Question

Calculate the enthalpy change of oxidation for the combustion of 1 mole of ethane gas (C₂H₆) in oxygen gas (O₂) to form carbon dioxide gas (CO₂) and water vapor (H₂O), given the following enthalpies of formation: ΔHf(C₂H₆) = -84.7 kJ/mol, ΔHf(CO₂) = -393.5 kJ/mol, ΔHf(H₂O) = -241.8 kJ/mol.

Answer 1

To calculate the enthalpy change of oxidation for the combustion of 1 mole of ethane gas, we need to consider the balanced chemical equation for the reaction:CH g  + 7/2 O g   2 CO g  + 3 HO g Now, we can use Hess's Law, which states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the products minus the sum of the enthalpy changes for the reactants:H =  Hf products  -  Hf reactants For this reaction, we have:H = [2  Hf CO  + 3  Hf HO ] - [Hf CH  + 7/2  Hf O ]Since the enthalpy of formation for an element in its standard state  such as O gas  is zero, the term for oxygen gas will not contribute to the calculation:H = [2   -393.5 kJ/mol  + 3   -241.8 kJ/mol ] - [ -84.7 kJ/mol ]H =  -787 kJ + -725.4 kJ  -  -84.7 kJ H =  -1512.4 kJ  + 84.7 kJH = -1427.7 kJ/molTherefore, the enthalpy change of oxidation for the combustion of 1 mole of ethane gas is -1427.7 kJ/mol.