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blogfast25
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Acids/bases: heats of neutralisation
This is a follow up on the back end of this thread here:
http://www.sciencemadness.org/talk/viewthread.php?tid=14164#...
... where it was suggested by some that the neutralisation of a weak acid with a strong base would produce less enthalpy than the neutralisation of a
strong acid with a strong base. I decided to put that assumption to the test.
So far we have the following data points:
1) The enthalpy of neutralisation of a strong acid with a strong base is approx. - 57.3 kJ/mol of water (literature).
2) The neutralisation of NaHCO3 with a strong acid is endothermic (literature).
3) The neutralisation of Na2CO3 with HCl appears experimentally neither much exothermic nor much endothermic.
4) The neutralisation enthalpy of acetic acid with NaOH was found experimentally to be about - 54.5 kJ/mol of water, so quite close to - 57.3 kJ/mol
of water. The strength of the acetic acid (distilled vinegar) will be experimentally verified by titration tomorrow.
Some new data points are:
5) The neutralisation enthalpy of NH4Cl (a very weak acid: pKa = 9.25) with NaOH (strong base) was experimentally found to be - 5.1 kJ/mol of water
and for (NH4)2SO4 (half a mol), - 5.0 kJ/mol of water, so much less than - 57.3 kJ/mol for strong bases with strong acids.
6) The neutralisation enthalpy of NH3 with HCl was experimentally found to be -9.7 kJ/mol of water, so also much lower than - 57.3 kJ/mol. The
strength of both the NH3 and HCl solutions have been verified with acid/base titrometry.
All in all quite inconclusive, I'd say. It'd be interesting to include some other alkanoic acids like 1-propanoic, 1-butanoic etc because we know
these decrease gradually in acid strength (Ka) as the aliphatic group increases in length...
[Edited on 31-7-2010 by blogfast25]
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12AX7
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It should vary with pKa because pKa is an indirect measure of energy. The neutralization of pH ~14 with pH ~ 7 should have half as much energy as to
pH ~0. And the neutralization of an alkyl metal (pKa ~ 50) in water certainly has more energy. I forget what the conversion is though.
Tim
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blogfast25
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Hi Tim,
I made that point here:
http://www.sciencemadness.org/talk/viewthread.php?tid=14164#...
For low Ka (< 1), the DG (delta G) of:
HA + H2O ---> H3O+ + A-
... is positive. Assuming no entropic effects, DH (delta H) for the dissociation reaction is then also positive. But for the acetic acid the data
point really doesn't fit that theory at all.
A bit more later...
[Edited on 1-8-2010 by blogfast25]
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blogfast25
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The acidity of my grade of 'distilled vinegar' was confirmed to be 4.65 w% of acetic acid, for '4.5 % acidity' advertised on the bottle, so the
molarity used in the calculation was correct.
I'm wondering whether there is some entropic effect that I'm overlooking:
in HA(aq) + H2O(l) ---> H3O+(aq) + A-(aq)
... which side is the most probable state, most chaotic, most entropic, left or right?
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Nicodem
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I would tend to believe that in acid/base reactions where non-fully dissociated species in water react it is the enthalpy of solvation changes that
contributes most. For example:
CH3COOH(aq) + OH<sup>-</sup>(aq) <=> CH3COO<sup>-</sup>(aq) + H2O
Acetic acid is solvated via H-bond (mostly as donor) but I think acetate should be more solvated. Most of the acetic acid is not dissociated, so it
takes some energy to break the O-H bond (pKa being the equilibrium constant). Perhaps the solvation changes compensate for the energy loss in O-H bond
breaking. Hydroxyde is highly solvated in water so it takes some energy to desolvate it, but this energy is already included in the -57.3 kJ/mol
enthalpy value.
CO3<sup>2-</sup>(aq) + 2H<sup>+</sup>(aq) <=> CO2(g) + H2O
Carbonate is strongly solvated while CO2 is not (leaves the reaction as gas), so a lot of energy is consumed in desolvating the carbonate anion. I
would not be surprised that all the energy from protonation is required for this, thus leaving the overall enthalpy close to zero. Protons are
extremely tightly solvated, but again this energy is already included in the -57.3 kJ/mol enthalpy value.
NH4<sup>+</sup> + OH<sup>-</sup>(aq) <=> NH3(aq) + H2O
The ammonium ion is surely more solvated than ammonia, so this change takes a lot of energy. Same with breaking the relatively strong N-H bond...
