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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: QuiteThePredicament on March 04, 2019, 12:00:54 PM
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In this double replacement reaction AgCl(aq) + NaCl(aq) :rarrow: AgCl(s) + Na+ + NO3- the Na+ and NO3- ions are held apart by the ion-dipole interactions between Na-O and NO3-H. On the other hand, Ag+ and Cl- ions have low electronegativity difference, thus come together to form the covalent AgCl precipitate.
However, I understand that the ion-dipole forces pulling the Ag+ and Cl- apart isn't strong enough because of their low polarity, but how are they attracted to each other? If it was caused by the force of the negative and positive electric fields of the ions, wouldn't they also interact with the dipole moments of oxygen and hydrogen surrounding them, staying in solution like Na+ and NO3- ions? Or is there no force to move them to each other on their own spontaneously, instead they have to collide with their own momentum, caused by someone stirring the solution?
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The logic looks fishy to me. Ionic salts can precipitate. Nor do I understand the "low polarity" of Ag+ and Cl-. One electron is the same charge on any ion. Where does this reasoning come from?
Besides that, I suppose you mean AgNO3 as a reactant.
Dissolution and precipitation are a matter of energy in the crystal and in the solution, as compared with the temperature. Or more cleanly, a matter of Gibbs energy. Considering only the energy in the solution is not enough.
And predictions based on qualitative properties or simple comparisons like the electronegativity are difficult. They use to fail in this case.
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Yes, AgNO3 is the reactant, mistyped it.
What you're saying is that, trying to predict solubility from how strong the ion-dipole interactions would be is wrong. What should be done, is to compare lattice and hydration energies and see if it's exothermic or endothermic. Only way to gather the energy values is by empirical evaluation, therefore trying to predict it is where the fault lies. Did I get it right?
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Don't forget entropy.
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What you're saying is that, trying to predict solubility from how strong the ion-dipole interactions would be is wrong. What should be done, is to compare lattice and hydration energies and see if it's exothermic or endothermic. Only way to gather the energy values is by empirical evaluation, therefore trying to predict it is where the fault lies. Did I get it right?
Yes, to the same reservation as Corribus pointed out, that G or µ determine equilibria, not E or H. A simple reasoning about it is that if energy leaves somewhere, it arrives elsewhere, so minimizing the energy can't be a driving force. It's the distribution of energy that counts.
Yes too, solubility is what determines a precipitation. The mere existence of double displacement tells that the ease of hydration of lone ions is not the whole picture, since all ions were dissolved prior to the precipitation. How well two (or possibly more) ions match to build a solid counts too.
I know no qualitative nor simple quantitative way to determine a solubility. Bigger ions are easier to hydrolyse, for instance (NO3)- where the charge is diffuse (like a charge has a smaller energy on a bigger sphere), but for a crystal, the energy depends on arrangement details which I don't imagine to evaluate by hand.
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In the precipitation step, the particles to be removed are part of the chemical reaction forming the precipitate. ... This is the same reaction type that you performed in the Experiment when the reaction between ions from two aqueous solutions produced a solid precipitate.
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What you're saying is that, trying to predict solubility from how strong the ion-dipole interactions would be is wrong. What should be done, is to compare lattice and hydration energies and see if it's exothermic or endothermic. Only way to gather the energy values is by empirical evaluation, therefore trying to predict it is where the fault lies. Did I get it right?
Yes, to the same reservation as Corribus pointed out, that G or µ determine equilibria, not E or H. A simple reasoning about it is that if energy leaves somewhere, it arrives elsewhere, so minimizing the energy can't be a driving force. It's the distribution of energy that counts.
Yes too, solubility is what determines a precipitation. The mere existence of double displacement tells that the ease of hydration of lone ions is not the whole picture, since all ions were dissolved prior to the precipitation. How well two (or possibly more) ions match to build a solid counts too.
I know no qualitative nor simple quantitative way to determine a solubility. Bigger ions are easier to hydrolyse, for instance (NO3)- where the charge is diffuse (like a charge has a smaller energy on a bigger sphere), but for a crystal, the energy depends on arrangement details which I don't imagine to evaluate by hand.
