Thermochemistry Challenge Problem - Chet's Paradox

[Chestermiller]

I have a reversible chemical reaction described by the balanced equation: $aA+bB=cC+dD$. I devise a reversible process to take a closed system containing these species (and its surroundings) from thermodynamic equilibrium state 1 to thermodynamic equilibrium state 2:

State 1: a moles of pure A and b moles of pure B at their standard states of 298 K and 1 bar

State 2: c moles of pure C and d moles of pure D at their standard states of 298 K and 1 bar

For each of the four species, I have data on the entropies of formation $S_f^0$ and the heats of formation at the standard states $H_f^0$. In the process, the four species can be considered to behave as ideal gases.

What is the change in entropy of my system for the reversible process I have devised?
What is the change in entropy of the surroundings for the reversible process I have devised?
What is the change in entropy of the universe for the reversible process I have devised?
What could the reversible process be that I have devised?
If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?
Why is the change in entropy of the surroundings not equal to minus the change in enthalpy of the system divided by the temperature?

Good luck. It took me 2 days to work this out.

 

[DrDu]

So, as nobody seems to answer after one day has passed I dare to post my solution:
Q:What is the change in entropy of my system for the reversible process I have devised?
$\Delta S= \sum_i S_{fi}^0\nu_i$, as S is a state function.

Q: What is the change in entropy of the surroundings for the reversible process I have devised?
$-\Delta H/T= -1/T\; \sum_i H_{fi}^0\nu_i$, as H is a state function.

Q: What is the change in entropy of the universe for the reversible process I have devised?
I have serious doubts that the entropy of the universe is a well defined physical quantity. :-)
But as the reaction is reversible, the change in total entropy of, say the system and a large enough environment is 0.

Q: What could the reversible process be that I have devised?
A: I would use a van't Hoff apparatus, i.e., mix the components via semipermeable membranes.

Q: If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?
Why is the change in entropy of the surroundings not equal to minus the change in enthalpy divided by the temperature?

The point is that before mixing the components have to be brought from standard pressure to the equilibrium partial pressure they have in the reaction chamber and the products back to standard pressure. In these steps entropy changes but H remains constant as for ideal gasses H(T,p)=H(T) . But as p isn't constant, neither is G, despite the expansion being reversible, i.e., $\Delta H=0 \ne T\Delta S$.
The change in entropy of each gas in this step is $nRT \ln (p/p_0)$. Although $\Delta H=0$, $Q\ne 0$, rather, $Q= \int p dV$, the volume work done during expansion which has to be compensated by a heat flux to keep internal energy and temperature constant.

Overall $\Delta G=-T\Delta S_\mathrm{mix}= -\sum_i n_i RT \ln(p_i/p_0)=-\xi RT \ln K$ where K is the equilibrium constant and $\xi =n_i/\nu_i$. This is also the mixing entropy which isn't compensated for by a change in enthalpy.

 

[Chestermiller]

So, as nobody seems to answer after one day has passed I dare to post my solution:
Q:What is the change in entropy of my system for the reversible process I have devised?
$\Delta S= \sum_i S_{fi}^0\nu_i$, as S is a state function.

Q: What is the change in entropy of the surroundings for the reversible process I have devised?
$-\Delta H/T= -1/T\; \sum_i H_{fi}^0\nu_i$, as H is a state function.

Q: What is the change in entropy of the universe for the reversible process I have devised?
I have serious doubts that the entropy of the universe is a well defined physical quantity. :-)
But as the reaction is reversible, the change in total entropy of, say the system and a large enough environment is 0.

Q: What could the reversible process be that I have devised?
A: I would use a van't Hoff apparatus, i.e., mix the components via semipermeable membranes.

Q: If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?
Why is the change in entropy of the surroundings not equal to minus the change in enthalpy divided by the temperature?

The point is that before mixing the components have to be brought from standard pressure to the equilibrium partial pressure they have in the reaction chamber and the products back to standard pressure. In these steps entropy changes but H remains constant as for ideal gasses H(T,p)=H(T) . But as p isn't constant, neither is G, despite the expansion being reversible, i.e., $\Delta H=0 \ne T\Delta S$.
The change in entropy of each gas in this step is $nRT \ln (p/p_0)$. Although $\Delta H=0$, $Q\ne 0$, rather, $Q= \int p dV$, the volume work done during expansion which has to be compensated by a heat flux to keep internal energy and temperature constant.

