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CHRIS25
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Trying to understand Bond energies
Totally new subject to learn. It all began when I asked myself - what does it take to break a bond? But in order to understand something in this I
asked why is NaOH in water an exothermic reaction, and herein lies the first of no doubt many relentless days of confusion.
One thing I read was that Bond Breaking is Endothermic and Bond Making is Exothermic, NaOH dissociates into Na and OH ions, it is an ionic compound
and has no bonds, but it is exothermic, no bonds are being broken. Therfore something else is happening. Na is not combining with anything so this is
not causing any heat to be given off (which I need to call energy)? So the culprit has to be the OH ion. Since we get an alkaline solution there has
to be a reaction somewhere. The OH therefore is NOT floating around in the water as I have always read for three years, it can't be, it has to be
doing something. If it was floating around doing nothing then there would be a neutral solution right? This is only logical really I think.
The OH is interacting with the H in the water to make the H3O so this is producing 'energy'? And the water is acting as a weak acid?
Lastly I was trying to understand how to calculate the bond energies here. My reference is this http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociatio...
But then I realised that this is about bond breaking, whereas I need to find energies for bond making? So this table http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_B... tells me that an H to H bond takes 432 Kj/mol of energy to combine, and that means
an endothermic reaction since it is a positive figure, this is where I am all tied up and need some guidance.
Ok answer found, one of the instructions was misleading a little, apparently the numbers can be either positive or negative depending on whether
energy is released or absorbed. So the H-H bond that is forming is a negative number, because bond making is always exothermic.
[Edited on 23-1-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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gdflp
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I believe that this is what you're looking for. In all aqueous solutions whether they have an acid or base dissolved in them the product of the
H3O+ concentration and the OH- concentration is equal to 1 x 10^-14M. In pure water, the concentrations
are equal and are both 1 x 10^-7M. When NaOH is added, some of the hydroxide ions react with the H3O+. For example if 40g of
solid sodium hydroxide is added to one liter of water, the OH- concentration becomes 1M, consequently the H3O+
concentration becomes 1 x 10^-14. This is a acid-base neutralization and releases an insignificantly small amount of energy. Another contributor is
the hydrogen bonding between the water molecules and the OH- ions. The main contributor however is the breaking of the ionic bonds between
Na+ and OH-.
[Edited on 1-23-2015 by gdflp]
[Edited on 1-23-2015 by gdflp]
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Molecular Manipulations
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Van der Waals forces is what you're looking for.
Water molecules are attracted to solid sodium hydroxide. Why? Water molecules are polar and sodium hydroxide is made of ions. These two are attracted
through the ion-dipole interaction, which is a van der Waals force.
The positive part of the water molecule (the hydrogens) attract the negative hydroxide ion while the negative oxygen part attracts the positive sodium
ion.
But why does sodium hydroxide dissolve exothermically while say ammonium nitrate dissolves endothermically? Because all ionic crystals have lattice
structures. When a solid lattice forms, energy is released. When a substance is dissolved, the lattice is pulled apart and destroyed - this requires
energy. This is where van der Waals comes in, there are three van der Waals forces at play here. The "solvent-solute attraction", the
"solvent-solvent attraction" and the "solute-solute attraction".
The relative strengths of each force are critical in determining whether it will dissolve endothermically, exothermically, or not at all.
When the solute-solvent attraction is greater than the solute-solute attraction the substance will dissolve exothermically.
So in a sense you're right about the breaking of bonds (the crystal lattice). But it gets way more complicated than that, I've tried to simplify it as
much as possible.
Does this answer your question?
[EDIT] It's got nothing to due with the self ionization of water, nor the hydroxide's reaction with the hydronium ions. Not everything is an
acid-base reaction!
[Edited on 23-1-2015 by Molecular Manipulations]
-The manipulator
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gdflp
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Quote: Originally posted by Molecular Manipulations  | It's got nothing to due with the self ionization of water, nor the hydroxide's reaction with the hydronium ions. Not everything is an acid-base
reaction!
[Edited on 23-1-2015 by Molecular Manipulations] |
I said that it was a small amount of energy, I didn't say that it was significant. And how is that not an acid-base reaction, hydronium ions are
reacting with hydroxide ions!
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Molecular Manipulations
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I was talking generally about the released energy and where it came from, which isn't an acid base reaction but several van der Waals forces, causing
sodium hydroxide's affinity to dissociate in water.
What you said is an acid-base reaction, but not the cause of the released energy. Even the heat from any acid-base reaction is also because
of van der Waals forces.
