Entropy in Reversible Reactions

[SuperSaiyan3]

Isn't the entropy (S) of REVERSIBLE reactions supposed to be ZERO?

That is what Kaplan was continuously preaching in my practice passages. Then when I came across a FL test, I got the answer to this question wrong.

CO2(g) + C(s) (equilibrium sign --> and <-- on top of each other) 2 CO(g)

What can be said about the value of &#8710;S° of the
reaction?
A. It is positive.
B. It is negative.
C. It is zero.
D. It cannot be determined from the information
given.

I chose C, since it was a reversible reaction, as all equilibrium reactions are.

If you can give your $0.02 on this, that would be great. Thanks.

---

Members don't see this ad.

 

[matth87]

First look at the # of moles. We see 2 moles going to 2 moles so we can't determine anything from that.

If 2 moles went to 1 mole entropy would be negative because it became more organized.

We can then look at it going from a (s) to a (g) as it changes phases the molecules gain kinetic energy and become more disordered. From that we could say the entropy is positive.

Thats my 2 cents. I would go with A

 

[SuperSaiyan3]

First look at the # of moles. We see 2 moles going to 2 moles so we can't determine anything from that.

If 2 moles went to 1 mole entropy would be negative because it became more organized.

We can then look at it going from a (s) to a (g) as it changes phases the molecules gain kinetic energy and become more disordered. From that we could say the entropy is positive.

Thats my 2 cents. I would go with A

but MattH, when would you apply that concept: "entropy change in reversible reactions is zero"? Would an equilibrium reaction qualify as a reversible reaction?

... and ya the answer is A. congrats.

 

[G1SG2]

but MattH, when would you apply that concept: "entropy change in reversible reactions is zero"? Would an equilibrium reaction qualify as a reversible reaction?

... and ya the answer is A. congrats.

Most reactions are reversible, even if it is to a small degree...the entropy change can't be 0, as that is the driving force of the reaction (creating more disorder). Unless there's something I'm missing here. Just look at which side contains more moles of gas.

 

[SuperSaiyan3]

ya that's what I thought as well. I guess it's just something that you'd apply to generic situations. Not when you have an equation laid out for you.

 

[neil100]

Alright here goes it!

The question asks delta S. Meaning change in entropy. So entropy of forward - entropy of reverse rxn.

The entropy change in forward rxn. is clearly positive since we are going from one mole of gas to two moles of gas and another reactant is a solid.

For the reverse, we are going from 2 moles of gas to one mole of gas and a solid hence it is NEGATIVE.

so, Delta S = + - (-) = + number. Hence the change in entropy is positive.

 

[matth87]

but MattH, when would you apply that concept: "entropy change in reversible reactions is zero"? Would an equilibrium reaction qualify as a reversible reaction?

... and ya the answer is A. congrats.

I have never heard that before. Sorry. I wouldn't know when to apply it.

General Rule of Thumb:
Breaking Bonds - Increases Disorder - Positive Entropy
Changing Phases - (s) --> (l) --> (g) - Increases Disorder - Positive Entropy

The opposite of these would be a Negative Entropy

 

[Podalarius]

A reversible thermodynamic process has an change in entropy of zero. They don't exist in real life, but are useful as a benchmark for efficiency of real processes. The best theoretical example is the isentropic expansion/compression of an ideal gas, along the curve pV^K = constant, where K = Cp/Cv [5/3 for a monoatomic gas, 7/5 for a diatomic gas].

A reversible chemical reaction just has an equilibrium condition, like the reaction described here.

 

[Podalarius]

Alright here goes it!

The question asks delta S. Meaning change in entropy. So entropy of forward - entropy of reverse rxn.

The entropy change in forward rxn. is clearly positive since we are going from one mole of gas to two moles of gas and another reactant is a solid.

For the reverse, we are going from 2 moles of gas to one mole of gas and a solid hence it is NEGATIVE.

so, Delta S = + - (-) = + number. Hence the change in entropy is positive.

That's the right idea, but it's not exactly right. The bold statement and the underlined statement are the exact same statement (and would be for any reversible reaction), so you'd just be doubling any effect going in one direction. Your definition of delta S is correct - it is the difference of the products and reactants, but that's calculated using the standard molar entropies and correcting for changes in volume/pressure/temperature.

Without that information, you can still use the statement in bold to qualitatively assess the system.

 

[wanderer]

Isn't the entropy (S) of REVERSIBLE reactions supposed to be ZERO?

That is what Kaplan was continuously preaching in my practice passages. Then when I came across a FL test, I got the answer to this question wrong.

CO2(g) + C(s) (equilibrium sign --> and <-- on top of each other) 2 CO(g)

What can be said about the value of &#8710;S° of the
reaction?
A. It is positive.
B. It is negative.
C. It is zero.
D. It cannot be determined from the information
given.

I chose C, since it was a reversible reaction, as all equilibrium reactions are.

If you can give your $0.02 on this, that would be great. Thanks.

The entropy of the universe is zero for reversible reactions (well not really, but I believe this is what you are getting stuck on), not for the system. Delta S here refers to the system, which is positive since you have more moles of gas on the right side.

 

[neil100]

The entropy of the universe is zero for reversible reactions (well not really, but I believe this is what you are getting stuck on), not for the system. Delta S here refers to the system, which is positive since you have more moles of gas on the right side.

👍 for the bold term.

 

[neil100]

That's the right idea, but it's not exactly right. The bold statement and the underlined statement are the exact same statement (and would be for any reversible reaction), so you'd just be doubling any effect going in one direction. Your definition of delta S is correct - it is the difference of the products and reactants, but that's calculated using the standard molar entropies and correcting for changes in volume/pressure/temperature.

Without that information, you can still use the statement in bold to qualitatively assess the system.

Well yes. We have no numbers to go by. You could make them up based on the # of moles and state of matter. All in all, it is a qualitative way to think about it because it's the only way in this case.