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

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)#

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.