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madchemistry
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I've made my calculus considering it as monoprotic 
We have both make another error: ammonia and formic acid are weak electrolyte, so we can't obtain data the way we followed! We have to use the
equation dG=-RTlnK with K=(Ka*Kb)/Kw
This way dG is 24,76kJ, a much more reasonable value 
[Edited on 1-3-2011 by madchemistry]
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blogfast25
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No, I’m sorry but your point about NH3 and HCOOH is incorrect. For the neutralisation of a weak base with an acid we find that the ΔH (not
ΔG!) is in fact the neutralisation enthalpy ΔH = - 57 kJ/mol for H+ + OH- -> H2O. In fact we find that the neutralisation of an acid with
a base (or vice versa) is almost completely independent from the nature of the acid and base. See for example a number of experiments carried out by
me here:
http://www.sciencemadness.org/talk/viewthread.php?tid=14247&...
… for the neutralisation enthalpy of acetic acid neutralised with NaOH.
BTW, the value you obtained for ΔG is positive: that would mean that the neutralisation of NH3 (aq) with HCOOH (aq) would simply not proceed!
That’s ABSURD (it does proceed with great vigour).
For the neutralisation of say, a weak base, with a strong acid, for example NH3 + HA, assume for this simplified example that HA is completely
dissociated:
HA + H2O --- > H3O+ + A- ([H3O+] . [A-] / [HA] is very large).
By contrast the base is weak and almost completely undissociated:
NH3 + H2O < --- > NH4+ + OH- ([NH4+] . [OH-] / [NH3] << 1)
Now we add the acid to the base solution and H3O+ + OH- --- > 2 H2O proceeds quickly and completely and a reaction enthalpy of – 57 kJ/mol of OH-
is released.
Of course this means that NH3 + H2O --- > NH4+ + OH- must also proceed from left to right (to maintain the equilibrium requirement). The reaction
has a ΔG = - RT lnKb which is of course positive (which is why the base doesn’t dissociate without forcing it to do so). – RT lnKb is rather
small (about + 27 kJ/mol) but it is made up entirely of entropy (times temperature), so that ΔH ≈ 0. The sum of the enthalpies of the
dissociation and the neutralisation reaction (H3O+ + OH- --- > 2 H2O) is thus approximately – 57 kJ/mol + 0 = -57 kJ/mol. Which is why the
reaction heat from acetic acid + NaOH is in fact the same as for HCl + NaOH, because the same thing holds true also for weak acids.
No, the neutralisation of 2 mol of NH3 in watery solution yields a reaction enthalpy of about 2 x -57 kJ/mol, probably slightly less…
Incidentally, in that same thread linked to above you’ll also find measured values for NH3 (aq) + HCl (aq) with heats of reaction again almost
indistinguishable from – 57 kJ/mol (allowing for inevitable measuring error, of course)…
[Edited on 1-3-2011 by blogfast25]
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madchemistry
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I asked my teacher and we will discuss the problem Friday. I wanted to kill him xD
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blogfast25
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Your reasoning that for:
HA + B --- > A- + BH+
... ΔG = - RT ln (KaKb/Kw) is correct, because the equilibrium constant for that reaction can be written as K = KaKb/Kw. But that ΔG is
decidedly negative and the reaction proceeds. However, IT IS NOT ΔH (= ΔG +TΔS), the value we’re actually looking
for.
We can see that ΔG = 0 for ln (KaKb/Kw) = 0 or KaKb/Kw = 1. From a certain degree of ‘weakness’ of the acid and base, they cannot neutralise
in water anymore.
[Edited on 3-3-2011 by blogfast25]
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madchemistry
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Now the results :
1) I was the only stupid that has spent time with the problem, all my mates didn't try, even though I'm just a guest because this is an advanced
course -.- ;
2) The prof said I can become a theorist xD For him was sufficient to consider the reduction of palladium and the oxidation od HCOOH to see that the
reaction was absolutely spontaneus and it wasn't necessary to do all the math we have done!
At the end, so much work for nothing xD
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blogfast25
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Well, I'm sure you've learned something from it. I did.
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madchemistry
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Absolutely me too 
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