Thermodynamics of chemical reactions

[Yash Agrawal]

TL;DR Summary

I have confusions regarding gibbs free energy change in chemical reactions

In chemical reactions generally ΔG < 0 , but if we were to consider a reversible path between pure reactants and products at 1 bar pressure , shouldn't the ΔG = 0 for every reaction ? and if it is due to non-pv work , I don't see any non pv work being done in reactions happing in a closed container

I'm just not sure how thermodynamics changes are calculated in case of chemical reactions

Also,
In electrochemical cell , we equate -nFE to ΔrG of reaction and so that is the maximum non-pv work obtained , so reaction should proceed reversibly to obtain maximum work , so does really reaction in cell occur reversibly ? I mean reversible reactions are only theoretical, So how could they happen ?

 

[Chestermiller]

Regarding your first question, even though the initial and final states of pure reactants and pure products, respectively, are at 1 bar and 298 K, and the reversible path between the initial state and the final state is all at constant temperature, the reversible path between the initial state and the final state is not at constant pressure. During the overall reversible path, the pressure of the pure reactants and the pure products must be varied. Even though the enthalpy of the pure reactants and the pure products are at constant enthalpy during these steps in the process, their entropies change. This causes the entropy change from the initial state to the final state to differ from $\Delta H^0/T^0$.

If you like, I will outline the overall (3 step) reversible process for you. This process involves a chemical reactor in which reactants are introduced and products are removed while the chemical reactor is operating at chemical equilibrium (quasi-statically).

 

[Yash Agrawal]

Regarding your first question, even though the initial and final states of pure reactants and pure products, respectively, are at 1 bar and 298 K, and the reversible path between the initial state and the final state is all at constant temperature, the reversible path between the initial state and the final state is not at constant pressure. During the overall reversible path, the pressure of the pure reactants and the pure products must be varied. Even though the enthalpy of the pure reactants and the pure products are at constant enthalpy during these steps in the process, their entropies change. This causes the entropy change from the initial state to the final state to differ from $\Delta H^0/T^0$.

If you like, I will outline the overall (3 step) reversible process for you. This process involves a chemical reactor in which reactants are introduced and products are removed while the chemical reactor is operating at chemical equilibrium (quasi-statically).

Yes , please outline it

 

[Yash Agrawal]

Regarding your first question, even though the initial and final states of pure reactants and pure products, respectively, are at 1 bar and 298 K, and the reversible path between the initial state and the final state is all at constant temperature, the reversible path between the initial state and the final state is not at constant pressure. During the overall reversible path, the pressure of the pure reactants and the pure products must be varied. Even though the enthalpy of the pure reactants and the pure products are at constant enthalpy during these steps in the process, their entropies change. This causes the entropy change from the initial state to the final state to differ from $\Delta H^0/T^0$.

If you like, I will outline the overall (3 step) reversible process for you. This process involves a chemical reactor in which reactants are introduced and products are removed while the chemical reactor is operating at chemical equilibrium (quasi-static

but ΔS of surroundings = - ΔH/T right ? also is it that heat absorbed by system in reversible path is not equal to ΔH of reaction ?

 

[Chestermiller]

but ΔS of surroundings = - ΔH/T right ?

No. For the reversible path, $\Delta S$ of the surroundings is not equal to $-\Delta H/T$ and $\Delta S$ of the system is not equal to $\Delta H/T$. The reversible path is not entirely at constant pressure, so $\Delta H$ is not equal to Q.

also is it that heat absorbed by system in reversible path is not equal to ΔH of reaction ?

Yes. This is the case.

 

[Chestermiller]

Yes , please outline it

The 2nd (middle) step in the reversible process consists of a continuous flow chemical reactor operating at constant temperature and pressure. The partial pressures of the reactants and products of this reactor are such that the reaction mixture is essentially at chemical equilibrium at all times.

The is not a batch chemical reactor. It is continuous flow, with reactants flowing into the reactor and products flowing out. Reactants are introduced into the reactor in stoichiometric proportions, and products are removed from the reactor in corresponding stoichiometric proportions. The rates of flow in and out are so slow that the reaction continuously has ample time to equilibrate, and never deviates significantly from the original equilibrium state.

In order to guarantee that the concentrations of the chemical species in the reactor do not change, the pure reactants must be introduced into the reactor through semi-permeable membranes at the same pressures as their partial pressures as in the reaction mixture, while pure products are removed through semi-permeable membranes at the same pressures as their partial pressures as in the reaction mixture.

