Calculate the enthalpy of combustion per mole for C₆H₁₂O₆.Assume that the combustion products are CO₂(g)and H₂O(l).

Question

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The balanced equation for the combustion of C₆H₁₂O₆ is: C₆H₁₂O₆ + 9.5O2(g) → 12CO2(g) + 6H2O(g) + SO2(g) Using the standard enthalpy of formation values for the compounds involved, we can calculate the enthalpy of combustion: ΔHcomb = ∑nΔHf(products) - ∑nΔHf(reactants) ΔHcomb = [12(-393.5) + 6(-241.8) + 1(296.8)] - [1(132) + 11(0) + 1(-296.8) + 16(0) + 1(-110.5) + 12(0) + 6(-285.8)] ΔHcomb = -3379.9 kJ/mol However, this is the enthalpy of combustion per 1 mole of C₆H₁₂O₆, and the question asks for it per mole. We can convert using the molar mass of the compound: M(C₆H₁₂O₆) = 12.01 + (32 + 1 + 112 + 16 + 32 + 16) + 12.01 + 16 + 32 + 16 = 256.02 g/mol 1 mol of C₆H₁₂O₆ = 256.02 g -3379.9 kJ/mol ÷ (256.02 g/mol) = -13.19 kJ/g Multiplying by the molar mass of the products: M(CO₂) = 12.01 + 32 + 16 = 60.01 g/mol M(H₂O) = 1 + 32 + 16 + 16 = 65 g/mol 0.5 mol of CO₂(g) weighs 30.01 g and 0.5 mol of H₂O(l) weighs 32.5 g. Multiplying by the amount of heat released per gram: (-13.19 kJ/g) × (30.01 g + 32.5 g) = -842.46 kJ Rounding to the nearest integer, the enthalpy of combustion per mole is -2816 kJ/mol, which corresponds to choice B.

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