-The manipulator
We are all here on earth to help others; what on earth the others are here for I don't know. -W. H. Auden
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CHRIS25
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Ok, this is interesting, ask myself a simple yet profound question, give myself weeks of work and cause an argument half way around other side of the
world. I have to learn about van de waals - ok; Enthalpy, now there is a new word! So as not to keep you guys may I ask one more thing. I was
originally asking myself the following: Apart from the concepts of ionic and covalency and the Pauline scale and electronegativity (we are talking
basic here), what other 'basic' concepts should I learn in order to be able to intelligently predict whether a reaction would occur to produce a new
compound?
A very quick look at enthalpy (have not yet read about van de waals) I was struck by this:
Breaking solute-solute attractions (endothermic), see for instance lattice energy in salts.
Breaking solvent-solvent attractions (endothermic), for instance that of hydrogen bonding
Forming solvent-solute attractions (exothermic), in solvation.
NaOH is NOT a bond. NaOH dissociates and this does NOT produce energy. Na floats around doing nothing. OH however qualifies for the Solvent -
Solute bond here above, No?
[Edited on 23-1-2015 by CHRIS25]
Ok, complicated now, Solvation, not what I thought it was, the Na ion is being surrounded by water molecules, there is an attraction here and that
qualifies as a solvent - solute I suppose?
[Edited on 23-1-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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Molecular Manipulations
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Yes the last thing you said is correct. There's nothing here that just "floats around".
NaOH is not a bond, per say, but there's still an attraction to itself from the equal and opposite charges, which cause the lattice structure,
breaking that is endothermic. This is the solute-solute attraction. But in this case the solvent-solute attraction is more exothermic than the
solute-solute attraction is endothermic, thus the total free energy is -, indicating an overall exothermic reaction.
-The manipulator
We are all here on earth to help others; what on earth the others are here for I don't know. -W. H. Auden
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gdflp
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In NaOH the O--H bond is covalent obviously, but the Na ion is also bonded to the OH ion. It's not covalent, but they're both charged ions being
attracted to each other which is the definition of an ionic bond. This requires energy to break, but the energy from the solvent-solute interactions
is greater than the energy required to break the ionic bond.
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CHRIS25
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Molecular and gd - thankyou for urging me forward a bit with the discussion, some helpful pointers here and a little confirmation was appreciated. A
side issue, not to be pedantic of course, but I did see the O-H as a covalent bond, and the Na-OH as an ionic bond, that was until within the context
of this ionization energy I read that ionic compounds do not have bonds between cation and anion, they do not share electrons. Now this seems
pedantic in its definition really, what is a bond what is not a bond, I mean what is important is that they are linked/bonded. So it was not the
first time I read this actually. Now I don't keep references to everything I read otherwise I would run out of disc space, so can not link here
unfortunately. But why would anyone write this. Rhetorically speaking, no question mark here..
Anyway thanks.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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Darkstar
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| Quote: Originally posted by CHRIS25 | | A side issue, not to be pedantic of course, but I did see the O-H as a covalent bond, and the Na-OH as an ionic bond, that was until within the
context of this ionization energy I read that ionic compounds do not have bonds between cation and anion, they do
not share electrons. Now this seems pedantic in its definition really, what is a bond what is not a bond, I mean what is important is
that they are linked/bonded. |
Oh, but they DO share those electrons! Despite what you're taught in school, "ionic bonds" and "covalent bonds" are actually the exact same bonds. The
atoms in a covalent bond are attracted to one another for the same reason the atoms in an ionic bond are. Both are caused by electrostatic attraction,
and both are the result of electrons becoming attracted to more than one nucleus simultaneously (and those nuclei becoming attracted back).
In reality, the oxygen and sodium atoms in NaOH are sharing those electrons to some degree. While the extreme difference in electronegativity
causes the electron density to end up primarily around oxygen, it isn't completely localized there, and is still being pulled towards sodium ever so
slightly. (technically, it's being pulled towards several sodium atoms) Truly ionic bonds, i.e. where the electrons are 100% localized and aren't
being shared what so ever, don't exist. A positively-charged nucleus is inevitably going to attract some electron density. There simply isn't
enough distance between the ions for that not to happen.
A bond is just said to be ionic when it's more ionic than it is covalent. In other words, "ionic bond" is another way to say "extremely polar
covalent bond."
