Entropy and spontaneity

[Hysteria24]

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Kaplan states that:

According to the second law of thermodynamics, all spontaneous processes proceed such that the entropy (ΔS) of the system plus its surroundings (entropy of the universe) increases (+ΔS).

This makes sense since disorder is favored.

However it then shows this equation:

ΔG = ΔH – TΔS

And when using the equation a spontaneous reaction must have a negative ΔG.

It is possible to have a negative ΔG (making a reaction spontaneous), along with a decreasing entropy (-ΔS). However that contradicts the first statement on this post.

If ΔH is negative and T is small enough, then ΔG can be negative, with ΔS negative.

ie. -3 = (-5) - (1)(-2)

Does anyone still follow, and care to shed some light on this for me?

 

[cuttiedentist]

Kaplan states that:

According to the second law of thermodynamics, all spontaneous processes proceed such that the entropy (ΔS) of the system plus its surroundings (entropy of the universe) increases (+ΔS).

This makes sense since disorder is favored.

However it then shows this equation:

ΔG = ΔH – TΔS

And when using the equation a spontaneous reaction must have a negative ΔG.

It is possible to have a negative ΔG (making a reaction spontaneous), along with a decreasing entropy (-ΔS). However that contradicts the first statement on this post.

If ΔH is negative and T is small enough, then ΔG can be negative, with ΔS negative.

ie. -3 = (-5) - (1)(-2)

Does anyone still follow, and care to shed some light on this for me?

im not sure if im right but this is how i think:
if the enthalpy of the rxn is negative enough to make G negative, it means that the rxn is very exothermic which means it release heat. releasing heat increase the entropy of the universe
correct me if im wrong

 

[poc91nc]

G = U + pV - TS

Internal energy is a function of Volume and Entropy.

Good P-chemists Have Studied Under Very Awesome Teachers!

 

[saaz55]

the way i think about it is. its all dependent on temperature when it comes to the gray pasrts of this equation. in ur example, the temperature u subbed in for is low compared to the energy needed to make it nonspontaneous, therefore the gibbs free energy stayed as a negative value. you will also find out that when the delta h and delta s values are positive you need a high temperature value to compensate for the positive delta h value, thus making it spontaneous.

i think kaplan has that table that tells you when the rxn is spontaneous or what temperatures are needed to make it spontaneous. if ever in doubt, you can always make up ur own numbers and plug and chug, if you have the time that is.

 

[predentn]

Kaplan states that:

According to the second law of thermodynamics, all spontaneous processes proceed such that the entropy (ΔS) of the system plus its surroundings (entropy of the universe) increases (+ΔS).

This makes sense since disorder is favored.

However it then shows this equation:

ΔG = ΔH – TΔS

And when using the equation a spontaneous reaction must have a negative ΔG.

It is possible to have a negative ΔG (making a reaction spontaneous), along with a decreasing entropy (-ΔS). However that contradicts the first statement on this post.

If ΔH is negative and T is small enough, then ΔG can be negative, with ΔS negative.

ie. -3 = (-5) - (1)(-2)

Does anyone still follow, and care to shed some light on this for me?

The entropy of the universe (system + surroundings) always increases

 

[Lonely Sol]

G = U + pV - TS

Internal energy is a function of Volume and Entropy.

Good P-chemists Have Studied Under Very Awesome Teachers!

You're right poc, Pchem is eliteness!!

Anyways, you should look at third law of thermodynamics, which states that:
"As a system approaches absolute zero of temperature, all processes cease and the entropy of the system approaches a minimum value."

just think about this law and apply it to the equation G = H-TS, it should make sense!

 

[Hysteria24]

i think kaplan has that table that tells you when the rxn is spontaneous or what temperatures are needed to make it spontaneous. if ever in doubt, you can always make up ur own numbers and plug and chug, if you have the time that is.

It does, and that is my point. This chart disobeys the stated second law of thermodynamics.

So I guess if the system is spontaneous with a negative entropy, then some part of the universe must increase in entropy to off set (like predentn emphasized).

Could this be the heat released from the reaction acting elsewhere, like cuttiedentist suggested?

 

[poc91nc]

see my response below

 

[Hysteria24]

Honestly, I don’t know what those formulas are trying to say.

