Calculating Entropy Change for N2O4 Equilibrium in Constant Pressure Vessel

[molly16]

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?

 

[Chestermiller]

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

 

[molly16]

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

 

[Chestermiller]

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

 

[molly16]

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