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Author: Subject: What about; two single molecule reaction?
maxpayne
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[*] posted on 5-8-2012 at 11:56
What about; two single molecule reaction?


As we all know this: H2SO4 + 2NaOH -> Na2SO4 + 2H2O

... my question is: What if single molecule of H2SO4 meets with single molecule NaOH, will they react? If not, why, when we all say that acids reacts with bases.

Also does probability allows such situation?
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Vargouille
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[*] posted on 5-8-2012 at 12:08


They will react to form sodium bisulfate.

H2SO4 -> H+ + HSO4-
NaOH -> OH- + Na+
H+ + OH- <=> H2O
Na+ + HSO4- <=> NaHSO4
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blogfast25
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[*] posted on 5-8-2012 at 12:50


maxpayne:



In watery medium there really are no ‘molecules’ like that.

NaOH is a stong base and dissociates completely in water:

NaOH(s) === > Na+(aq) + OH-(aq)

H2SO4 is a strong acid and also more or less completely dissociates in water:

H2SO4(l) + H2O(l) === > H3O+(aq) + HSO4 -(aq)

When both solution are mixed together the only chemical reaction taking place is:

H3O+(aq) + OH-(aq) === > 2H2O (l)

The sodium and sulphate ions play no part whatsoever. They’re so-called ‘spectator ions’.




[Edited on 5-8-2012 by blogfast25]

[Edited on 6-8-2012 by blogfast25]




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maxpayne
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[*] posted on 6-8-2012 at 07:06


blogfast25, thank you very much, if I only had a teacher like you back then ;)

Let's keep up on this: NaOH(s) === > Na+(aq) + OH-(aq)

This dissociation is what I suspected it was happening, but please, tell me how this ions look like in solution, are they really free or bonded somehow and from where the energy came, that is needed for dissociation??

This energy question is my big ???.
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blogfast25
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[*] posted on 6-8-2012 at 08:00


They’re solvated to some degree: surrounded by water molecules that are loosely ‘bound’ to them because H2O is a permanent dipole.

When dissolving an ionic solvent, the energy needed is basically governed by Gibbs Law: ΔG = ΔH – TΔS. ΔH is the Enthalpy of Solution and is basically the energy to break up the ionic lattice, it is always positive (ENDOthermic). ΔS is the Entropy change, it is always positive (solutions are more disordered than ionic lattice) and thus - TΔS is always negative. Whether an ionic compound is soluble in water basically depends on whether ΔG is negative or not (negative = soluble).

But there’s a complication, at least in many cases. When solvated ions are formed, usually solvation energy is released and as a result the dissolution of many ionic compounds is EXOthermic. Interesting examples are both NaOH and H2SO4, the dissolution process of which in both cases releases considerable amounts of Enthalpy: the solutions heat up as you make them.

In reality we cannot distinguish between the energy needed to break up the lattice and energy released due to solvation. These overall energies are determined experimentally.




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maxpayne
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[*] posted on 6-8-2012 at 10:52


Very, very interesting.

It is a large subject to study this Gibbs free energy, and I have too many questions now to ask here about energy origins, etc. So let's get back to that spectator ions.

You actually said before that sodium and sulfate ions "watch" the reaction in solution of other "relatively free" ions. So, it seems to me that sodium sulfate begins to exist only when there is no water and it is wrong to say that when I mix before mentioned solutions, I get Na2SO4. Only when I remove the water (evaporating) I produced Na2SO4?
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blogfast25
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[*] posted on 6-8-2012 at 11:40


Yes, that's quite correct. Although we tend to see a sodium sulphate solution as a, erm... sodium sulphate solution, more modern interpretations tend to see the solute (sodium sulphate)/solvent (water) system as a system in its own right with its own properties.

Certainly a sodium sulphate solution doesn't share many properties with the 'pure substance', Na2SO4(s),of which in itself different (solid) hydrates exist (each with their own properties, such as MP and BP). These hydrates can co-exist in equilibrium with sodium sulphate solutions of various concentrations (see sodium sulphate - water phase diagrams).

[Edited on 6-8-2012 by blogfast25]




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maxpayne
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[*] posted on 6-8-2012 at 12:36


blogfast25, thank you for your time and effort, I see you have quite knowledge in this "invisible chemistry", and you certainly did answered as a professional. I did learned something useful now, and it seems that things are not always as they look at the first glance.

[Edited on 6-8-2012 by maxpayne]
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blogfast25
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[*] posted on 6-8-2012 at 12:57


You're welcome. There are plenty good texts available online about this subject. Search and ye shall find.



