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Heat of reaction!

madchemistry - 26-2-2011 at 01:23

Hi everybody! I'm an italian student of chemistry, and I need your help

In the lab the teacher reduced a solution of tetrammine palladium with formic acid. He started with a yellow solution containing ammonia and the complex and began do drop HCl. Suddenly a yellow precipitate formed and disappeared. When the precipitate became hard to dissolve, he stopped dropping HCl and started with HCOOH. The reaction was very very slow at the beginning, but after about 30 minutes every drop of the acid caused a very huge amount of gas to be formed. At the end the resulting solution was very clear and incolor with palladium metallic precipitate. Now the problem is: how to work out how many heat was involved in the reduction?? I have not found any data on the web or in my books, so I have no idea!
Hel please

I tought I may do something with the reduction potential, but I do not really know how to solve this problem

[Edited on 26-2-2011 by madchemistry]

[Edited on 26-2-2011 by madchemistry]

madchemistry - 26-2-2011 at 08:13

I don't need the exact value, but I hope someone can just show me the way

Magpie - 26-2-2011 at 09:47

Give us the reaction equation, showing all reactants and products. Then find the heats of formation of those reactants and products. Maybe we can help here. Then it is just simple math: addition and subtraction.

kmno4 - 26-2-2011 at 10:11

[Pd(NH3)2]Cl2 + HCOOH -> Pd + 2 NH4Cl + CO2
(for heat estimation there is no need to use ionic form of this equation)

blogfast25 - 26-2-2011 at 11:12

Quote: Originally posted by kmno4

[Pd(NH3)2]Cl2 + HCOOH -> Pd + 2 NH4Cl + CO2
(for heat estimation there is no need to use ionic form of this equation)

Rewrite as:

[Pd(NH3)2](2+) (aq) + 2 Cl(-) (aq) + HCOOH (aq) -> Pd (s) + 2 NH4(+) (aq) + 2 Cl(-) (aq) + CO2 (g)

Or (Cl(-) is a spectator ion):

[Pd(NH3)2](2+) (aq) + HCOOH (aq) -> Pd (s) + 2 NH4(+) (aq) + CO2 (g)

Break down as (for Hess’ Law):

{I} [Pd(NH3)2](2+) (aq) -> Pd(2+) (aq) + 2 NH3 (aq)

{II} 2 NH3 (aq) + HCOOH -> 2 NH4(+) (aq) + COO(2-) (aq)

{III} Pd(2+) (aq) + COO(2-) (aq) -> Pd (s) + CO2 (g)

The heat of reaction (HoR) of {I} can be estimated from the complexing constant K = ([Pd2+].[NH3]^2)/ [Pd(NH3)2](2+)] and Nernst (Free Energy).

The HoR of {II} is small and probably about half the HoR of the neutralisation energy of ammonia and acetic acid.

The HoR of {III} can be estimated from the cell potential of the redox reaction and Nernst (Free Energy for redox reactions).

Add up (acc. Hess) for reasonable estimate of overall HoR. Not simple but definitely doable… With Pd being so 'noble', this value is likely to be 'large', as per observation.

[Edited on 26-2-2011 by blogfast25]

madchemistry - 26-2-2011 at 12:54

The initial complex is Pd(NH3)4, HCl just react with excess ammonia but the dichloro complex doesn't precipitate. But I think this is not really important.
My problem is that following the way proposed, for the first and the last semi-reaction I can find the variation in Free Energy, but how can I estimate from this the heat without information on entropy? The transition in the last equation bring to the formation of a solid compound, so I think entropy is not neglictible

hissingnoise - 26-2-2011 at 13:27

Quote:

The transition in the last equation bring to the formation of a solid compound, so I think entropy is not neglictible

WTF! Newly arrived and already name-dropping . . .

madchemistry - 26-2-2011 at 13:56

What? I'm not Engish so don't bother if I made some (many) grammar errors

[Edited on 26-2-2011 by madchemistry]

kmno4 - 26-2-2011 at 15:45

Why do you ask about entropy/free energy ?
To estimate heat of reaction you only need to know enthalpies of formation of reagents.
I gave "[Pd(NH3)2]Cl2" formula because this compound is well known and it is possible that thermodynamic data (deltaH) for it are available.

madchemistry - 26-2-2011 at 23:32

Because using the method described by blogfast I can obtain Free Energy (dG), not entalpy!

kmno4 - 27-2-2011 at 00:42

Quote: Originally posted by madchemistry

Because using the method described by blogfast I can obtain Free Energy (dG), not entalpy!

