# 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?

## $\setminus \textsf{\text{Sorry if I have asked this before!}}$ The reaction is as follows: $\setminus \texttt{N O \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \setminus \leftrightarrow N {O}_{2} \left(g\right)}$

##### 1 Answer
Aug 5, 2018

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

However, the $\Delta {G}_{f}^{\circ}$ of ${\text{NO}}_{2} \left(g\right)$ is incorrect, and it should be $\text{51.3 kJ/mol}$, 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,

${K}_{P} = 1.53 \times {10}^{6}$

What are the implied units of ${K}_{P}$? Hint:

${K}_{P} = {P}_{N {O}_{2}} / \left({P}_{N O} {P}_{{O}_{2}}^{1 / 2}\right)$

And instead of what we did here, we are to calculate ${K}_{P}$ from $\Delta {G}_{r x n}^{\circ}$.

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

$\Delta {G}_{r x n}^{\circ} = {\sum}_{P} {n}_{P} \Delta {G}_{f , P}^{\circ} - {\sum}_{R} {n}_{R} \Delta {G}_{f , R}^{\circ}$

where $\Delta {G}_{f}^{\circ}$ is the change in Gibbs' free energy of reaction at ${25}^{\circ} \text{C}$ and $\text{1 atm}$ in $\text{kJ/mol}$, and $P$ and $R$ are products and reactants. $n$ is the mols of substance.

Hence, since the reaction is

${\text{NO"(g) + 1/2"O"_2(g) rightleftharpoons "NO}}_{2} \left(g\right)$,

we just have

$\Delta {G}_{r x n}^{\circ} = \left[{\text{1 mol" cdot "51.3 kJ/mol NO"_2] - ["1 mol" cdot "86.6 kJ/mol NO" + 1/2 "mol O"_2 cdot "0 kJ/mol O}}_{2}\right]$

$=$ $- \text{35.3 kJ}$

$= - {\text{35.3 kJ/mol NO}}_{2} \left(g\right)$

We know that for gas-phase reactions at equilibrium,

$\Delta {G}_{r x n}^{\circ} = - R T \ln {K}_{P}$

and for aqueous reactions at equilibrium,

$\Delta {G}_{r x n}^{\circ} = - R T \ln {K}_{c}$

And therefore, for this spontaneous gas-phase reaction,

$\textcolor{b l u e}{{K}_{P}} = {e}^{- \Delta {G}_{r x n}^{\circ} / R T}$

$= {e}^{- \left(- \text{35.3 kJ/mol")//("0.008314 kJ/mol"cdot"K" cdot "298.15 K}\right)}$

$= \textcolor{b l u e}{1.53 \times {10}^{6}}$