My original question is just that the second law of thermodynamics states that a spontaneous reaction (and surroundings) will have a positive entropy. But using the formulas I can create a hypothetical spontaneous reaction with a negative entropy.

That’s my problem.

 

[TechDoc]

I hope this helps.

if dH < 0 (Exothermic) and (dS < 0) then the reaction is spontaneous Only at low temps, if |T dS| < |dH|

 

[predentn]

Honestly, I don’t know what those formulas are trying to say.

My original question is just that the second law of thermodynamics states that a spontaneous reaction (and surroundings) will have a positive entropy. But using the formulas I can create a hypothetical spontaneous reaction with a negative entropy.

That’s my problem.

&#916;S(universe)=&#916;S(system) + &#916;S(surroundings)

&#916;G=&#916;H-T&#916;S(sys)

If &#916;S(sys)<0 then &#916;S(surr) is >0 and the law is NOT broken.

&#916;S(UNIVERSE) is always positive.

 

[jay47]

i think what they are tring to say is that although the entropy of the system may decrease, this doesn't mean that the reaction can't be spontaneous or that the entropy of the universe decreases. The last post said it best when they said that if the entropy of the system decreases, the entropy of the universe increases. Its all about the big picture and not so much about spontaniety. (I don't understand the calc up there but I'm sure it's right).

 

[poc91nc]

You can create "hypothetical" situations all you want...that's all they are...hypothetical...the REALITY is that the sign of gibb's free energy is dictated by the sign of entropy. Again, there is no contradiction.

Here is a deriviation I found in my gchem book:

delta S(univ) = delta S(sys) + delta S(surr)

At constant pressure, delta S(surr) = - delta H(sys)/T

delta S(univ) = delta S(sys) - delta H(sys)/T

- T delta S(univ) = delta H(sys) - T delta S(sys)

Delta G(sys) = delta H(sys) - T delta S(sys)

-T delta S(univ) = delta H(sys) - T delta S(sys) = delta G(sys)

See how they're related? Again...there is no contradiction. What you have proposed is merely hypothetical...the deriviation shows that your hypothetical situation isn't true...

Again...it's all about ENTROPY.

Another way of looking at it is from the definition of G

G = H - TS

dG = dH - TdS - SdT

Enthalpy is a function of pressure and entropy. However, spontaneity w/ regard to gibbs free energy is at constant T and P. Therefore, again...it's all about entropy.

If you still don't see that there isn't a contradiction...you should find a deriviation of Gibbs free energy in your chemistry textbook.

 

[JeffChou]

Kaplan states that:

According to the second law of thermodynamics, all spontaneous processes proceed such that the entropy (&#916;S) of the system plus its surroundings (entropy of the universe) increases (+&#916;S).

This makes sense since disorder is favored.

However it then shows this equation:

&#916;G = &#916;H &#8211; T&#916;S

And when using the equation a spontaneous reaction must have a negative &#916;G.

It is possible to have a negative &#916;G (making a reaction spontaneous), along with a decreasing entropy (-&#916;S). However that contradicts the first statement on this post.

If &#916;H is negative and T is small enough, then &#916;G can be negative, with &#916;S negative.

ie. -3 = (-5) - (1)(-2)

Does anyone still follow, and care to shed some light on this for me?

The problem that I see with that statement is that it says "i.e. the entropy of the universe" right after it says entropy of the surroundings. The equation that follows (&#916;Suniv = &#916;Ssys + &#916;Ssur) makes an important distinction between the entropy of the universe and the entropy of the system. When you look at the equation, it makes more sense. A restatement of the second law of thermodynamics from my text book reads: "In any spontaneous process there is always an increase in the entropy of the universe." That is, &#916;Suniv must be positive, but the entropies of the system or the surroundings can be negative (just not both). I think the Kaplan book may have just may have made a mistake.

Your confusion lies in the &#916;S's. In this equation: &#916;G = &#916;H &#8211; T&#916;S, the &#916;S is the entropy of the system, not of the surroundings, nor of the universe. As a side note, when there is no subscript, the &#916;S usually refers to the entropy of the system. This is the negative entropy value you have come up with.

In summary, what we have are two ways to determine spontaneity. Gibb's free energy is used when we have a reaction at an unchanging temperature and pressure (a requirement better understood if you look at the derivation). We must use &#916;Ssys (entropy of the system) to find spontaneity for Gibb's.

