1st law of thermodynamics with chemical reaction?

In summary, the author is saying that the equations for a closed system involving chemical reaction should be written in terms of differentials, not sums. The first law for an isolated system should be written in terms of sums instead.
  • #1
Amin2014
113
3
##dU = dw + dq ##

vs

##dU = dw + dq + µdN##

Which equation do we apply to a closed system involving chemical reaction? According to textbooks, the first equation holds for any closed system in the absence of fields and kinetic energy. However, later chapters use the second equation for closed systems involving irreversible chemical reactions. This doesn't make sense to me because the first law for surroundings gives:

##dU_{surr}= -dw - dq##

Combined with ##dU = dw + dq + µdN## we have :

## dU_{univ} = dU + dU_{surr} = µdN ## ! Which contradicts the statement that energy is conserved for an isolated system.
 
Science news on Phys.org
  • #2
That should be
$$dU = dw + dq + \sum_i µ_i dN_i$$

If the system is closed, what do the two statements of the first law tell you about ##\sum_i µ_i dN_i##?
 
  • #3
DrClaude said:
That should be
$$dU = dw + dq + \sum_i µ_i dN_i$$

If the system is closed, what do the two statements of the first law tell you about ##\sum_i µ_i dN_i##?

ok I get it you're saying ##\sum_i µ_i dN_i## is zero for all closed systems.

Except that, ##\sum_i µ_i dN_i=0## is used for reversible reactions and equilibrium. It's actually used as the criterion for equilibrium. How could it be the criterion for equilibrium if it holds for reversible and irreversible reactions alike?
 
Last edited:
  • #4
Amin2014 said:
ok I get it you're saying ##\sum_i µ_i dN_i## is zero for all closed systems.
Yes. If the system is closed, then not all ##dN_i## will have the same sign, but everything will conspire to keep ##\sum_i µ_i dN_i = 0##. Note that this can be complicated to calculate, as ##\mu## is a function of ##P## and ##T##, which can change due to the reaction.
 
  • #5
Whenever I see a thermodynamic equation written in terms of differentials, I automatically think that it must apply only to a reversible process. That doesn't seem to be the notation they use in your book. But, I would never write the first law using differentials dq and dw, since q and w apply to any path, both reversible and irreversible. The correct equation for an open system is $$dU=TdS-PdV+\sum{\mu_i dN_i}$$This applies to the differential change in state between two closely neighboring thermodynamic equilibrium states of a system.

For a closed system, even with chemical reaction, the correct equation is always $$\Delta U=Q-W$$where the changes in U take into account the changes in the number of moles of the various chemical species present in the mixture.
 
  • #6
Chestermiller said:
Whenever I see a thermodynamic equation written in terms of differentials, I automatically think that it must apply only to a reversible process. That doesn't seem to be the notation they use in your book. But, I would never write the first law using differentials dq and dw, since q and w apply to any path, both reversible and irreversible. The correct equation for an open system is $$dU=TdS-PdV+\sum{\mu_i dN_i}$$This applies to the differential change in state between two closely neighboring thermodynamic equilibrium states of a system.

For a closed system, even with chemical reaction, the correct equation is always $$\Delta U=Q-W$$where the changes in U take into account the changes in the number of moles of the various chemical species present in the mixture.

To quote from "physical chemistry, 6th edition by Ira Levine":
$$dU=TdS-PdV+\sum{\mu_i dN_i}$$
$$dH=TdS+VdP+\sum{\mu_i dN_i}$$
$$dA=-SdT-PdV+\sum{\mu_i dN_i}$$
$$dG=-SdT+Vdp+\sum{\mu_i dN_i}$$
These equations are the extensions of the Gibbs equations to processes involving exchange of matter with surroundings or irreversible composition changes."

Are you saying the author is wrong to consider these equations applicable to irreversible change? Not even after integrating?

P.S. Nice to see your smile again ;)
 
  • #7
Amin2014 said:
To quote from "physical chemistry, 6th edition by Ira Levine":
$$dU=TdS-PdV+\sum{\mu_i dN_i}$$
$$dH=TdS+VdP+\sum{\mu_i dN_i}$$
$$dA=-SdT-PdV+\sum{\mu_i dN_i}$$
$$dG=-SdT+Vdp+\sum{\mu_i dN_i}$$
These equations are the extensions of the Gibbs equations to processes involving exchange of matter with surroundings or irreversible composition changes."

Are you saying the author is wrong to consider these equations applicable to irreversible change? Not even after integrating?

