How to Find Equilibrium Pressures for NO₂ and N₂O₄ at 298 K?

[Youngster]

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

Homework Equations

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.

 

[Chestermiller]

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.