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### Topic: Attraction between species during precipitate formation  (Read 2566 times)

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#### mjc123

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• Mole Snacks: +245/-11 ##### Re: Attraction between species during precipitate formation
« Reply #15 on: March 13, 2019, 05:50:09 AM »
I. The tabulated empirical data refer to a specific pressure and temperature. In that sense they are constant. Whatever pressure and temperature you're working at, ΔH°f(298K, 1 atm) has the same value. But the heat of formation at another temperature will not be the same (see below). This is important because when you use the equation for ΔG at a particular temperature, you must use the ΔH and ΔS values for that temperature, not 298K; e.g.
ΔG°(400K) = ΔH°(400K) - 400*ΔS°(400K)
Over a small temperature range, ΔH and ΔS usually don't vary much, so it is a reasonable approximation to use the 298K values at (say) 310K, but probably not at 400K.
These considerations apply equally to ΔG°, ΔH° and ΔS°, not just ΔG°. You can find tabulated values of  ΔG°f(298K, 1 atm). But you can define a standard state for G, H and S at any temperature. There is no such thing as THE standard state.
II. Enthalpy and entropy vary with temperature, depending on the heat capacity. dH/dT = Cp and dS/dT = Cp/T. So if reactants and products have different heat capacities, ΔH and ΔS vary with temperature.
So dΔH/dT = ΔCp, where ΔCp = ΣCp(products) - ΣCp(reactants).
dΔS/dT = ΔCp/T
(In fact it is a bit more complicated, as Cp itself varies with temperature.)
(Note that to a first approximation this doesn't affect ΔG, as dΔG/dT = dΔH/dT - TdΔS/dT - ΔS = ΔCp - ΔCp - ΔS = ΔS.)
Another thing that affects ΔH° and ΔS° is phase changes; e.g. if a reactant has a melting point of 300K, then below this temperature the standard state is solid, and above 300K the standard state is liquid, so there is a step change in H due to the heat of fusion.

#### QuiteThePredicament

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« Reply #16 on: March 13, 2019, 01:00:48 PM »
I. The tabulated empirical data refer to a specific pressure and temperature. In that sense they are constant. Whatever pressure and temperature you're working at, ΔH°f(298K, 1 atm) has the same value. But the heat of formation at another temperature will not be the same (see below). This is important because when you use the equation for ΔG at a particular temperature, you must use the ΔH and ΔS values for that temperature, not 298K; e.g.
ΔG°(400K) = ΔH°(400K) - 400*ΔS°(400K)
Over a small temperature range, ΔH and ΔS usually don't vary much, so it is a reasonable approximation to use the 298K values at (say) 310K, but probably not at 400K.
These considerations apply equally to ΔG°, ΔH° and ΔS°, not just ΔG°. You can find tabulated values of  ΔG°f(298K, 1 atm). But you can define a standard state for G, H and S at any temperature. There is no such thing as THE standard state.
II. Enthalpy and entropy vary with temperature, depending on the heat capacity. dH/dT = Cp and dS/dT = Cp/T. So if reactants and products have different heat capacities, ΔH and ΔS vary with temperature.
So dΔH/dT = ΔCp, where ΔCp = ΣCp(products) - ΣCp(reactants).
dΔS/dT = ΔCp/T
(In fact it is a bit more complicated, as Cp itself varies with temperature.)
(Note that to a first approximation this doesn't affect ΔG, as dΔG/dT = dΔH/dT - TdΔS/dT - ΔS = ΔCp - ΔCp - ΔS = ΔS.)
Another thing that affects ΔH° and ΔS° is phase changes; e.g. if a reactant has a melting point of 300K, then below this temperature the standard state is solid, and above 300K the standard state is liquid, so there is a step change in H due to the heat of fusion.

So, the ΔG° refers to the ΔG at the conditions which the reaction takes place in? When an equilibrium is established at, let's say 500K and 2 atm, the ΔG° in lnK = -ΔG°/RT would be equal to the Gibbs free energy change of the reaction at 500K and 2 atm? In that case, why isn't it just written as ΔG? What exactly is the difference between ΔG, and ΔG°?

