Calculate the standard free energy change for the combustion of one mole of methane using the values for standard free energies of formation of the products and reactants. The sign of the standard free energy change allows chemists to predict if the reaction is spontaneous or not under standard conditions and whether it is product-favored or reactant-favored at equilibrium.

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.

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