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

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Discussion Overview

The discussion revolves around the calculation of the equilibrium constant for the reaction 2NO2(g)⇌N2O4(g) at a temperature of 670K. Participants explore the methodology used to derive the equilibrium constant from the standard Gibbs free energy change.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The original poster calculated ΔG°rxn as -2.8 kJ and derived K=1.65 using the equation ΔG°rxn = -RT*lnK.
  • One participant confirmed the ΔG°rxn value but suggested a different value for the gas constant (8.315 J K-1 mol-1) without believing it to be the source of the error.
  • Another participant questioned the temperature at which the ΔG°rxn was determined, implying it may not have been at 670K.
  • The original poster later acknowledged forgetting to account for the temperature and recalculated ΔG° at 670K, resulting in a new equilibrium constant of K=1.34e-5, which they stated was correct.

Areas of Agreement / Disagreement

There is no consensus on the initial calculation of the equilibrium constant, as participants identified potential errors and corrections, but the discussion reflects multiple viewpoints on the methodology and assumptions involved.

Contextual Notes

Participants noted the importance of ensuring that the standard Gibbs free energy change is calculated at the correct temperature, which was a point of confusion. The discussion also highlights the dependency on specific values from tables and the implications of using different constants.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in thermodynamics, particularly in calculating equilibrium constants and understanding the implications of temperature on Gibbs free energy.

littlebearrrr
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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 :)
 
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littlebearrrr said:
1) Using tabulated free energies of formation, I obtained ΔG°rxn=-2.8 kJ=-2.8x10^3 J.
I get the same answer.

littlebearrrr said:
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.

littlebearrrr said:
3) I took e^ of both sides. K=1.65
This is where things go bad! Could you show your calculation?
 
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In your table, what temperature is that ΔGo at? I bet it's not at 670 K.

Chet
 
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@ 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. :)
 

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