#### mjc123

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« Reply #17 on: March 13, 2019, 02:26:37 PM »
No, it refers to standard conditions at that temperature. When an equilibrium is established at 500K and  2 atm, ΔG° is the Gibbs free energy change at 500K and 1 atm. (Assuming 1 atm is your standard pressure. And note that this means a partial pressure of 1 atm for each gaseous reagent, not a total pressure of 1 atm.) ΔG is the Gibbs free energy change at the actual conditions. They are related by (as you have already quoted)
ΔG = ΔG° + RTlnQ
where Q is the reaction quotient.
When the actual conditions are at equilibrium, ΔG = 0 and Q = K. This assumes adjusting the equilibrium (pressures, concentrations) at a given temperature. The reason is that it is simple to adjust for pressure by the relation μ = μ° + RT ln(P/P°), and that is how the above equation for ΔG is derived. (μ is the chemical potential - the partial molar Gibbs free energy. Have you come across it?) Adjusting for temperature is more difficult because of the reasons we have discussed. In practice, of course, one often has to take a tabulated value at (say) 298K, and adjust it to the reaction temperature using the heat capacities. But that doesn't give a nice neat equation like ΔG = ΔG° + RTlnQ. So we calculate ΔG° at the reaction temperature and apply this equation, rather than having a complicated equation using ΔG°(298K). The important thing is that in the equation ΔG° = RT lnK, ΔG° is at the reaction temperature.

#### QuiteThePredicament

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• Mole Snacks: +1/-0 ##### Re: Attraction between species during precipitate formation
« Reply #18 on: March 13, 2019, 07:16:22 PM »
No, it refers to standard conditions at that temperature. When an equilibrium is established at 500K and  2 atm, ΔG° is the Gibbs free energy change at 500K and 1 atm. (Assuming 1 atm is your standard pressure. And note that this means a partial pressure of 1 atm for each gaseous reagent, not a total pressure of 1 atm.) ΔG is the Gibbs free energy change at the actual conditions. They are related by (as you have already quoted)
ΔG = ΔG° + RTlnQ
where Q is the reaction quotient.
When the actual conditions are at equilibrium, ΔG = 0 and Q = K. This assumes adjusting the equilibrium (pressures, concentrations) at a given temperature. The reason is that it is simple to adjust for pressure by the relation μ = μ° + RT ln(P/P°), and that is how the above equation for ΔG is derived. (μ is the chemical potential - the partial molar Gibbs free energy. Have you come across it?) Adjusting for temperature is more difficult because of the reasons we have discussed. In practice, of course, one often has to take a tabulated value at (say) 298K, and adjust it to the reaction temperature using the heat capacities. But that doesn't give a nice neat equation like ΔG = ΔG° + RTlnQ. So we calculate ΔG° at the reaction temperature and apply this equation, rather than having a complicated equation using ΔG°(298K). The important thing is that in the equation ΔG° = RT lnK, ΔG° is at the reaction temperature.

I now realize that the standard conditions do not define a set temperature, I got it mixed up with the STP. It determines the values for when the reaction is run at 1 atm pressure, and the concentration of solutes is 1M. That being the case, we can use the same ΔG° value we measured in those conditions, for the same reaction with different pressures and concentrations, as long as we keep temperature and other variables constant. For example, let's say we've determined the ΔG for a reaction at 500K, 1 atm and 1M concentrated solutes called it the ΔG°. This value can be used to determine the K of the same reaction that is run at 500K. The ΔG stays the same for the same reaction run at 500K 1 atm solutes at 2M, 500K 2 atm solutes at 1 M, 500K 19 atm solutes at 7M,  etc... And we can pick one of these at random, measure the ΔG value of it, and use it as the ΔG° as well. There'd be no difference. This explains also why any pressure can be chosen to be the standard condition, because it doesn't change with it. Is all of this correct, and if it is, does it hold true for every ΔX° value, like the ΔH° and ΔS°, just like it does for the ΔG°?