Overall $\Delta G=-T\Delta S_\mathrm{mix}= -\sum_i n_i RT \ln(p_i/p_0)=-\xi RT \ln K$ where K is the equilibrium constant and $\xi =n_i/\nu_i$. This is also the mixing entropy which isn't compensated for by a change in enthalpy.

This is all entirely correct, except for the answer to question 2. The change in entropy of the surroundings for this reversible process is minus the change in entropy of the system. In your answer to question 4, you showed why this is so. The change in entropy between the entrance and exit of the equilibrium box is equal to $1/T\; \sum_i H_{fi}^0\nu_i$, since $\Delta G=0$. But there is additional entropy change at constant enthalpy in the first and last steps.

The paradox for me was "if the pressure is constant, how could the standard change in entropy not be equal to the change in enthalpy divided by T?" What you showed (and what I figured out) was that the initial and final steps in the process were not at constant pressure.

By the way, you required way less time than I did figuring all this out. Congrats.

 

[DrDu]

This is all entirely correct, except for the answer to question 2. The change in entropy of the surroundings for this reversible process is minus the change in entropy of the system.

Thank's Jeff,
you are obviously right. I mixed this up.

 

[Nidum]

Just an engineers curiosity - is it possible to represent this type of process on an enthalpy/entropy diagram ?

 

[Chestermiller]

Just an engineers curiosity - is it possible to represent this type of process on an enthalpy/entropy diagram ?

It can be done for the total enthalpy and entropy. But, it won't be unique, except for the initial and final end points. This is because the intermediate values will depend on the choice of chemical equilibrium state in the van't Hoff reactor box.

 

[Nidum]

Interesting . Thanks .

 

[Mark Harder]

Perhaps it's just a misunderstanding of what you mean by "equilibrium", but I can't think of a chemical example in which there can be 2 states consisting of pure reactants on the one hand and pure products on the other, and each is in equilibrium. Either the equilibrium state lies between these two extremes, or either pure products or pure reactants are equilibrium states. For example, 2moles of H₂O is an equilibrium state, but 2H₂ + O₂ is not.

 

[Chestermiller]

Perhaps it's just a misunderstanding of what you mean by "equilibrium", but I can't think of a chemical example in which there can be 2 states consisting of pure reactants on the one hand and pure products on the other, and each is in equilibrium. Either the equilibrium state lies between these two extremes, or either pure products or pure reactants are equilibrium states. For example, 2moles of H₂O is an equilibrium state, but 2H₂ + O₂ is not.

I guess it is misunderstanding. I said that in the initial thermodynamic equilibrium state, the reactants were pure and, in the final equilibrium state, the products were pure. Certainly, a pure substance can be in a thermodynamic equilibrium state. In my question, I didn't say anything about the process for transitioning from the first state to the second state.

 

[mfb]

They have to be separated, otherwise they cannot be in thermodynamic equilibrium, but the problem statement doesn't exclude that option.

 

[I like Serena]

Now I'm confused.
If 2 moles of $H_2$ and 1 mole of $O_2$ are perfectly mixed (at maximum entropy), they are in thermal equilibrium are they not?

 

[DrDu]

Now I'm confused.
If 2 moles of $H_2$ and 1 mole of $O_2$ are perfectly mixed (at maximum entropy), they are in thermal equilibrium are they not?

In thermal equilibrium maybe but not necessarily in thermodynamic equilibrium as they can react with each other.
At normal temperatures and without a catalyst, this reaction takes very long time, so you may rather call this a quasi-equilibrium.

 

[Chestermiller]

Now I'm confused.
If 2 moles of $H_2$ and 1 mole of $O_2$ are perfectly mixed (at maximum entropy), they are in thermal equilibrium are they not?

When I originally posed this problem, I made the conscious decision not to bother mentioning that, in the initial and final states, the pure reactants and the pure products are in separate containers. Apparently this has caused confusion. Sorry about that, chief. Anyway, that's what I meant.

 

[Amin2014]

I have a reversible chemical reaction described by the balanced equation: $aA+bB=cC+dD$. I devise a reversible process to take a closed system containing these species (and its surroundings) from thermodynamic equilibrium state 1 to thermodynamic equilibrium state 2:

State 1: a moles of pure A and b moles of pure B at their standard states of 298 K and 1 bar

State 2: c moles of pure C and d moles of pure D at their standard states of 298 K and 1 bar

For each of the four species, I have data on the entropies of formation $S_f^0$ and the heats of formation at the standard states $H_f^0$. In the process, the four species can be considered to behave as ideal gases.