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CHRIS25
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Yep, I was looking through my notes and can quite clearly understand why you are right. The ionic and covalent bond is a grey area and separation of
the definition seems just for convenience, I get that. So much to learn here. Trouble with learning form the internet is that I always source at
least 10 different references and make sure that everyone is agreeing, and for a long time I was under the impression that ionic and covalent were
static separations with a definitive cut-off point. So the water breaks apart the NaOH and the water itself is breaking apart into H and O ions and
this is endothermic, then he attraction between the OH and the Na to the surrounding water molecules via their + and - charge release energy leading
to an exothermic ending.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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| Quote: Originally posted by CHRIS25 | | Totally new subject to learn. It all began when I asked myself - what does it take to break a bond? But in order to understand something in this I
asked why is NaOH in water an exothermic reaction, and herein lies the first of no doubt many relentless days of confusion. |
The question 'what is a chemical bond?' is one that has pre-occupied scientists for the last 200 years or so and has only been resolved with the
advent of Quantum Mechanics (aka 'wave mechanics') in the early part of the 20th century. It lies at the heart of chemistry, which decribes
essentially the breaking and making of chemical bonds, known also as chemical reactions.
Forget for a minute the finer points re. covalent v. ionic bonds.
To try and understand what chemical bonds are, start from the simplest type known: the H-H bond, aka the σ bond (sigma bond), which holds together
diatomic hydrogen (H<sub>2</sub> or 'hydrogen', to you and me). With its many variations on the H-H theme, this type of bonds probably
accounts for 75 % or more of all chemical bonding.
But to understand how σ bonds work you need to understand firstly how electrons reside in a single, free hydrogen atom. In chemical/physical jargon:
what is the "atomic orbital" (no, NOT 'orbit'!) of that lone electron that makes up a hydrogen atom (together with the proton nucleus, of course!)?
A single, free hydrogen atom is a proton as nucleus and a single electron moving in the proton's central electrical field. Due to its almost
unimaginably small size this system is a so-called Quantum System and its properties are described by the Schrödinger Equation, which describes the
moving electron as a wave rather than as a particle.
From free hydrogen's quantum mechanical properties can also be deduced that a related quantum system comprised of 2 protons and 2 electrons
also exists, due to the formation of a σ bond : H-H, where the hyphen represents the σ bond and the two H the two protons. While the atomic orbital
of free hydrogen atoms contain only 1 electron, the molecular orbital of H<sub>2</sub> contains 2 electrons 'residing' mainly in the
space between the 2 protons (H).
Simple σ bonds hold together many so-called homodiatomic molecules like all the halogens (X<sub>2</sub> .
But most compounds containing more than one type of atom and made of single bonds only are made of σ bonds only, for example all of the alkanes
(C<sub>n</sub>H<sub>2n+2</sub>.
As regards the bond energy, in the case of a simple H-H sigma bond, if we represent an free hydrogen atom as H* then the bond energy is simply the
Reaction Enthalpy (per mol of reaction product) for:
2 H*(g) === > H<sub>2</sub>(g)
<b>For a layman's level overview of these concepts I recommend studying (no less!) the following web pages:</b>
http://www.chemguide.co.uk/atommenu.html
Other types of basic chemical bonds like p bonds, π bonds, ionic bonding etc, become relatively easy to understand once the principle of the σ bond
is understood.
Good luck with your studies!
[Edited on 27-1-2015 by blogfast25]
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aga
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I found this to be an interesting visualisation of the s,p and d orbitals :-
https://www.youtube.com/watch?v=K-jNgq16jEY
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blogfast25
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Definitely one of the better videos on orbitals. Good find!
Careful with false idolatry though: the images are useful but don't tell the whole story, not by a long shot.
Knowledge question: what does this:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mecha...
... have to do with atomic orbitals?
For 10 points.
[Edited on 26-1-2015 by blogfast25]
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Darkstar
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I'm a lot stronger in chemistry than I am in physics (especially quantum mechanics), but I'll give it a try. Like the particle in the box, an electron
around the nucleus of an atom is in a bound state. Its attraction to the positively-charged protons creates a potential energy well, localizing the
electron as a wave packet and giving it quantized energy levels. The lowest possible energy of the electron is thus n=1, as its energy cannot be zero
since restricting the electron to a single point in space and time would require infinite momentum. Like the particle in the box, the probability of
finding the electron in some region of space changes as its energy increases, as do the number of regions where there's a zero probability of it being
found.
However, unlike the particle in the box, the electron isn't trapped in an infinite potential well, and can escape given a high enough energy state.