#### mjc123

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« Reply #19 on: March 14, 2019, 07:07:06 AM »
Quote
The ΔG stays the same for the same reaction run at 500K 1 atm solutes at 2M, 500K 2 atm solutes at 1 M, 500K 19 atm solutes at 7M,  etc...
No, ΔG varies with pressure/concentration. We can choose any of these as our standard conditions, but then everything must be referred to that standard. Strictly, Q should be written
Q = (PC/P°)(PD/P°)/(PA/P°)(PB/P°)       for a reaction A + B C + D
Then ΔG = ΔG° + RTlnQ  is independent of the standard P° chosen, as changes in ΔG° are balanced by changes in Q
Note that for equal number of moles of gas on each side, P° cancels and ΔG° and Q are independent of P°, but this is not generally true.

#### QuiteThePredicament

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• Mole Snacks: +1/-0 ##### Re: Attraction between species during precipitate formation
« Reply #20 on: March 14, 2019, 01:06:49 PM »
Quote
The ΔG stays the same for the same reaction run at 500K 1 atm solutes at 2M, 500K 2 atm solutes at 1 M, 500K 19 atm solutes at 7M,  etc...
No, ΔG varies with pressure/concentration. We can choose any of these as our standard conditions, but then everything must be referred to that standard. Strictly, Q should be written
Q = (PC/P°)(PD/P°)/(PA/P°)(PB/P°)       for a reaction A + B C + D
Then ΔG = ΔG° + RTlnQ  is independent of the standard P° chosen, as changes in ΔG° are balanced by changes in Q
Note that for equal number of moles of gas on each side, P° cancels and ΔG° and Q are independent of P°, but this is not generally true.

The ΔG is the change of free energy of the reaction, ΔG° is the change of free energy of the reaction that is run at the same temperature, but at a arbitrarily chosen pressure/conc that is picked as the standard state. T term is the temperature that the reaction is run at, and the Q is the (products)/(reactants), adjusted for the standard pressure and conc as you stated.
Let's say I measured the ΔG for the reaction run at 500K 1 atm 2M and called it the standard state. This makes that value the ΔG° for all the reactions that are run at 500K, i.e. 500K 1 atm 3M, 500K 2 atm 1M?
Separate question, let's say there is a reaction that I run at 400K 2atm and 3M. I chose the 400K 1atm 1M as my standard state, determined the ΔG value for it and marked it as the ΔG°, that I'm going to use for the aforementioned 400K 2atm 3M. Then I find the Q value for the reaction at those conditions, and use the formula to get the ΔG(400K 2atm 3M). After this, I keep the reaction conditions the same, but change my definition of the standard state. I now set it at 400K, 7atm, 9M. I find the ΔG, call it the ΔG°, find the Q that is adjusted for my new definition of the standard, and use the formula again to calculate the ΔG°(400K 2atm 3M). This ΔG value that I find at the end would be the same as the one that I found before, right?

#### mjc123

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« Reply #21 on: March 14, 2019, 02:01:11 PM »
Quote
Let's say I measured the ΔG for the reaction run at 500K 1 atm 2M and called it the standard state. This makes that value the ΔG° for all the reactions that are run at 500K, i.e. 500K 1 atm 3M, 500K 2 atm 1M?
Yes
Quote
Separate question, let's say there is a reaction that I run at 400K 2atm and 3M. I chose the 400K 1atm 1M as my standard state, determined the ΔG value for it and marked it as the ΔG°, that I'm going to use for the aforementioned 400K 2atm 3M. Then I find the Q value for the reaction at those conditions, and use the formula to get the ΔG(400K 2atm 3M). After this, I keep the reaction conditions the same, but change my definition of the standard state. I now set it at 400K, 7atm, 9M. I find the ΔG, call it the ΔG°, find the Q that is adjusted for my new definition of the standard, and use the formula again to calculate the ΔG°(400K 2atm 3M). This ΔG value that I find at the end would be the same as the one that I found before, right?
Yes, it should be. The ΔG° that I have bolded should be ΔG, but otherwise that's correct.

