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Attraction between species during precipitate formation
naughtydelhi:
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.
QuiteThePredicament:
--- Quote from: Enthalpy on March 07, 2019, 07:00:10 AM ---
--- Quote from: QuiteThePredicament on March 05, 2019, 06:43:40 PM ---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?
--- End quote ---
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.
--- End quote ---
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?
mjc123:
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?
QuiteThePredicament:
--- Quote from: mjc123 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?
--- End quote ---
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.
mjc123:
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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