Entropy Change at Boiling Point of Chloroform

[ACLerok]

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?

 

[gravenewworld]

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.

 

[ACLerok]

Alright, thanks alot! 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 alot!

 

[gravenewworld]

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.