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Enthalpy of formation of C2H4, CO2 and H2O are 52, −394 and −286 kJ/mol respectively (Given) The reaction is C2H4 + 3O2 → 2CO2 + 2H2O change in enthalpy, ΔH = ΔHproducts − ΔHreactants = 2 × (−394) + 2 × (−286) − (52 + 0) = −1412 kJ/mol

1. (a) −1412 kJ/mol
2. (b) −1412 kJ/mol
3. (c) −1412 kJ/mol
4. (d) −1412 kJ/mol

Correct answer: (a) −1412 kJ/mol

Solution

The enthalpy change is calculated correctly using the given formula and data. The result is −1412 kJ/mol, which matches option (a).

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