Answer:
The heat of combustion of liquid butane is: -2636 kJ/mol
Explanation:
This problem is solved by using Hess law of heats summation:
We have the heat of combustion for gaseous butane:
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
and we want the heat of combustion for liquid butane.
What we need to do is add the heat of vaporization for liquid butane to this equation:
C₄H₁₀ (l) ⇒ C₄H₁₀ (g) ΔHºvap = 22.44 kJ/mol
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
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C₄H₁₀ (l) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)
So,
ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol
