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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 withdH -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.