#### QuiteThePredicament

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« Reply #22 on: March 14, 2019, 03:25:55 PM »
Quote
Let's say I measured the ΔG for the reaction run at 500K 1 atm 2M and called it the standard state. This makes that value the ΔG° for all the reactions that are run at 500K, i.e. 500K 1 atm 3M, 500K 2 atm 1M?
Yes
Quote
Separate question, let's say there is a reaction that I run at 400K 2atm and 3M. I chose the 400K 1atm 1M as my standard state, determined the ΔG value for it and marked it as the ΔG°, that I'm going to use for the aforementioned 400K 2atm 3M. Then I find the Q value for the reaction at those conditions, and use the formula to get the ΔG(400K 2atm 3M). After this, I keep the reaction conditions the same, but change my definition of the standard state. I now set it at 400K, 7atm, 9M. I find the ΔG, call it the ΔG°, find the Q that is adjusted for my new definition of the standard, and use the formula again to calculate the ΔG°(400K 2atm 3M). This ΔG value that I find at the end would be the same as the one that I found before, right?
Yes, it should be. The ΔG° that I have bolded should be ΔG, but otherwise that's correct.

Typo, meant to write ΔG.
Ultimately, when calculating the K for a reaction using the lnK = -ΔG°/RT, I only have to find the ΔG value of that reaction at the same temperature, doesn't matter what the pressure/conc is, and use it as the ΔG°. For a reaction taking place at 400K 1atm 2M, I can take the ΔG of the same reaction running at 400K 2 atm 3M or 400K 4 atm 1M, and use it as the ΔG° for this reaction, and it'd yield the same K value?

#### mjc123

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« Reply #23 on: March 15, 2019, 07:09:58 AM »
Not generally; see what I wrote about Q on post#19. Same applies to K (= value of Q at equilibrium). Pressures or concentrations must be referenced to the standard state, and therefore the value of K will vary with the standard state chosen, unless the number of moles is equal on both sides of the equation. More precisely
K(P°2) = K(P°1) * (P°1/P°2)Δn where Δn is the change in number of moles.
The value of K will always be constant at a given temperature as long as you use the same standard state for reference, whatever the actual reaction conditions.

#### QuiteThePredicament

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• Mole Snacks: +1/-0 ##### Re: Attraction between species during precipitate formation
« Reply #24 on: March 15, 2019, 10:43:00 AM »
Not generally; see what I wrote about Q on post#19. Same applies to K (= value of Q at equilibrium). Pressures or concentrations must be referenced to the standard state, and therefore the value of K will vary with the standard state chosen, unless the number of moles is equal on both sides of the equation. More precisely
K(P°2) = K(P°1) * (P°1/P°2)Δn where Δn is the change in number of moles.
The value of K will always be constant at a given temperature as long as you use the same standard state for reference, whatever the actual reaction conditions.

The ΔG, recorded at the conditions which are defined as the standard state can be used as ΔG° to calculate all K values for the same reaction that take place within the same temperature, but changing the definition would require me to adjust the K for the new ones. Like using a 2m long stick to measure the length of a road, calling it x units, then use a 3m one, and that y. I'd have to adjust the values I get when switching from x to y. The length of the road can be described at K, which would stay the same when I keep everything the same, but change my method of measuring.
Measuring different roads with the x units (Same reaction and same temp K, but at different pressure/conc since I can only change the quantity that my x method measures, for this to be consistent, changing temp would be same as changing the curvature of the road which'd mess up everything) would also yield the same results as long as I adjust everything else to comply with the new meter to x ratio. Using this analogy to simplify it has helped me understand. Is it correct though?

#### mjc123

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« Reply #25 on: March 15, 2019, 12:30:48 PM »
If I understand you correctly, yes.

#### QuiteThePredicament

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« Reply #26 on: March 15, 2019, 01:25:34 PM »
Nice, that was all. Thanks for clarifying.