The combustion of one mole of butane(C₄H₁₀) takes place at 298K and 1atm.After combustion CO₂(g) and H₂O(l) are produced and 2878.7kJ/mol of heat is liberated. Compute the standard enthalpy of formation of CO₂(g) and H₂O(l) are -393.5 kJ/mol and -285.8kJ/mol respectively.

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Solution

Combustion reaction of butane:
C₄H₁₀ (g) + $\frac{13}{2}$ O₂ (g) $\to$ 4CO₂ (g) + 5H₂O (l)

$Δ_{r} H ° = Σ Δ_{f} H ° (p r o d u c t s) - Σ Δ_{f} H ° (r e a c \tan t s) Δ_{r} H ° = 4 [Δ_{f} H ° (C O_{2})] + 5 [Δ_{f} H ° (H_{2} O)] - [Δ_{f} H ° (C_{4} H_{10})] - \frac{13}{2} [Δ_{f} H ° (O_{2})]$

Given that:
Heat of reaction = -2878.7 kJ/mol (negative sign is taken because combustion is always exothermic)
Heat of formation of CO₂ = -393.5 kJ/mol
Heat of formation of H₂O = -285.8 kJ/mol
Heat of formation of O₂ = 0

-2878.7 = 4(-393.5) + 5(-285.8) - $Δ_{f} H ° (C_{4} H_{10})$ - 0
$Δ_{f} H ° (C_{4} H_{10})$ = -1574 - 1429 + 2878.7
$Δ_{f} H ° (C_{4} H_{10})$ = -124.3 kJ/mol

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