How can I calculate the enthalpy of a reaction?

Since enthalpy is a state function, we only need to know the beginning and end states of the reaction, and we can figure out the standard enthalpy for the reaction.

Standard enthalpies of formation #DeltaH_f^@# are tabulated in Thermodynamics tables you should be able to find in your textbook appendix.

With those, we can construct the following equation basically looking at the enthalpy required to form each component of the reaction (Enthalpy of formation) and finding the difference between the beginning and end states:

#\mathbf(DeltaH_"rxn"^@ = sum_(P) n_PDeltaH_(f,P)^@ - sum_(R) n_RDeltaH_(f,R)^@)#

where:

• #n# is the number of #"mol"#s of reactant #R# or product #P#. You could also call it the stoichiometric coefficient, since those are relative to each other, as are #"mol"#s.
• #sum# means add (within parentheses).

So, if we take this reaction as an example (using data from here):

#color(green)("CH"_4(g) + 2"O"_2(g) -> "CO"_2(g) + 2"H"_2"O"(g))#

#DeltaH_(f,"CH"_4(g))^@ = "-74.9 kJ/mol"#

#DeltaH_(f,"O"_2(g))^@ = "0 kJ/mol"#

#DeltaH_(f,"CO"_2(g))^@ = "-393.5 kJ/mol"#

#DeltaH_(f,"H"_2"O"(g))^@ = "-241.8 kJ/mol"#

We would be able to calculate the enthalpy for the reaction at standard conditions of #25^@ "C"# and #"1 bar"#:

#color(blue)(DeltaH_"rxn"^@) = sum_(P) n_PDeltaH_(f,P)^@ - sum_(R) n_RDeltaH_(f,R)^@#

#= [n_("CO"_2)DeltaH_(f,"CO"_2(g))^@ + n_("H"_2O)DeltaH_(f,"H"_2"O"(g))^@] - [n_("CH"_4)DeltaH_(f,"CH"_4(g))^@ + n_("O"_2)DeltaH_(f,"O"_2(g))^@]#

#= [(1xx"-393.5 kJ/mol") + (2xx"-241.8 kJ/mol")] - [(1xx"-74.9 kJ/mol") + (2xx"0 kJ/mol")]#

#=# #color(blue)("-802.2 kJ/mol")#

And then if you wanted to calculate #DeltaH_"rxn"# for the same reaction under the same conditions with a different amount of methane, since #DeltaH_"rxn"^@# is defined on a per-mol basis (you could also call it normalized to #"1 mol"#), simply multiply #DeltaH_"rxn"^@# by the quantity:

#"mass of Methane (Limiting Reagent) used"/"mass of 1 mol of Methane (Limiting Reagent)"#

...and that scales the standard reaction down to your reaction (again, if under the same reaction conditions as the standard reaction, and only differing in the amount of limiting reagent used).