Using data from Appendix 4, calculate AG for the reaction below at the following conditions: 35°C, PH25 = 1.5 x 104 atm; Ps02 = 2.0 x 10² atm; PH20 = 2.5 x 10² atm. %3D 2H,S (e) + SO2 (8) → 35 mombic (s) + 2H2O (g)
Using data from Appendix 4, calculate AG for the reaction below at the following conditions: 35°C, PH25 = 1.5 x 104 atm; Ps02 = 2.0 x 10² atm; PH20 = 2.5 x 10² atm. %3D 2H,S (e) + SO2 (8) → 35 mombic (s) + 2H2O (g)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Transcribed Image Text:**Appendix Four: Thermodynamic Data for Various Substances**
This appendix provides extensive thermodynamic data for various substances, including the changes in enthalpy (ΔH°), Gibbs free energy (ΔG°), and entropy (ΔS°) under standard conditions. Each table row lists a substance, its state, and the corresponding thermodynamic values.
### Key Columns in the Table
- **Substance and State**: Indicates the chemical substance and its physical state. For example, BaO(s) refers to solid Barium Oxide.
- **ΔH° (kJ/mol)**: Change in enthalpy under standard conditions, measured in kilojoules per mole. This represents the heat absorbed or released during a reaction.
- **ΔG° (kJ/mol)**: Gibbs free energy change under standard conditions, measured in kilojoules per mole. This indicates the spontaneity of a reaction; a negative ΔG° suggests the reaction is spontaneous.
- **ΔS° (J/K·mol)**: Change in entropy under standard conditions, measured in joules per Kelvin per mole. This represents the disorder or randomness of a system.
### Example Data Entries
#### Barium (continued)
- **BaSO4(s)**:
- **ΔH°**: -1465 kJ/mol
- **ΔG°**: -1353 kJ/mol
- **ΔS°**: 132 J/K·mol
#### Carbon
- **C(s) (graphite)**:
- **ΔH°**: 0 kJ/mol
- **ΔG°**: 0 kJ/mol
- **ΔS°**: 6 J/K·mol
- **C(s) (diamond)**:
- **ΔH°**: -10.5 kJ/mol
- **ΔG°**: -3 kJ/mol
- **ΔS°**: 198 J/K·mol
#### Chlorine
- **Cl2(g)**:
- **ΔH°**: 0 kJ/mol
- **ΔG°**: 0 kJ/mol
- **ΔS°**: 223 J/K·mol
- **Cl2(aq)**:
- **ΔH°**: -23 kJ/mol
- **ΔG
![### Gibbs Free Energy Calculation for a Chemical Reaction
**Problem Statement:**
Using data from Appendix 4, calculate ΔG for the reaction below at the following conditions:
* Temperature: 35°C
* Partial pressures:
- \( P_{H_2S} = 1.5 \times 10^{-4} \) atm
- \( P_{SO_2} = 2.0 \times 10^{-2} \) atm
- \( P_{H_2O} = 2.5 \times 10^{-2} \) atm
**Chemical Reaction:**
\[
2H_2S_{(g)} + SO_2_{(g)} \rightarrow 3S_{\text{rhombic (s)}} + 2H_2O_{(g)}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c13c2cd-7092-4d54-837b-d1e36936f099%2F460f9efe-437f-4ac2-a9b2-5359fed47711%2Fp37ezyb.jpeg&w=3840&q=75)
Transcribed Image Text:### Gibbs Free Energy Calculation for a Chemical Reaction
**Problem Statement:**
Using data from Appendix 4, calculate ΔG for the reaction below at the following conditions:
* Temperature: 35°C
* Partial pressures:
- \( P_{H_2S} = 1.5 \times 10^{-4} \) atm
- \( P_{SO_2} = 2.0 \times 10^{-2} \) atm
- \( P_{H_2O} = 2.5 \times 10^{-2} \) atm
**Chemical Reaction:**
\[
2H_2S_{(g)} + SO_2_{(g)} \rightarrow 3S_{\text{rhombic (s)}} + 2H_2O_{(g)}
\]
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