# What is the difference between enthalpy of formation, combustion, solution, and neutralization?

Nov 6, 2016

The equation(s) are similar, but the context is clearly different:

• Enthalpy of formation is the enthalpy for a formation reaction, and requires that the reactants are all in their standard state. That means they must be in their natural state at ${25}^{\circ} \text{C}$ and $\text{1 atm}$, such as "C"("graphite"), $\text{Al} \left(s\right)$, ${\text{H}}_{2} \left(g\right)$, ${\text{F}}_{2} \left(g\right)$, etc.
• Enthalpy of combustion is the enthalpy for the combustion reaction of a specified compound.
• Enthalpy of solution is the enthalpy for dissolving a compound into solution, which one could write as a reaction.
• Enthalpy of neutralization is the enthalpy for a neutralization reaction.

The similarity is that they can all be categorized under $\Delta {H}_{\text{rxn}}$. They are just different types of reactions.

These tend to be done at constant pressure, like in a coffee-cup calorimeter. By definition it means:

$\Delta {H}_{\text{rxn}} = {q}_{P}$,

if both in $\text{kJ}$, where ${q}_{P}$ is the heat evolved or absorbed during the reaction at constant pressure (an open-air system).

When you want the units in $\text{kJ/mol}$, you divide by the $\text{mol}$s of:

• the important compound for formation and dissolution/solvation reactions (involving one main compound).
• the limiting reactant in combustion and neutralization reactions (those involving more than one reactant).

As a result, you get the following equation for different contexts that are all assumed to be at ${25}^{\circ} \text{C}$ and $\text{1 atm}$:

ENTHALPY OF FORMATION EXAMPLE

$\boldsymbol{\Delta {\overline{H}}_{f , \text{NH"_4"Cl"(s))^@ = (q_"rxn")/(n_("NH"_4Cl} \left(s\right)}}$

for the standard reaction given by:

$\frac{1}{2} \text{N"_2(g) + 2"H"_2(g) + 1/2"Cl"_2(g) -> "NH"_4"Cl} \left(s\right)$

since the enthalpy of reaction is the difference in the sums of the $\Delta {H}_{f}^{\circ}$ of the products minus the sums of the $\Delta {H}_{f}^{\circ}$ of the reactants, but $\Delta {H}_{f}^{\circ} = 0$ for all the reactants in their natural/elemental states.

What I just said is:

$\Delta {H}_{\text{rxn}}^{\circ} = {\sum}_{P} {n}_{P} \Delta {\overline{H}}_{f , P}^{\circ} - {\cancel{{\sum}_{R} {n}_{R} \Delta {\overline{H}}_{f , R}^{\circ}}}^{0}$

when all reactants are in their elemental state. That means with only one product, $\Delta {H}_{\text{rxn}}^{\circ} = \Delta {H}_{f}^{\circ}$ of the product in $\text{kJ}$, and therefore, $\Delta {\overline{H}}_{\text{rxn}}^{\circ} = \Delta {\overline{H}}_{f}^{\circ}$ in $\text{kJ/mol}$.

ENTHALPY OF COMBUSTION EXAMPLE

bb(DeltabarH_(c,"CH"_4(g))^@ = (q_"rxn")/(n_("CH"_4(g))))

for the combustion reaction given by:

$\text{CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O} \left(g\right)$

ENTHALPY OF SOLUTION/SOLVATION EXAMPLE

$\boldsymbol{\Delta {\overline{H}}_{{\text{solv","NH"_3(g))^@ = (q_"rxn")/(n_("NH}}_{3} \left(g\right)}}$

for the dissolution process given by:

${\text{NH"_3(g) stackrel("H"_2"O"(l)" ")(->) "NH}}_{3} \left(a q\right)$

ENTHALPY OF NEUTRALIZATION EXAMPLE

bb(DeltabarH_("neut","NH"_4"Cl"(aq))^@ = (q_"rxn")/(n_("Limiting Reactant")))

for the neutralization (acid-base) reaction given by:

$\text{NH"_3(aq) + "HCl"(aq) -> "NH"_4"Cl} \left(a q\right)$

Again, note that they are all DeltabarH_"rxn" = q_"rxn"/(n_"something"). That is because the units on the left and right are both $\text{kJ/mol}$. Keeping track of your units can help you reason out why this is the equation.