Final answer:
The enthalpy of combustion of ethylene gas (C2H4) at 298 K and 1 atmospheric pressure, calculated using the enthalpies of formation for the reactants and products, is -1323.3 kJ/mol.
Step-by-step explanation:
To calculate the enthalpy of combustion of ethylene gas (C2H4), we must consider the balanced chemical equation for its combustion:
• C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)
The enthalpy change for the reaction can be calculated using the enthalpies of formation of the reactants and products:
• ΔHcombustion = [∑ΔHf products] - [∑ΔHf reactants]
Using the given enthalpies of formation:
• CO2: -393.5 kJ/mol
• H2O: -241.8 kJ/mol
• C2H4: +52.3 kJ/mol
The enthalpy of combustion is:
• ΔHcombustion = [2(-393.5 kJ/mol) + 2(-241.8 kJ/mol)] - [+52.3 kJ/mol]
• ΔHcombustion = [(-787.0) + (-483.6)] - [52.3]
• ΔHcombustion = -1323.3 kJ/mol
Therefore, the enthalpy of combustion of ethylene gas at 298 K and 1 atmospheric pressure is -1323.3 kJ/mol.