P.S. Nice to see your smile again ;)
What I'm saying is that I interpret these relationships differently. These relationships apply to any pair of closely neighboring thermodynamic equilibrium states of a system, irrespective of how tortuous, extensive, and possibly irreversible the actual process path was that brought about the change between these two differentially separated equilibrium states. For more on this, see the very nice answer provided by GeorgioP in another online forum: https://physics.stackexchange.com/questions/455029/relations-derived-from-ist-law-of-thermodynamics
 
  • #8
Chestermiller said:
What I'm saying is that I interpret these relationships differently. These relationships apply to any pair of closely neighboring thermodynamic equilibrium states of a system, irrespective of how tortuous, extensive, and possibly irreversible the actual process path was that brought about the change between these two differentially separated equilibrium states. For more on this, see the very nice answer provided by GeorgioP in another online forum: https://physics.stackexchange.com/questions/455029/relations-derived-from-ist-law-of-thermodynamics

Chet, I understand what you are saying. The way I see it, differential equations involving state functions (such as ##dU=TdS-PdV##) apply only to reversible paths. However, if we are able to integrate them between two equilibrium states, the resulting relation will hold between those states for all paths, including irreversible ones. But how does any of this relate to my question?
Consider a closed system involving an irreversible chemical reaction. Are you saying that we cannot envisage a reversible path connecting the same initial and final states, and for this reason we cannot combine the two equations to yield ##\sum{\mu_i dN_i}## =0 between those two states?
 
  • #9
Amin2014 said:
Chet, I understand what you are saying. The way I see it, differential equations involving state functions (such as ##dU=TdS-PdV##) apply only to reversible paths. However, if we are able to integrate them between two equilibrium states, the resulting relation will hold between those states for all paths, including irreversible ones. But how does any of this relate to my question?
Consider a closed system involving an irreversible chemical reaction. Are you saying that we cannot envisage a reversible path connecting the same initial and final states, and for this reason we cannot combine the two equations to yield ##\sum{\mu_i dN_i}## =0 between those two states?
Who says we can't envisage a reversible path connect the same initial and final states?
 
  • #10
Amin2014 said:
These equations are the extensions of the Gibbs equations to processes involving exchange of matter with surroundings or irreversible composition changes."

In his book “The Principles of Chemical Equilibrium” Kenneth Denbigh discusses the context in chapter 2.7 “The chemical potential”.
 
  • #11
Chestermiller said:
Who says we can't envisage a reversible path connect the same initial and final states?
For a chemical reaction, occurring irreversibly in a closed environment, can we come up with a reversible path that connects the initial and final states of the system?
If not, then my question is resolved. I just looked at my old notes and realized this was my resolution to this same question I had come up with some years ago. If however, you can think of a reversible path connecting the exact same initial and final conditions, then we can conclude ∑ μdN = 0 for the irreversible and reversible paths alike, which makes no sense to me since we are using ∑ μdN = 0 as the very criterion for achieving equilibrium! How could that be a valid criteria if it holds for reversible and irreversible paths alike?
 
Last edited:
  • #12
Amin2014 said:
For a chemical reaction, occurring irreversibly in a closed environment, can we come up with a reversible path that connects the initial and final states of the system?
If not, then my question is resolved. I just looked at my old notes and realized this was my resolution to this same question I had come up with some years ago. If however, you can think of a reversible path connecting the exact same initial and final conditions, then we can conclude ∑ μdN = 0 for the irreversible and reversible paths alike, which makes no sense to me since we are using ∑ μdN = 0 as the very criterion for achieving equilibrium! How could that be a valid criteria if it holds for reversible and irreversible paths alike?
The equation ##\sum{\mu dN}=0## applies only to small deviations from the thermodynamic equilibrium state at constant temperature and pressure.

If you want to devise a reversible path between initially pure reactants and finally pure products, you can easily do so using a Van't Hoff equilibrium box. Are you familiar with this device?
 
  • #13
I finally managed to sort things out in my head:

Assuming the absence of kinetic and potential energy changes, the only restriction on this equation ##dU = dw + dq## is for the system to be closed ; As you can see the equation is quite general. However, we need to keep in mind that ##dw## accounts for ALL types of work, including chemical work. Some books will write this as ##dU = dw + dw_*+dq## where ##dw_*## represents non P - V forms of work.

Next we have ## dU = TdS - PdV ## which holds true for all processes, reversible or not, provided that the initial and final states are equilibrium states and there is no work other than P _V work. So chemical work must be zero (##\sum{\mu dN}=0##) for the equation to apply.

Finally we have ##dU = TdS - PdV + \sum{\mu dN}## which holds true for all processes, reversible or not, provided the only forms of work are P - V and chemical work.

Chestermiller said:
Who says we can't envisage a reversible path connect the same initial and final states?
In the case of a closed system involving irreversible chemical change, we cannot find a reversible path connecting the initial and final states that doesn't involve chemical work, thus we cannot apply ##dU = TdS - PdV## to this situation.

