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Chapter 19: Problem 84

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

Expert verified
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 \( P_{CO_2}^{(400)} = K_{400} \) and at \( 180 ^\circ C \), we find \( P_{CO_2}^{(180)} = K_{180} \). Thus, we can determine the equilibrium pressure of CO2 at both temperatures.

Step by step solution

01

Gather Gibbs free energies of formation

Using data from Appendix C, we find the Gibbs free energies of formation (鈭咷f) at 298 K for the following substances: - \( PbCO_3(s): 鈭咷f(PbCO_3) = -699.9\,\text{k} \text{J/mol} \) - \( PbO(s): 鈭咷f(PbO) = -217.3\,\text{k} \text{J/mol} \) - \( CO_2 (g): 鈭咷f(CO_2) = -394.4\,\text{k} \text{J/mol} \)
02

Calculate the equilibrium constant at each temperature

Using the Gibbs free energies of formation, we can calculate the Gibbs free energy change (鈭咷) for the reaction at 298 K by the following equation: \( 鈭咷 = 鈭咷_{PbO} + 鈭咷_{CO_2} - 鈭咷_{PbCO_3} \) Now, we can calculate the equilibrium constant (K) at each temperature by using the Van't Hoff equation: \( K = e^{-\frac{鈭咷}{R \times T}} \), where \( R = 8.314 \,\text{J/mol K} \) is the gas constant and T is the temperature. (a) For \( 400^\circ C = 673 \,\text{K} \): \( K_{400} = e^{-\frac{鈭咷}{R \times 673}} \) (b) For \( 180^\circ C = 453 \,\text{K} \): \( K_{180} = e^{-\frac{鈭咷}{R \times 453}} \)
03

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

The equilibrium constant expression for the reaction is: \( K = \frac{P_{CO_2}}{1} \) because the stoichiometric coefficients are 1 and the reactants are in solid form. Hence, we can calculate the equilibrium pressure of CO2: (a) At \( 400^\circ C \): \( P_{CO_2}^{(400)} = K_{400} \) (b) At \( 180^\circ C \): \( P_{CO_2}^{(180)} = K_{180} \) By calculating these values, we will find the equilibrium pressure of CO2 at both 400 掳 C and 180 掳 C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( \Delta G \), can be calculated using the equation:\[ \Delta G = \Delta H - T \Delta S \]where:
  • \( \Delta H \) is the change in enthalpy.
  • \( T \) is the temperature in Kelvin.
  • \( \Delta S \) is the change in entropy.
If \( \Delta G \) is negative, the reaction is spontaneous. If it's positive, the reaction is non-spontaneous. In this exercise, we calculated \( \Delta G \) 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:\[ K = P_{CO_2} \]The equilibrium constant varies with temperature, and that's why we calculate it at different temperatures like 400 掳C and 180 掳C in this exercise. By using the Van't Hoff equation:\[ K = e^{-\frac{\Delta G}{R \times T}} \]we could find \( K \) at different temperatures where:
  • \( \Delta G \) 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 \( CO_2 \), 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 \( CO_2 \) in the equilibrium.

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Most popular questions from this chapter

Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from \(300 \mathrm{~K}\) to \(200 \mathrm{~K}\). (a) Describe how this change might be carried out while keeping the volume of the gas constant. (b) Describe how it might be carried out while keeping the pressure of the gas constant. (c) Does the change in \(\Delta E\) depend on the particular pathway taken to carry out this change of state? Explain.

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

(a) In a chemical reaction two gases combine to form a solid. What do you expect for the sign of \(\Delta S ?\) (b) How does the entropy of the system change in the processes described in Exercise \(19.12 ?\)

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example, from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?\) (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure 5.10 that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.
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