Good luck.
No comment.

hissingnoise - 27-2-2011 at 05:32

Quote:

[Edited on 26-2-2011 by madchemistry]

Your English is fine, madchemistry . . .

madchemistry - 27-2-2011 at 07:14

If I say something stupid please tell me, I'm here to learn...
However, turninng back to the problem:

I found this data:
1)HCOOH + 2H+ +2e- --->C + 2H2O E=0,621
2)C02 + 4H+ + 4e- --------> C + 2H2O E=0,207
3)Pd(2+)-----Pd E=0,987
4)dfG(NH4+)=-79,31kJ
5)dfG(NH3) (aq) =-26,50 kJ
6)dfG(Hform) (aq)=-351
7)dfG(form-) (aq) =-351

Now I can write:
(1)+(3)-(2)= Pd(2+) +HCOOH ------>CO2 + Pd +2H+ E=0,621-0,207+0,987=1,407V
dG(1)=-nFE=-271.55
dG for formation of complex from ammonia and Pd(2+) is
dG(2)=(RTlnK)/1000 kJ (I consider K for formation);
dG for neutralization of ammonia with H+ and formic acid is
dG(3)=4*[(4)-(5)] (dG is equal for HCOOH and HCOO-) =-211,24 kJ

so the total dG for the reaction:
Pd(NH3)4 (2)+ + 3HCOOH -------> Pd + CO2 + 2HCOO- + 4NH4+
must be dG(tot)=-482,79 + (1/1000)RT ln K

I found that lnK for the (en) complex is 27,3 , so considering that complex with en are far more stable than that with ammonia, I can ipotize that k is about 20.
This way I obtain dG=-433,26
But this valour is not too high?? Where am I making a mistake?

[Edited on 27-2-2011 by madchemistry]

[Edited on 27-2-2011 by madchemistry]

blogfast25 - 27-2-2011 at 08:54

Madchemistry:

Admittedly I haven’t checked your latest calculation (although roughly you seem on the right track).

My suggestion was to calculate ΔG for steps {I}, {II} and {III} and add them up according to Hess. You are right bringing up that ΔG isn’t the actual heat of reaction (ΔH) because ΔG = ΔH - T ΔS. And the ΔG calculated from an equilibrium constant (or a cell potential) includes the entropic term.

But, but, but. For reactions {I} and {II} the entropic terms are likely to be quite small (in the case of {II} the ΔH itself can frankly be neglected: the heat coming off the neutralisation of a weak base (NH3) with a relatively weak acid (formic acid) is really quite small).

For reaction {III}, where a gas is formed, ΔS is likely to be considerable but can be easily estimated: simply look up the entopic value of CO2 (look up in NIST database) at 298 K. ΔS for reaction {III} is then approximately equal to that value, and by subtracting T ΔS from the ΔG for reaction {III} you get its ΔH.

In summary: for reaction {I}, ignore the ΔS term as about 0. Ignore {II} altogether, its heat is small. For {III}, estimate ΔG from the cell potential, then subtract from it T ΔG(CO2). Add up for a reasonable estimate.

Alternatively, measure ΔH by running the reaction in a calorimeter. You MAY (note emphasis) even find that the reaction is endothermic: the neutralisations of carbonates and bicarbonates with strong acid e.g. are endothermic due to the evolution of… CO2!

[Edited on 27-2-2011 by blogfast25]

blogfast25 - 27-2-2011 at 09:05

Quote: Originally posted by kmno4

Why do you ask about entropy/free energy ?
To estimate heat of reaction you only need to know enthalpies of formation of reagents.
I gave "[Pd(NH3)2]Cl2" formula because this compound is well known and it is possible that thermodynamic data (deltaH) for it are available.

Bullshit. The heats of formation that are tabled are for ionic substances usually for the 298 K solids. In solution these values are almost irrelevant.

Or when calculating the neutralisation heat of NaOH do you simply ‘add up’ the HoFs of the reagents???

In reality that would work for NaOH (s) + HCl (g) (for instance) -> NaCl (s) + H2O (l).

But not for Na+ (aq) + OH- (aq) + H3O+ (aq) + Cl- (aq) -> Na+ (aq) + Cl- (aq) 2 H2O (l). Obviously.

Next time you want to be scathing, engage brain before keyboard.

Look at what you wrote above:

Quote: Originally posted by kmno4

[(for heat estimation there is no need to use ionic form of this equation)

Oh, no?

The mistake you’re making is colossal and most sixth graders wouldn’t make it.

