# Calculating Equilibrium Pressures

1. Sep 25, 2013

### Youngster

1. The problem statement, all variables and given/known data

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

2. Relevant equations

dG = VdP - SdT

3. 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.

2. Sep 25, 2013

### Staff: Mentor

How is the equilibrium constant expressed in terms of the partial pressures of NO2 and N2O4? Also note that the standard heat of reaction and standard entropy change of reaction are evaluated at constant pressure (1 atm), not constant volume. So the problem you have been working on is really problem 2. You have yet to address problem 1.