I see. Alright, one last question. An exothermic dissolution should always be spontaneous. Increasing the temperature during an exothermic dissolution will stop and then shift the equilibrium away from dissolution, but when we look at ΔG=ΔH-ΔTS, the ΔTS will increase while the others stay the same, lowering the already negative Gibbs energy even further. If the Gibbs is below zero, why does the reaction stop and turn the other way?
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If you think that lnK = -ΔG°/RT where K is the equilibrium constant
then lnK = -ΔH°/RT + ΔS°/R
Now, what affects the variation of K with temperature?
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If you think that lnK = -ΔG°/RT where K is the equilibrium constant
then lnK = -ΔH°/RT + ΔS°/R
Now, what affects the variation of K with temperature?
I got it this way, is this correct?
ΔG = ΔG° + RT lnQ
Q=[products]
lnQ = ΔH/RT - ΔS/R - ΔG°/RT
ΔH and ΔG° should be constant at every temperature so at standart temp this can be written
ΔG° = ΔH - 273ΔS
ΔS is positive for dissolution, thus ΔH > ΔG°
Q=e^((ΔH - ΔG°)/RT - ΔS/R) and we know that ΔS/R is constant, and ΔH - ΔG° is a positive term.
As T approaches infinity, limit of Q will e^(-ΔS/R) and as T approaches 0+ the limit will be infinity. Proving that concentration of products decrease as temperature increases.
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ΔG° is NOT constant with temperature. This is a very common mistake in equilibrium questions.
You are seriously considering the limits of solubility as T approaches 0 and infinity? (Note that - if it were relevant - ΔH and ΔS are not constant over the range T = 0 - ∞. We generally assume they are constant over a relatively small practical T range.)
It's much simpler than this. You have
ln K = -ΔH°/RT + ΔS°/R
The only term on the RHS that is temperature dependent (assuming, as a first approximation, that ΔH° and ΔS° are constant) is the first. Differentiating,
d(lnK)/dT = ΔH°/RT2
Thus whether K increases or decreases with temperature depends on the sign of ΔH°, not on ΔS°.
I'm also not convinced you fully grasp the difference between ΔG and ΔG°. You correctly say
ΔG = ΔG° + RTlnQ
At equilibrium, ΔG = 0 and Q = K, which gives us
ΔG° = -RTlnK
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ΔG° is NOT constant with temperature. This is a very common mistake in equilibrium questions.
You are seriously considering the limits of solubility as T approaches 0 and infinity? (Note that - if it were relevant - ΔH and ΔS are not constant over the range T = 0 - ∞. We generally assume they are constant over a relatively small practical T range.)
It's much simpler than this. You have
ln K = -ΔH°/RT + ΔS°/R
The only term on the RHS that is temperature dependent (assuming, as a first approximation, that ΔH° and ΔS° are constant) is the first. Differentiating,
d(lnK)/dT = ΔH°/RT2
Thus whether K increases or decreases with temperature depends on the sign of ΔH°, not on ΔS°.
I'm also not convinced you fully grasp the difference between ΔG and ΔG°. You correctly say
ΔG = ΔG° + RTlnQ
At equilibrium, ΔG = 0 and Q = K, which gives us
ΔG° = -RTlnK
I. At standart temperature and pressure, ΔG should be equal to ΔG° and ΔG° = ΔH°-TΔS°. T value should be constant and 298K. Is it the ΔH° and ΔS° changing with temperature, making ΔG° a T related function? If so, what's their relation with Temperature , and if not, what is it that's making the ΔG° vary with temperature?
II. I thought that when we heat up the solution, the reaction would start and we couldn't use the ΔG° = -RTlnK formula during that because the state of equilibrium wouldn't be established until it ended. But now I see that we can set up the formula for the next point of equilibrium, i.e we stop increasing the temp and reaction completes, satisfying the new K with more reactants and less products. I get it now. This is correct, yes?
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I. There is not one "standard condition". ΔG° refers to standard conditions at a particular temperature. There is a different value for ΔG° for each temperature. You cannot use ΔG°(298K) for other temperatures. What makes ΔG° vary with temperature is the T term in ΔG° = ΔH°-TΔS°. (Note that the variation of ΔG° with temperature depends on ΔS°, but the variation of ΔG°/RT with temperature depends on ΔH°. That's what often confuses people.)
