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Standard entropy of a liquid at melting temperature

  1. Jun 2, 2014 #1
    I am presented a review of data which gives:
    vapour pressures of a liquid have been measured and fit to the following equation:
    Log10 (mmHg) = -3571/T + 8.999
    The melting point has been determined to be 392.7 K.
    A Cp value given for the liquid is 250 J/mol K
    and the ΔSvap is 117 J/mol K

    The standard entropy S°(392.7) for the liquid is calculated, although I am unable to recalculate it correctly for verification.


    T1= 392.7 K, P1 = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point).
    T2= 298.15 K, P2= 1.052E-3 mmHg

    From the Clausius Clapeyron:

    ln(0.8045/1.052E-3)= ΔH/R*(1/298.15 K - 1/392.7 K)

    ΔH/R = 8221.89 K-1
    ΔH = 68361 J/mol

    ΔH/T = ΔS = 68361/298.15 = 229.29 J/mol K

    However the answer is S°(392.7) = 350.60 J/mol K.

    I tried adding the ΔSvap to the derived value (229.29+117.2) , however that only gives me 346.5 J/mol K
     
  2. jcsd
  3. Jun 2, 2014 #2
    Is the standard reference state of the material a vapor at 298K and a hypothetical pressure of 1 atm?

    Chet
     
  4. Jun 3, 2014 #3
    at 298K it is a solid
     
  5. Jun 3, 2014 #4
    Is 298K and 1 atm (solid) the standard reference state for this material? The vapor pressure relationship suggests that the heat of vaporization is constant, but yet there is a heat capacity given, implying that the heat of vaporization is not constant? Is there any value given for the heat of fusion? The entropy of vaporization...is that from liquid to vapor? At what temperature? There doesn't seem to be enough information given to solve this problem.

    Chet
     
  6. Jun 3, 2014 #5
    Yes this is the standard reference state of the material. Heat of vaporization is not constant.
    The heat of fusion is unknown (no measured data).

    The entropy of vaporization is from liquid to vapour at 460 K.

    There is also vapour pressure data for the solid-gas; Log10 (mmHg) = -3571/T + 18.279, with entropy of sublimation ΔSsub = 295 J/ mol K at 377 K.
     
  7. Jun 3, 2014 #6
    Is there any heat capacity data on the solid between 298 and 377?
    Chet
     
  8. Jun 4, 2014 #7
    yes, it is 155 + 0.086T - 1.8[itex]\times[/itex]106 T-2 (J/mol K)
     
  9. Jun 5, 2014 #8
    OK. You're starting out with solid at 298 and 1 atm, and your final state is liquid at 392.7 and 1 atm. You need to dream up a sequence of reversible processes that takes you from state 1 to state 2, and you need to calculate the change in entropy for each of these steps. The first two steps might be:

    1. solid at 298 and 1 atm ---> solid at 377 and 1 atm.
    2. solid at 377 and 1 atm ----> solid at 377 and equilibrium sublimation pressure at 377

    You need to be able to get the change in entropy over each step of the entire sequence.

    Can you think of the additional steps to complete the sequence?

    Chet
     
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