Asked by Steve
Determine the Enthalpy of combustion of ethyne (C2H2) using the following enthalpies of formation;
Data:
DHf°(H2O)= -285.9kj/mol
DHf°(2H2)= +226.7Kj/mol
DHf°(CO2)= -393.5Kj/mol
Equation; C2H2--------->CO2 + H2O
Data:
DHf°(H2O)= -285.9kj/mol
DHf°(2H2)= +226.7Kj/mol
DHf°(CO2)= -393.5Kj/mol
Equation; C2H2--------->CO2 + H2O
Answers
Answered by
DrBob222
Balance the equation.
2C2H2 + 5O2 ==>4CO2 + 2H2O
dHfo rxn = (n*dHfo products)-(n*dHfo reactanta). That will give you dHfo for the rxn which is for 2 mols C2H2. Divide by 2 for kJ/mol.
2C2H2 + 5O2 ==>4CO2 + 2H2O
dHfo rxn = (n*dHfo products)-(n*dHfo reactanta). That will give you dHfo for the rxn which is for 2 mols C2H2. Divide by 2 for kJ/mol.
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