Question

Octane (C8H18) undergoes combustion according to the following thermochemical equation:
2 C8H18(l) + 25 O2(g) ---> 16 CO2(g) + 18 H2O(l) DHo = -11,200 kJ
Given : DHfo [CO2(g)] = -393.5 kJ/mol and DHfo[H2O(l)] = -285.8 kJ/mol.
Calculate the enthalpy of formation of one mole of liquid octane.

1) -6296 kJ
2. ) -120 kJ
3. ) -11,440 kJ
4. ) -5144 kJ
5. ) -240 kJ

I calculated 2 to be correct is it?

Answer 1

Since the heat of formation for 2 moles of octane is 11,200, you divide by 2 to get 5600kJ for 1 mole. According to the equation above, 1 mole of octane will give you 8 moles of CO2 and 9 moles of H2O. The heat of formation for O2 is zero because it naturally exists and has a bond energy of 0. Recall that heat of formation is=
heat of formation of products - heat of formation of reactants
So you get:
[8(393.5) + 9(285.8)] - [enthalpy of reaction]=-240kJ / 2 =>
The answer is b) -120kJ

Answer 2

Octane (C8H18) undergoes combustion according to the following equation. How many grams of octane are needed to produce 9620.5 kJ of heat?(Given M(C8H18) = 114 g/mol) %3D