Which of the options could be characteristic of a rxn with + deltaH & + deltaS?

[FROGGBUSTER]

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Which of the options could be characteristic of a rxn with + deltaH & + deltaS?

A. The reaction is spontaneous
B. The reaction is nonspontaneous
C. The reaction is at equilibrium
D. The reaction is exothermic.
E. Two of the above


The answer is E, and the solutions indicate that A & B could be both true, which I agree with.

But what about C? The reaction is at equilibrium when deltaG = 0, and it's possible for delta G to be zero with a + deltaH & + deltaS.

Am I thinking about this incorrectly, or is this a mistake?

 

[crax]

the obvious response would be dG has no bearing on rate. the midpoint between spontaneous and non spontaneous is dG = 0, this doesn't seem to contribute to being in equilibrium between products and reactants.

 

[rah08e]

Which of the options could be characteristic of a rxn with + deltaH & + deltaS?

A. The reaction is spontaneous
B. The reaction is nonspontaneous
C. The reaction is at equilibrium
D. The reaction is exothermic.
E. Two of the above


The answer is E, and the solutions indicate that A & B could be both true, which I agree with.

But what about C? The reaction is at equilibrium when deltaG = 0, and it's possible for delta G to be zero with a + deltaH & + deltaS.

Am I thinking about this incorrectly, or is this a mistake?

If a rxn has a +DeltaH and +DeltaS, it will only be non-spontaneous at low temps right? And I'm still confused about the exothermic thing. I thought that if a reaction has a -DeltaG its exothermic?

 

[LetsGo2DSchool]

If a rxn has a +DeltaH and +DeltaS, it will only be non-spontaneous at low temps right? And I'm still confused about the exothermic thing. I thought that if a reaction has a -DeltaG its exothermic?

Exothermic/endothermic is directly related to delta H. Delta G is about free energy and thus, spontaneous/nonspontaneous.

 

[flin5845]

If a rxn has a +DeltaH and +DeltaS, it will only be non-spontaneous at low temps right? And I'm still confused about the exothermic thing. I thought that if a reaction has a -DeltaG its exothermic?

you are getting dG and dH mixed up.

dG tells you if a reaction is spontaneous or non-spontaneous
dH tells you if a reaction is endothermic or exothermic
dS tells you the "disorder" of a reaction

So think about the equation. dG= dH- TdS

If dG is negative, dH could be positive (endothermic) and if dS was positive, then under HIGH temps, the reaction could be spontaneous.

 

[flin5845]

Which of the options could be characteristic of a rxn with + deltaH & + deltaS?

A. The reaction is spontaneous
B. The reaction is nonspontaneous
C. The reaction is at equilibrium
D. The reaction is exothermic.
E. Two of the above


The answer is E, and the solutions indicate that A & B could be both true, which I agree with.

But what about C? The reaction is at equilibrium when deltaG = 0, and it's possible for delta G to be zero with a + deltaH & + deltaS.

Am I thinking about this incorrectly, or is this a mistake?

I would say it is possible.

if dG=0 then dH=TdS. I am just picking random numbers but if dH was 2 and dS was 2 and temp was 1 then 2=1(2) and they are equal.

I am not sure if you can think about it that way, but it satisfies the answer. But then again I do not know if you can "assume" this.

 

[FROGGBUSTER]

I would say it is possible.

if dG=0 then dH=TdS. I am just picking random numbers but if dH was 2 and dS was 2 and temp was 1 then 2=1(2) and they are equal.

I am not sure if you can think about it that way, but it satisfies the answer. But then again I do not know if you can "assume" this.

Yeah that was my thought process too. I don't know how it's an assumption though, delta G = 0 at equilibrium so it seems to be pretty cut and dry to me that it could work. I guess most of you are agreeing?

 

[flin5845]

There was another post about this, and they said that at EQ deltaS=0. Since the forward rate= reverse rate, the NET change in entropy is zero. But the question states that dS is positive, we can not say that the rxn is at EQ.

 

[FROGGBUSTER]

There was another post about this, and they said that at EQ deltaS=0. Since the forward rate= reverse rate, the NET change in entropy is zero. But the question states that dS is positive and therefore we can not say that the rxn is at EQ.