Take the example of the neutralisation of a SOLUTION of NaOH with a strong acid. By using the HoF @ 298K for SOLID NaOH you simply forget that most of the enthalpy of NaOH’s HoF is already accounted for when you dissolved the stuff into water. The same is true for the acid. In reality the sodium and chlorine ions play the role of spectator ions that take no part in the reaction and no lattice energy of the NaCl is released EITHER. The only reaction that takes place is H3O+ (aq) + OH- (aq) -> 2 H2O (l). So when calculating heats of reaction in solution of ionic substances THERE IS EVERY NEED TO USE THE IONIC FORM OF THIS EQUATION.

[Edited on 27-2-2011 by blogfast25]

blogfast25 - 27-2-2011 at 11:09

Madchemist’s calculation of ΔG = - 271.5 kJ/mol for the redox reaction:

Pd(2+) (aq) + COO(2-) (aq) -> Pd (s) + CO2 (g)

… appears correct to me.

Now we assume that both sides of that equation have about the same amount of entropy EXCEPT for the released CO2 then we can assume ΔS = ΔS,CO2 = 214 J/mol.K (NOTE: J NOT kJ) and ΔH = ΔG + TΔS ≈ - 208 kJ/mol (@ 298 K) for that specific reaction. Quite a whopper! It would basically cause a steam explosion.

This would be slightly increased (negatively) still by the small neutralisation heat of NH3 + formic acid, which I’ll neglect.

But Pd(NH3)2(2+) (aq) -> Pd(2+) + 2 NH3(aq) is likely to be strongly endothermic because the equilibrium constant K = [Pd(NH3)2 (2+)]/([Pd(2+)].[NH3]^2]) is positive: we know the complex forms spontaneously upon mixing Pd(2+) (aq) and NH3 (aq).

Thus ΔG = - RTlnK = - 0.00831477 (kJ/mol.K) x 298 (K) x lnK = - 2.4777 (kJ/mol) x lnK must be strongly negative (for the formation of the complex).

I’ve not found an actual complexation constant for this complex. This table here:

http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm

… gives an idea of the magnitude of the complexation constants of some really stable complexes.

If we assume (for argument’s sake) that the decomplexation free energy of the complex was about + 208 kJ/mol and that the entropic term of decomplexation is weak, then:

– 208 = - 2.4777 lnK or K = 2.9.E+36, at which point the reaction would neither be endo nor exothermic (but would still proceed because of the escaping CO2). At smaller (and probably more realistic) values for K the overall reaction would be exothermic, depending on the size of K. Interestingly, from experimental calorimetry the value of K could thus be very roughly estimated, allowing of course for some of the assumptions made…

[Edited on 27-2-2011 by blogfast25]

madchemistry - 27-2-2011 at 11:30

Quote: Originally posted by blogfast25

This would be slightly increased (negatively) still by the small neutralisation heat of NH3 + formic acid, which I’ll neglect.

You're right and I also thought the same, but trying to do the math I obtain a VERY large value!

I have that the heat of reaction is that of NH3 + H+ --->NH4.
HoF for NH4+ (aq) is-79.3, for NH3(aq)=-26,50 and for NH3(gas)=-16.4.
If I consider that NH3 is aq dG(neu)=-52,8kJ, for NH3(gas) dG(neu)=-62,9. We have to multiply by four and so it is obtained a value that is almost that of the reduction reaction O.o I controlled various source and data are OK! Probably I'm doin some really stupid error but I didn't manageto work out what it is!

Quote: Originally posted by blogfast25

– 208 = - 2.4777 lnK or K = 2.9.E+36, at which point the reaction would neither be endo nor exothermic (but would still proceed because of the escaping CO2).
[Edited on 27-2-2011 by blogfast25]

Why do you put equal the dGs? Shouldn't it be necessary to sum the value for reduction of Pd(2+) with that of dissociation of Pd(NH3)4 (2+) ?

[Edited on 27-2-2011 by madchemistry]

blogfast25 - 27-2-2011 at 12:30

Madchemist:

The value ΔH = - 208 kJ/mol is far too high to be the value of the overall reaction (and of course it’s the estimated ΔH for the redox reaction ALONE). Work out just how much water you could bring to the boil (from RT) with 208 kJ of heat!

It also turns out that due to lack of concrete value for the complexation constant of the ammonia-palladium complex we can’t actually estimate its ΔG (and thus its ΔH).

So instead I’m playing the devil’s advocate: I’m assuming that the heat of reaction of the reaction you observed is actually quite small (if it wasn’t quite small the reactor would have started to boil or worse – that didn’t happen, ergo I’m roughly right on this).

