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The combustion of octane is a common reaction in automobile engines.

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. 163 kJ/mole

B. -163 kJ/mole

C. 333 kJ/mole

D. -333 kJ/mole


This question asks you to determine the change in enthalpyfor a net reaction given the enthalpy change for two other reactions. Hess’s Law tells us that the sum of two reactions gives us the enthalpy change for the net reaction: ΔH1 + ΔH2 = ΔH3. Since the net reaction has H2O (l) as a product, reaction 2 must be reversed and the sign of the enthalpy change negated (-ΔH2). The sum of the reaction enthalpies for the net reaction is as follows:

ΔH3 = ΔH1 + (-ΔH2)

ΔH3 = -250 kJ/mole + (-88 kJ/mole)

ΔH3 = -333 kJ/mole


Therefore, the change in enthalpy (ΔH3) for the net reaction above is -333 kJ/mole, or answer choice D. Furthermore, the combustion of a fuel is a highly exothermic process making answer choice A and C incorrect.

Topics #chemistry #enthalpy #mcat #octane