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

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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),
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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