You can also look at this as entropic changes given that it is about changes in the overall (dis)order of the system, but from the perspective of
individual ions it is all about how much energy it takes to break the solvation bonds with the solvent or how much energy is gained by forming such
bonds. Obviously the overall reaction enthalphy is the sum of all these changes plus the enthalpy of the H<sup>+</sup>(aq) +
OH<sup>-</sup>(aq) reaction.
Edit: Mostly changed the interpretation of the acetic acid + hydroxide reaction as the previous made no sense. I still think a lot does not make sense
which is not that strange as it is mostly guesswork. The point being that the answer most likely is connected with the solvation enthalpies.
[Edited on 2/8/2010 by Nicodem]
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blogfast25
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@ Nicodem:
Yes, solvation or other energies must play a part in weak acid/base neutralisations, as it can be shown easily that even in the case of the weakest
possible monoprotonic acid 1 mol of water is formed per mol of acid neutralised. So by rights the neutralisation energy for H3O+ + OH- ---> 2 H2O
of -57.3 kJ/mol of H2O must be liberated, whatever the monoprotonic acid. Endothermic effects must offset that.
The case of NH4+/NH3 is particularly confusing, at least to moi. Most ammonium salts I've tested are lightly endothermic when you dissolve
them in water (at least the chloride and sulphate are). NH3 itself though is strongly exothermic. Wouldn't one on these grounds expect the
neutralisation of NH4+ salts to be strongly exothermic? I mean, where does the dissolution energy of the NH3 go???
These things keep me awake at night! ;-)
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woelen
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I think that splitting the proton from NH4(+) ions takes a lot of energy, which compensates for the solvation energy, released by dissolving NH3 in
water. On the other hand, dissolving NH3 in water is not that exothermic, I tried a few times and it is much less exothermic than e.g. dissolving
gaseous HCl in water.
[Edited on 2-8-10 by woelen]
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blogfast25
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Good point, Woelen.
To get this absolutely right we probably would to have construct something akin to the Born Haber Cycle but for watery reactions, taking into account all energy consumptions and releases.
As regards the exothermicity of dissolving NH3 in water, in one patent I found the released energy is recovered to use as energy for a stripper
section. In the stripper section a process solution is thermally stripped of dissolved NH3... Can't be that low if it's worth recovering. The NH3
cycled between solutions, acting as a NaF solubility suppressant.
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Nicodem
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Quote: Originally posted by blogfast25  | | Yes, solvation or other energies must play a part in weak acid/base neutralisations, as it can be shown easily that even in the case of the weakest
possible monoprotonic acid 1 mol of water is formed per mol of acid neutralised. So by rights the neutralisation energy for H3O+ + OH- ---> 2 H2O
of -57.3 kJ/mol of H2O must be liberated, whatever the monoprotonic acid. Endothermic effects must offset that. |
I think what you keep neglecting is the energy required to deprotonate weak acids. HCl(aq), AcOH(aq) or NH4<sup>+</sup> can not give the
same amount of heat upon reaction with OH<sup>-</sup> even if you could somehow eliminate all enthalpy changes resulting from ions
solvations and desolvations. The energy required for full dissociation (deprotonation energy) of these three acids is very different (given by pKa)
and this energy must be detracted from the -57.3 kJ/mol value you give for the H2O "association". In the case of strong acids which are more or less
fully dissociated in water this energy is close to zero, but with weak acids like acetic or ammonium it is certainly the major factor. I think that
only in reactions such as the neutralisation of (hydrogen)carbonates (where the strongly solvated anion changes into a species that leaves the
solution) is the solvation change is the prevalent factor explaining the overall enthalpy.
| Quote: | | The case of NH4+/NH3 is particularly confusing, at least to moi. Most ammonium salts I've tested are lightly endothermic when you dissolve
them in water (at least the chloride and sulphate are). NH3 itself though is strongly exothermic. Wouldn't one on these grounds expect the
neutralisation of NH4+ salts to be strongly exothermic? I mean, where does the dissolution energy of the NH3 go??? |
The enthalpy of dissolution is no direct indication of the solvation enthalpy, certainly not so in dissolving salts where the enthalpy is the
difference between solvation and lattice energy. OK, the dissolution of NH3 is more directly connected with the solvation, but I think the most direct
indication of the solvation energy would be the dilution enthalpy. Anyway, the solvation enthalpy of the formed NH3 goes in the overall enthalpy, thus
it is consumed in the desolvation of the ammonium ions and N-H bond breaking (as Woelen noted above). Obviously this energy per self would not be
enough, but since there is also the H2O "association" energy released, the overall reaction is what it is, a very slightly exothermic neutralisation.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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blogfast25
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To be sure, a second reading of the reaction enthalpy of HAc (aq) + NaOH (aq) ---> NaAc (aq) + H2O (l) (acetic acid) was obtained as -55.5 kJ/mol,
making the average value -55.0 kJ/mol.
