Use standard enthalpies of formation to calculate ΔHᵣₓₙ for the following reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Given the standard enthalpies of formation: CH₄(g) = -74.8 kJ/mol, O₂(g) = 0 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(g) = -241.8 kJ/mol.

• A. -571.6 kJ/mol
• B. -802.3 kJ/mol
• C. -890.3 kJ/mol
• D. -483.6 kJ/mol

Answer: (B)

Verified step by step guidance

1. Identify the given reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g).
2. Recall the formula for calculating the standard enthalpy change of a reaction (ΔHᵣₓₙ) using standard enthalpies of formation: ΔHᵣₓₙ = ΣΔHᵢ(products) - ΣΔHᵢ(reactants).
3. List the standard enthalpies of formation for each compound: CH₄(g) = -74.8 kJ/mol, O₂(g) = 0 kJ/mol, CO₂(g) = -393.5 kJ/mol, H₂O(g) = -241.8 kJ/mol.
4. Calculate the total enthalpy of the products: ΔH(products) = [1 * ΔH(CO₂) + 2 * ΔH(H₂O)] = [1 * (-393.5) + 2 * (-241.8)] kJ/mol.
5. Calculate the total enthalpy of the reactants: ΔH(reactants) = [1 * ΔH(CH₄) + 2 * ΔH(O₂)] = [1 * (-74.8) + 2 * (0)] kJ/mol. Then, use the formula to find ΔHᵣₓₙ: ΔHᵣₓₙ = ΔH(products) - ΔH(reactants).