Question

The combustion of ethene, C₂H₄ , occurs via the reaction

C₂H₄(g)+3O₂(g)→2CO₂(g)+2H₂O(g)

with heat of formation values given by the following table:

Substance ΔH∘f
(kJ/mol )
C₂H₄ (g) 52.47
CO₂(g) − 393.5
H₂O(g) − 241.8

Answer 1

The heat of combustion for 1 mole of ethanol is calculated using the given standard enthalpies of formation for ethanol, carbon dioxide, and water. The result is -1367 kJ/mol, which indicates the energy released when ethanol combusts.

Step-by-step explanation:

The student is asking about the heat of combustion for ethanol (C2H5OH). In chemistry, the calculation of heat of combustion generally involves using the enthalpies of formation to determine the total energy change during a chemical reaction. The heat of combustion indicates the amount of energy released when one mole of a substance is completely burned in oxygen to form water and carbon dioxide.

To calculate the heat of combustion for ethanol, we can use the standard enthalpies of formation provided in the question:

• Ethanol, C2H5OH(l), − 278 kJ/mol
• Water, H2O(l), − 286 kJ/mol
• Carbon Dioxide, CO2(g), − 394 kJ/mol

The balanced chemical equation for the combustion of ethanol is:

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

From this equation, we can calculate the heat of combustion using the formula:

ΔHcombustion = ∑ΔHf (products) - ∑ΔHf (reactants)

By plugging in the values, we see that the heat of combustion for ethanol is:

(2 × -394 kJ/mol + 3 × -286 kJ/mol) - (1 × -278 kJ/mol) = − 1367 kJ/mol.

Therefore, the heat of combustion for 1 mole of ethanol is − 1367 kJ/mol, indicating that this amount of energy is released during the process.