What is the change in entropy of my system for the reversible process I have devised?
What is the change in entropy of the surroundings for the reversible process I have devised?
What is the change in entropy of the universe for the reversible process I have devised?
What could the reversible process be that I have devised?
If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?
Why is the change in entropy of the surroundings not equal to minus the change in enthalpy of the system divided by the temperature?

Good luck. It took me 2 days to work this out.

I have only just read the question and have a question myself. How are you able to fully carry out the reaction in a closed system? Some of the reactants will still be left.

 

[Chestermiller]

I have only just read the question and have a question myself. How are you able to fully carry out the reaction in a closed system? Some of the reactants will still be left.

You are underestimating the complexity of the reversible process required to convert pure A and pure B to pure C and pure D. Like I asked you in the other thread, are you familiar with the concept to the Van't Hoff equilibrium box?

 

[Amin2014]

What is the change in entropy of my system for the reversible process I have devised?

## ΔS = (Sc + Sd) - (Sa +Sb)

What is the change in entropy of the surroundings for the reversible process I have devised?

ΔS (surr) = (Sa + Sb) - (Sc + Sd)

What is the change in entropy of the universe for the reversible process I have devised?

Zero

What could the reversible process be that I have devised?

I don't know. But you seem to have non P-V work going on. Maybe it's being used to keep the products separate from reactants so as to make the reaction complete?

If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?

Because ΔH = q is only valid when you don't have non P-V work

Why is the change in entropy of the surroundings not equal to minus the change in enthalpy of the system divided by the temperature?

Same reason as previous question

Good luck. It took me 2 days to work this out.[/QUOTE]

 

[Chestermiller]

## ΔS = (Sc + Sd) - (Sa +Sb)

ΔS (surr) = (Sa + Sb) - (Sc + Sd)

Zero

I don't know. But you seem to have non P-V work going on. Maybe it's being used to keep the products separate from reactants so as to make the reaction complete?

Because ΔH = q is only valid when you don't have non P-V work

Same reason as previous question

Good luck. It took me 2 days to work this out.

[/QUOTE]
To see what the reversible process actually looks like for a closed system, see pages 506-507 of Smith and van Ness, Introduction to Chemical Engineering Thermodynamics.

 

[Amin2014]

You are underestimating the complexity of the reversible process required to convert pure A and pure B to pure C and pure D. Like I asked you in the other thread, are you familiar with the concept to the Van't Hoff equilibrium box?

No I am not familiar with no dude's box lol

 

[Amin2014]

To see what the reversible process actually looks like for a closed system, see pages 506-507 of Smith and van Ness, Introduction to Chemical Engineering Thermodynamics.

I will read into it. Surely the short answer to the last two questions is because your process involves non P-V work?

Δ H = W* + TΔS
Δ G = W*

 

[Chestermiller]

I will read into it. Surely the short answer to the last two questions is because your process involves non P-V work?

Δ H = W* + TΔS
Δ G = W*

Why don’t you wait and read it before you reach any conclusions?

 

[James Pelezo]

In your question 'If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?' is the reader to presume the system is not at equilibrium but in transition to the reversible state or is it already at equilibrium, but in a closed system where expansion or compression is not a factor? I'm thinking if the process is in transition then ΔH ≠ TΔS (spontaneity factors would be in play) or, if already at equilibrium then ΔG = 0 = ΔH - TΔS => ΔH = TΔS which contradicts the question's objective. Very interesting questions. Appreciate any clarifications you are willing to share. Thx

 

[Chestermiller]

In your question 'If the pressures in states 1 and 2 are the same, why is the change in entropy of the system not equal to the change in enthalpy divided by the temperature?' is the reader to presume the system is not at equilibrium but in transition to the reversible state or is it already at equilibrium, but in a closed system where expansion or compression is not a factor? I'm thinking if the process is in transition then ΔH ≠ TΔS (spontaneity factors would be in play) or, if already at equilibrium then ΔG = 0 = ΔH - TΔS => ΔH = TΔS which contradicts the question's objective. Very interesting questions. Appreciate any clarifications you are willing to share. Thx

The initial state and the final state are thermodynamic equilibrium states. In between, there is a reversible process that transitions the system from the initial thermodynamic equilibrium state to the final thermodynamic equilibrium state. You will note that DrDu was able to figure out the answer to my question.

 

[James Pelezo]

Thank you.