What is the standard enthalpy of formation of SO2(g)?

Well, we can always check our answer. To check, it should be #(-296.81 pm 0.20)# #"kJ/mol"#. You should use NIST more often.

I actually got #-"310.17 kJ/mol"# though.

Supposedly, the #DeltaH_(rxn)^@# is #-"171.2 kJ"# for this reaction as-written:

#2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g)#

You have to look up #DeltaH_f^@# for #"SO"_3(g)# first. I got #-"395.77 kJ/mol"# from NIST.

Now, #DeltaH_(rxn)^@# can be calculated from tabulated enthalpies of formation.

#DeltaH_(rxn)^@ = sum_P n_P DeltaH_(f,P)^@ - sum_R n_R DeltaH_(f,R)^@#

where #P# stands for products, #R# for reactants, #n# for the mols of stuff, and #DeltaH_f^@# is the enthalpy of forming 1 mol of the substance from its elements in their elemental states at #25^@ "C"# and #"1 bar"#.

We should know that #DeltaH_(f,O_2(g))^@ = "0 kJ/mol"#... no energy is needed to form something by doing nothing to it. We have #"O"_2(g)# in the atmosphere naturally.

#-"171.2 kJ" = ["2 mol" xx -"395.77 kJ/mol"] - ["2 mol" xx DeltaH_(f,SO_2(g))^@ + "1 mol" xx "0 kJ/mol"]#

Resolving the units, we obtain:

#-"171.2 kJ" = -"791.54 kJ" - 2DeltaH_(f,SO_2(g))^@ "kJ"#

And so:

#color(blue)(DeltaH_(f,SO_2(g))^@ = -"310.17 kJ/mol")#

Apparently, this is rather off from what NIST has (#4.5%# error). I wouldn't trust that scientist's measuring strategies. :)