By now equating the estimated ΔH of the redox reaction with the negative of the ΔH of the decomplexation reaction, I can ascertain roughly the conditions in which the overall reaction really doesn’t generate much heat at all, as observed. In particular the order of magnitude of the complexation constant K, order of magnitude which is roughly in line with the complexation constants of some similar, highly stable complexes.

This DOES NOT give you the actual value of ΔH for the overall reaction, but it does allow some nifty conclusions to be reached and that is better than NOTHING AT ALL.

Now let me see if I can find an estimated HoR value for reaction {II}...

Edit: in fact I’m definitely a bit wrong on that aspect: the heat of reaction (ΔH, not ΔG) of {II} is likely to be in the order of about - 100 kJ/mol (of ammonium formiate), so not as small and negligible as I first stated. To the – 208 kJ/mol of {I} has thus to be added another – 100 kJ/mol from {II}, making it say about -300 kJ/mol of Pd reduced, for {II} +{III}.

To illustrate my point above, it can easily be shown that if the reaction had been carried out to completion using a 1 M solution of the palladium complex that for a ΔH ≈ -300 kJ the temperature of the water would have been increased by about ΔT =72 C!

This has then to be ‘offset’ roughly by the enthalpy needed to decomplexate the Pd(NH3)2(2+) (reaction {I}), so – 300 ≈ - 2.4777 lnK or K ≈ 4.E+52. That would roughly be the condition for the reaction to be enthalpically neutral (ΔH ≈ 0), in reality the reaction is likely to be somewhat exothermic because K will be smaller than that.

[Edited on 27-2-2011 by blogfast25]

madchemistry - 27-2-2011 at 13:17

Uff, there must be some points we are not considering

Now I understand perfectly what you have done and it is absolutely reasonable, but that value of K is undoubtely out of any order! It is too high if you see generally the order of magnitude of other stability constant of complexes, and overall I found for [Pd(en)2](2+) a value of ln K of 2*10^27, the amino complex can't be more stable!
Tomorrow I will show the problem to another professor (I will meet the prof who has given me this problem only wednesday) and perhaps he can find the error

blogfast25 - 27-2-2011 at 13:49

Quote: Originally posted by madchemistry

Uff, there must be some points we are not considering

In that table I linked to there are complexes with far higher Ks. But even at 2E+27 that makes for a massive 155 kJ (2.477 ln(2*10^27)! No, the order of magnitudes are right… At 300 - 155 = 145 kJ enthalpy, a 1 M solution would heat up by about 35 C and a 0.1 M solution by about 3.5 C. The ball park is correct. Without a precise value for K there’s nothing further that can be done here…

By all means ask your professor. I'm curious as to the exact value of HoR for this oxidation reaction because I measured the oxidation enthaply of some watery oxidation reactions in the past...

[Edited on 27-2-2011 by blogfast25]

madchemistry - 28-2-2011 at 00:20

Quote: Originally posted by blogfast25

Edit: in fact I’m definitely a bit wrong on that aspect: the heat of reaction (ΔH, not ΔG) of {II} is likely to be in the order of about - 100 kJ/mol (of ammonium formiate), so not as small and negligible as I first stated. To the – 208 kJ/mol of {I} has thus to be added another – 100 kJ/mol from {II}, making it say about -300 kJ/mol of Pd reduced, for {II} +{III}.

[Edited on 27-2-2011 by blogfast25]

How do you calcolate this dH? I founf for NH4+ entalpy of formation is -132,51 kJ, for ammonia (aq) is -80,28 kJ and so dH=4*(-132,51+80,20)= 4*(-52,23)=208,92
I multiply by four because we have four mol that have to be neutralised.

blogfast25 - 28-2-2011 at 08:18

Madchemist:

Rewrite:

2 NH3 (aq) + HOOCH (aq) -> 2 NH4+ (aq) COO(2-) (aq)

As:

2 NH3 (aq) + 2 H2O (l) -> 2NH4+ (aq) + 2 OH- (aq)

HOOCH (aq) + 2 H2O (l) -> COO(2-) (aq) + 2 H3O+ (aq)

2 H3O+ (aq) + 2 OH- (aq) -> 4 H2O (l)

And add up.

The reaction {II} is essentially the neutralisation of 2 moles of OH- (aq) with 2 moles of H3O+ (aq). Experiment shows that entropic effects play hardly a role and that the neutralisation of a watery solution of a weak base with a watery solution of a weak acid yields ΔH almost equal to the heat of neutralisation – 57 kJ/mol H2O (strong acid + strong base). Times 2 (for 2 mol): ΔH {II} ≈ -100 kJ/mol of Pd.

