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

We need to calculate the enthalpy change for the reaction:
C2H2 + 5/2 O2 → 2 CO2 + H2O
The reaction involves breaking the bonds in C2H2 and O2 and forming bonds in CO2 and H2O. Therefore, the enthalpy change can be calculated as follows:
Enthalpy change = ∑(BE of bonds broken) - ∑(BE of bonds formed)
BE(C-C) + BE(C-H) + 5/2 BE(O=O) = (2 × BE(C=O)) + BE(O-H)
= (0) + (2 × 802 kJ/mol) + 462 kJ/mol - (456 kJ/mol + 5/2 × 499 kJ/mol)
= -1010 kJ/mol
The negative sign indicates that the reaction is exothermic, so the correct answer is D) -1010 kJ/mol.

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

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