Calculate the Gibbs free energy (in kJ) at 25 degrees C of the following reaction using the data in the table. 1A +1B + 4C - 4D + 3E AHF°(kJ/mol) sOU/mol/K) A -82.84 121.97 B 91.68 C -89.37 170.11 12.15 D -88.02 41.54 E 64.03 32.33
Calculate the Gibbs free energy (in kJ) at 25 degrees C of the following reaction using the data in the table. 1A +1B + 4C - 4D + 3E AHF°(kJ/mol) sOU/mol/K) A -82.84 121.97 B 91.68 C -89.37 170.11 12.15 D -88.02 41.54 E 64.03 32.33
Introductory Circuit Analysis (13th Edition)
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Question
11.
![**Gibbs Free Energy Calculation for a Chemical Reaction**
**Objective:**
Calculate the Gibbs free energy (in kJ) at 25 degrees Celsius for the following chemical reaction using the data provided in the table.
**Chemical Reaction:**
\[ 1A + 1B + 4C \rightarrow 4D + 3E \]
**Data Table:**
| Species | \(\Delta H_f^0\) (kJ/mol) | \(S^0\) (J/mol·K) |
|---------|----------------|--------------|
| A | -82.84 | 121.97 |
| B | 91.68 | 170.11 |
| C | -89.37 | 12.15 |
| D | -88.02 | 41.54 |
| E | 64.03 | 32.33 |
**Explanation:**
- \(\Delta H_f^0\) represents the standard enthalpy of formation for each species.
- \(S^0\) denotes the standard entropy for each species.
- The given values are used to calculate the Gibbs free energy change (\(\Delta G\)) using the equation:
\[
\Delta G = \Delta H - T\Delta S
\]
Where:
- \(\Delta H = \Sigma (\Delta H_f^0 \, \text{products}) - \Sigma (\Delta H_f^0 \, \text{reactants})\)
- \(\Delta S = \Sigma (S^0 \, \text{products}) - \Sigma (S^0 \, \text{reactants})\)
- \(T\) is the temperature in Kelvin (25°C = 298 K).
**Instructions:**
1. Calculate \(\Delta H\) using the given enthalpies of formation.
2. Calculate \(\Delta S\) using the given entropies.
3. Use the Gibbs free energy equation to find \(\Delta G\).
This calculation will indicate the spontaneity of the reaction under standard conditions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a00ba11-4727-4380-bbb0-88ea42161338%2Fd32ebb52-1b83-4d27-bd05-aaa79ebda0af%2Fe7ado7.jpeg&w=3840&q=75)
Transcribed Image Text:**Gibbs Free Energy Calculation for a Chemical Reaction**
**Objective:**
Calculate the Gibbs free energy (in kJ) at 25 degrees Celsius for the following chemical reaction using the data provided in the table.
**Chemical Reaction:**
\[ 1A + 1B + 4C \rightarrow 4D + 3E \]
**Data Table:**
| Species | \(\Delta H_f^0\) (kJ/mol) | \(S^0\) (J/mol·K) |
|---------|----------------|--------------|
| A | -82.84 | 121.97 |
| B | 91.68 | 170.11 |
| C | -89.37 | 12.15 |
| D | -88.02 | 41.54 |
| E | 64.03 | 32.33 |
**Explanation:**
- \(\Delta H_f^0\) represents the standard enthalpy of formation for each species.
- \(S^0\) denotes the standard entropy for each species.
- The given values are used to calculate the Gibbs free energy change (\(\Delta G\)) using the equation:
\[
\Delta G = \Delta H - T\Delta S
\]
Where:
- \(\Delta H = \Sigma (\Delta H_f^0 \, \text{products}) - \Sigma (\Delta H_f^0 \, \text{reactants})\)
- \(\Delta S = \Sigma (S^0 \, \text{products}) - \Sigma (S^0 \, \text{reactants})\)
- \(T\) is the temperature in Kelvin (25°C = 298 K).
**Instructions:**
1. Calculate \(\Delta H\) using the given enthalpies of formation.
2. Calculate \(\Delta S\) using the given entropies.
3. Use the Gibbs free energy equation to find \(\Delta G\).
This calculation will indicate the spontaneity of the reaction under standard conditions.
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