Answer:
The standard free energy of combustion of 1 mole of methane = -801.11 kJ
The negative sign shows that this reaction is spontaneous under standard conditions.
The negative sign on the standard free energy also means this combustion reaction is product-favoured at equilibrium.
Explanation:
The chemical reaction for the combustion of methane is given by
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
The standard free energy of formation for the reactants and products as obtained from literature include
For CH₄, ΔG⁰ = -50.50 kJ/mol
For O₂, ΔG⁰ = 0 kJ/mol
For CO₂, ΔG⁰ = -394.39 kJ/mol
For H₂O(g), ΔG⁰ = -228.61 kJ/mol
ΔG(combustion) = ΔG(products) - ΔG(reactants)
ΔG(products) = (1×-394.39) + (2×-228.61) = -851.61 kJ/mol
ΔG(reactants) = (1×-50.50) + (2×0) = -50.50 kJ/mol
ΔG(combustion) = ΔG(products) - ΔG(reactants)
ΔG(combustion) = -851.61 - (-50.5) = -801.11 kJ/mol
Since we're calculating for 1 mole of methane, ΔG(combustion) = -801.11 kJ
- A negative sign on the standard free energy means that the reaction is spontaneous under standard conditions.
- A positive sign indicates a non-spontaneous reaction.
- A Gibb's free energy of 0 indicates that the reaction is at equilibrium.
- A negative sign on the standard free energy also means that if the reaction reaches equilibrium, it will be product favoured.
- A positive sign on the standard free energy means that the reaction is reactant-favoured at equilibrium.
Hope this Helps!!!