So what we have here is a continuous flow chemical reactor that is operating reversibly (i.e., essentially at equilibrium).

OK so far? If so, I will next proceed to analyze this 2nd step in the reversible path using the open system (continuous flow) versions of the 1st and 2nd laws of thermodynamics.

 

[Yash Agrawal]

got it , so we are going from pure reactants (initially in a cylinder containing pure reactants connected to reactor by SPM) to pure products (in cylinders connected to reactor on the other side) ?

 

[Yash Agrawal]

No. For the reversible path, $\Delta S$ of the surroundings is not equal to $-\Delta H/T$ and $\Delta S$ of the system is not equal to $\Delta H/T$. The reversible path is not entirely at constant pressure, so $\Delta H$ is not equal to Q.

Yes. This is the case.

Yes got it , as pressure is not constant throughout we can't equate q = ΔH , can you confirm that for actual reaction taking place (not this reversible pathway) ΔS of surroundings = -ΔH/T but ΔS of system not equal to -ΔH/T (indeed equal to what it comes out in reversible pathway)

 

[Chestermiller]

Yes got it , as pressure is not constant throughout we can't equate q = ΔH , can you confirm that for actual reaction taking place (not this reversible pathway) ΔS of surroundings = -ΔH/T but ΔS of system not equal to -ΔH/T (indeed equal to what it comes out in reversible pathway)

No. For any actual path employing a single constant temperature reservoir at temperature T, $\Delta S$ of surroundings =-Q/T, but $\Delta S$ of system = $\frac{Q}{T}+\sigma$, where $\sigma$ is equal to the entropy generation as a result of irreversibility.

 

[Yash Agrawal]

No. For any actual path employing a single constant temperature reservoir at temperature T, $\Delta S$ of surroundings =-Q/T, but $\Delta S$ of system = $\frac{Q}{T}+\sigma$, where $\sigma$ is equal to the entropy generation as a result of irreversibility.

but In actual irreversible path at constant P and T , Q = ΔH ? My reasoning : ΔH = ΔU + ΔPV , applying it for initial and final states (internal pressure is same initially and finally only , while external pressure is constant throughout) , ΔH = pΔV = -W = Q .

 

[Chestermiller]

but In actual irreversible path at constant P and T , Q = ΔH ? My reasoning : ΔH = ΔU + ΔPV , applying it for initial and final states (internal pressure is same initially and finally only , while external pressure is constant throughout) , ΔH = pΔV = -W = Q .

That is only if the externally applied pressure is held constant, and equal to the initial pressure of the gas in a closed system.

 

[Yash Agrawal]

That is only if the externally applied pressure is held constant, and equal to the initial pressure of the gas in a closed system.

That's what I am assuming , Is there any other way of maintaining constant pressure irreversibly ? When it is mentioned for example that combustion of methane happens at 1atm , that means external pressure is 1atm throughout (that's how I have understood it till now )

 

[Chestermiller]

That's what I am assuming , Is there any other way of maintaining constant pressure irreversibly ? When it is mentioned for example that combustion of methane happens at 1atm , that means external pressure is 1atm throughout (that's how I have understood it till now )

You can assume anything you want for an irreversible path, but to get the change in G (and the standard change in G ) between two states, you need a reversible path. For the standard change in G, there is no reversible path at constant P and constant T.

 

[Yash Agrawal]

You can assume anything you want for an irreversible path, but to get the change in G (and the standard change in G ) between two states, you need a reversible path. For the standard change in G, there is no reversible path at constant P and constant T.

Yes right , no problem with that. I just want to ask that when ΔH for a reaction is given, it is equal to heat released/absorbed when reaction happen in a closed container at constant external pressure and temperature , right ?

 

[Chestermiller]

Yes right , no problem with that. I just want to ask that when ΔH for a reaction is given, it is equal to heat released/absorbed when reaction happen in a closed container at constant external pressure and temperature , right ?

Only if the external pressure is held equal to the initial pressure of the gas.

 

[Chestermiller]

You are asking the same kind of questions (and my responses are the same) as another member who I interacted with in this thread: https://www.physicsforums.com/threa...ctrolytic-galvanic-cell.1002849/#post-6490318
Please read this thread over carefully. There is a wealth of valuable information in it of interest to you.

 

[Yash Agrawal]

Only if the external pressure is held equal to the initial pressure of the gas.

Yes that is necessary otherwise how initial state would be in equilibrium.