| Quote: Originally posted by Nicodem |
I think what you keep neglecting is the energy required to deprotonate weak acids. HCl(aq), AcOH(aq) or NH4<sup>+</sup> can not give the
same amount of heat upon reaction with OH<sup>-</sup> even if you could somehow eliminate all enthalpy changes resulting from ions
solvations and desolvations. The energy required for full dissociation (deprotonation energy) of these three acids is very different (given by pKa)
and this energy must be detracted from the -57.3 kJ/mol value you give for the H2O "association". In the case of strong acids which are more or less
fully dissociated in water this energy is close to zero, but with weak acids like acetic or ammonium it is certainly the major factor. I think that
only in reactions such as the neutralisation of (hydrogen)carbonates (where the strongly solvated anion changes into a species that leaves the
solution) is the solvation change is the prevalent factor explaining the overall enthalpy.
(Snip...)
The enthalpy of dissolution is no direct indication of the solvation enthalpy, certainly not so in dissolving salts where the enthalpy is the
difference between solvation and lattice energy. OK, the dissolution of NH3 is more directly connected with the solvation, but I think the most direct
indication of the solvation energy would be the dilution enthalpy. Anyway, the solvation enthalpy of the formed NH3 goes in the overall enthalpy, thus
it is consumed in the desolvation of the ammonium ions and N-H bond breaking (as Woelen noted above). Obviously this energy per self would not be
enough, but since there is also the H2O "association" energy released, the overall reaction is what it is, a very slightly exothermic neutralisation.
|
Regarding the first part I think Nicodem is being a little unfair to me, in the sense that I took the deprotonation reaction into account right from
the start. But it doesn't explain everything IMHO. Let me reiterate my initial thoughts on the reaction enthalpies for the neutralisations of weak
acids with strong bases:
1) Each reaction produces one mol of water (for monoprotonic acids) and thus - 57.3 kJ/mol of water MUST be released.
2) The Gibbs Free Energy of the deprotonation (index d) reaction (for weak acids i.e. K < 1): HA + H2O ---> H3O+ + A- is positive because:
ΔGd = -RTlnK and K < 1
For instance for acetic acid pKa = 4.76, Ka = 10^-4.76. This value has to be divided by the water constant of 55.55 mol/l, so K = 3.13 10^-7
and so isothermal ΔGd = 8.31 J/ K mol x 298 K x ln (3.13 10^-7) = + 37.1 kJ/mol
Of course ΔGd = ΔHd - TΔSd. We can't assume T ΔSd ≈ 0 because then ΔHd ≈ + 37.1 kJ/mol and that doesn't fit -55.0
= - 57.3 + 37.1. So either we assume ΔGd is mostly made of entropy or other enthalpies must play a part.
Incidentally, if for arguments sake we assumed NO other enthalpies play a part, it can the be shown that ΔSd = - 0.115 kJ/K mol, which would
suggest the dissociated acid to be less disorderly than the non-dissociated form and that doesn't seem to make a lot of sense to me. Take KCl
for instance: endothermic when dissolved because the state K+ (aq) + Cl- (aq) + H2O (l) is more disordered than the state KCl (s) + H2O (l).
So I do believe solvation energies play a part. But how precisely?
Incidentally, for the reaction NH4+ (aq) + OH- (aq) ---> NH3 (aq) + H2O (l)
with Ka = 10^-9.25 and K = 1.01 10^-11, @ 298 K, ΔGd = + 62.7 kJ/mol and that doesn't fit either.
[Edited on 3-8-2010 by blogfast25]
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Nicodem
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While reading your calculations it occurred to me that I was quite wrong about solvation enthalpy and that this phenomenon actually only explains the
(bi)carbonate neutralization. The reason is in that the (de)solvation enthalpies are already included in the dissociation enthalpy! This becomes
obvious if you for example read the dissociation equilibrium reaction:
AcOH(aq) + H2O <=> AcO<sup>-</sup>(aq) + H3O<sup>-</sup>(aq)
Obviously here the energy change between solvated acetic acid and solvated acetate anion is already included and these are also the starting and
ending species in the neutralization of acetic acid.