We also must have at least either the &#916;Ssur or &#916;Ssys positive to get a positive &#916;Suniv and, expectedly, a spontaneous reaction.

If, perchance, you have Zumdahl and Zumdahl's Chemistry, you will find the chapter on Spontanaeity, Entropy, and Free Energy quite helpful.

Hope that helps.
Jeff

 

[ucla2134]

Kaplan states that:

According to the second law of thermodynamics, all spontaneous processes proceed such that the entropy (?S) of the system plus its surroundings (entropy of the universe) increases (+?S).

This makes sense since disorder is favored.

However it then shows this equation:

?G = ?H &#8211; T?S

And when using the equation a spontaneous reaction must have a negative ?G.

It is possible to have a negative ?G (making a reaction spontaneous), along with a decreasing entropy (-?S). However that contradicts the first statement on this post.

If ?H is negative and T is small enough, then ?G can be negative, with ?S negative.

ie. -3 = (-5) - (1)(-2)

Does anyone still follow, and care to shed some light on this for me?

I use water to understand this concept:

H20 (l)<=>H20(g)

delta Ssurr+delta Ssys = delta Suniverse>=0
delta G = delta H &#8211; delta S.
deltaS system is positive when water liquid converted to gas. Even if delta S system is positive, the reaction cannot occur spontaneously when delta H is positive and at low Temperature.
deltaS system is negative when gas converted to liquid. Even if delta S system is negative, the reaction can still occur spontaneously when delta H is negative and at low Temperature.

Some people on previously post says it all depends on Temperature, but look at the equation, it depend both on the delta H system, T and delta S system.

 

[catinthehat1234]

I have a question...TBR asks "DeltaS for a gas-to-solid phase change is always"...and the answer is positive at all temperatures and pressures. This makes sense to me, because the order is increasing going from a gas to a liquid (disorder is decreasing). But another answer offers that it could be zero, negative, or positive depending on the temperature. I had a gut feeling that this was wrong, but I'm not sure why, considering that entropy depends on temperature? Thanks in advance!

 

[tommyngu]

I have a question...TBR asks "DeltaS for a gas-to-solid phase change is always"...and the answer is positive at all temperatures and pressures. This makes sense to me, because the order is increasing going from a gas to a liquid (disorder is decreasing). But another answer offers that it could be zero, negative, or positive depending on the temperature. I had a gut feeling that this was wrong, but I'm not sure why, considering that entropy depends on temperature? Thanks in advance!

Isn't a gas to solid transition exothermic with a negative entropy change?
For phase changes, entropy will decrease going from gas to liquid, liquid to solid, or gas to solid.
(and entropy would increase in the reverse direction)
As for the second part of your question, are you sure it's referring to entropy? It sounds more like it's referring to Gibb's.

 

[catinthehat1234]

Isn't a gas to solid transition exothermic with a negative entropy change?
For phase changes, entropy will decrease going from gas to liquid, liquid to solid, or gas to solid.
(and entropy would increase in the reverse direction)
As for the second part of your question, are you sure it's referring to entropy? It sounds more like it's referring to Gibb's.

Oh, yes, I meant to say negative at all temperatures and pressures. I mean the answer choice is "can be zero, positive, or negative, depending on the temperature"...so I'm wondering, essentially, if entropy is dependent on temperature?

 

[Lnguyen7]

Entropy is dependent on temperature. Entropy is increased when temperature is raised because of greater molecular motion.

 

[catinthehat1234]

Entropy is dependent on temperature. Entropy is increased when temperature is raised because of greater molecular motion.

Alright, but then why would the answer be "negative at all temperatures and pressures"?

 

[Lnguyen7]

Alright, but then why would the answer be "negative at all temperatures and pressures"?

First problem: "DeltaS for a gas-to-solid phase change is always"...and the answer is negative at all temperatures and pressures.

This statement really depends on what they meant by "all temperature". If "all temperature" meant ANY temperature, then the statement is invalid. What if the temperature was high enough such that no deposition occurred? This could result in a positive delta S. However, if all temperatures meant the range of temperature where deposition took place then deltaS is negative at all temperature in this range only.

Second problem: "DeltaS for a gas-to-solid phase change is always"... could be zero, negative, or positive depending on the temperature.
Third law says the entropy of a perfect crystalline substance is zero at 0 K. +/- delta S explained above.