One thing you might want to check is the values of the half-potentials for {III}. Your reasoning is sound there but I didn’t check those values.

[Edited on 28-2-2011 by blogfast25]

madchemistry - 28-2-2011 at 13:17

Ok, thanks! I asked my physical chemistry teacher and she said that the way proposed is correct. Now on wednesday I will see my inorganic teacher and I will ask him the correct solution!
However, thank you very much for your patience and you disponibility

[Edited on 28-2-2011 by madchemistry]

blogfast25 - 28-2-2011 at 14:07

disponibilita (It.) = availability (En.) I lived in Milan for 3 1/2 wonderful years...

No quitting from here w/o filling me in on the value, d'accordo caro?

Ci vediamo!

[Edited on 28-2-2011 by blogfast25]

blogfast25 - 1-3-2011 at 07:36

Madchemist:

Turns out there is one glaring error in the scheme: formic acid is a monoprotic, not a biprotic acid: HCOOH (a) + H2O (l) -> H3O+ (aq) + HCOO- (aq).

But as luck would have it changes little to the ΔH calculation!

The partial reactions become:

{I} [Pd(NH3)2](2+) (aq) -> Pd(2+) (aq) + 2 NH3 (aq)

As before.

{II} NH3 (aq) + HCOOH (aq) -> NH4+ (aq) + HCOO- (aq) which is a neutralisation reaction as shown above.

And NH3 (aq) + H2O (l) -> NH4+ (aq) + OH- (aq)

And H3O+ (aq) + OH- (aq) -> 2 H2O (l) which both form the neutralisation of the second NH3.

So there are two neutralisations, as above, with total ΔH ≈ -100 kJ/mol of Pd.

{III} Pd(2+) (aq) + HCOO- (aq) + H2O (l) -> Pd (s) + CO2 (g) + H3O+ (aq) for the redox reaction.

Add {I} + {II} + {III} up to:

Pd(NH3)2(2+) (aq) + HCOOH (aq) -> Pd (s) + 2 NH4+ (aq) + CO2 (g)

With the two Cl- (aq) spectator ions we started off with:

Pd(NH3)2Cl2 (aq) + HCOOH (aq) -> Pd (s) + 2 NH4Cl (aq) + CO2 (g)

Assuming the value of cell potential of {III} remains the same as you calculated nothing has really changed in terms of ΔH calculation! But this is the correct reaction scheme based on actual formic acid…

We can further simplify (I’ll leave out the suffixes that indicate ‘state’). Start from the redox reaction you defined:

Pd2+ + HCOOH -> Pd + CO2 + 2 H+

Then Pd(NH3)2(2+) -> Pd2+ + 2 NH3

Now add:

2 NH3 + 2 H2O -> 2 NH4+ + 2 OH-
2 H+ + 2OH- -> 2 H2O

Together they make 2 neutralisations of the weak base NH3, with ΔH ≈ -100 kJ.

Add up and add Cl- spectator ions:

Pd(NH3)2Cl2 (aq) + HCOOH (aq) -> Pd (s) + 2 NH4Cl (aq) + CO2 (g)

The overall heat of reaction can be estimated using Hess as:

ΔH ≈ - 208 kJ/mol - 100 kJ/mol + RTlnK = -308 kJ/mol + RTlnK

With K the complexation constant (and assuming entropic effects are small, so ΔH ≈ ΔG). In reality there is an entropic effect for Pd(NH3)2(2+) -> Pd2+ + 2 NH3 because the right hand side is more disordered, thus ΔS > 0, but it’s reasonable to expect this value to be small… Assuming the calculated cell potential to be roughly correct, we can also expect the reduction of uncomplexed Pd2+ with formic acid to be seriously exothermic, presumably this is why the experiment was carried out using the ammonium complex, rather than with ‘naked’ Pd2+.

[Edited on 1-3-2011 by blogfast25]

madchemistry - 1-3-2011 at 11:20

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]

blogfast25 - 1-3-2011 at 13:24

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]

madchemistry - 2-3-2011 at 10:28

I asked my teacher and we will discuss the problem Friday. I wanted to kill him xD

blogfast25 - 3-3-2011 at 08:10

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]

madchemistry - 4-3-2011 at 06:26

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

blogfast25 - 4-3-2011 at 07:34

Well, I'm sure you've learned something from it. I did.

madchemistry - 4-3-2011 at 07:46

Quote: Originally posted by blogfast25

Well, I'm sure you've learned something from it. I did.

Absolutely me too