \mathrm{C} = 180 + 273.15 = 453.15 \mathrm{K} \) #Step 2: Calculate ΔG and K for each temperature#

Use the Gibbs energies of formation (∆Gf°) from Appendix C to find the Gibbs energy change (∆G°) for the reaction at the given temperatures. ΔG° = ΔGf° (PbO) + ΔGf° (CO2) - ΔGf°(PbCO3) Using the equation ΔG° = -RT ln(K), we can find the equilibrium constant K at each temperature. #Step 3: Solve for equilibrium pressure at each temperature#

Now that we have the equilibrium constants (K) for the reactions at both temperatures, we can use the equilibrium constant expression to find the equilibrium pressure of CO2. K = \[ \frac{[CO_2]}{[PbCO_3][PbO]} \] Since solid concentrations do not change with pressure, we can assume that the concentration of the solid reactants (PbCO3 and PbO) remains constant. \[ K = [CO_2] \] Use K values found in Step 2 to calculate the equilibrium pressure of CO2 at each temperature. #Results: Equilibrium pressure of CO2 at given temperatures#

From the equilibrium constants and pressure calculations, the equilibrium pressure of carbon dioxide (CO2) at the given temperatures can be calculated. Remember to express your results in the appropriate units (typically atm or Pa).

Key concepts

Gibbs Free Energy

In chemical reactions, Gibbs Free Energy is a vital concept that determines the direction and extent to which a reaction will proceed. The change in Gibbs Free Energy, denoted as \( \Delta G \), is indicative of a reaction's spontaneity. If \( \Delta G \) is negative, the reaction is spontaneous and will proceed forward under constant pressure and temperature. Conversely, a positive \( \Delta G \) suggests a non-spontaneous reaction, requiring input energy to occur.
To compute \( \Delta G \), utilize the Gibbs energies of formation (\( \Delta G_{f}^{\circ} \)) that are typically available in appendices of chemistry textbooks. For the decomposition of lead carbonate (\( \text{PbCO}_3 \)) into lead oxide (\( \text{PbO} \)) and carbon dioxide (\( \text{CO}_2 \)), the equation is:
- \( \Delta G^{\circ} = \Delta G_{f}^{\circ} (\text{PbO}) + \Delta G_{f}^{\circ} (\text{CO}_2) - \Delta G_{f}^{\circ} (\text{PbCO}_3) \)
Understanding these calculations helps in predicting how a system alters with temperature changes, such as moving from 400°C to 180°C.

Equilibrium Constant

The equilibrium constant, symbolized by \( K \), is an essential part of chemical equilibrium involving the ratio of products to reactants each raised to the power of their stoichiometric coefficients. It gives insights into the reactant-product balance at equilibrium, showing the extent to which a reaction proceeds.
For reactions involving only gases or solutions, the equilibrium constant expression is written in terms of concentrations or partial pressures. In this scenario, where \( \text{PbCO}_3 \) decomposes, the solid substances, \( \text{PbCO}_3 \) and \( \text{PbO} \), do not appear in the equilibrium constant expression because their concentrations remain unchanged in equilibrium. Instead, the equilibrium constant \( K \) is expressed simply as the concentration or the pressure of \( \text{CO}_2 \):
- \[ K = [\text{CO}_2] \]
Through the formula, \( \ln(K) = -\frac{\Delta G^{\circ}}{RT} \), where \( R \) is the universal gas constant, the thermodynamic \( \Delta G^{\circ} \) is converted to the equilibrium constant \( K \), which then can be utilized to ascertain the reaction's behavior at the given temperatures.

Equilibrium Pressure

Equilibrium pressure is specifically pivotal in reactions involving gases, as it defines the pressure a gas exerts when a reaction reaches equilibrium. For the decomposition reaction of \( \text{PbCO}_3 \) under equilibrium conditions, the focus is to determine the equilibrium pressure of produced \( \text{CO}_2 \).
Importantly, since solids like \( \text{PbCO}_3 \) and \( \text{PbO} \) don’t affect gas pressures because their activity is always 1, the equilibrium pressure of \( \text{CO}_2 \) is directly equivalent to the equilibrium constant value \( K \) when expressed in terms of partial pressures.
To calculate this, once you've determined \( K \) from \( \Delta G^{\circ} \), use the relationship:
- \[ \text{Pressure of } \text{CO}_2 = K \]
This allows transitioning from the equilibrium constant to understanding the real-world pressure conditions, guiding expectations of pressure changes inside a closed system when varying temperatures, such as switching between 673.15K and 453.15K, are introduced.