What Could Be Wrong with My Equilibrium Constant Calculation at 670K?

[littlebearrrr]

Homework Statement

Estimate the value of the equilibrium constant at 670K for the following reaction: 2NO2(g)⇌N2O4(g)

Homework Equations

ΔG°rxn = -RT*lnK

The Attempt at a Solution

This should be an easy problem. But for some reason, MasteringChem won't accept it, so I'm wondering if I did anything wrong. Here's what I did:

1) Using tabulated free energies of formation, I obtained ΔG°rxn=-2.8 kJ=-2.8x10^3 J.
2) I then divided this by 8.314 J/molK multiplied by 670K (multiplied by -1, which cancels).
3) I took e^ of both sides. K=1.65

Anyone know what I'm doing wrong? Just a small hint will suffice. Thank you in advance :)

 

[DrClaude]

1) Using tabulated free energies of formation, I obtained ΔG°rxn=-2.8 kJ=-2.8x10^3 J.

I get the same answer.

2) I then divided this by 8.314 J/molK multiplied by 670K (multiplied by -1, which cancels).

I would've used 8.315 J K^-1 mol^-1, but I don't think that's the problem here.

3) I took e^ of both sides. K=1.65

This is where things go bad! Could you show your calculation?

 

[Chestermiller]

In your table, what temperature is that ΔG^o at? I bet it's not at 670 K.

Chet

 

[littlebearrrr]

@ DrClaude and Chestermiller:

Sorry for the late response. Thank you for reviewing my post. To DrClaude: I forgot to account for the temperature in my calculations, as noted by Chestermiller. I calculated the std. change in enthalpy and entropy, and used the values to find ΔG^o at 670K. Then I plugged it into the equation from my first post to get a new answer of K=1.34e-5 (which turned out to be correct). Thank you both once again for looking at my post, and Chestermiller for offering that helpful bit of advice. :)