Calculate the enthalpy of combustion of ethylene gas to form CO2, and H2Oat 298K and 1 atm pressure. the enthalpies of formation of CO2, H2O and C2H4 are -393.7, -241.8 and +52.3 kJ per moles respectively.
Answers
The enthalpy of combustion of ethylene gas to form CO₂, and H₂O at S.T.P. conditions is, - 1218.70 kJ / mole.
• The combustion of ethylene to form CO₂ and H₂O can be represented by the following equation :
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
• The temperature and pressure conditions under which the reaction takes place are 298 K and 1 atm respectively. These are the standard conditions of temperature and pressure (S.T.P.).
• The enthalpy of combustion of a compound at S.T.P. is given by the formula :
∆H°c = Sum of ∆H°f of all products - Sum of ∆H°f of all reactants -(i)
where ∆Hc° denotes the enthalpy of combustion of a compound,
and ∆Hf ° denotes the enthalpy of formation of a compound.
• Given,
∆Hf ° of C₂H₄ = 52.3 kJ / mol
∆Hf ° of CO₂ = - 393.7 kJ / mol
∆Hf ° of H₂O = - 241.8 kJ / mol
• The enthalpies of each reactant and product should be multiplied by their stoichiometric coefficient or number of moles mentioned in the equation.
• Here, the reactants are C₂H₄ and O₂, and the products are CO₂ and H₂O. However, the heat of formation of O₂ is taken as zero, because oxygen freely occurs in atmosphere and involves no energy in formation.
• Putting the values of enthalpies in equation (i), we get,
∆Hc° (C₂H₄) = (52.3 kJ / mol + 0) - (2 × 393.7 kJ / mol + 2 × 241.8 kJ / mol)
=> ∆Hc° (C₂H₄) = (52.3 kJ / mol) - (787.4 kJ / mole + 483.6 kJ / mol)
=> ∆Hc° (C₂H₄) = 52.3 kJ / mol - 1271.0 kJ / mol
=> ∆Hc° (C₂H₄) = - 1218.70 kJ / mol (Answer)