Question

Consider the 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) ${180}^{\circ }\mathrm{C}$.

Short answer

Using the given data and the Van't Hoff equation, we can calculate the equilibrium constant at each temperature and subsequently determine the equilibrium pressure of CO2. At ${400}^{\circ }C$, we find $P_{C{O}_2}^{(400)}=K_{400}$ and at ${180}^{\circ }C$, we find $P_{C{O}_2}^{(180)}=K_{180}$. Thus, we can determine the equilibrium pressure of CO2 at both temperatures.

Step-by-step solution

Gather Gibbs free energies of formation

Using data from Appendix C, we find the Gibbs free energies of formation (∆Gf) at 298 K for the following substances: - $PbC{O}_3(s):∆Gf(PbC{O}_3)=-699.9\text{k}\text{J/mol}$ - $PbO(s):∆Gf(PbO)=-217.3\text{k}\text{J/mol}$ - $C{O}_2(g):∆Gf(C{O}_2)=-394.4\text{k}\text{J/mol}$

Calculate the equilibrium constant at each temperature

Using the Gibbs free energies of formation, we can calculate the Gibbs free energy change (∆G) for the reaction at 298 K by the following equation: $∆G=∆G_{PbO}+∆G_{C{O}_2}-∆G_{PbC{O}_3}$ Now, we can calculate the equilibrium constant (K) at each temperature by using the Van't Hoff equation: $K=e^{-\frac{∆G}{R\times T}}$, where $R=8.314\text{J/mol K}$ is the gas constant and T is the temperature. (a) For ${400}^{\circ }C=673\text{K}$: $K_{400}=e^{-\frac{∆G}{R\times 673}}$ (b) For ${180}^{\circ }C=453\text{K}$: $K_{180}=e^{-\frac{∆G}{R\times 453}}$

Use the equilibrium constant expression to solve for the equilibrium pressure of CO2

The equilibrium constant expression for the reaction is: $K=\frac{P_{C{O}_2}}{1}$ because the stoichiometric coefficients are 1 and the reactants are in solid form. Hence, we can calculate the equilibrium pressure of CO2: (a) At ${400}^{\circ }C$: $P_{C{O}_2}^{(400)}=K_{400}$ (b) At ${180}^{\circ }C$: $P_{C{O}_2}^{(180)}=K_{180}$ By calculating these values, we will find the equilibrium pressure of CO2 at both 400 ° C and 180 ° C.

Key concepts

Gibbs Free Energy

Gibbs Free Energy is a crucial concept in thermodynamics. It helps predict whether a chemical reaction will occur spontaneously. In simple terms, it's an energy measure that takes into account both enthalpy (heat content) and entropy (disorder or randomness) of a system. For a reaction at constant pressure and temperature, the change in Gibbs Free Energy, noted as $\mathrm{\Delta }G$, can be calculated using the equation:$$\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$$where:

• $\mathrm{\Delta }H$ is the change in enthalpy.
• $T$ is the temperature in Kelvin.
• $\mathrm{\Delta }S$ is the change in entropy.

If $\mathrm{\Delta }G$ is negative, the reaction is spontaneous. If it's positive, the reaction is non-spontaneous. In this exercise, we calculated $\mathrm{\Delta }G$ using the Gibbs Free Energies of formation. This allows us to further determine the equilibrium constant, showing the reaction's favorability and extent.

Equilibrium Constant

The equilibrium constant, $K$, is key in understanding the extent to which a reaction proceeds. It is derived from the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients. For the reaction given, it was simplified since solids were involved, resulting in:$$K=P_{C{O}_2}$$The equilibrium constant varies with temperature, and that's why we calculate it at different temperatures like 400 °C and 180 °C in this exercise. By using the Van't Hoff equation:$$K=e^{-\frac{\mathrm{\Delta }G}{R\times T}}$$we could find $K$ at different temperatures where:

• $\mathrm{\Delta }G$ is the change in Gibbs Free Energy.
• $R=8.314\text{J/mol K}$ is the universal gas constant.
• $T$ is the temperature in Kelvin.

Thus, $K$ provides a crucial link between Gibbs Free Energy and the equilibrium position of a reaction, indicating how significantly the products are favored over the reactants at a particular temperature.

Le Chatelier's Principle

Le Chatelier's Principle is a useful tool to predict how a system at equilibrium reacts to changes in conditions. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance. This principle is particularly valuable in industrial chemistry for optimizing reactions. When considering the effect of pressure and temperature as in this exercise, Le Chatelier's Principle helps us understand the system's response without intricate calculations.

• Raising the temperature of an endothermic reaction will shift the equilibrium towards the products to absorb the added heat.
• Conversely, cooling an exothermic reaction will shift equilibrium towards product formation.
• In gas-phase reactions like the one involving $C{O}_2$, increasing pressure (by decreasing volume) will shift the equilibrium towards the side with fewer moles of gas.

While not explicitly used in solving this exercise, Le Chatelier's Principle strengthens the conceptual understanding of how temperature influences the pressure of $C{O}_2$ in the equilibrium.