Hess’s law type 2

The second way to apply Hess’s law uses standard enthalpies of formation.

formation reaction

a reaction in which a compound is formed from its elements

C_(s) + O_2(g) → CO_2(g), $\mathrm{\Delta }H=-393.5\text{kJ}$

standard enthalpy of formation ($\mathrm{\Delta }{H}_{\text{f}}$)

the molar enthalpy for a formation reaction (expressed per mole of product produced) where the elements are in their standard states and the reaction occurs at SATP (25 ºC and 101.3 kPa)

$\mathrm{\Delta }{H}_{\text{f}}=-393.5\text{kJ/mol}$ CO_2(g)

We can calculate the enthalpy change of a reaction using the individual $\mathrm{\Delta }{H}_{\text{f}}$ values of the products and reactants:

$\mathrm{\Delta }H=\sum n\mathrm{\Delta }{H}_{\text{f(products)}}-\sum n\mathrm{\Delta }{H}_{\text{f(reactants)}}\text{.}$

Example

What is the enthalpy change for CH_4(g) + 2 O_2(g) → CO_2(g) + 2 H₂O_(l)?

Begin with the formula given above. Substitute moles and standard enthalpies of formation for each product and reactant:

$\mathrm{\Delta }H=[(1\text{mol})(-393.5\text{kJ/mol})+(2\text{mol})(-285.8\text{kJ/mol})]$ $-[(1\text{mol})(-74.4\text{kJ/mol})+0]\text{.}$

This evaluates to −890.7 kJ. Notice that the answer is in kJ, not kJ/mol.