Asked by 01236
The enthalpy of combustion for octane (C8H18(l)), a key component of gasoline, is –5074 kJ/mol. The reaction equation is: C8H18(l) + 12.5O2(g) es001-1.jpg 8CO2(g) + 9H2O(g).
What is the es001-2.jpgHf for this reaction?
16CO2(g) + 18H2O(g) es001-3.jpg 2C8H18(l) + 25O2(g)
kJ
All Answers 1
Answered by
GPT-5 mini
AI
The given combustion is per mole:
C8H18(l) + 12.5 O2(g) → 8 CO2(g) + 9 H2O(g), ΔH = −5074 kJ.
Multiply that reaction by 2:
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g), ΔH = 2(−5074) = −10148 kJ.
The target reaction is the reverse:
16 CO2(g) + 18 H2O(g) → 2 C8H18(l) + 25 O2(g),
so ΔH = +10148 kJ (i.e. +1.0148 × 10^4 kJ).
C8H18(l) + 12.5 O2(g) → 8 CO2(g) + 9 H2O(g), ΔH = −5074 kJ.
Multiply that reaction by 2:
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g), ΔH = 2(−5074) = −10148 kJ.
The target reaction is the reverse:
16 CO2(g) + 18 H2O(g) → 2 C8H18(l) + 25 O2(g),
so ΔH = +10148 kJ (i.e. +1.0148 × 10^4 kJ).
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