Calculate the equilibrium constant, K, at 25.0 °C for each of the reactions. Use the thermodynamic information provided in the table. The hydrogenation of acetylene to ethane. C,H,(g) +2H,(g) →C,H(g) K = x10 TOOLS Compound CH₂Cl(g) CH₂(g) C₂H₂(g) C₂H6 (g) Cl₂(g) H₂(g) H₂O(1) HCl(g) HNO₂ (aq) NO(g) NO₂ (g) AG; (kJ. mol-¹) 48.50 -50.72 209.2 -32.82 0.00 0.00 -237.13 -95.30 -111.25 86.55 51.31
Calculate the equilibrium constant, K, at 25.0 °C for each of the reactions. Use the thermodynamic information provided in the table. The hydrogenation of acetylene to ethane. C,H,(g) +2H,(g) →C,H(g) K = x10 TOOLS Compound CH₂Cl(g) CH₂(g) C₂H₂(g) C₂H6 (g) Cl₂(g) H₂(g) H₂O(1) HCl(g) HNO₂ (aq) NO(g) NO₂ (g) AG; (kJ. mol-¹) 48.50 -50.72 209.2 -32.82 0.00 0.00 -237.13 -95.30 -111.25 86.55 51.31
Chemistry
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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
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![### Calculating the Equilibrium Constant for the Hydrogenation of Acetylene
#### Problem Description
Calculate the equilibrium constant, \( K \), at 25.0 °C for the hydrogenation of acetylene to ethane, using the thermodynamic information provided in the table.
#### Chemical Reaction
The balanced chemical equation for the hydrogenation of acetylene (C₂H₂) to ethane (C₂H₆) is as follows:
\[ \text{C}_2\text{H}_2 (g) + 2 \text{H}_2 (g) \rightleftharpoons \text{C}_2\text{H}_6 (g) \]
#### Required Formula
The equilibrium constant \( K \) can be calculated from the standard Gibbs free energy change \( \Delta G^\circ \) using the formula:
\[ \Delta G^\circ = -RT \ln K \]
Where:
- \( R \) is the gas constant, 8.314 J/(mol·K)
- \( T \) is the temperature in Kelvin (298.15 K for 25.0 °C)
#### Thermodynamic Data
The following table provides the standard Gibbs free energy of formation \( \Delta G_f^\circ \) for each compound involved in the reaction:
| Compound | \( \Delta G_f^\circ \) (kJ · mol⁻¹) |
|----------------|--------------------------------------|
| CH₃Cl(g) | 48.50 |
| CH₄(g) | -50.72 |
| C₂H₂(g) | 209.2 |
| C₂H₆(g) | -32.82 |
| Cl₂(g) | 0.00 |
| H₂(g) | 0.00 |
| H₂O(l) | -237.13 |
| HCl(g) | -95.30 |
| HNO₃(aq) | -111.25 |
| NO(g) | 86.55 |
| NO₂(g) | 51.31 |
#### Steps to Calculate \( \Delta G^\circ \) for the Reaction
1. **Identify Products and Reactants:**
- Products: C₂H₆(g)
- Reactants: C₂H₂(g),](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F623e3454-0888-4b1f-b867-4e9e54aaab3f%2F07ab382b-7134-426d-85c6-e372cc226fc9%2Ftw98pxm_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Equilibrium Constant for the Hydrogenation of Acetylene
#### Problem Description
Calculate the equilibrium constant, \( K \), at 25.0 °C for the hydrogenation of acetylene to ethane, using the thermodynamic information provided in the table.
#### Chemical Reaction
The balanced chemical equation for the hydrogenation of acetylene (C₂H₂) to ethane (C₂H₆) is as follows:
\[ \text{C}_2\text{H}_2 (g) + 2 \text{H}_2 (g) \rightleftharpoons \text{C}_2\text{H}_6 (g) \]
#### Required Formula
The equilibrium constant \( K \) can be calculated from the standard Gibbs free energy change \( \Delta G^\circ \) using the formula:
\[ \Delta G^\circ = -RT \ln K \]
Where:
- \( R \) is the gas constant, 8.314 J/(mol·K)
- \( T \) is the temperature in Kelvin (298.15 K for 25.0 °C)
#### Thermodynamic Data
The following table provides the standard Gibbs free energy of formation \( \Delta G_f^\circ \) for each compound involved in the reaction:
| Compound | \( \Delta G_f^\circ \) (kJ · mol⁻¹) |
|----------------|--------------------------------------|
| CH₃Cl(g) | 48.50 |
| CH₄(g) | -50.72 |
| C₂H₂(g) | 209.2 |
| C₂H₆(g) | -32.82 |
| Cl₂(g) | 0.00 |
| H₂(g) | 0.00 |
| H₂O(l) | -237.13 |
| HCl(g) | -95.30 |
| HNO₃(aq) | -111.25 |
| NO(g) | 86.55 |
| NO₂(g) | 51.31 |
#### Steps to Calculate \( \Delta G^\circ \) for the Reaction
1. **Identify Products and Reactants:**
- Products: C₂H₆(g)
- Reactants: C₂H₂(g),
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