Question

So using Hess's law with dodecane, how would you calculate the heat of combustion using -352.1 as heat of formation?

Answer 1

Ans. Balanced Reaction:      C₁₂H₂₆(l) + 18.5 O₂(g) --------> 12CO₂(g) + 13H₂O(g)

Using Hess’s Law, the standard enthalpy of reaction, dHrxn is given by summation of standard enthalpy of formation of products subtracted by summation of standard enthalpy of formation of reactants.

Using Hess Law,

dHrxn = [12 x dH_f CO₂(g) + 13 x dH_f H₂O(l)] – [ dHf C₁₂H₂₆(l) – 9.5 x [dH_f O₂(g)]

Or, dHrxn = [12 x (- 393.5 kJ/mol) + 13 x (- 285.8 kJ/mol)] - [ - 352.1 kJ/mol – 0 ]

Or, dHrxn = (- 4722 kJ mol^-1 - 3715.4 kJ mol^-1) + 352.1 kJ mol^-1

Or, dHrxn = -8437.4 kJ mol^-1 + 352.1 kJ mol^-1

Hence, dHrxn = -8085.3 kJ/ mol

Hence, standard heat of combustion (dHrxn) of dodecane = -8085.3 kJ/ mol

Note: dH_f H₂O(l) = -285.8 kJ/ mol.            dH_f H₂O(g) = -241.8 kJ/mol

The standard state of water is liquid.

# If the non-standard condition dH_f H₂O(g) = -241.8 kJ/mol is used, then-

Or, dHrxn = [12 x (- 393.5 kJ/mol) + 13 x (- 241.8 kJ/mol)] - [ - 352.1 kJ/mol – 0 ]

Hence, dHrxn = -7513.3 kJ/ mol