Btw, chemical work is just another form of work, albeit at the microscopic level. The surroundings does work on the system, which causes internal change (structural rearrangement) in the components of the system. This work is saved as potential energy in the molecules. So as the surroundings loses energy (by doing work) the system gains it, abiding to the law of conservation of energy for an isolated system.
 
Last edited:
  • #14
Amin2014 said:
I finally managed to sort things out in my head:

Assuming the absence of kinetic and potential energy changes, the only restriction on this equation ##dU = dw + dq## is for the system to be closed ; As you can see the equation is quite general. However, we need to keep in mind that ##dw## accounts for ALL types of work, including chemical work. Some books will write this as ##dU = dw + dw_*+dq## where ##dw_*## represents non P - V forms of work.

Next we have ## dU = TdS - PdV ## which holds true for all processes, reversible or not, provided that the initial and final states are equilibrium states and there is no work other than P _V work. So chemical work must be zero (##\sum{\mu dN}=0##) for the equation to apply.

Finally we have ##dU = TdS - PdV + \sum{\mu dN}## which holds true for all processes, reversible or not, provided the only forms of work are P - V and chemical work.To clarIn the case of a closed system involving irreversible chemical change, we cannot find a reversible path connecting the initial and final states that doesn't involve chemical work, thus we cannot apply ##dU = TdS - PdV## to this situation.

Btw, chemical work is just another form of work, albeit at the microscopic level. The surroundings does work on the system, which causes internal change (structural rearrangement) in the components of the system. This work is saved as potential energy in the molecules. So as the surroundings loses energy (by doing work) the system gains it, abiding to the law of conservation of energy for an isolated system.
Usually, in the case of chemical reactions, we are working with the Gibbs free energy G, not the internal energy U. And, irrespective of whether a reaction is reversible or not, we can determine a reversible path between the same two end states (assuming they are thermodynamic equilibrium states) for which we can determine the change in G.
 
  • #15
Chestermiller said:
Usually, in the case of chemical reactions, we are working with the Gibbs free energy G, not the internal energy U. And, irrespective of whether a reaction is reversible or not, we can determine a reversible path between the same two end states (assuming they are thermodynamic equilibrium states) for which we can determine the change in G.

Yes, but not a path without any chemical work. This point was initially puzzling to me and I wasn't able to pin point the source of my confusion.

To quote the 5th point in link you reffered me to: " 5. "##TdS = dU + PdV ## This equation holds good for any process, reversible or irreversible, undergone by a closed system."

Unfortunately this type of misrepresentation is present among many thermodynamics texts which can lead to confusion. The equation does not hold when we have work other than P-V, such as chemical work.
 
  • #16
Amin2014 said:
Yes, but not a path without any chemical work. This point was initially puzzling to me and I wasn't able to pin point the source of my confusion.

To quote the 5th point in link you reffered me to: " 5. "##TdS = dU + PdV ## This equation holds good for any process, reversible or irreversible, undergone by a closed system."

Unfortunately this type of misrepresentation is present among many thermodynamics texts which can lead to confusion. The equation does not hold when we have work other than P-V, such as chemical work.
What is your definition of chemical work?
 
  • #17
Chestermiller said:
What is your definition of chemical work?
The mathematical definition would be:
Chemical work = ##∑ μ dN##

Basically it's the work done by the surroundings at the microscopic level to get a chemical reaction or phase change to occur in the system under conditions of constant temperature and pressure (or constant temperature and volume).

Update: Actually,I have an update on this but I would like to hear your opinion first.
 
Last edited:
  • #18
Amin2014 said:
The mathematical definition would be:
Chemical work = ##∑ μ dN##

Basically it's the work done by the surroundings at the microscopic level to get a chemical reaction or phase change to occur in the system under conditions of constant temperature and pressure (or constant temperature and volume).

Update: Actually,I have an update on this but I would like to hear your opinion first.
I would call that the change in Gibbs Free Energy or the change in Helmholtz Free Energy, not the change in internal energy (unless it is at constant volume and entropy).
 
  • #19
Chestermiller said:
I would call that the change in Gibbs Free Energy or the change in Helmholtz Free Energy, not the change in internal energy (unless it is at constant volume and entropy).
Yes but the point is, when you can't apply ##dU=TdS-PdV## to a process, you can't apply any of the other similar Gibbs equations such as ## dG = VdP - SdT## either.
 
  • #20
Consider the reaction A + B ---> C to take place in a closed environment. Let the initial coordinates be T1 and P1, with T2 and P2 as the final coordinates. The reaction takes place irreversibly.

