MCAT Chemistry Question – Combustion of Octane
- by
- Jan 07, 2017
- MCAT Question of the Day
1) 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g) ΔH1 = -250 kJ/mole
2) 18 H2O (l) → 18 H2O (g) ΔH2 = 88 kJ/mole
Net Reaction:
3) 2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (l) ΔH3 = ?
Given ΔH1 and ΔH2, what is ΔH3 for the net reaction above?
A. 162 kJ/mole
B. -162 kJ/mole
C. 338 kJ/mole
D. -338 kJ/mole
Click for Explanation
ΔH3 = ΔH1 + (-ΔH2)
ΔH3 = -250 kJ/mole + (-88 kJ/mole)
ΔH3 = -338 kJ/mole
Therefore, the change in enthalpy (ΔH3) for the net reaction above is -338 kJ/mole, or answer choice D. Furthermore, the combustion of a fuel is a highly exothermic process, making answer choices A and C incorrect.
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