# Problem: Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g). Calculate Δ˚Hrxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is –146.8 kJ/mol{{ m{kJ}}}/{{ m{mol}}}.)

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###### FREE Expert Solution

Balanced reaction:

C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g)

Recall that ΔH˚rxn can be calculated from the enthalpy of formation (ΔH˚f) of the reactants and products involved:

$\overline{){\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{products}}{\mathbf{-}}{\mathbf{∆}}{\mathbf{H}}{{\mathbf{°}}}_{\mathbf{f}\mathbf{,}\mathbf{reactants}}}$ ###### Problem Details
Pentane (C5H12) is a component of gasoline that burns according to the following balanced equation: C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g). Calculate Δ˚Hrxn for this reaction using standard enthalpies of formation. (The standard enthalpy of formation of liquid pentane is –146.8 kJ/mol.)