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
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
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F264afef6-5fb8-47a1-a0f2-19efba1e1114%2Fa2f7e39d-ac0d-48ed-9093-0343ea844214%2Fo26nj8l_processed.jpeg&w=3840&q=75)
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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