Do you have a link to that thread?

 

[flin5845]

http://forums.studentdoctor.net/showthread.php?t=556869

 

[flin5845]

Here is what wiki has to say,

http://en.wikipedia.org/wiki/Thermodynamic_equilibrium


For a completely isolated system, ΔS = 0 at equilibrium.

 

[HannibalLecter]

Ok frogger just for your benefit and I am strictly looking out for you on this:

Stick to the F8cking graph dude, if you start looking at the deltaG = deltaH ..... equation for this, I think you will be wasting a lot of time and brain power on the test that takes like 2 seconds to solve. It literally took me 15 secs to solve this problem.

Just memorize the four quadrants and you will be golden. This is how in depth the real DAT is and don't try to memorize extraordinary steps to solve your answer man.

P.S. I know that equation is important to know for the test but don't use it on this problem.

 

[FROGGBUSTER]

Here is what wiki has to say,

http://en.wikipedia.org/wiki/Thermodynamic_equilibrium


For a completely isolated system, ΔS = 0 at equilibrium.

Thanks for looking into this for me. I think Kaplan may be wrong here though, especially because it doesn't mention anything about it being a compleely isolated system.

The solution references delta S = 0, but now I'm thinking it just meant delta G=0.

Ok frogger just for your benefit and I am strictly looking out for you on this:

Stick to the F8cking graph dude, if you start looking at the deltaG = deltaH ..... equation for this, I think you will be wasting a lot of time and brain power on the test that takes like 2 seconds to solve. It literally took me 15 secs to solve this problem.

Just memorize the four quadrants and you will be golden. This is how in depth the real DAT is and don't try to memorize extraordinary steps to solve your answer man.

P.S. I know that equation is important to know for the test but don't use it on this problem.

Thanks, I know where you're coming from but I think it's important to ask as many questions as you can when you're studying for the test.

 

[flin5845]

So I was really confused when I found that other thread about ΔS=0, because I never really looked at it that way. So I asked Chad about that question, and he said that C is in fact FALSE.

Here is what Chad said:

That is indeed correct. If a rxn has reached equilibrium, then ΔG=0, ΔH=0, and ΔS=0 at that point. We commonly only look at ΔG=0 for equilibrium but the other two are just as true.

Hope this helps!

I never realized that at equilibrium they are all equal to zero.

 

[FROGGBUSTER]

So I was really confused when I found that other thread about ΔS=0, because I never really looked at it that way. So I asked Chad about that question, and he said that C is in fact FALSE.

Here is what Chad said:

That is indeed correct. If a rxn has reached equilibrium, then ΔG=0, ΔH=0, and ΔS=0 at that point. We commonly only look at ΔG=0 for equilibrium but the other two are just as true.

Hope this helps!

I never realized that at equilibrium they are all equal to zero.

Awesome stuff, thanks flin.

 

[budda10000]

Which of the options could be characteristic of a rxn with + deltaH & + deltaS?

A. The reaction is spontaneous
B. The reaction is nonspontaneous
C. The reaction is at equilibrium
D. The reaction is exothermic.
E. Two of the above


The answer is E, and the solutions indicate that A & B could be both true, which I agree with.

But what about C? The reaction is at equilibrium when deltaG = 0, and it's possible for delta G to be zero with a + deltaH & + deltaS.

Am I thinking about this incorrectly, or is this a mistake?

The reaction is endolthermic and it could be spon. at high temp and nonspon at low temps.

 

[wired202808]

Ok frogger just for your benefit and I am strictly looking out for you on this:

Stick to the F8cking graph dude, if you start looking at the deltaG = deltaH ..... equation for this, I think you will be wasting a lot of time and brain power on the test that takes like 2 seconds to solve. It literally took me 15 secs to solve this problem.

Just memorize the four quadrants and you will be golden. This is how in depth the real DAT is and don't try to memorize extraordinary steps to solve your answer man.

P.S. I know that equation is important to know for the test but don't use it on this problem.

what chart are you referring to?