Consider the decomposition of calcium carbonate: AH = 177.8 kJ/mol CACO3(s) Ca0(s) + CO2(8) As° = 160.5 J/K• mol Calculate the equilibrium pressure of CO, at 25°C and 773°C. Assume the AH and AS remain constant over the temperature range. at 25°C Pco x 10 atm %3D (Enter your answer in scientific notation) at 773°C Pco2 atm %3D

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Thermodynamics of Calcium Carbonate Decomposition**

In this exercise, we will explore the decomposition reaction of calcium carbonate (CaCO₃) and calculate the equilibrium pressure of carbon dioxide (CO₂) at different temperatures.

**Reaction:**

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

**Given Thermodynamic Data:**

- Enthalpy change (\(\Delta H^{\circ}\)): 177.8 kJ/mol
- Entropy change (\(\Delta S^{\circ}\)): 160.5 J/K·mol

**Problem Statement:**

Compute the equilibrium pressure of CO₂ at:
- \(25^{\circ}C\)
- \(773^{\circ}C\)

Assumptions: \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are constant over the temperature range.

**Calculations:**

- At \(25^{\circ}C\):
  
  \[ P_{CO_2} = \boxed{\phantom{answer}} \times 10^{\boxed{\phantom{answer}}} \text{ atm} \]

  *(Please provide the answer in scientific notation)*

- At \(773^{\circ}C\):

  \[ P_{CO_2} = \boxed{\phantom{answer}} \text{ atm} \]

Feel free to plug in your calculated values for a deeper understanding of the application of thermodynamic principles in chemical equilibria!
Transcribed Image Text:**Thermodynamics of Calcium Carbonate Decomposition** In this exercise, we will explore the decomposition reaction of calcium carbonate (CaCO₃) and calculate the equilibrium pressure of carbon dioxide (CO₂) at different temperatures. **Reaction:** \[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \] **Given Thermodynamic Data:** - Enthalpy change (\(\Delta H^{\circ}\)): 177.8 kJ/mol - Entropy change (\(\Delta S^{\circ}\)): 160.5 J/K·mol **Problem Statement:** Compute the equilibrium pressure of CO₂ at: - \(25^{\circ}C\) - \(773^{\circ}C\) Assumptions: \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are constant over the temperature range. **Calculations:** - At \(25^{\circ}C\): \[ P_{CO_2} = \boxed{\phantom{answer}} \times 10^{\boxed{\phantom{answer}}} \text{ atm} \] *(Please provide the answer in scientific notation)* - At \(773^{\circ}C\): \[ P_{CO_2} = \boxed{\phantom{answer}} \text{ atm} \] Feel free to plug in your calculated values for a deeper understanding of the application of thermodynamic principles in chemical equilibria!
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