(Bi)carbonates are an exception because the dissociation equilibrium reaction and the neutralization reaction are not complementary:
HCO3<sup>-</sup>(aq) + H2O <=> H2CO3(aq) + OH<sup>-</sup>(aq)
The subsequent release of CO2 is not included in the equilibrium described by pKa1 and pKa2, so the enthalpy of this missing reaction needs to be
included in the neutralization enthalpy:
H2CO3(aq) <=> CO2(g) + H2O
If you think entropic factors in these reactions are large, then maybe you can take some time and do a few measurements of AcOH + NaOH reaction at 3
or 4 different temperatures so that we can calculate its entropy. Better waste a bit of time playing with your calorimeter than being unable to sleep.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
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blogfast25
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| Quote: Originally posted by Nicodem |
If you think entropic factors in these reactions are large, then maybe you can take some time and do a few measurements of AcOH + NaOH reaction at 3
or 4 different temperatures so that we can calculate its entropy. Better waste a bit of time playing with your calorimeter than being unable to sleep.
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That's a nice idea. Perhaps for ammonium sulphate: for ammonium neutralisations the discrepancy seems to be the largest. Not to mention that I've
plenty ammonium sulphate... Perhaps 0 C, 25 C, 50 and 70 C or for starters 0 C and 75 C, just to see.
Your concern for my insomnia is touching... : -) In all 'seriousness'. I just think these are interesting little problems worth investigating...
I do see one potential complication: the enthalpy of neutralisation (H3O+ + OH-) may be temperature dependent. Normally making that correction for
small temperature differences can be done mathematically but what's the heat capacity of a mol of H3O+ (LOL)? Or one mol of OH-? It'd be cautious to
measure the enthalpy of neutralisation at at least two temps, just to be sure...
[Edited on 3-8-2010 by blogfast25]
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woelen
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Use dilute solutions, at most 3 mol/l. Then the heat capacity of the solution roughly is equal to the heat capacity of water. Of course, your
temperature changes are less and harder to obtain accurately, but at least you do not have to worry about problems with wrong heat capacity.
On the other hand, if you use measurements at different concentrations (e.g. 1.5 mol/, 3 mol/l, 6 mol/l) then you can determine the heat capacity of
the other than water compounds in the solution. You can do that by measuring the change of temperature for a certain solution. If you take a solution
with exactly twice the concentration, then you expect a change of temperature, exactly twice as large (of course you also must take into account the
heat capacity of your vessel, so you might have somewhat less than 2 times). Any deviation can be used to compute the effect on heat capacity of the
dissolved chemicals. Repeat measurements in order to reduce the effect of random (non-systematic) errors.
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blogfast25
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Thanks for the advice, Woelen, but it didn't really teach me anything I didn't already know. I've been using concentrations around 1 M for most
experiments. I've always assumed the heat capacity of the solution is Cp, water but I know more precise methods are available. Regarding replicas,
repeating each experiment with a 1 l 'calorimeter' (and old thermos flask) would be costly but I do have about 3 repeats and they show the method is
remarkably reproducible if care is taken... Some 20 experiments have been conducted so far and this afternoon 2 more will be conducted.
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blogfast25
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First up, the cold neutralisation of 1 mol of NH4+ (1/2 mol (NH4)2SO4) with 1 mol NaOH.
As a side observation, dissolving 1/2 a mol of ammonium sulphate in 800 ml of water caused the solution to cool down by about 1 C, so negligibly
endothermic.
By contrast, dissolving 1 mol of NaOH in 200 ml of water the temperature rise was about 52 C, corresponding to about 44 kJ/mol released in dissolution
enthalpy!
The neutralisation was carried out at T = 1.6 C and the measured enthalpy of reaction ΔH = -6.7 kJ/mol of water which isn't significantly
different from the RT value of -5.1 kJ/mol.
This experiment will now be repeated at about 70 - 80 C.
Second, the neutralisation enthalpy of NaOH + HCl at T = 71 C. I found a value of about - 55.8 kJ/mol of water, so not significantly different from
the RT literature value of - 57.3 kJ/mol.
