Question

Give the ΔH value for the combustion of ethanol as shown in the reaction C2H5OH(g)+3O2(g)⟶Δ2CO2(g)+3H2O(g)+1278 kJ .

Express your answer using four significant figures. If the value is positive, do not include the + sign in your answer.

Answer 1

The enthalpy change for the combustion of ethanol is -510 kJ/mol.

Step-by-step explanation:

The enthalpy change for the combustion of ethanol can be calculated using the enthalpies of formation of the reactants and products. The balanced equation for the combustion of ethanol is C₂H₅OH(g) + 3O₂(g) → 2CO₂(g) + 3H₂O(g), and the given enthalpies of formation are: C₂H₅OH(I) = -278 kJ/mol, CO₂(g) = -394 kJ/mol, and H₂O(I) = -286 kJ/mol.

To calculate the enthalpy change, we can use the following equation:

ΔH = Σ(n * ΔHf(products)) - Σ(n * ΔHf(reactants))

ΔH = (2 * -394 kJ/mol + 3 * -286 kJ/mol) - (-278 kJ/mol)

ΔH = -788 kJ/mol - (-278 kJ/mol)

ΔH = -510 kJ/mol

Therefore, the ΔH value for the combustion of ethanol is -510 kJ/mol. Since the value is negative, we do not include the + sign in the answer.