When the oxide of generic metal M is heated at 25.0 °C, a negligible amount of M is produced. MO₂ (s) M(s) + O₂(g) AG= 286.2- kJ mol When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium. Include physical states and represent graphite as C(s). chemical equation: 108.2 Incorrect What is the thermodynamic equilbrium constant for the coupled reaction? K = 9.248 ×1018
When the oxide of generic metal M is heated at 25.0 °C, a negligible amount of M is produced. MO₂ (s) M(s) + O₂(g) AG= 286.2- kJ mol When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. What is the chemical equation of this coupled process? Show that the reaction is in equilibrium. Include physical states and represent graphite as C(s). chemical equation: 108.2 Incorrect What is the thermodynamic equilbrium constant for the coupled reaction? K = 9.248 ×1018
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Coupled Chemical Reactions and Thermodynamic Equilibrium
#### Introduction
When the oxide of a generic metal \( \text{M} \) is heated at \( 25.0^\circ \text{C} \), a negligible amount of metal \( \text{M} \) is produced. The reaction is represented by the following chemical equation:
\[ \text{MO}_2(s) \rightleftharpoons \text{M}(s) + \text{O}_2(g) \hspace{2cm} \Delta G^\circ = 286.2 \, \frac{\text{kJ}}{\text{mol}} \]
This reaction, when coupled with the conversion of graphite to carbon dioxide, becomes spontaneous.
#### Coupled Chemical Equation
To find the chemical equation for the coupled process, we combine the given reaction with the combustion of graphite (graphite is represented as \( \text{C}(s) \)):
\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \]
Combining these two reactions:
\[ \text{MO}_2(s) + \text{C}(s) \rightleftharpoons \text{M}(s) + \text{CO}_2(g) \]
This combined reaction represents the overall process in equilibrium.
#### Thermodynamic Equilibrium Constant
The thermodynamic equilibrium constant \( K \) for the coupled reaction can be calculated from the Gibbs free energy change (\( \Delta G^\circ \)) for the coupled reaction. The relationship is given by:
\[ K = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvins.
The given equilibrium constant for the coupled reaction is:
\[ K = 9.248 \times 10^{18} \]
This high value of the equilibrium constant indicates that at equilibrium, the reaction heavily favors the formation of products \( \text{M}(s) \) and \( \text{CO}_2(g) \).
### Conclusion
By coupling the reduction of metal oxide with the combustion of graphite, the reaction becomes thermodynamically spontaneous. This explains how favorable conditions can be achieved to produce metal \( \text{M} \) from its oxide at standard temperatures.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f33cc2f-0221-41a8-a3a2-cc96bc7960a2%2F6e398f86-99df-4e2d-acde-76c7e1606038%2Feurmqd_processed.png&w=3840&q=75)
Transcribed Image Text:### Coupled Chemical Reactions and Thermodynamic Equilibrium
#### Introduction
When the oxide of a generic metal \( \text{M} \) is heated at \( 25.0^\circ \text{C} \), a negligible amount of metal \( \text{M} \) is produced. The reaction is represented by the following chemical equation:
\[ \text{MO}_2(s) \rightleftharpoons \text{M}(s) + \text{O}_2(g) \hspace{2cm} \Delta G^\circ = 286.2 \, \frac{\text{kJ}}{\text{mol}} \]
This reaction, when coupled with the conversion of graphite to carbon dioxide, becomes spontaneous.
#### Coupled Chemical Equation
To find the chemical equation for the coupled process, we combine the given reaction with the combustion of graphite (graphite is represented as \( \text{C}(s) \)):
\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \]
Combining these two reactions:
\[ \text{MO}_2(s) + \text{C}(s) \rightleftharpoons \text{M}(s) + \text{CO}_2(g) \]
This combined reaction represents the overall process in equilibrium.
#### Thermodynamic Equilibrium Constant
The thermodynamic equilibrium constant \( K \) for the coupled reaction can be calculated from the Gibbs free energy change (\( \Delta G^\circ \)) for the coupled reaction. The relationship is given by:
\[ K = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvins.
The given equilibrium constant for the coupled reaction is:
\[ K = 9.248 \times 10^{18} \]
This high value of the equilibrium constant indicates that at equilibrium, the reaction heavily favors the formation of products \( \text{M}(s) \) and \( \text{CO}_2(g) \).
### Conclusion
By coupling the reduction of metal oxide with the combustion of graphite, the reaction becomes thermodynamically spontaneous. This explains how favorable conditions can be achieved to produce metal \( \text{M} \) from its oxide at standard temperatures.
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