- #1

Youngster

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## Homework Statement

For the reaction

2 NO[itex]_{2}[/itex](g) [itex]\leftrightarrow[/itex] N[itex]_{2}[/itex]O[itex]_{4}[/itex](g)

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

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

*dH -TdS*, with all three values available in the problem.

So using the following relation between K[itex]_{eq}[/itex] and free energy:

ΔG°[itex]_{rxn}[/itex] = -RTln(K[itex]_{eq}[/itex])

Solving for K[itex]_{eq}[/itex], I obtain 6.973

So from this point, how do I calculate the partial pressures of NO[itex]_{2}[/itex] and N[itex]_{2}[/itex]O[itex]_{4}[/itex]? 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[itex]_{rxn}[/itex] goes to zero.