One might reason that, by means of constructing a reversible path connecting the above two end states, we could apply ##dU = TdS - PdV## to calculate the changes in internal energy (or ##dG = VdP - SdT## for Gibbs Free energy). However, the validity of these equations is limited to processes that don't involve P-V work, and it is impossible to fully traverse the path from the initial to final states without introducing non P - V work. Therefore, under the conditions of irreversible composition change, we cannot apply ##dU =TdS - PdV## or similar equations to calculate changes in potential functions, which negates the claims of textbooks that these equations can be used for any process in a closed system. (see picture)

Getting from State 2
A+B =C.png
 

Attachments

  • A+B =C.png
    A+B =C.png
    4.5 KB · Views: 590
  • #21
If you are saying that in a closed system involving chemical reaction, the equations ##dU=TdS-PdV+\sum{\mu dN}## and ##dG=-SdT+VdP+\sum{\mu dN}## describe the differences in U and G between two closely neighboring thermodynamic equilibrium states, I agree. Certainly, U=U(S,V,N's) and G=G(T,P,N's). But, for a closed system that experiences a finite change in U between two thermodynamic equilibrium states separated by finite changes in the parameters, irrespective of whether the intermediate process is reversible or irreversible, $$\Delta U=Q-W$$where the change in U includes the effects of the changes in the species concentrations. Are we in agreement on this?
 
  • #22
Chestermiller said:
If you are saying that in a closed system involving chemical reaction, the equations ##dU=TdS-PdV+\sum{\mu dN}## and ##dG=-SdT+VdP+\sum{\mu dN}## describe the differences in U and G between two closely neighboring thermodynamic equilibrium states, I agree. Certainly, U=U(S,V,N's) and G=G(T,P,N's).
Provided we don't have non P-V work, then yes.

Chestermiller said:
But, for a closed system that experiences a finite change in U between two thermodynamic equilibrium states separated by finite changes in the parameters, irrespective of whether the intermediate process is reversible or irreversible, $$\Delta U=Q-W$$where the change in U includes the effects of the changes in the species concentrations. Are we in agreement on this?
Provided W includes all forms of work, YESSS!

Also I am saying the fifth point in this link is false:
https://physics.stackexchange.com/questions/455029/relations-derived-from-ist-law-of-thermodynamics
we cannot apply the equation to processes which involve non P-V work. This includes any reversible paths connecting the end states of an irreversible chemical reaction in a closed system.
 
Last edited:
  • #25
Amin2014 said:
##dU = dw + dq ##

vs

##dU = dw + dq + µdN##

Which equation do we apply to a closed system involving chemical reaction?
For a closed system, only the first equation is correct.
 
  • #26
Amin2014 said:
For a chemical reaction, occurring irreversibly in a closed environment, can we come up with a reversible path that connects the initial and final states of the system?
Of course we can, but the reversible path may involve open systems.
 

1. What is the 1st law of thermodynamics?

The 1st law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another.

2. How does the 1st law of thermodynamics apply to chemical reactions?

In chemical reactions, the 1st law of thermodynamics means that the total energy of the reactants must equal the total energy of the products. This is because energy is conserved, so any energy lost or gained during the reaction must be accounted for.

3. Can the 1st law of thermodynamics be violated?

No, the 1st law of thermodynamics is a fundamental law of physics and has been consistently observed to hold true. It is a principle that cannot be violated.

4. How does the 1st law of thermodynamics relate to enthalpy?

Enthalpy is a measure of the total energy of a system, including both its internal energy and any work done by or on the system. The 1st law of thermodynamics applies to enthalpy because it states that the total energy of a system must remain constant.

5. Can the 1st law of thermodynamics be applied to open systems?

Yes, the 1st law of thermodynamics can be applied to open systems, which are systems that can exchange matter and energy with their surroundings. In these systems, the total energy must still be conserved, but the energy can be transferred in or out of the system through matter exchange.

Similar threads

I Reversible / Irreversible Fundamental Equation
    • Thermodynamics
Replies
3
Views
799
I Why is ##dQ = dH## still valid for chemical reaction (Callen)?
    • Thermodynamics
Replies
31
Views
1K
I Graphical difference between adiabatic and isothermal processes.
    • Thermodynamics
Replies
1
Views
642
Question about the 2nd law of thermodynamics (heat loss)
    • Introductory Physics Homework Help
Replies
1
Views
870
What is ##Tds## in this equation : ##dU = Tds - PdV##
    • Introductory Physics Homework Help
Replies
14
Views
702
Why does the entropy of a closed system remain constant in a reversible process?
    • Thermodynamics
Replies
10
Views
1K
B Change in entropy of reversible isothermal process
    • Thermodynamics
Replies
11
Views
327
B Entropy change for reversible and irreversible processes
    • Thermodynamics
Replies
12
Views
1K
I Application of the first law of thermodynamics
    • Thermodynamics
2
Replies
46
Views
2K
Thermal Physics: Reversible processes
    • Thermodynamics
Replies
19
Views
2K

Hot Threads

Recent Insights

Back
Top