Standard Entropy for Liquid at Melting Point

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SUMMARY

The standard entropy of a liquid at its melting point of 392.7 K was calculated using the Clausius-Clapeyron equation. The vapor pressure of the liquid at this temperature was determined to be 0.8045 mmHg. The calculated ΔH was found to be 68361 J/mol, leading to an initial entropy value of 174.07 J/mol K. However, this value was identified as approximately half of the expected result, prompting the consideration of additional variables such as the specific heat capacity (Cp) of 250 J/mol K and the entropy of vaporization at 117 J/mol K.

PREREQUISITES
  • Understanding of the Clausius-Clapeyron equation
  • Familiarity with thermodynamic concepts such as ΔH and ΔS
  • Knowledge of vapor pressure calculations
  • Basic principles of entropy and its calculation
NEXT STEPS
  • Review the Clausius-Clapeyron equation in detail
  • Study the relationship between heat capacity (Cp) and entropy
  • Investigate the concept of entropy of vaporization
  • Explore advanced thermodynamic calculations involving phase transitions
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Chemistry students, thermodynamics researchers, and professionals involved in phase transition studies will benefit from this discussion.

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Homework Statement


The vapour pressures of a liquid have been measured and fit to the following equation:
Log10 (mmHg) = -3571/T + 6.124
The melting point has been determined to be 392.7 K.
Calculate the standard entropy of the liquid at the melting point.


Homework Equations





The Attempt at a Solution

 
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My attempt was as follows:
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
ΔH = 68361 J/mol

ΔH/T = ΔS = 68361/392.7 = 174.07 J/mol K

However the answer is about twice this value and a Cp value given for the liquid is 250 J/mol K and the entropy of vaporisation is 117 J/mol K (2 additonal variables)
 

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