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?

Md Ariful Islam

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Md Ariful Islam

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Just like we did , 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:
(pg. 8)
(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 , we are to calculate $K_P$ from $DeltaG_(rxn)^@$.

As , 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,

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

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

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

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