Calculating Entropy Change for N2O4 Equilibrium in Constant Pressure Vessel

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In summary, the conversation discusses the process of a constant pressure vessel containing 1.0 mol of N2O4 at 1 bar and 298 K reaching equilibrium. The gibbs energy and enthalpy of formation, as well as the entropy values for N2O4 and NO2, are given. The solution involves calculating the final equilibrium state and using the free energies of formation to find the equilibrium constant.
  • #1
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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  • #2
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
 
  • #3
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:
  • #4
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
 
  • #5
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
 

Related to Calculating Entropy Change for N2O4 Equilibrium in Constant Pressure Vessel

What is entropy of surroundings?

Entropy of surroundings refers to the measure of the degree of randomness or disorder in the environment outside of a system. It is a thermodynamic property that describes the energy dispersal or the amount of unavailable energy in a system.

How is entropy of surroundings related to the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that the entropy of a closed system will always increase over time. This also means that the entropy of surroundings will also increase as the system becomes more disordered, leading to a decrease in the amount of usable energy.

What factors affect the entropy of surroundings?

The entropy of surroundings is affected by the heat transfer, work, and matter exchange between the system and the surroundings. The greater the heat transfer and work done by the system, the higher the entropy of surroundings will be. Additionally, the number of particles and their arrangement also play a role in determining the entropy of surroundings.

How is the entropy of surroundings different from the entropy of a system?

The entropy of a system refers to the amount of disorder within the system itself, while the entropy of surroundings refers to the amount of disorder in the environment outside of the system. The two are related, but they are distinct thermodynamic properties.

How can the entropy of surroundings be calculated?

The entropy of surroundings can be calculated using the equation ΔS = Q/T, where Q is the heat transferred between the system and surroundings, and T is the temperature in Kelvin. This equation is based on the definition of entropy as a measure of energy dispersal over a range of temperatures.

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