Question

Octane (C8H18) undergoes combustion according to the following thermochemical equation:

2C8H18(l) + 25O2(g)     16CO2(g) + 18H2O(l)       ∆H°rxn = −10,800 kJ/mol

Given that ∆H°f[CO2(g)] = −394 kJ/mol and ∆H°f[H2O(l)] = −286 kJ/mol, calculate the standard enthalpy of formation of octane.
   a. −326 kJ/mol
   b.326 kJ/mol
   c.210 kJ/mol
   d.-218 kJ/mol

Answer 1

Given

2C8H18(l) + 25O2(g) --> 16CO2(g) + 18H2O(l)     ∆H°rxn = −10,800 kJ/mol

The delta H of reaction = Sum of ∆H°f of products - Sum of ∆H°f reactants

So

∆H°rxn = −10,800 kJ/mol = [16X∆H°f CO2 + 18X ∆H°f H2O] - [2X∆H°f C8H18 + 25∆H°f O2]

Note: ∆H°f of elements in their native state = 0 so ∆H°f O2 = 0

−10,800 kJ/mol = [16 X (-394) + 18 ( -286)] - [ 2X∆H°f C8H18]

     −10,800    = -11452 - [2X∆H°f C8H18]

652 = - 2X∆H°f C8H18

∆H°f C8H18 = -326 KJ / mole