Enthalpy of combustion of Graphite, Hydrogen and Ethyl alcohol are -393.5 KJ/ mole -286 KJ/ mole and -1368 KJ/mole respectively. Standard
enthalpy of formation of ethyl alcohol is ____KJ/ mole.

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-554

-277

-189.5

-138.5

answer is C.

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Detailed Solution

Required equation is

$2 C (graphite) + 3 H_{2} (g) + \frac{1}{2} O_{2} (g) \to C_{2} H_{5} OH (l)$

$C (graphite) + O_{2} (g) \to {CO}_{2} (g), Δ H = - 393.5 KJ . . . . (1)$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H = - 286 K J . . . . . (2)$

$C_{2} H_{5} OH (l) + 3 O_{2} (g) \to 2 {CO}_{2} (g) + 3 H_{2} O (l),$

$Δ H = - 1368 K J \dots . (3)$

Equation 1 x 2 + equation 2 x 3 - equation 3 = Required equation

$[2 (- 393.5) + 3 (- 286) - (- 1368)] K J / mole$

$= [- 787 - 858 + 1368] KJ / mole = - 277 KJ / mole$

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