What is the Entropy of Vaporization for Water at 100 Degrees Celsius?

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SUMMARY

The entropy of vaporization (∆S) for water at 100 degrees Celsius can be calculated using the formula ∆S = ∆H / T. In this discussion, the enthalpy of vaporization (∆H) was determined to be 46.2 kJ/mol, which converts to 46200 J/mol. By substituting the temperature of 373.15 K into the equation, the resulting entropy of vaporization is calculated to be 123.8 J/(mol*K), indicating that the value of ∆S must be less than this threshold for the process to be spontaneous.

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  • Understanding of thermodynamic concepts such as enthalpy (∆H) and entropy (∆S).
  • Familiarity with the ideal gas law and phase changes in substances.
  • Basic knowledge of temperature conversions, particularly from Celsius to Kelvin.
  • Proficiency in algebraic manipulation of thermodynamic equations.
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  • Study the derivation and implications of the Gibbs free energy equation (∆G = ∆H - T∆S).
  • Learn about phase transitions and their thermodynamic properties in detail.
  • Explore the concept of spontaneity in chemical reactions and its relation to entropy.
  • Investigate the effects of temperature on the entropy of various substances.
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Homework Statement


"From the boiling point you found of ∆H(vaporization) at a 100 degrees celsius, find ∆S(vaporization) at 100 degrees celsius from the formula given (see eqn.)
This is for water.

Homework Equations



∆H-T∆S<0

The Attempt at a Solution



From the task before, I found ∆H(vap.) at 100 deg. celsius to be 372.1 K. Used the formula, and it gave me;

∆S>∆H/ T. That gave me ∆S>124.8 J/(mol*K)

Doesnt make much sense to me, is this correct or wrong?
 
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P.s. I meant ∆H(vap.) at 100 deg. celsius to be 46.2 kJ/ mol
Edit:

Tried: ∆H-T∆S<0

Foumd ∆H to be 46200 J/mol, and T=373,15 K. Gives me that;

46200J/mol - 373,15K*∆S<0

123.8 J/(mol*K)>∆S.

Still does not make any sense! Really confused.
 
Last edited:

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