Answer :
To determine the temperature at which the given reaction becomes spontaneous, we need to analyze the Gibbs free energy changes and the entropy change associated with the reaction.
Given the reaction:
[tex]\[ 2 \text{SOCl}_2(g) \rightarrow 2 \text{SCl}_2(g) + \text{O}_2(g) \][/tex]
And the provided data:
- [tex]\(\Delta S^{\circ} = 134 \frac{J}{K}\)[/tex]
- [tex]\(\Delta G_f^{\circ}(\text{SOCl}_2) = -202.85 \frac{kJ}{mol}\)[/tex]
- [tex]\(\Delta G_f^{\circ}(\text{SCl}_2) = -83.22 \frac{kJ}{mol}\)[/tex]
First, we need to convert the Gibbs free energy of formation values from kJ to J:
[tex]\[ \Delta G_f^{\circ}(\text{SOCl}_2) = -202.85 \frac{kJ}{mol} \times 1000 \frac{J}{kJ} = -202850 \frac{J}{mol} \][/tex]
[tex]\[ \Delta G_f^{\circ}(\text{SCl}_2) = -83.22 \frac{kJ}{mol} \times 1000 \frac{J}{kJ} = -83220 \frac{J}{mol} \][/tex]
Next, calculate the Gibbs free energy change for the reactants and products:
- For the reactants ([tex]\(2 \text{SOCl}_2\)[/tex]):
[tex]\[ \Delta G_{\text{reactants}} = 2 \times \Delta G_f^{\circ}(\text{SOCl}_2) = 2 \times (-202850 \frac{J}{mol}) = -405700 \frac{J}{mol} \][/tex]
- For the products ([tex]\(2 \text{SCl}_2 + \text{O}_2\)[/tex]):
[tex]\(\Delta G_f^{\circ}\)[/tex] for [tex]\( \text{O}_2(g) \)[/tex] in its standard state is zero, so:
[tex]\[ \Delta G_{\text{products}} = 2 \times \Delta G_f^{\circ}(\text{SCl}_2) = 2 \times (-83220 \frac{J}{mol}) = -166440 \frac{J}{mol} \][/tex]
Now, calculate the Gibbs free energy change for the reaction ([tex]\(\Delta G_{\text{rxn}}\)[/tex]):
[tex]\[ \Delta G_{\text{rxn}} = \Delta G_{\text{products}} - \Delta G_{\text{reactants}} \][/tex]
[tex]\[ \Delta G_{\text{rxn}} = -166440 \frac{J}{mol} - (-405700 \frac{J}{mol}) = 239260 \frac{J}{mol} \][/tex]
To determine the temperature at which the reaction becomes spontaneous, we use the relationship between Gibbs free energy ([tex]\(\Delta G\)[/tex]), enthalpy ([tex]\(\Delta H\)[/tex]), and entropy ([tex]\(\Delta S\)[/tex]):
[tex]\[ \Delta G_{\text{rxn}} = \Delta H_{\text{rxn}} - T \Delta S_{\text{rxn}} \][/tex]
For the reaction to be spontaneous, [tex]\(\Delta G_{\text{rxn}} < 0\)[/tex]. Therefore, we need to find the temperature at which [tex]\(\Delta G_{\text{rxn}}\)[/tex] becomes zero:
[tex]\[ 0 = \Delta H_{\text{rxn}} - T \Delta S_{\text{rxn}} \][/tex]
[tex]\[ T_{\text{spontaneous}} = \frac{\Delta H_{\text{rxn}}}{\Delta S_{\text{rxn}}} \][/tex]
Since [tex]\(\Delta G_{\text{rxn}}\)[/tex] at 298 K is known, we can use this to find [tex]\(\Delta H_{\text{rxn}}\)[/tex]:
[tex]\[ \Delta G_{\text{rxn}} = \Delta H_{\text{rxn}} - 298 K \times \Delta S^{\circ} \][/tex]
Solving for [tex]\(\Delta H_{\text{rxn}}\)[/tex]:
[tex]\[ \Delta H_{\text{rxn}} = \Delta G_{\text{rxn}} + 298 K \times \Delta S^{\circ} \][/tex]
But we can solve directly using the given delta_G_rxn (at T=0):
[tex]\[ T_{\text{spontaneous}} = \frac{\Delta G_{\text{rxn}}}{\Delta S^{\circ}} = \frac{239260 \frac{J}{mol}}{134 \frac{J}{K \cdot mol}} \][/tex]
[tex]\[ T_{\text{spontaneous}} \approx 1785.52 K \][/tex]
Therefore, the temperature at which the reaction becomes spontaneous is approximately 1785.52 K.