I think this is due to the two temperature corrections that would normally have to be applied to go from ΔH, 298 to ΔH, T cancel each other
because the heat capacity of H3O+ (aq) + OH- (aq) + H2O (l) is essentially the same as that of H2O (l).
I may repeat this experiment at near O C.
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Nicodem
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Blogfast, it is nice that you take so much time for these measurments and it is interesting to follow your results, but with all the effort invested I
can't do otherwise but to remind you that the proper way to measure enthalpies is to use an electric heating element incorporated in the
calorimeter/thermos to introduce heat (I^2×R×t) until the same ΔT is obtained as obtained in the reaction. Just measuring the ΔT with a
thermometer and then using water's Cp is certainly not properly scientific. Though, at least you get similar errors by using similar approximations,
but the numbers, even if good for comparison, are not reliable and can not be compared with literature values.
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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blogfast25
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That's a rather curious comment, Nicodem. "Not properly scientific"? Very sloppy formulation: if I had a penny for every undergrad student that had
determined the neutralisation enthalpy or the latent heat of ice in a chemistry of physics practicum, using the method I use, I'd have a tidy sum.
Your comment also potentially rubbishes much of what many experimenters on SM are doing. To judge the 'scientificness' of a method solely by its
accuracy is contentious: Bang!, goes the Hadron collider experiment...
Measuring the enthalpies by means of coulocalorimety bypasses the problem of Cp,H2O not being temperature independent but only a good operator with
good kit will get that method to work more precisely than the one I use.
It is not a question of "Just measuring the ΔT with a thermometer and then using water's Cp" either. Let me explain with an example, carried out
this morning: the neutralisation of 0.5 mol NaOH in 500 g of water with 0.5 mol of HCl in 500 g of water. In fact no ΔT is determined.
The calorific value k of the flask had previously independently determined to be 86 J/K.
The NaOH solution loaded into the flask and closed, while I prepared the HCl solution from standardised HCl stock, was swirled for about 2 minutes.
Temp. of the NaOH sol. + flask = 24.4 C (closed flask)
Temp. of the HCl sol. = 17.8 C
After mixing and stabilising (closed flask) = 28.2 C
Enthalpy content of NaOH + flask, relative to O C; H1 = (4.1813 x 500 + 86) x 24.4 = 53.1 kJ
Enthalpy content of HCl, relative to 0 C; H2 = 4.1813 x 500 x 17.8 = 37.2 kJ
Enthalpy of content and flask after mixing and reaction, relative to 0 C; H3 = (4.1813 x 1000 + 86) x 28.2 = 120.3 kJ
Reaction enthalpy ΔH = H3 - H2 - H1 = 30.0 kJ, estimated neutralisation enthalpy = - 60.0 kJ/mol of water, for -57.3 kJ/mol listed. Yesterday I
obtained - 62.7 kJ/mol but for one mol and using quite concentrated HCl.
Yes, it does rely on constant Cp but in most experiments the actual temperature does orbit around 298 K, bar a few exceptions.
I think more interesting criticisms would include:
* measurement of mass by volume
* neglect of heat capacity of solutes
* quality of used reagents
* possibly the use of too concentrated solutions, thereby introducing inadvertently dilution enthalpies.
Instead of being a bit all over the place, I will now actually concentrate on the systems HAc + NaOH and HCl + NaOH, addressing some of the concerns
in that list...
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Nicodem
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Well, sorry, I did not mean to badly criticize your method. Actually I think you are doing a wonderful job. What I wanted to say, was that your way of
measuring the enthalpy might be good for preliminary experiments, but I do not think it is a good idea to invest so much time and effort into
measurements, but no effort in using a more reliable method. Your way of doing it does indeed give relatively good results, but how sure are we the
+/- few kJ/mol are good enough? For example, are you sure your measurements will give a good enough data trend to calculate the entropy? It might end
up with you doing days of experiments just to realize the data is too scattered to draw a enthalpy vs. temperature plot.
Incorporating a heating element inside your thermos can be easy, so perhaps you should consider doing it. All you need is a multimeter for measuring
the current and potential across a resistor/heater. You can use an electronics resistor for a heater (just insulate the wires). I never heard of
undergraduate students using anything else in their practice but this kind of control. This is the method we used at the university when I was a
student. And that is how we measured the calorimeter constant at different temperatures.
