Question

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?

`\sf{"Sorry if I have asked this before!"}`

The reaction is as follows:
`\tt{NO(g)+1/2O_2(g)\harrNO_2(g)}`

Answer 1

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)`