Standard Entropy for Liquid at Melting Point

AI Thread Summary
The discussion focuses on calculating the standard entropy of a liquid at its melting point, determined to be 392.7 K. The vapor pressure equation provided is used to find the pressure of the liquid at this temperature, resulting in a value of 0.8045 mmHg. Using the Clausius-Clapeyron equation, the enthalpy change (ΔH) is calculated as 68361 J/mol, leading to an initial entropy value (ΔS) of 174.07 J/mol K. However, this calculated entropy is noted to be approximately half of the expected value, 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. The discussion highlights the need to incorporate these factors for an accurate calculation of standard entropy.
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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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