Thermochemistry Challenge Problem - Chet's Paradox

In summary: This is the essence of a reversible process. The change in entropy of the surroundings is just a convenient bookkeeping device to keep track of the flow of heat into or out of the system.In summary, the reversible process devised involves mixing the components using a van't Hoff apparatus. The change in entropy of the system can be calculated using the entropies of formation and the number of moles of each species. The change in entropy of the surroundings is equal to the negative change in entropy of the system. The change in entropy of the universe is assumed to be 0 since the reaction is reversible. The mixing
  • #1
Chestermiller
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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.
 
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  • #2
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.
 
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  • #3
DrDu said:
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.
 
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  • #4
Chestermiller said:
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.
 
  • #5
Just an engineers curiosity - is it possible to represent this type of process on an enthalpy/entropy diagram ?
 
  • #6
Nidum said:
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.
 
  • #7
Interesting . Thanks .
 
  • #8
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.
 
  • #9
Mark Harder said:
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.
 
  • #10
They have to be separated, otherwise they cannot be in thermodynamic equilibrium, but the problem statement doesn't exclude that option.
 
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  • #11
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?
 
  • #12
I like Serena said:
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.
 
  • #13
I like Serena said:
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.
 
  • #14
Chestermiller said:
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.
 
  • #15
Amin2014 said:
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?
 
  • #16
Chestermiller said:
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]
 
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  • #17
Amin2014 said:
## 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.
 
  • #18
Chestermiller said:
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
 
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  • #19
Chestermiller said:
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*
 
  • #20
Amin2014 said:
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?
 
  • #21
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
 
  • #22
James Pelezo said:
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.
 
  • #23
Thank you.
 

1. What is the Thermochemistry Challenge Problem - Chet's Paradox?

The Thermochemistry Challenge Problem - Chet's Paradox is a thought-provoking problem in the field of thermochemistry that challenges our understanding of energy conservation and the second law of thermodynamics.

2. Who is Chet and why is this paradox named after him?

Chet is a fictional character created by scientists to present this paradox. The name "Chet" is derived from the word "chetana", which means consciousness in Sanskrit. The paradox revolves around the idea of consciousness and how it relates to energy conservation.

3. What is the paradox presented in Chet's Paradox?

The paradox is as follows: Chet is standing on a frictionless surface and is holding a weight in his hand. He first raises the weight to a certain height and then releases it. According to the law of conservation of energy, the potential energy of the weight should be converted into kinetic energy as it falls to the ground. However, Chet can also choose to focus his consciousness on the weight, causing it to remain suspended in mid-air without any change in its potential energy. This seems to defy the law of conservation of energy.

4. How do scientists explain Chet's Paradox?

Scientists have proposed various explanations for Chet's Paradox, but the most widely accepted one is the concept of "virtual work". This concept states that when Chet focuses his consciousness on the weight, he is actually performing a type of work on it, which cancels out the potential energy that would have been converted into kinetic energy. This allows for the law of conservation of energy to be upheld.

5. Why is Chet's Paradox important in the field of thermochemistry?

Chet's Paradox challenges our understanding of energy conservation and the second law of thermodynamics, which are fundamental principles in the study of thermochemistry. It also highlights the complex relationship between consciousness and energy, and the need for further research in this area. Understanding and solving this paradox could lead to advancements in our understanding of energy transfer and conservation in various systems.

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