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Topic: Attraction between species during precipitate formation  (Read 1495 times)

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Offline QuiteThePredicament

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Attraction between species during precipitate formation
« on: March 04, 2019, 12:00:54 PM »
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?

Offline Enthalpy

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Re: Attraction between species during precipitate formation
« Reply #1 on: March 05, 2019, 04:02:01 PM »
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.

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #2 on: March 05, 2019, 06:43:40 PM »
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?

Offline Corribus

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Re: Attraction between species during precipitate formation
« Reply #3 on: March 05, 2019, 10:59:30 PM »
Don't forget entropy.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Enthalpy

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Re: Attraction between species during precipitate formation
« Reply #4 on: March 07, 2019, 07:00:10 AM »
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.

Offline naughtydelhi

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Re: Attraction between species during precipitate formation
« Reply #5 on: March 09, 2019, 04:57:09 AM »
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.

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #6 on: March 09, 2019, 09:25:14 AM »
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?
« Last Edit: March 09, 2019, 10:31:26 AM by QuiteThePredicament »

Offline mjc123

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Re: Attraction between species during precipitate formation
« Reply #7 on: March 11, 2019, 05:33:15 AM »
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?

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #8 on: March 11, 2019, 03:07:13 PM »
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.   

Offline mjc123

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Re: Attraction between species during precipitate formation
« Reply #9 on: March 12, 2019, 05:45:48 AM »
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

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #10 on: March 12, 2019, 08:17:29 AM »
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?         

Offline mjc123

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Re: Attraction between species during precipitate formation
« Reply #11 on: March 12, 2019, 10:06:59 AM »
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.

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #12 on: March 12, 2019, 12:20:16 PM »
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?

Offline mjc123

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Re: Attraction between species during precipitate formation
« Reply #13 on: March 12, 2019, 01:49:58 PM »
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.)

Offline QuiteThePredicament

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Re: Attraction between species during precipitate formation
« Reply #14 on: March 12, 2019, 02:32:33 PM »
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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