# Question #0ff54

May 10, 2015

The standard enthalpy of formation for a compound expresses the change in enthalpy (it can be positive or negative) when 1 mole of that compound is formed from the most stable form of its elements (at 1 atm and 273.15 K).

Standard enthalpies of formation are thus calculated for reactions that produce 1 mole of a specific compound. For example, you could write the balanced chemical equation for your synthesis reaction like this

${N}_{2 \left(g\right)} + 2 {O}_{2 \left(g\right)} \to 2 N {O}_{2 \left(g\right)}$

This reaction will also have a change in enthalpy, $\Delta {H}_{\text{rxn}}$, but it not be equal to the standard enthalpy of formation, $\Delta {H}_{\text{f}}^{\circ}$, because it produces 2 moles of nitrogen dioxide, instead of only 1 mole.

To get the reaction for which $\Delta {H}_{\text{rxn" = DeltaH_"f}}^{\circ}$ you need to divide the above equation by 2, in order to get only 1 mole of the product

$\text{1/2} {N}_{2 \left(g\right)} + {O}_{2 \left(g\right)} \to N {O}_{2 \left(g\right)}$

That's how you can determine the reaction for which the change in enthalpy is equal to the standard enthalpy of formation for any compound.

Here's another example. Assume you want the reaction that produces $\Delta {H}_{\text{f}}^{\circ}$ for iron (II) sulfide, $F e S$. You start from

$8 F e + {S}_{8} \to 8 F e S$

Notice that you produce 8 moles of iron (II) sulfide, but you only need to produce 1 mole $\to$ divide all the coefficients by 8

$F e + \text{1/8} {S}_{8} \to F e S$

This reaction has $\Delta {H}_{\text{rxn" = DeltaH_"f}}^{\circ}$.