FeaturedChallenge Thermochemistry Challenge Problem - Chet's Paradox

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1. May 4, 2017

Staff: Mentor

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.

Last edited: May 5, 2017
2. May 5, 2017

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.

3. May 5, 2017

Staff: Mentor

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.

Last edited: May 5, 2017
4. May 6, 2017

DrDu

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

5. May 6, 2017

Nidum

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

6. May 6, 2017

Staff: Mentor

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. May 6, 2017

Nidum

Interesting . Thanks .

8. May 10, 2017

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.

9. May 10, 2017

Staff: Mentor

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. May 10, 2017

Staff: Mentor

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

11. May 10, 2017

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?

12. May 10, 2017

DrDu

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. May 10, 2017

Staff: Mentor

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.