[Electrolize]

Calculate Kp for the reaction:
N₂O₄ (g) -> 2NO₂ (g)
at 24°C, given that ΔG° = 329.83 J/mole N₂O₄.

[Jorriss]

You can get Kp in terms of Kc and Kc in terms of Gibbs free energy, maybe?

[Electrolize]

You can get Kp in terms of Kc and Kc in terms of Gibbs free energy, maybe?

I thought that too but the problem is that, what is given is ΔG° = 329.83 J/mole N₂O₄ for the half-reaction and not the overall, so it doesn't work...

[jsrozner]

 :delta:G =  :delta:G^o + RTlnK

I'm pretty sure you can consider K_c to = K_p
in which case you just substitute the values in, taking care to use 24⁰C = 297K.

and its at equilbrium so  :delta:G = 0.

I thought that too but the problem is that, what is given is ΔG° = 329.83 J/mole N₂O₄ for the half-reaction and not the overall, so it doesn't work...

I don't really understand what you're trying to say, but really the given info means that for every time the reaction occurs (for every mole of N2O4 consumed) the standard change in free energy is 329.83 J. It's not for a half reaction. I'm fairly certain this is just a straightforward plug into the equation.

[Electrolize]

:delta:G =  :delta:G⁰ + RTlnK

I'm pretty sure you can consider K_c to = K_p
in which case you just substitute the values in, taking care to use 24⁰C = 297K.

and its at equilbrium so  :delta:G = 0.

I thought that too but the problem is that, what is given is ΔG° = 329.83 J/mole N₂O₄ for the half-reaction and not the overall, so it doesn't work...

I don't really understand what you're trying to say, but really the given info means that for every time the reaction occurs (for every mole of N2O4 consumed) the standard change in free energy is 329.83 J. It's not for a half reaction. I'm fairly certain this is just a straightforward plug into the equation.

Ok so I have come to the same conclusion. Thank you for chiming in.