If #\tt{\DeltaG°_(f,NO_2(g))=31.3" kJ"//"mol"}# and #\tt{\DeltaG°_(f,NO(g))=86.6" kJ"//"mol"}# at 298 K, then calculate #\tt{K_p}# at 298 K?

Just like we did here, this problem just reverses the situation.

However, the #DeltaG_f^@# of #"NO"_2(g)# is incorrect, and it should be #"51.3 kJ/mol"#, as shown by two sources here:
http://sunny.moorparkcollege.edu/~dfranke/chemistry_1B/thermodynamics.pdf (pg. 8)
http://gchem.ac.nctu.edu.tw/file.php/1/OldExam/Chem_1/Old-Final/Chem1_Final-99-Ans.pdf (pg. 4)

Once we fixed that,

#K_P = 1.53 xx 10^6#

What are the implied units of #K_P#? Hint:

#K_P = P_(NO_2)/(P_(NO)P_(O_2)^(1//2))#

And instead of what we did here, we are to calculate #K_P# from #DeltaG_(rxn)^@#.

As in this question, we again use Gibbs' free energy of formations:

#DeltaG_(rxn)^@ = sum_P n_P DeltaG_(f,P)^@ - sum_R n_RDeltaG_(f,R)^@#

where #DeltaG_f^@# is the change in Gibbs' free energy of reaction at #25^@ "C"# and #"1 atm"# in #"kJ/mol"#, and #P# and #R# are products and reactants. #n# is the mols of substance.

Hence, since the reaction is

#"NO"(g) + 1/2"O"_2(g) rightleftharpoons "NO"_2(g)#,

we just have

#DeltaG_(rxn)^@ = ["1 mol" cdot "51.3 kJ/mol NO"_2] - ["1 mol" cdot "86.6 kJ/mol NO" + 1/2 "mol O"_2 cdot "0 kJ/mol O"_2]#

#=# #-"35.3 kJ"#

#= -"35.3 kJ/mol NO"_2(g)#

We know that for gas-phase reactions at equilibrium,

#DeltaG_(rxn)^@ = -RTlnK_P#

and for aqueous reactions at equilibrium,

#DeltaG_(rxn)^@ = -RTlnK_c#

And therefore, for this spontaneous gas-phase reaction,

#color(blue)(K_P) = e^(-DeltaG_(rxn)^@//RT)#

#= e^(-(-"35.3 kJ/mol")//("0.008314 kJ/mol"cdot"K" cdot "298.15 K"))#

#= color(blue)(1.53 xx 10^6)#