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bbartlog
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[*] posted on 7-8-2012 at 06:22


Although the case of H2SO4 + NaOH is pretty straightforward, I think there are some other ones that are less so. For example

Al + Cl2 --> ??
Li + O2 --> ??





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blogfast25
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[*] posted on 7-8-2012 at 06:24


Quote: Originally posted by bbartlog  
Although the case of H2SO4 + NaOH is pretty straightforward, I think there are some other ones that are less so. For example

Al + Cl2 --> ??
Li + O2 --> ??



Sure but direct union of elements are a different class altogether!




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barley81
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[*] posted on 7-8-2012 at 09:42


Blogfast, enthalpy of solution and entropy of solution do not have to be positive! Calcium chloride dissolves exothermically, while ammonium nitrate dissolves endothermically (enthalpy of solution for calcium chloride is negative; enthalpy of solution for ammonium nitrate is positive).


Entropy of solution need not be positive. Hydrocarbons have negative entropy of solution in water:
http://www.mpcfaculty.net/mark_bishop/solubility_entropy.htm
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maxpayne
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[*] posted on 7-8-2012 at 12:12


I think that blogfast already mentioned complication, and without further thinking, my sense tells me that mathematical equation do not explain everything, especially in Gibbs law. For me, it is just a fancy equilibrium math, nothing else.
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blogfast25
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[*] posted on 7-8-2012 at 12:36


Quote: Originally posted by barley81  
Blogfast, enthalpy of solution and entropy of solution do not have to be positive! Calcium chloride dissolves exothermically, while ammonium nitrate dissolves endothermically (enthalpy of solution for calcium chloride is negative; enthalpy of solution for ammonium nitrate is positive).


Entropy of solution need not be positive. Hydrocarbons have negative entropy of solution in water:
http://www.mpcfaculty.net/mark_bishop/solubility_entropy.htm



I never said anything of the sort: you badly read what I wrote.

ΔG has to be negative for dissolution to occur, not necessarily ΔH.

When dissolution is endothermic that’s because TΔS > ΔH. Even endothermic dissolution means ΔG < 0. And ΔS is always positive.

Where dissolution is exothermic, that's caused by solvation energy (always negative).

Re. your link, nice page but sorry, but I don’t believe that:

<i>”Overall, the attractions in the system after hexane and other hydrocarbon molecules move into the water are approximately equivalent in strength to the attractions in the separate substances. For this reason, little energy is absorbed or evolved when a small amount of a hydrocarbon is dissolved in water. To explain why only very small amounts of hydrocarbons such ashexane dissolve in water, therefore, we must look at the change in the entropy of the system. It is not obvious, but when hexane molecules move into the water layer, the particles in the new arrangement created areactually less dispersed (lower entropy) than the separate liquids. The natural tendency toward greater dispersal favors the separate hexane and water and keeps them from mixing.”</i>

This argument is pulled by the hair. Far more likely is that the hydrogen bonds in water cannot be easily broken by the hydrocarbon. But a weak solution of hexane in water still has higher entropy than the separate fluids, undissolved in each other.


[Edited on 7-8-2012 by blogfast25]




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blogfast25
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[*] posted on 7-8-2012 at 12:39


Quote: Originally posted by maxpayne  
For me, it is just a fancy equilibrium math, nothing else.


No Sir. This thermodynamic principle underpins much of chemistry, including theory on what and what doesn't dissolve in each other.




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maxpayne
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[*] posted on 7-8-2012 at 12:47



Code:
No Sir. This thermodynamic principle underpins much of chemistry, including theory on what and what doesn't dissolve in each other.


Then explain to me origins of the energy needed for dissociation. Equation uses only quantities, it does not explain themselves.
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blogfast25
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[*] posted on 7-8-2012 at 12:57


This energy is known as lattice energy or coulombic energy. Imagine the ions of NaCl (Na+ and Cl-) to be at infinite distance from each other (at atomic scale even infinite isn’t very large). These ions are attracted by electrostatic attraction (coulombic attraction) and thus have Potential Energy. When allowed to move towards each other this potential energy is converted into heat (enthalpy).

But when dissolving the lattice, the exact opposite must happen and we must expend energy (enthalpy) to rip the lattice apart. Without solvation energy the dissociation energy is therefore closely related to the lattice energy. Exact values for molar lattice energies are known and tabled, for many binary ionic compounds.

The lattice energy usually makes up the most part of the Enthalpy of Formation of (solid!) binary ionic compounds, see Born Haber Cycle:

http://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle


[Edited on 7-8-2012 by blogfast25]

[Edited on 7-8-2012 by blogfast25]




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