Entropy Change at Boiling Point of Chloroform

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Discussion Overview

The discussion revolves around calculating the entropy change associated with the phase transition of chloroform from liquid to gas at its boiling point, as well as a related problem involving the condensation of methanol. The scope includes theoretical calculations and applications of thermodynamic principles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the heat of vaporization of chloroform and seeks help in applying it to calculate the entropy change for vaporization at the boiling point.
  • Another participant clarifies the state of chloroform at equilibrium and explains the relationship between Gibbs free energy, enthalpy, and entropy, suggesting a formula to calculate the change in entropy.
  • A third participant introduces a different problem involving the condensation of methanol and shares their calculated entropy change, seeking assistance with a subsequent question about the entropy of the liquid phase.
  • A later reply confirms the calculated entropy change for methanol and provides a method to find the entropy of the liquid using the relationship between the entropies of products and reactants.

Areas of Agreement / Disagreement

Participants generally agree on the methods to calculate entropy changes for the problems presented, but the discussion includes multiple distinct problems, and no consensus is reached on the specific calculations for the second problem involving methanol.

Contextual Notes

The calculations depend on the definitions of thermodynamic quantities and the specific conditions of the phase transitions. The assumptions regarding equilibrium and temperature are critical to the discussions but are not fully explored in every detail.

Who May Find This Useful

This discussion may be useful for students preparing for exams in thermodynamics or those interested in phase transitions and entropy calculations in chemistry and physics.

ACLerok
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The heat of vaporization of chloroform at its boiling point (61.2 degrees C) is 29.6*10^3 J/mol. What is the entropy change when 1.2 mol CHCl3 vaporizes at its boiling point?

I realize I am supposed to use the definition of entropy which is S=q/T but I don't know how the 1.2 mol of CHCl3 plays a part. Anyone care to help?
 
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To make things easier, it is best to describe what the state is of the chloroform at the beginning and at the end.
CHCl3(l,61.2 C, 1 atm) <------> (equilibrium) CHCl3(g, 61.2 C, 1 atm)


Since it is an equilibrium we are talking about DG=0 (delta G). You also know DG=DH-DS(T). So 0=DH-DS(T) or change in entropy=DH/T. In this case DH is just equal to DH_vap=29.6kJ/mol. So DS=(29.6kJ/mol)/342.1K . You can do the arithmetic. As you will see DS is in kJ/Kmol. DS is usually written in J/K so multiply the answer you get by your 1.2 mol.

A more interesting question would have been if your chloroform started at a different temperature than at the boiling point and the gas was heated to a temperature hotter than 61.2 C.
 
Alright, thanks a lot! I have one more problem before my exam in the morning.

The enthalpy change when liquid methanol, CH3OH, vaporizes at 25 degrees C is 37.4 kJ/mol. What is the entropy change when 1.5 mol of vapor in quilibrium with liquid condenses to liquid at 25 degrees C?

I was able to do this and i got -188.

Then it says that the entropy of the vapor at 25C is 252 J/molK. What is the entropy of the liquid at this temperature?

How do I do this second part? Thanks a lot!
 
Yes -188 J/K is right. You know DS for the entire process is -188J/K. But DS=Sum of S for Products-Sum S Reactants. So -188J/k=(1.5mol)(x)-(1.5mol)(252J/Kmol). X is the entropy of the liquid. Solve for x.
 

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