f enthalpies of formation of $C_{2} H_{4} (g), {CO}_{2} (g) and H_{2} O (l)$ at $25^{\circ} C$ and 1atm pressure be 52, -394 and $- 286 kJ {mol}^{- 1}$ respectively, the enthalpy of combustion of $C_{2} H_{4} (g)$ will be

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$+ 1412 kJ {mol}^{- 1}$

$- 141.2 kJ {mol}^{- 1}$

$+ 141.2 kJ {mol}^{- 1}$

$- 1412 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

$C_{2} H_{4} + 3 O_{2} \to 2 {CO}_{2} + 2 H_{2} O$

${ΔH}_{reaction} = [2 \times {ΔH}_{f}^{\circ} ({CO}_{2}) + 2 \times {ΔH}_{f}^{\circ} (H_{2} O)]$ $- [{ΔH}_{f}^{\circ} (C_{2} H_{4}) + 3 \times {ΔH}_{f}^{\circ} (O_{2})]$

$= [2 (- 394) + 2 (- 286)] - [52 + 0] = - 1412 kJ .$

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