 

[Yash Agrawal]

You are asking the same kind of questions (and my responses are the same) as another member who I interacted with in this thread: https://www.physicsforums.com/threa...ctrolytic-galvanic-cell.1002849/#post-6490318
Please read this thread over carefully. There is a wealth of valuable information in it of interest to you.

Sure

 

[Yash Agrawal]

You are asking the same kind of questions (and my responses are the same) as another member who I interacted with in this thread: https://www.physicsforums.com/threa...ctrolytic-galvanic-cell.1002849/#post-6490318
Please read this thread over carefully. There is a wealth of valuable information in it of interest to you.

been though it carefully several times , understood most of the things . But for galvanic cell part , can you please elaborate how reaction happen reversibly at constant Pressure ? I mean for ideal gas reaction , in reversible pathway we define 3 steps in which step 1 and 3 are responsible for extra entropy generation other than ΔH/T in 2nd step , and entropy from 1st and 3rd step comes R.ln(K) , so you are claiming that in electrochemical reactions this extra entropy is zero (now it is due to non-pv work which is responsible for making ΔG not zero ), but for that k = 1 for every electrochemical reaction , which is certainly not the case.

Or my question would be solved if you could derive ** ΔG(standard) = -RT.ln(K) ** for electrochemical reactions (like you derived for ideal gas reactions by writing total entropy for three step reversible process )

Just curious to know how electrochemical reactions happens reversibly at constant pressure (unlike ideal gas reactions )

 

[Chestermiller]

In the case of the ideal gas reactions, the reaction is made to proceed reversibly using the Equilibrium Box. In the case of an electrolytic cell, the reaction is made to proceed reversibly (at essentially zero reaction rate and current) by applying a voltage to the cell opposite to the way that the current wants to flow. Without applying this voltage (say, by short circuiting the cell), the reaction will proceed at finite rate and the process will be irreversible.

 

[Yash Agrawal]

In the case of the ideal gas reactions, the reaction is made to proceed reversibly using the Equilibrium Box. In the case of an electrolytic cell, the reaction is made to proceed reversibly (at essentially zero reaction rate and current) by applying a voltage to the cell opposite to the way that the current wants to flow. Without applying this voltage (say, by short circuiting the cell), the reaction will proceed at finite rate and the process will be irreversible.

So if we were to connect a voltmeter to two electrodes of a galvanic cell , the reading it show experimentally is different from what we calculate from E = -ΔG/nF ?

 

[Chestermiller]

So if we were to connect a voltmeter to two electrodes of a galvanic cell , the reading it show experimentally is different from what we calculate from E = -ΔG/nF ?

No. My understanding is that this is the voltage we will show with no current flowing.

 

[Yash Agrawal]

No. My understanding is that this is the voltage we will show with no current flo

No. My understanding is that this is the voltage we will show with no current flowing.

This 'No' is for showing same value or not ?

 

[Chestermiller]

With this voltage applied, allowing a tiny current to flow and do work until 1 Faraday of charge has transferred, we will have carried out the reaction reversibly with a free energy change of $\Delta G$.

 

[Yash Agrawal]

With this voltage applied, allowing a tiny current to flow and do work until 1 Faraday of charge has transferred, we will have carried out the reaction reversibly with a free energy change of $\Delta G$.

If no opposing voltage is applied , but voltmeter has very high resistance (which generally it has) , then also current would be very less , then do we obtain reading as -ΔG/nF ?

 

[Chestermiller]

If no opposing voltage is applied , but voltmeter has very high resistance (which generally it has) , then also current would be very less , then do we obtain reading as -ΔG/nF ?

Yes, but all the energy would be dissipated in the resister,

 

[Yash Agrawal]

Yes, but all the energy would be dissipated in the resister,

Yes , Thanks. Then why students are even made to calculate them if that value of emf is not correct in practical circumstances where we geniunely need to use the cell ?

 

[Chestermiller]

Yes , Thanks. Then why students are even made to calculate them if that value of emf is not correct in practical circumstances where we geniunely need to use the cell ?

I don't know how to answer this. In practical situations, the EMF is still very close to the ideal reversible potential even when minor irreveribilities are present. In the case of the high resister, all the irreversibility occurs outside the cell, and doesn't significantly affect the potential. The resistor is an irreversible feature of the surroundings. It's like a frictional resistance outside a gas-cylinder arrangement.

 

[Yash Agrawal]

In practical situations, the EMF is still very close to the ideal reversible potential

I think so