II. If the reaction doesn't start, what use is the formula? The equation tells you how the position of equilibrium varies with temperature. If the system is not at equilibrium, it will adjust until Q = K. Whether that means shifting to the left or right depends on where the system is relative to the equilibrium state. So if you have a saturated solution at one temperature, and you increase the temperature, if the dissolution is exothermic the solubility will decrease, and solid will precipitate.
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I. There is not one "standard condition". ΔG° refers to standard conditions at a particular temperature. There is a different value for ΔG° for each temperature. You cannot use ΔG°(298K) for other temperatures. What makes ΔG° vary with temperature is the T term in ΔG° = ΔH°-TΔS°. (Note that the variation of ΔG° with temperature depends on ΔS°, but the variation of ΔG°/RT with temperature depends on ΔH°. That's what often confuses people.)
II. If the reaction doesn't start, what use is the formula? The equation tells you how the position of equilibrium varies with temperature. If the system is not at equilibrium, it will adjust until Q = K. Whether that means shifting to the left or right depends on where the system is relative to the equilibrium state. So if you have a saturated solution at one temperature, and you increase the temperature, if the dissolution is exothermic the solubility will decrease, and solid will precipitate.
Standart conditions are 298K and 1 atm, ΔH° and ΔS° are constant because they're measured at these exact standart T and P values. But ΔG° is not, it differs with temperature. Is it because it's only defined for the standart pressure (1 atm)? So that it stays constant throughout pressure changes?
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298K and 1 atm are commonly used standard conditions, but they are not absolute. You can define whatever standard conditions you like, as long as you are consistent in applying them. If you define ΔG° at a given temperature and pressure, then if you change the pressure at that temperature, ΔG will change but ΔG° will not. However, if you change the temperature, ΔG° will change. If we assume ΔH° and ΔS° do not vary with temperature (actually they do a bit, but often they can be treated as constant over a small range of temperature), then ΔG° varies due to the relation ΔG° = ΔH° - TΔS°.
(I am not now talking about things like "standard heats of formation" that you can find tabulated, usually referring to 298K and 1 atm. I'm talking about the value of ΔG° that you use in the relation ΔG° = RT lnK, and that is ΔG° at the temperature for which you want to determine K. K then reflects how the equilibrium pressures (or concentrations) differ from the standard pressure/conc. at that temperature.)
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298K and 1 atm are commonly used standard conditions, but they are not absolute. You can define whatever standard conditions you like, as long as you are consistent in applying them. If you define ΔG° at a given temperature and pressure, then if you change the pressure at that temperature, ΔG will change but ΔG° will not. However, if you change the temperature, ΔG° will change. If we assume ΔH° and ΔS° do not vary with temperature (actually they do a bit, but often they can be treated as constant over a small range of temperature), then ΔG° varies due to the relation ΔG° = ΔH° - TΔS°.
(I am not now talking about things like "standard heats of formation" that you can find tabulated, usually referring to 298K and 1 atm. I'm talking about the value of ΔG° that you use in the relation ΔG° = RT lnK, and that is ΔG° at the temperature for which you want to determine K. K then reflects how the equilibrium pressures (or concentrations) differ from the standard pressure/conc. at that temperature.)
I. Then the reason ΔG° differs from the tabulated empirical data (e.g ΔH°f) is that, the empirical data you mentioned is measured at both a certain pressure and temperature, thus it stays constant (or a slight, negligible change) throughout both T and P changes, while ΔG it's measured only at a certain pressure that is chosen as the standard, but there is NO standard temperature it is measured at, making it a function with the variable T. Did I get it right this time?
II. You said that ΔH° and ΔS° does in fact slightly vary with respect to T, how does that occur?
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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.
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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°?
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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.
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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°?
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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 ::equil:: 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.
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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 ::equil:: 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?
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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
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.
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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
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?
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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.
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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?
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If I understand you correctly, yes.
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Nice, that was all. Thanks for clarifying.