[Edited on 27-1-2015 by Darkstar]
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CHRIS25
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Allow me to translate the above for a 2% share in Darkstar's points: electrons in the outermost shells of atoms are partially screened from the
positively protons on the inner shells. This distance from the protons coupled with the number of electrons on the inner shells is called the binding
energy (total energy + kinetic energy is E 'psi' from Walker-schrodingers equation). The energy required to remove an electron from its atom and
leave therefore an ion in solution is the Negative ionization energy. The positive or opposite is represented by the pauling scale where this is the
strength of an atom to pull electrons away from other atoms. (Not going well is it....... ) Well I am trying desperately to find a relationship
between, or a connection, between all these ionization energies, pauling scale, gibbs free energy figures: As isolated concepts they are
understandable, as mathematical formulas they are understandable to a degree, the reasonings are not difficult to assimilate, but and this a big but,
they remain at this moment (after three days) in isolation of one another and this makes it difficult to see how they come together to form a cohesive
understanding of the actual physical process of two chemicals reacting. So I am drawing a NaCl + H2SO4 and then apply all the above to every atom in
that reaction in order to understand this visually. Numbers and delta signs and Kl/mole on paper are useless without the practicality of relating all
of this to physically pouring salt into sulphuric acid for example. So I guess I don't deserve even 2%.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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The particle (electron) in a box can be seen as essentially a simply analogy of a hydrogen atom.
The time-independent Schrödinger Equation treats the moving electron as a standing (matter) wave, with ψ(x,y,z) the amplitude of the wave
distributed over space (x,y,z). At any point in space, ψ<sup>2</sup>(x,y,z) ['psi squared'] represents the probability of finding the
electron at location (x,y,z).
The electron in a box is bound by the walls of the box, hence it is confined to the box. In the case of a hydrogen atom the electron is bound to the
vicinity of the nucleus by the electrostatic attraction between the proton (nucleus) and the electron. Both are examples of 'bound electrons'.
Solving the Schrodinger equation (SE) yields the various eigen functions ψ<sub>1</sub>, ψ<sub>2</sub>,
ψ<sub>3</sub>, etc for n = 1, 2, 3, etc (with n the 'quantum number') and the energy levels E1, E2, E3, etc associated with them.
Solving the SE for a particle in a box is much easier than for a hydrogen atom because the box is one dimensional and because there is no Potential
Energy term but the solutions for both show strong similarities nonetheless: both systems are quantised, wavy and with non-zero Energy ground states
(E<sub>1</sub> > 0).
That electrons are present in atoms as waves is relevant because waves are additive and that is, very simply put, what happens when molecular orbitals
are formed: the atomic orbitals involved are 'added up'.
[Edited on 27-1-2015 by blogfast25]
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blogfast25
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Chris:
No mate, you need to learn this step by step. You're conflating all sorts of related concepts and end up with 'word salad'.
Patience.
Let's take a look at this:
| Quote: Originally posted by CHRIS25 | | Totally new subject to learn. It all began when I asked myself - what does it take to break a bond? But in order to understand something in this I
asked why is NaOH in water an exothermic reaction, and herein lies the first of no doubt many relentless days of confusion. |
You are referring to the fact that dissolving NaOH in water is strongly exothermic and asking why that is, right?
Solid NaOH is a three dimensional lattice made up of equal amounts of positively (+1) charged sodium ions and negatively charged (-1) hydroxide ions.
The electrostatic attraction forces of these ions are what we call the ionic bonds and what holds the crystal together.
To dissolve this lattice all these bonds have to be broken and that costs a very appreciable amount of energy ('enthalpy'). So why is this
dissolution so exothermic and not endothermic?
Because something else happens at the same time of the dissolution: solvation.
When the sodium and hydroxide ions enter the solution they interact strongly with the solvent, in this case water.
Water molecules are so-called permanent dipoles. Imagine a water molecule a bit like a boomerang shaped object, with the oxygen atom at the bend and
the two hydrogen atoms at the ends. Due to strong difference in electronegativity between O and H, the covalent H-O bonds are polarised, with a
partial negative charge residing at the oxygen atom and a (smaller) partial charge residing at the hydrogens.
When sodium ions enter the water, water molecules orient themselves in the electrical field of that ion, so that the oxygen side faces the sodium ions
and the hydrogen side away from them.
For the hydroxide ions and water molecules the situation is similar but opposite.
It is this orienting of water molecules vis-a-vis the sodium and hydroxide ions that causes large amounts of heat to be released, much larger than the
energy needed to pull apart the lattice itself.
The overall process is thus strongly exothermic, due to this solvation enthalpy.