Given the reaction:
[tex]\[ 2 \text{SOCl}_2(g) \rightarrow 2 \text{SCl}_2(g) + \text{O}_2(g) \][/tex]
And the provided data:
- [tex]\(\Delta S^{\circ} = 134 \frac{J}{K}\)[/tex]
- [tex]\(\Delta G_f^{\circ}(\text{SOCl}_2) = -202.85 \frac{kJ}{mol}\)[/tex]
- [tex]\(\Delta G_f^{\circ}(\text{SCl}_2) = -83.22 \frac{kJ}{mol}\)[/tex]
First, we need to convert the Gibbs free energy of formation values from kJ to J:
[tex]\[ \Delta G_f^{\circ}(\text{SOCl}_2) = -202.85 \frac{kJ}{mol} \times 1000 \frac{J}{kJ} = -202850 \frac{J}{mol} \][/tex]
[tex]\[ \Delta G_f^{\circ}(\text{SCl}_2) = -83.22 \frac{kJ}{mol} \times 1000 \frac{J}{kJ} = -83220 \frac{J}{mol} \][/tex]
Next, calculate the Gibbs free energy change for the reactants and products:
- For the reactants ([tex]\(2 \text{SOCl}_2\)[/tex]):
[tex]\[ \Delta G_{\text{reactants}} = 2 \times \Delta G_f^{\circ}(\text{SOCl}_2) = 2 \times (-202850 \frac{J}{mol}) = -405700 \frac{J}{mol} \][/tex]
- For the products ([tex]\(2 \text{SCl}_2 + \text{O}_2\)[/tex]):
[tex]\(\Delta G_f^{\circ}\)[/tex] for [tex]\( \text{O}_2(g) \)[/tex] in its standard state is zero, so:
[tex]\[ \Delta G_{\text{products}} = 2 \times \Delta G_f^{\circ}(\text{SCl}_2) = 2 \times (-83220 \frac{J}{mol}) = -166440 \frac{J}{mol} \][/tex]
Now, calculate the Gibbs free energy change for the reaction ([tex]\(\Delta G_{\text{rxn}}\)[/tex]):
[tex]\[ \Delta G_{\text{rxn}} = \Delta G_{\text{products}} - \Delta G_{\text{reactants}} \][/tex]
[tex]\[ \Delta G_{\text{rxn}} = -166440 \frac{J}{mol} - (-405700 \frac{J}{mol}) = 239260 \frac{J}{mol} \][/tex]
To determine the temperature at which the reaction becomes spontaneous, we use the relationship between Gibbs free energy ([tex]\(\Delta G\)[/tex]), enthalpy ([tex]\(\Delta H\)[/tex]), and entropy ([tex]\(\Delta S\)[/tex]):
[tex]\[ \Delta G_{\text{rxn}} = \Delta H_{\text{rxn}} - T \Delta S_{\text{rxn}} \][/tex]
For the reaction to be spontaneous, [tex]\(\Delta G_{\text{rxn}} < 0\)[/tex]. Therefore, we need to find the temperature at which [tex]\(\Delta G_{\text{rxn}}\)[/tex] becomes zero:
[tex]\[ 0 = \Delta H_{\text{rxn}} - T \Delta S_{\text{rxn}} \][/tex]
[tex]\[ T_{\text{spontaneous}} = \frac{\Delta H_{\text{rxn}}}{\Delta S_{\text{rxn}}} \][/tex]
Since [tex]\(\Delta G_{\text{rxn}}\)[/tex] at 298 K is known, we can use this to find [tex]\(\Delta H_{\text{rxn}}\)[/tex]:
[tex]\[ \Delta G_{\text{rxn}} = \Delta H_{\text{rxn}} - 298 K \times \Delta S^{\circ} \][/tex]
Solving for [tex]\(\Delta H_{\text{rxn}}\)[/tex]:
[tex]\[ \Delta H_{\text{rxn}} = \Delta G_{\text{rxn}} + 298 K \times \Delta S^{\circ} \][/tex]
But we can solve directly using the given delta_G_rxn (at T=0):
[tex]\[ T_{\text{spontaneous}} = \frac{\Delta G_{\text{rxn}}}{\Delta S^{\circ}} = \frac{239260 \frac{J}{mol}}{134 \frac{J}{K \cdot mol}} \][/tex]
[tex]\[ T_{\text{spontaneous}} \approx 1785.52 K \][/tex]
Therefore, the temperature at which the reaction becomes spontaneous is approximately 1785.52 K.
