Most positive value of Delta S

[Awuah29]

Ok, here another chem question!
Why is the answer here #3 ? We going here from liquid to gas (exothermic)
Delta H (-) Delta S (+) spontaneous at all temp
Delta H (+) Delta S (-) Nonspontaneous at all temp ???

How do I tackle this question here ?
Thanks

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[llopop]

Okay first thing entropy is always increasing in the universe. As far as an individual reaction count the #of moles on each side. Whenever there is an increase in the #of moles there is a positive value for "Delta S". Another concept in this problem is that liquids have a lower state of order therefore going from a liquid to a gas would be a positive increase in "Delta S." The only answer that has both of these concepts is #3.

As far as your other question a +H,-S will be spontaneous under low temps. -H,+S will be spontaneous under all temps. Hope that helped

 

[Lonely Sol]

As far as your other question a +H,-S will be spontaneous under low temps. -H,+S will be spontaneous under all temps. Hope that helped

I always thought +H, -S = non spontaneous at all temp and
-H and -S, spontaneous only at low temp.

Someone correct me if I am wrong

 

[llopop]

My bad i think I wrote that out wrong. If G=H-TS then -H,-S would occur under low temps, +H,-S would not occur under any temps and -H,+S would occur at all temps.

 

[jackbauer!]

Okay first thing entropy is always increasing in the universe. As far as an individual reaction count the #of moles on each side. Whenever there is an increase in the #of moles there is a positive value for "Delta S". Another concept in this problem is that liquids have a lower state of order therefore going from a liquid to a gas would be a positive increase in "Delta S." The only answer that has both of these concepts is #3.

As far as your other question a +H,-S will be spontaneous under low temps. -H,+S will be spontaneous under all temps. Hope that helped

not necessarily true. you must count the # of moles of gas on each side.

d = (# moles of gas as products) - (# moles of gas as reactants)

if d > 0, then d(S) is + (i.e. higher entropy, less order)
if d < 0, then d(S) is - (i.e. lower entropy, more order)
if d = 0, then must look at # of moles of solids & liquids

hope this helps, jB!

 

[Awuah29]

not necessarily true. you must count the # of moles of gas on each side.

d = (# moles of gas as products) - (# moles of gas as reactants)

if d > 0, then d(S) is + (i.e. higher entropy, less order)
if d < 0, then d(S) is - (i.e. lower entropy, more order)
if d = 0, then must look at # of moles of solids & liquids

hope this helps, jB!

thanks Jack !

 

[jdmsamurai]

when i think of delta S think of total destruction and chaos...the world wants to be chaotic (less ordered)...if you have a book on the shelf...it wants to fall down into a pile......so if you tear apart a solid into a liquid...you increase the entropy...if you tear apart a liquid into a gas you increase the entropy because now this molecules are more chaotic......and if you tear two moles of gas into 3 moles of gas its more chaotic....imagine tearing apart a car into all its pieces....thats chaotic....aka increase of S

 

[BodybldgDoc]

going from liquid ( more ordered state ) ---> to gas ( less ordered state) increases the entropy. If you look at the answer choices only #3 does this.