Q.At 33K, $N_2O_4$ is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Step 1. $N_2O_4\rightleftharpoons2NO_2$; starting with 1 mole $N_2O_4$ and dissociation $\alpha=0.5$: total moles $=1+\alpha=1.5$; mole fraction $N_2O_4=(1-\alpha)/(1+\alpha)=0.5/1.5=1/3$; mole fraction $NO_2=2\alpha/(1+\alpha)=1/1.5=2/3$.

Step 2. At total pressure 1 atm, partial pressures equal the mole fractions: $p(NO_2)=2/3$, $p(N_2O_4)=1/3$. $K_p=\dfrac{p(NO_2)^2}{p(N_2O_4)}=\dfrac{(2/3)^2}{1/3}=\dfrac{4/9}{1/3}=\dfrac{4}{3}\approx1.333$. …

Convert the 50% dissociation into mole fractions/partial pressures to find $K_p$, then apply $\Delta …

• Silently 'correcting' the given 33 K to a more physically typical temperature (e.g. 330 K) -- the number is used exactly as printed in the source, and the METHOD (finding $K_p$ from % dissociation, then $\Delta G^0=-RT\ln K_p$) i …