# How can I do Hess's law with fractions?

Feb 14, 2016

Usually with ease.

#### Explanation:

Let's take a simple one, formation of water:

$2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right)$ $\Delta {H}^{\circ} = - 572 \cdot k J \cdot m o {l}^{-} 1$.

Now you have been given an enthalpy value for the reaction. What does it mean? It means that $572$ $k J$ are evolved PER MOLE OF REACTION AS WRITTEN! Since its per mole of $\text{REACTION}$ as written, we know that for each 2 moles of dihydrogen that combine with 1 mole of dioxygen to form 2 moles of water, $572$ $k J$ of energy ARE EVOLVED.

If I wanted to represent the reaction a bit more economically I could write:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$; $\Delta {H}^{\circ} = - 286 \cdot k J \cdot m o {l}^{-} 1$.

Quite clearly the same energy transfer occurs, but here only the 1 mole of product results (and of course 1/2 the energy). This last reaction represents (by definition) DeltaH_f""^@, ${H}_{2} O$; the enthalpy of formation of water. Of course, I am free to use the 1/2 coefficient.

So, what's the message? That enthalpies are extensive quantities, and depend on the amounts of reactants and products. And when you do heat summations with Hess' law, you add and subtract together the actual numerical quantities of reactions, and double/treble/halve the enthalpies associated with them if required.