Estimate the enthalpy change for the combustion of one mole of acetylene, C₂H₂, to form carbon dioxide and water vapor.
BE(C-H) = 456 kJ/mol
BE(C $\equiv$ C) = 962 kJ/mol
BE(O=O) = 499 kJ/mol
BE(C=O) = 802 kJ/mol
BE(O-H) = 462 kJ/mol

Question

Estimate the enthalpy change for the combustion of one mole of acetylene, C₂H₂, to form carbon dioxide and water vapor.
BE(C-H) = 456 kJ/mol
BE(C

\equiv

C) = 962 kJ/mol
BE(O=O) = 499 kJ/mol
BE(C=O) = 802 kJ/mol
BE(O-H) = 462 kJ/mol

Quizplus · Accepted Answer

The balanced chemical equation for the combustion of acetylene is:

2C2H2 + 5O2 -> 4CO2 + 2H2O

The enthalpy change can be calculated using the bond enthalpies of the reactants and products:

Reactants:
4xC-H bonds x 456kJ/mol = 1824 kJ/mol
1xCC bond x 962 kJ/mol = 962 kJ/mol
5xO=O bonds x 499 kJ/mol = 2495 kJ/mol

Products:
8xC=O bonds x 802 kJ/mol = 6416 kJ/mol
4xO-H bonds x 462 kJ/mol = 1848 kJ/mol

ΔH = ΣBE(bonds broken) - ΣBE(bonds formed)
ΔH = (1824 kJ/mol + 962 kJ/mol + 2495 kJ/mol) - (6416 kJ/mol + 1848 kJ/mol)
ΔH = - 1271 kJ/mol or - 1.271MJmol-1

Therefore, the enthalpy change for the combustion of one mole of acetylene is -1010 kJ/mol (rounded to the nearest 10 kJ/mol). The answer is D.

Estimate the Enthalpy Change for the Combustion of One Mole ≡\equiv≡

Estimate the Enthalpy Change for the Combustion of One Mole $\equiv$