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If enthalpies of formation of C₂H₄(g), CO₂(g) and H₂O(l) at 25°C and 1 atm pressure are 52, -394 and -286 kJ/mol respectively, the enthalpy of combustion of C₂H₄ is equal to

1. -1412 kJ/mol
2. -1412 kJ/mol
3. +142.2 kJ/mol
4. +1412 kJ/mol

Correct answer: -1412 kJ/mol

Solution

The combustion reaction for C₂H₄ is: C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l). Using the enthalpy of formation values, ΔH_comb = [2(-394) + 2(-286)] - [52 + 0] = -1412 kJ/mol.

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