Question

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

Answer 1

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^@ "C"` and `"1 atm"`, such as `"C"("graphite")`, `"Al"(s)`, `"H"_2(g)`, `"F"_2(g)`, 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 `DeltaH_"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:

`DeltaH_"rxn" = q_P`,

if both in `"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 `"kJ/mol"`, you divide by the `"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^@ "C"` and `"1 atm"`:

ENTHALPY OF FORMATION EXAMPLE

`bb(DeltabarH_(f,"NH"_4"Cl"(s))^@ = (q_"rxn")/(n_("NH"_4Cl"(s)))`

for the standard reaction given by:

`1/2"N"_2(g) + 2"H"_2(g) + 1/2"Cl"_2(g) -> "NH"_4"Cl"(s)`

since the enthalpy of reaction is the difference in the sums of the `DeltaH_f^@` of the products minus the sums of the `DeltaH_f^@` of the reactants, but `DeltaH_f^@ = 0` for all the reactants in their natural/elemental states.

What I just said is:

`DeltaH_"rxn"^@ = sum_P n_P DeltabarH_(f,P)^@ - cancel(sum_R n_R DeltabarH_(f,R)^@)^(0)`

when all reactants are in their elemental state. That means with only one product, `DeltaH_"rxn"^@ = DeltaH_f^@` of the product in `"kJ"`, and therefore, `DeltabarH_"rxn"^@ = DeltabarH_f^@` in `"kJ/mol"`.

ENTHALPY OF COMBUSTION EXAMPLE

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

for the combustion reaction given by:

`"CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(g)`

ENTHALPY OF SOLUTION/SOLVATION EXAMPLE

`bb(DeltabarH_("solv","NH"_3(g))^@ = (q_"rxn")/(n_("NH"_3(g)))`

for the dissolution process given by:

`"NH"_3(g) stackrel("H"_2"O"(l)" ")(->) "NH"_3(aq)`

ENTHALPY OF NEUTRALIZATION EXAMPLE

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

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

`"NH"_3(aq) + "HCl"(aq) -> "NH"_4"Cl"(aq)`

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