delta G equilibrium

[joonkimdds]

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Which of the following could be characteristic of a
reaction with a positive ΔH and a positive ΔS?
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 becuz both A and B are right.
I want to know why C cannot be the right answer.

In equilibrium, delta G = 0
that means delta H and Tdelta S are equal to each other.
if we assume that T = 1, that makes delta H = delta S.
so if delta H is 50, then delta S is 50, and both of them are positive.

The solution says that in equilibrium, delta S should be Zero.
I don't understand it.
delta G should be zero but I don't know why delta S should be zero too.

 

[FuzzyChicken]

At equilibrium there is a dynamic forward and reverse process happening equally at the same time, thus, there is no net disorder occurring and therefore entropy is zero.

I dunno if this helps, but that is how I think of it.

 

[tranv117]

At equilibrium there is a dynamic forward and reverse process happening equally at the same time, thus, there is no net disorder occurring and therefore entropy is zero.

I dunno if this helps, but that is how I think of it.

At equilibrium The Net Free Energy (deltaG) is equal to zero. This means that deltaH=TdeltaS. Im not sure about this, but I remember there being some question where it asks about equilibrium also. Something like at equilibrium, deltaH/TdeltaS=1 since theyre equal. i dont get how deltaS can be 0, this would make the above equation equal to 0 or be undefined... To make this statement true, even if deltaH and deltaS are postive, the T is the factor that determines whether or not the reaction is at equilibrium. The example of T=1 is kind of hard to explain because T= 1K is very close to 0K which would mean that the entropy is 0.

I understand that at equilibrium the reaction proceeding forward and backwards is constant. But it doesnt mean that there are equal amounts of reactants going forward as products going backwards. quick example...4 moles of reactant forwards and 2 moles of product backwards. The net change in entropy is constant (won't change) BUT doesn't mean it is equal to zero...

Anyone else care to explain this?

 

[GCT]

There is no net reaction at equilibrium ; enthalpy and entropy both describe an overall process.

 

[joonkimdds]

At equilibrium there is a dynamic forward and reverse process happening equally at the same time, thus, there is no net disorder occurring and therefore entropy is zero.

I dunno if this helps, but that is how I think of it.

if we think that delta S is zero,
delta G = delta H - TdeltaS
delta G = delta H
0 = delta H (at equilibrium)

I don't think that's how it works though.


Just like what tranv117 said, I saw many questions saying
deltaH/TdeltaS = 1 because they should be the same.
if we set all of them equal to zero
zero/zero = supposed to be undefined in mathmetical term, not 1.

 

[joonkimdds]

let's take a vote.
yay or nay?

 

[hersheyslove]

if we think that delta S is zero,
delta G = delta H - TdeltaS
delta G = delta H
0 = delta H (at equilibrium)

I don't think that's how it works though.


Just like what tranv117 said, I saw many questions saying
deltaH/TdeltaS = 1 because they should be the same.
if we set all of them equal to zero
zero/zero = supposed to be undefined in mathmetical term, not 1.

Your analogy does not make any sense at all since you cannot set them 0 yourself. You have to learn the definition of delta G.
As to the questions, it cannot be at the equilibrium because the delta H and delta S have opposite sign, which is goes against the definition of the the delta G at equilibrium.

 

[tranv117]

Anyways, its just like the answer explained. At equilibrium, deltaS=0. so reaction cannot be at equilibrium since we're told deltaS is positive. Just something you gotta know.

 

[joonkimdds]

Your analogy does not make any sense at all since you cannot set them 0 yourself. You have to learn the definition of delta G.
As to the questions, it cannot be at the equilibrium because the delta H and delta S have opposite sign, which is goes against the definition of the the delta G at equilibrium.

I am not the one who set them equal Zero. it's the Kaplan.
So are you agreeing with me that Kaplan is wrong?