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Entropy of surroundings

  1. Apr 22, 2014 #1
    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?
     
  2. jcsd
  3. Apr 22, 2014 #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
     
  4. Apr 22, 2014 #3
    How would I do that?
    I tried an ICE table but I dont 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: Apr 22, 2014
  5. Apr 22, 2014 #4
    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
     
  6. Apr 22, 2014 #5
    Ohhh ok I see, thanks
     
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