Calculating Entropy Change for N2O4 Equilibrium in Constant Pressure Vessel

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Discussion Overview

The discussion revolves around calculating the entropy change for the surroundings during the equilibrium of N2O4 and NO2 in a constant pressure vessel. Participants explore the necessary steps to determine the final equilibrium state and the corresponding entropy change, focusing on thermodynamic principles and calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the initial conditions and thermodynamic data for N2O4 and NO2, along with an attempt to calculate the reaction's enthalpy and Gibbs energy changes.
  • Another participant suggests determining the final equilibrium state by calculating the number of moles of each species present at equilibrium.
  • A participant expresses uncertainty about how to proceed without the equilibrium constant, indicating an attempt to use an ICE table for calculations.
  • It is noted that the equilibrium constant can be derived from the free energies of formation, prompting further inquiry into this method.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to calculate the equilibrium state, as there is uncertainty regarding the availability of the equilibrium constant and how to derive it from the given data.

Contextual Notes

The discussion highlights limitations in the provided data, particularly the absence of the equilibrium constant, which is necessary for completing the calculations. There are also unresolved steps in the mathematical reasoning related to the entropy change.

molly16
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1.0 mol of N2O4 placed in a constant pressure vessel at P = 1bar and T = 298 K. The system is allowed to slowly (reversibly) come to equilibrium. Given gibbs energy of formation, enthalpy of formation and entropy (the values are below) calculate the entropy change to the surroundings.

N2O4:
gibbs energy of formation = 99.8 kJ/mol
Enthalpy of formation =11.1 kJ/mol
Entropy = 304.3 J/mol K

NO2 :
gibbs energy of formation = 51.3 kJ/mol
enthalpy of formation = 33.2 kJ/mol
entropy = 240.1 J/mol K

N2O4 (g) <=> 2NO2 (g)

Attempt at a solution:

So first I found the enthalpy and free energies of the reaction
delta Hrxn = 2(33.2) - 11.1 = 55.3 kJ/mol
delta Grxn = 2(51.3) - 99.8 = 2.8 kJ/mol

since the surroundings are at constant pressure I know:
delta Ssurroundings = qsurroundigs/T = delta H surroundings/ T

but I'm not sure where to go from here. Can anyone help?
 
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You need to start out by figuring out the final equilibrium state, in terms of the final number of moles of the two species present.

Chet
 
Chestermiller said:
You need to start out by figuring out the final equilibrium state, in terms of the final number of moles of the two species present.

Chet

How would I do that?
I tried an ICE table but I don't have the value for the equilibrium constant
-----------N204------------2 NO2
I -------(1.0 mol)---------(0 mol)
C ---------(-x)---------------(x)
E-------(1.0 - x)------------(x)

So I'm stuck
 
Last edited:
molly16 said:
How would I do that?
I tried an ICE table but I don't have the value for the equilibrium constant
-----------N204------------2 NO2
I -------(1.0 mol)---------(0 mol)
C ---------(-x)---------------(x)
E-------(1.0 - x)------------(x)

So I'm stuck
You are supposed to be able to calculate the equilibrium constant from the free energies of formation. Do you know how to do that?

Chet
 
Chestermiller said:
You are supposed to be able to calculate the equilibrium constant from the free energies of formation. Do you know how to do that?

Chet

Ohhh ok I see, thanks
 

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