[Edited on 27-1-2015 by blogfast25]
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CHRIS25
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"It is this orienting of water molecules vis-a-vis the sodium and hydroxide ions that causes large amounts of heat to be released, much larger than
the energy needed to pull apart the lattice itself."
Yes Ok I get that I understand, but step by step now, the standard enthalpy of formation for NaOH is -411 KJ/mole. Water does not have any? because
it is an elemental and it had no reactants to produce it (in this context only). Therefore the difference between the enthalpy of the reactants and
the products is in this instance 'Zero', and the way I am looking at this reaction, the maths seems tell me that there is no positive or negative
enthalpy happening. On top of this where is the enthalpy for pulling the NaOH apart? It must be in the H2O, in which case there is an endothermic
reaction here since energy Must be absorbed surely? So where is the positive enthalpy data for this sort of thing or am I completely off the rails
here? I have watched about 7 videos on this subject and it's easy for say NaCl and H2SO4, NaCl has -411 and the acid -814, Na2SO4 -1387 and HCl (2x
92 aq) is -184 KJ/mole. All this calculates to -346 KJ/mole energy released (exothermic). Why can I not calculate this for NaOH in water?
[Edited on 27-1-2015 by CHRIS25]
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
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The Standard Enthalpy of Formation of NaOH is the Reaction Enthalpy for:
Na(s) + 1/2 O2(g) + 1/2 H2(g) === > NaOH(s) at STP.
But that value has NOTHING to do with the dissolution energy because the NaOH is neither being formed nor destroyed, only transformed (dissolved).
The dissolution energy is basically the sum of two energies:
1. the energy needed to pull apart the entire lattice. It's an endotherm, say ΔH<sub>1</sub>
2. the solvation energy. That's an exotherm, say ΔH<sub>2</sub>
The overall energy released or consumed by the sum of these two processes is simply say ΔH = ΔH<sub>1</sub> +
ΔH<sub>2</sub>. The second term is more negative than the first one is positive, so ΔH < 0. Exothermic.
Of course water has a value for Standard Heat of Formation [- 285.8 kJ/mol] but it has nothing to do with this issue either: no water molecules are
being created or destroyed.
For the most part heats of dissolution can not be calculated from theory, only determined by experiment.
[Edited on 27-1-2015 by blogfast25]
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CHRIS25
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Oh! My brain has been severely endothermonized and is in a state of positive entropy ....... At this moment in time all I can figure out is that if the energy released in forming new bonds is greater than the energy
contained in the reactants then this points to a strong bond, the inverse then is also true. So that if the enthalpy of the formation for the
reactants is lower than the enthalpy of the products the product has a strong bond. The gibbs free energy is a measure of entropy and enthalpy added
together where temperature is multiplied by the entropy. But while I understand what entropy is, (ie, disorder and order briefly, the videos explain
it well) what they fail to explain is the practicality of what these figures really mean in practical terms. IE. I look at aluminium chloride and see
the KJ/mole for enthalpy, entropy and Gibbs free energy. The enthalpy I have cracked, that at least I can visulaize as a practical help in looking at
a balanced equation and doing an experiment, the Gibbs free energy is supposed to be the difference between the enthalpy and entropy, but that
explains nothing about its purpose in a practical experiment even though the theory makes sense the practical down to earth process of viewing a
balanced equation in this light has me lost, and the entropy makes sense in theory BUT is just darn confusing when applied to an experiment. I have
pages of notes and diagrams, but I fail to grasp its purpose when looking at a balanced equation and asking what might happen if I mix these two
reactants.
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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aga
Forum Drunkard
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Sooo ...
a common noob error is to try to work out enthalpy from the wrong involved species then ?
This is sounding kinda familiar.
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CHRIS25
National Hazard
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| Quote: Originally posted by aga | Sooo ...
a common noob error is to try to work out enthalpy from the wrong involved species then ?
This is sounding kinda familiar. |
Are there enough us to start a group then?
‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some
Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)
Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)
The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by
precision and law. (me)
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blogfast25
International Hazard
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Location: Neverland
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Chris:
At this point in time don't try to relate bond strength/bond energy to related concepts like change of Gibbs Free Energy during a reaction just yet.
Crawling > walking > running, remember?
For now try and understand the workings of σ and p bonds as well as you can.
Then we can try and move forward.
BTW, as Einstein once said: 'Education by Utoob vids doesn't work'
Or in any case, that's what he WOULD have said had he been with us today! 
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blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
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You need a focus group first!
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