Question

The standard entropy change for 1.00 moles of Ar as it evaporates at -185.7 °C (∆H°-185.7 = 1558 cal mol-1) is?

1. 17.8 J mol-1K-1 2. 87.5 J mol-1K-1 3. 15.0 J mol-1K-1
4. 46.0 J mol-1K-1 5. 74.5 J mol-1K-1 6. none of the previous answers

I thought I would use deltaG=deltaH-TdeltaS, but I don't have deltaG, what do I do, or is deltaG zero because it is standard? Thank you. Or do I just used a table of entropy values to calculate this because it is standard?

Answer 1

I think I can answer your question but your phrasing is bad. You don't ask a question, really, and that word "is" is out of place.
Delta G is zero but not because it is standard but because it is in equilibrium at the boiling point (I assume that is the b.p. for Ar). Delta G is always zero when the system is at equilibrium such as the b.p. or m.p.

Answer 2

Sorry, but I didn't write the question, my teacher did.
I did deltaH/T=deltaS, and I got 17.81cal/Kmol
Is the conversion factor for cal to J 4.1858? If so I used that and got 74.5 (answer 5?)