Standard enthalpy of formation

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy from the formation of 1 mole of the compound from its constituent elements, with all substances in their standard states at 101.3 kPa and 298 K. Its symbol is ΔH^o_f. The superscript theta (zero) on this symbol indicates that the process has been carried out under standard conditions.

For example, the standard enthalpy of formation of carbon dioxide would be the enthalpy of the following reaction under the conditions above:

C_(s,graphite) + O_{2 (g)} → CO_{2 (g)}

Note that all elements are written in their standard states, and one mole of product is formed. This is true for all enthalpies of formation.

The standard enthalpy of formation is measured in units of energy per amount of substance. Most are defined in kilojoules per mole (kJ mol⁻¹).

All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.

Now let's use these with Hess' law to determine the reaction enthalpy for the following reaction.

According to Hess' law, we could chose to break down the reactants to their constituent elements in their standard state and then reconstruct them into the products.

 From Hess' law, we have, ΔH₁^o = ΔH₂^o + ΔH₃^o

Lets explore ΔH₂^o + ΔH₃^o a bit further.

In step 2, we broke the reactants (methane and oxygen) into their constituent elements at standard state.  This is the reverse of the formation of these reactants so the total enthalpy change is simply the negative of the sum of the standard enthalpy of formation for the reactants (multiplied, of course by the approprite coefficients to give us the correct stoichiometry of the desired reaction).  Thus,

 ΔH= - [sum of Std enthalpies of formation of reactants]

ΔH= -ΣΔH_f^o (reactants)

In step 3, we took elements in their standard state and formed the products.  Thus, the  total enthalpy change is simply the sum of the standard enthalpy of formation for the products (multiplied, of course by the approprite coefficients to give us the correct stoichiometry of the desired reaction). Thus, 

ΔH = + sum of Std enthalpies of formation of products

ΔH = ΣΔH_f^o(products)

So now, we can write:

For the combustion of methane we're dealing with here, we can write

          ΔH = [ΔH_f^o (CO₂) + 2 ΔH_f^o  (H₂O)] - [ΔH_f^o  (CH₄) + 2 ΔH_f^o  (O₂)]

We can look up the values for Heat of formation in any standard reference book.  

           ΔH = [(-393.5 kJ) + 2 (-285.8 kJ)] - [(-74.7 kJ) + 2 (0)]

           ΔH = -890.4 kJ

In general, we can write:

Note that the coefficients n used here are unitless because they represent mole ratios, not numbers of moles. This allows the final entalpy change values to have units of kJ/mol, as they must have.  Note that if one or more of the reactants or products are not in their standard state, we simply can remove the superscript 'not' from the final enthalpy change symbol and the same equation will still work.