# Calculating Equilibrium Pressures

## Homework Statement

For the reaction

2 NO$_{2}$(g) $\leftrightarrow$ N$_{2}$O$_{4}$(g)

ΔH°$_{rxn}$ = -57.2 kJ and ΔS°$_{rxn}$ = -175.8 $J/K$

A 1 L container is initially filled with 1 atm of NO$_{2}$ at 298 K. Find the equilibrium pressures of NO$_{2}$ and N$_{2}$O$_{4}$ at 298 K if

a. volume is held constant

b. pressure is held constant

dG = VdP - SdT

## The Attempt at a Solution

So using the fundamental thermodynamic equation above, I found that ΔG = -4811.6 J. SdT goes to zero since the temperature doesn't appear to change. VdP can be substituted withdH -TdS, with all three values available in the problem.

So using the following relation between K$_{eq}$ and free energy:

ΔG°$_{rxn}$ = -RTln(K$_{eq}$)

Solving for K$_{eq}$, I obtain 6.973

So from this point, how do I calculate the partial pressures of NO$_{2}$ and N$_{2}$O$_{4}$? Can I stop at 6.973?

As for the second question, I find that if pressure is held constant, the equilibrium constant is just 1, since ΔG$_{rxn}$ goes to zero.