# g12 chem enthalpy change benzene question?

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- micatkieLv 72 months agoFavourite answer
According to the given data, we obtain the following three thermochemical equations:

6CO₂(g) + 3H₂O(ℓ) → C₆H₆(ℓ) + (15/2)O₂(g) ΔH°₁ = -(1/2)(-6534) = +3267 kJ/mol

6C(s) + 6O₂(g) → 6CO₂(g) ΔH°₂ = 6(-393.5) = -2361 kJ/mol

3H₂(g) + (3/2)O₂(g) → 3H₂O(ℓ) ΔH°₃ = 3(-285.8) = -857.4 kJ/mol

Add the above three thermochemical equations, and cancel 6CO₂(g), 3H₂O(ℓ) and (15/2)O₂(g) on the both side:

6C(s) + 3H₂(g) → C₆H₆(ℓ) ΔHf°[C₆H₆(ℓ)]

ΔHf°[C₆H₆(ℓ)]

= ΔH°₁ + ΔH°₂ + ΔH°₃

= (+3267) + (-2361) + (-857.4) kJ/mol

= +48.6 kJ/mol

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