Here are a few references that you might find interesting:
http://dx.doi.org/10.1016/0021-9614(77)90198-7
http://dx.doi.org/10.1021/ja01122a008
http://dx.doi.org/10.1021/ed045p57
http://dx.doi.org/10.1039/CT8885300865 (very old!)
http://dx.doi.org/10.1039/CT8875100593 (very old!)
I only read the abstracts, so I'm unsure if you can find anything interesting there (also, I do not have access to the first one).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
unworthy of the scientist. Some of the greatest men of science have publicly repudiated a theory which earlier they hotly defended. In this lies their
scientific temper, not in the scientific defense of the theory. - Weston La Barre (Ghost Dance, 1972)
Read the The ScienceMadness Guidelines!
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blogfast25
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Nicodem:
Yes, nothing I can disagree with there. But I'll pass on the amperocalorimetric determinations for now. I'll check out these links, thanks...
What I think is needed is an experiment design that concentrates on determining the difference between the heats of neutralisation of the systems HAc
+ NaOH and HCl + NaOH, in my particular conditions, rather than try and determine absolute values for the enthalpies. After all it's the 'missing
energy' that concerns me here. And signal to noise ratio of course needs to be determined, at least roughly. If that difference could be established
with roughly known accuracy then repeating the design at different temperatures should reveal whether entropic effects are in play or not.
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blogfast25
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The following experiment was carried out at RT (between 20 and 21 C initial temperature):
Two standardised solutions of 0.2 M nominal were prepared, one HAc (acetic acid), one HCl.
One solution of 0.2 M NaOH nominal but about 5 % stronger than the acid solutions was prepared.
An amount, precisely weighed, of about 200 g of the acid solutions was reacted with a slight excess of precisely weighed NaOH solution (also about 200
g). The precise amounts of acid neutralised where thus known for each run. These amount were 0.03 - 0.04 mol.
Temperatures were determined as outlined above and the heats of neutralisation were obtained. Three runs were recorded for each acid. The results are
as follows:
1) For the heat of neutralisation of 0.2 M HCl with 0.2 M NaOH the values -56.4; - 56.1; -58.5 kJ/mol were obtained with an average of -57.0 kJ/mol
and SD = 1.6 kJ/mol.
2) For the heat of neutralisation of 0.2 M HAc with 0.2 M NaOH the values -42.1; - 38.4; -39.8 kJ/mol were obtained with an average of -40.1 kJ/mol
and SD = 1.9 kJ/mol.
Note that there's more spread in values than I anticipated and that for HCl + NaOH I'm now quite close to the literature listed value of -57.3 kJ/mol.
The real value of -57.3 kJ/mol can probably only be determined by determining the limit for conc. ---> 0.
The value for HAc + NaOH is considerably less than previously obtained in more concentrated solutions (0.8 M) and the more reliable value is the one
in more diluted conditions (I feel) of -40.1 kJ/mol.
Pooling together both SDs (they are not significantly different), the relative difference between the heats of neutralisation of HCl and HAc is -57.0
- (-40.1) = -16.9 kJ/mol +/- 3.5 kJ/mol.
That difference is higher than previously estimated but still doesn't account for the Free Energy change for the deprotonation of HAc, estimated to be
+37.1 kJ/mol, at least not if we consider most of that Free Energy to be enthalpy (but we don't know that, of course).
As Nicodem suggested I will now rerun some of the experiments at higher temperature, as significant changes in the results would indicate that
TΔS may play some part. From preliminary results with HCl + NaOH reported above, as well as from theory, I will assume the neutralisation of HCl
+ NaOH isn't temperature sensitive.
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kmno4
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Your value of heat of neutralisation of CH3COOH with NaOH solution is in error.
I do not know what, but you do something wrongly.
Literature values for neutralisation of ~0,3 M solutions are:
NaOH - HCl : 57,5 kJ/mol
NaOH - CH3COOH : 56,1 kJ/mol
Values found on some web page are 57,1 and 55,9 kJ/mol and also here http://dx.doi.org/10.1021/ja01378a011 (worth of reading)
BTW.
Thermodynamic data for reaction
CH3COOH(aq) -> CH3COO(-)(aq) + H3O(+)(aq)
are (in kcal/mol, rewritten from book by G.Kortum) :
ΔG = 6,49 ( = -RTlnKa) ; ΔH = -0,09 ; TΔS = - 6,58
[Edited on 11-8-2010 by kmno4]
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entropy51
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blogfast25, how are you standardizing your solutions of acid and base? What standard acids and bases are you using for comparison? What indicator
are you using for your titrations? Are you checking the pH in some way after your calorimetry to verify that you had equivalent amounts of acid and
base? Thanks!
