If enthalpies of formation for $C_{2} H_{4 (g)}, {CO}_{2 (g)} and H_{2} O_{(l)}$ at 25°C and 1 atm pressure are 52, - 394 and – 286 kJ/mol respectively, then enthalpy of combustion of C₂H₄(g) will be

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– 141.2 kJ / mol

+ 1412 kJ / mol

+ 141.2 kJ / mol

– 1412 kJ / mol

answer is D.

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Detailed Solution

$C_{2} H_{4} + 3 O_{2} \to 2 {CO}_{2} + 2 H_{2} O {ΔH}^{\circ} = {ΔH}^{\circ}_{products} - {ΔH}^{\circ}_{reactants}$
$\begin{array}{l} = 2 \times (- 394) + 2 \times (- 286) - (52 + 0) \\ = - 1412 kJ / mol \end{array}$

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