Answer :
To solve this problem, let's determine the equilibrium constant ([tex]\( K \)[/tex]) for the given reaction at [tex]\( 25^{\circ}C \)[/tex].
Given the reaction:
[tex]\[ CH_4(g) + 2H_2O(g) \rightarrow CO_2(g) + 4H_2(g) \][/tex]
We need to follow these steps:
1. Calculate the standard Gibbs free energy change ([tex]\( \Delta G^{\circ}_{\text{reaction}} \)[/tex]).
The standard Gibbs free energy change for a reaction is given by:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \Sigma \Delta G_f^{\circ}(\text{products}) - \Sigma \Delta G_f^{\circ}(\text{reactants}) \][/tex]
Using the Gibbs free energy of formation ([tex]\( \Delta G_f^{\circ} \)[/tex]) values from the table:
- For [tex]\( CO_2(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -394.4 \)[/tex] kJ/mol
- For [tex]\( H_2(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = 0 \)[/tex] kJ/mol
- For [tex]\( CH_4(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -50.81 \)[/tex] kJ/mol
- For [tex]\( H_2O(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -228.6 \)[/tex] kJ/mol
Substituting these values into the equation:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \left( \Delta G_f^{\circ}(CO_2) + 4 \times \Delta G_f^{\circ}(H_2) \right) - \left( \Delta G_f^{\circ}(CH_4) + 2 \times \Delta G_f^{\circ}(H_2O) \right) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \left( -394.4 + 4 \times 0 \right) - \left( -50.81 + 2 \times -228.6 \right) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -394.4 - (-50.81 - 457.2) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -394.4 + 50.81 + 457.2 \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113.61 \ \text{kJ/mol} \][/tex]
2. Convert [tex]\( \Delta G^{\circ}_{\text{reaction}} \)[/tex] into joules:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113.61 \ \text{kJ/mol} \times 1000 \ (\text{J/kJ}) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113610 \ \text{J/mol} \][/tex]
3. Calculate the equilibrium constant [tex]\( K \)[/tex]:
The relation between Gibbs free energy change and the equilibrium constant is given by:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -RT \ln K \][/tex]
Where:
- [tex]\( R \)[/tex] is the universal gas constant ([tex]\( 8.314 \ \text{J/K} \cdot \text{mol} \)[/tex])
- [tex]\( T \)[/tex] is the temperature in Kelvin ([tex]\( 25^{\circ}C = 298.15 \ \text{K} \)[/tex])
Rearranging the equation for [tex]\( K \)[/tex]:
[tex]\[ K = \exp\left( -\frac{\Delta G^{\circ}_{\text{reaction}}}{RT} \right) \][/tex]
[tex]\[ K = \exp\left( -\frac{113610 \ \text{J/mol}}{8.314 \ \text{J/K} \cdot \text{mol} \times 298.15 \ \text{K}} \right) \][/tex]
[tex]\[ K = \exp\left( -\frac{113610}{2476.9} \right) \][/tex]
[tex]\[ K = \exp\left( -45.867 \right) \][/tex]
[tex]\[ K = 1.2453218972075848 \times 10^{-20} \][/tex]
So, the equilibrium constant [tex]\( K \)[/tex] for the reaction is:
[tex]\[ K = 1.245 \times 10^{-20} \][/tex]
The correct answer from the multiple choice options is:
[tex]\[ \boxed{1.2 \times 10^{-20}} \][/tex]
Given the reaction:
[tex]\[ CH_4(g) + 2H_2O(g) \rightarrow CO_2(g) + 4H_2(g) \][/tex]
We need to follow these steps:
1. Calculate the standard Gibbs free energy change ([tex]\( \Delta G^{\circ}_{\text{reaction}} \)[/tex]).
The standard Gibbs free energy change for a reaction is given by:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \Sigma \Delta G_f^{\circ}(\text{products}) - \Sigma \Delta G_f^{\circ}(\text{reactants}) \][/tex]
Using the Gibbs free energy of formation ([tex]\( \Delta G_f^{\circ} \)[/tex]) values from the table:
- For [tex]\( CO_2(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -394.4 \)[/tex] kJ/mol
- For [tex]\( H_2(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = 0 \)[/tex] kJ/mol
- For [tex]\( CH_4(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -50.81 \)[/tex] kJ/mol
- For [tex]\( H_2O(g) \)[/tex]: [tex]\( \Delta G_f^{\circ} = -228.6 \)[/tex] kJ/mol
Substituting these values into the equation:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \left( \Delta G_f^{\circ}(CO_2) + 4 \times \Delta G_f^{\circ}(H_2) \right) - \left( \Delta G_f^{\circ}(CH_4) + 2 \times \Delta G_f^{\circ}(H_2O) \right) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = \left( -394.4 + 4 \times 0 \right) - \left( -50.81 + 2 \times -228.6 \right) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -394.4 - (-50.81 - 457.2) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -394.4 + 50.81 + 457.2 \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113.61 \ \text{kJ/mol} \][/tex]
2. Convert [tex]\( \Delta G^{\circ}_{\text{reaction}} \)[/tex] into joules:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113.61 \ \text{kJ/mol} \times 1000 \ (\text{J/kJ}) \][/tex]
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = 113610 \ \text{J/mol} \][/tex]
3. Calculate the equilibrium constant [tex]\( K \)[/tex]:
The relation between Gibbs free energy change and the equilibrium constant is given by:
[tex]\[ \Delta G^{\circ}_{\text{reaction}} = -RT \ln K \][/tex]
Where:
- [tex]\( R \)[/tex] is the universal gas constant ([tex]\( 8.314 \ \text{J/K} \cdot \text{mol} \)[/tex])
- [tex]\( T \)[/tex] is the temperature in Kelvin ([tex]\( 25^{\circ}C = 298.15 \ \text{K} \)[/tex])
Rearranging the equation for [tex]\( K \)[/tex]:
[tex]\[ K = \exp\left( -\frac{\Delta G^{\circ}_{\text{reaction}}}{RT} \right) \][/tex]
[tex]\[ K = \exp\left( -\frac{113610 \ \text{J/mol}}{8.314 \ \text{J/K} \cdot \text{mol} \times 298.15 \ \text{K}} \right) \][/tex]
[tex]\[ K = \exp\left( -\frac{113610}{2476.9} \right) \][/tex]
[tex]\[ K = \exp\left( -45.867 \right) \][/tex]
[tex]\[ K = 1.2453218972075848 \times 10^{-20} \][/tex]
So, the equilibrium constant [tex]\( K \)[/tex] for the reaction is:
[tex]\[ K = 1.245 \times 10^{-20} \][/tex]
The correct answer from the multiple choice options is:
[tex]\[ \boxed{1.2 \times 10^{-20}} \][/tex]
