Answer :
To find the change in molar entropy (\( \Delta S_{\text{m}} \)) and the change in molar enthalpy (\( \Delta H_{\text{m}} \)) for the given reaction, we'll use the equation:
\[ \Delta S_{\text{m}} = \sum{n \Delta S_{\text{products}}} - \sum{m \Delta S_{\text{reactants}}} \]
\[ \Delta H_{\text{m}} = \sum{n \Delta H_{\text{products}}} - \sum{m \Delta H_{\text{reactants}}} \]
Given:
- \( \Delta S_{\text{formation}} \) for A = -100 J/molK
- \( \Delta S_{\text{formation}} \) for B = 200 J/molK
- \( \Delta H_{\text{formation}} \) for A = -100 kJ/mol
- \( \Delta H_{\text{formation}} \) for B = 100 kJ/mol
The reaction is \( 2A \rightarrow B \), so:
- \( n = 1 \) for B (products)
- \( m = 2 \) for A (reactants)
Calculating \( \Delta S_{\text{m}} \):
\[ \Delta S_{\text{m}} = (1 \times 200 \, \text{J/molK}) - (2 \times -100 \, \text{J/molK}) \]
\[ \Delta S_{\text{m}} = 200 \, \text{J/molK} + 200 \, \text{J/molK} \]
\[ \Delta S_{\text{m}} = 400 \, \text{J/molK} \]
Calculating \( \Delta H_{\text{m}} \):
\[ \Delta H_{\text{m}} = (1 \times 100 \, \text{kJ/mol}) - (2 \times -100 \, \text{kJ/mol}) \]
\[ \Delta H_{\text{m}} = 100 \, \text{kJ/mol} + 200 \, \text{kJ/mol} \]
\[ \Delta H_{\text{m}} = 300 \, \text{kJ/mol} \]
Now, to find the change in Gibbs free energy (\( \Delta G \)) at constant temperature and pressure, we use the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Given:
- Temperature (T) = 25°C = 298 K
- Pressure = 1.00 bar
\[ \Delta G = 300 \, \text{kJ/mol} - (298 \, \text{K} \times 400 \, \text{J/molK}) \]
\[ \Delta G = 300 \, \text{kJ/mol} - 119.2 \, \text{kJ/mol} \]
\[ \Delta G = 180.8 \, \text{kJ/mol} \]
So, the change in Gibbs free energy (\( \Delta G \)) at 25°C and 1.00 bar pressure for the reaction \( 2A \rightarrow B \) is \( 180.8 \, \text{kJ/mol} \).
\[ \Delta S_{\text{m}} = \sum{n \Delta S_{\text{products}}} - \sum{m \Delta S_{\text{reactants}}} \]
\[ \Delta H_{\text{m}} = \sum{n \Delta H_{\text{products}}} - \sum{m \Delta H_{\text{reactants}}} \]
Given:
- \( \Delta S_{\text{formation}} \) for A = -100 J/molK
- \( \Delta S_{\text{formation}} \) for B = 200 J/molK
- \( \Delta H_{\text{formation}} \) for A = -100 kJ/mol
- \( \Delta H_{\text{formation}} \) for B = 100 kJ/mol
The reaction is \( 2A \rightarrow B \), so:
- \( n = 1 \) for B (products)
- \( m = 2 \) for A (reactants)
Calculating \( \Delta S_{\text{m}} \):
\[ \Delta S_{\text{m}} = (1 \times 200 \, \text{J/molK}) - (2 \times -100 \, \text{J/molK}) \]
\[ \Delta S_{\text{m}} = 200 \, \text{J/molK} + 200 \, \text{J/molK} \]
\[ \Delta S_{\text{m}} = 400 \, \text{J/molK} \]
Calculating \( \Delta H_{\text{m}} \):
\[ \Delta H_{\text{m}} = (1 \times 100 \, \text{kJ/mol}) - (2 \times -100 \, \text{kJ/mol}) \]
\[ \Delta H_{\text{m}} = 100 \, \text{kJ/mol} + 200 \, \text{kJ/mol} \]
\[ \Delta H_{\text{m}} = 300 \, \text{kJ/mol} \]
Now, to find the change in Gibbs free energy (\( \Delta G \)) at constant temperature and pressure, we use the equation:
\[ \Delta G = \Delta H - T \Delta S \]
Given:
- Temperature (T) = 25°C = 298 K
- Pressure = 1.00 bar
\[ \Delta G = 300 \, \text{kJ/mol} - (298 \, \text{K} \times 400 \, \text{J/molK}) \]
\[ \Delta G = 300 \, \text{kJ/mol} - 119.2 \, \text{kJ/mol} \]
\[ \Delta G = 180.8 \, \text{kJ/mol} \]
So, the change in Gibbs free energy (\( \Delta G \)) at 25°C and 1.00 bar pressure for the reaction \( 2A \rightarrow B \) is \( 180.8 \, \text{kJ/mol} \).
