Calculating Entropy Change for N2O4 Equilibrium in Constant Pressure Vessel

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The discussion revolves around calculating the entropy change for a system involving 1.0 mol of N2O4 in a constant pressure vessel at 1 bar and 298 K as it reaches equilibrium. Participants highlight the need to determine the final equilibrium state by calculating the equilibrium constant from the Gibbs free energies of formation for N2O4 and NO2. The calculated reaction enthalpy and Gibbs energy changes are 55.3 kJ/mol and 2.8 kJ/mol, respectively. An ICE table is suggested to track the changes in moles of N2O4 and NO2, but the lack of the equilibrium constant poses a challenge. The conversation emphasizes the importance of calculating the equilibrium constant to proceed with the entropy change calculations.
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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
 
Thread 'Confusion regarding a chemical kinetics problem'

Thread 'Confusion regarding a chemical kinetics problem'

TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
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