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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) \(180^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The equilibrium pressure of CO鈧 in the system at \(400^{\circ}\mathrm{C}\) is approximately 1.48 脳 10鈦烩伌 bar, and at \(180^{\circ}\mathrm{C}\), it is approximately 1.53 脳 10鈦宦 bar.

Step by step solution

01

Find the standard Gibbs free energy change of the reaction

To find the standard Gibbs free energy change, 鈭咷掳, of the reaction, we can use the equation: 鈭咷掳 = 鈭咷f掳(products) - 鈭咷f掳(reactants) Using the data in Appendix C, we find the following values for the standard Gibbs free energy of formation, 鈭咷f掳, at 298 K, 鈭咷f掳(PbCO鈧) = -699.5 kJ/mol 鈭咷f掳(PbO) = -218.4 kJ/mol 鈭咷f掳(CO鈧) = -394.4 kJ/mol Now, substitute the values into the equation, 鈭咷掳 = (-218.4 - (-394.4)) - (-699.5) = 175.5 kJ/mol
02

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, 鈭咷掳, and temperature, T: K = \(\exp\left(\frac{-\Delta G^{\circ}}{RT}\right)\) Where R is the universal gas constant (8.314 J/mol鈰匥). We need to find the equilibrium constant, K, at the two given temperatures, (a) 400掳C and (b) 180掳C. Since the given 鈭咷f掳 values are at 298 K, we will use the same value of 鈭咷掳 for both cases as an approximation. (a) For 400掳C: T = 400 + 273.15 = 673.15 K K = \(\exp\left(\frac{-175500}{8.314\times673.15}\right)\) K 鈮 1.48 脳 10鈦烩伌 (b) For 180掳C: T = 180 + 273.15 = 453.15 K K = \(\exp\left(\frac{-175500}{8.314\times453.15}\right)\) K 鈮 1.53 脳 10鈦宦
03

Find the equilibrium pressure of CO鈧 using the expression for K

The reaction involves one mole of a gaseous product (CO鈧) and no gaseous reactants. Therefore, the expression for the equilibrium constant, K, is: K = (P_CO鈧)/P掳 Where P_CO鈧 is the equilibrium pressure of CO鈧, and P掳 is the standard pressure (1 bar). Now, we can solve for the equilibrium pressure of CO鈧 at the given temperatures. (a) For 400掳C: 1.48 脳 10鈦烩伌 = (P_CO鈧)/1 bar P_CO鈧 = 1.48 脳 10鈦烩伌 bar (b) For 180掳C: 1.53 脳 10鈦宦 = (P_CO鈧)/1 bar P_CO鈧 = 1.53 脳 10鈦宦 bar Thus, the equilibrium pressure of CO鈧 in the system at 400掳C is about 1.48 脳 10鈦烩伌 bar, and at 180掳C, it is about 1.53 脳 10鈦宦 bar.

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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, 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鈥檚 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鈰匥.
  • \( T \) 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掳C and 180掳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.

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Most popular questions from this chapter

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) is a liquid hydrocarbon at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{6} \mathrm{H}_{12}(l)\) to form \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\)

For the majority of the compounds listed in Appendix C, the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ}\). (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) \(\mathrm{An}\) exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

A particular reaction is spontaneous at \(450 \mathrm{~K}\). The enthalpy change for the reaction is \(+34.5 \mathrm{~kJ}\). What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

(a) How can we calculate \(\Delta S\) foran isothermal process? (b) Does \(\Delta S\) for a process depend on the path taken from the initial to the final state of the system? Explain.

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(\mathrm{~s})\) to \(\mathrm{I}_{2}(\mathrm{~g})\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as WebElements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

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