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blogfast25
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| Quote: Originally posted by kmno4 | Your value of heat of neutralisation of CH3COOH with NaOH solution is in error.
I do not know what, but you do something wrongly.
Literature values for neutralisation of ~0,3 M solutions are:
NaOH - HCl : 57,5 kJ/mol
NaOH - CH3COOH : 56,1 kJ/mol
Values found on some web page are 57,1 and 55,9 kJ/mol and also here http://dx.doi.org/10.1021/ja01378a011 (worth of reading)
BTW.
Thermodynamic data for reaction
CH3COOH(aq) -> CH3COO(-)(aq) + H3O(+)(aq)
are (in kcal/mol, rewritten from book by G.Kortum) :
ΔG = 6,49 ( = -RTlnKa) ; ΔH = -0,09 ; TΔS = - 6,58
[Edited on 11-8-2010 by kmno4] |
If you look a little higher up, I found two vales for HAc neutralisation of -54.5 and -55.5 kJ/mol, using 0.8 M HAc. But the latest results, at 0.2 M
were much lower. That part will now be repeated as I can see nothing particularly wrong with the raw data.
As regards the calculation of ΔG by G.Kortum, are you sure you're copying correctly? That part has been elaborated on higher up: ΔG = -RT
lnK, not Ka, with K = [H3O+][Ac-]/[HAc][H2O] for the reaction:
CH3COOH(aq) + H2O(l) -> CH3COO(-)(aq) + H3O(+)(aq)
For acid/base calculations [H2O] can be incorporated in Ka but for thermocalcs you need the real thing: K, not Ka.
I'm intrigued about the values for ΔH and TΔS, where does he get them from?
| Quote: Originally posted by entropy51 | | blogfast25, how are you standardizing your solutions of acid and base? What standard acids and bases are you using for comparison? What indicator
are you using for your titrations? Are you checking the pH in some way after your calorimetry to verify that you had equivalent amounts of acid and
base? Thanks! |
A fundamental standard was created as a solution from a known quantity of pure anhydrous Na2CO3 (washing soda, filtered and very carefully
recrystallised, then dehydrated for 1 h @ 110 C (till constant weight)). Used immediately to prepare the standard solution, 0.1 N with known titre,
using a calibrated measuring flask.
Against it is titrated a HCl solution, using a 0.1 N HCl solution as titrant and determining the last end point by means of pH meter (direct pH
readings, continuously throughout titration) to determine solution's titre. The pH jump is used (throughout all titrations) to determine end point. It
is very sharp in all cases. I use a 25 ml volumetric burette.
Against the standardised HCl solution is then titrated a 0.1 N NaOH solution, using pH meter as end point determinator.
Standardised NaOH and HCl solutions are the used to calibrate HAc and HCl 0.2 M solutions, using the pH meter.
I accept there are more accurate ways of doing this but having no pure oxalic acid and no commercially standardised solutions, I have to 'make
do'. Suggestions are most welcome.
Final pH of neutralisations is always checked: a small excess of base is used to ensure full neutralisation of the acid.
As it happens I prepared a new sample of pure anhydrous Na2CO3 today and will verify the titre of the master solutions of NaOH and HCl tomorrow.
[Edited on 11-8-2010 by blogfast25]
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entropy51
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| Quote: Originally posted by blogfast25 | | I accept there are more accurate ways of doing this but having no pure oxalic acid and no commercially standardised solutions, I have to 'make
do'. Suggestions are most welcome. |
It sounds as if you're being fairly rigorous in your
standardizations! The only suggestion I would make is that I try to compare to more than a single standard, and when I do this I often find that the
Na2CO3 is the outlier.
I recrystallize technical grade oxalic acid several times for a standard. It seems to work better for me than Na2CO3, but that may just be me.
I also use accurate dilutions of the constant boiling HCl which I find to probably be the most accurate standard for me. This one is not too
difficult if you have distillation equipment and an accurate balance.
Keep up the good work!
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blogfast25
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entropy51:
I have had some problems with Na2CO3 in the past, particularly with turbid solutions which turned even more turbid overnight! I seem to have that
under control now.
Explain your constant boiling HCl method, please...
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