[GeLe5000]

Hello.

Can someone confirm my idea that the following sentence can be misleading.

A + B ----------> C + H₂O (endergonic in standard conditions)

"In an aqueous environment, like in living systems, the exergonic direction, thermodynamically privileged, is hydrolysis." (translated from "Construire une cellule" Ch. De Duve, = "Blueprint for a cell. The nature and origin of life")

 It doesn't mean that in the presence of a large amount of water, very little of C will be produced for thermodynamic reasons. On the contrary, it's for kinetic reasons, a water "mass effect" that will strongly reverse the reaction through hydrolysis :
v_reverse = k x (D) x (H₂O) is huge compared to v_forward.


Thank you for your ideas.

[magician4]

I can't confirm.

"endergonic" clearly is a thermodynamic statement, not a kinetic one.

furthermore, I think your reasoning is wrong: kinetics in, for example ,  esterification / hydrolysis is of second order, if memory serves. so, speed will be proportional to the concentration of water to the power of one at best, meaning: proportional to approx. 55 mol/L
looks like a huge number to you? well, you forgot 'bout the k here, which well will reach from very huge numbers to the very minuscule, compared to "55", and easily might overcompensate anything of influence those 55 M might have to offer.

so , you can't judge a total reaction speed by cherrypicking the isolated influence of just one component: by chance, this might give you realities results, but more often than not, you will fail.


regards

Ingo

[GeLe5000]

Thank you very much.

I conclude that I must keep exploring the literature in order to understand (perhaps) why it's so often mentioned that in the presence of water (55 M/L), a condensation reaction is not feasible for thermodynamic reasons. Because what seems strange to me is the fact that this concentration doesn't appear in the calculation of K_eq = (C) / (A) x (B). Thus ΔG is independent of water concentrations.

[magician4]

(...) is the fact that this concentration doesn't appear in the calculation of K_eq = (C) / (A) x (B).

I don't know how you possibly could come to this expression.
for example, a well known constant is about the esterification of acetic acid with ethanol:

C₂H₅OH + CH₃COOH  CH₃COOC₂H₅ + H₂O

where of course the water does show up in the respective LMA expression:

K ≈ 4 = [CH₃COOC₂H₅] * [H₂O] / ( [C₂H₅OH] * [CH₃COOH] )

hence, in a waterbased environment, with let's say c₀ (ethanol, acetic acid) = 0.1 M and water at a ~ constant 55 M , you'd have like 7.2 * 10^-4 M of ester in equilibrium

as you see, this reaction is feasible - though the net gain is small.*⁾

However, talking about the origin of life  / the "chemical evolution" part , all these considerations about condensation at / near room temperature might well be beside the point, as for example the Miller/Urey experiment teaches us.
  It may well be that those fundamental molecules of life didn't come into being in a lukewarm pool of water, but with thunder and lightning, erupting volcano's lava meeting the sea, and in the superheated brine in black smokers, instead.

... and this changes everything

Thus ΔG is independent of water concentrations.

as shown above, it isn't

regards

Ingo


*⁾
however, at times even small concentrations might well be high enough to create a protective environment. think about soap, and how little soap is required to produce micelles

[GeLe5000]

Thank you again.

I must say that it's the first time that I see the calculation of a product's concentration implying the concentration of water (0.00072 = 4 / 55 x 0.1 x 0.1).

In biochemistry, they say for example :

glucose + Pi --------> glucose-6-P + H2O

K = glucose-6-P / (glucose) x (Pi) = 0.0039

"Notice that H2O is not included in this expression, as its concentration (55,5 M) is assumed to remain unchanged by the reaction" (Lehninger's "Principles of Biochemistry")

I'll try to find if H2O concentration is actually implied in the calculation of 0.0039, if the other concentrations at equilbrium are available somewhere.

It would mean that's not written but taken into account in the result. It's a surprise...

[magician4]

you can "include" one constant ( here: the conc. of water, 55 M ) in the original  LMA constant

(as we often do with Ka, Kb values for waterbased solutions, for example)

hence , the K given in that "Lehninger expression" in question is not the K that belongs to the "full" ( i.e. including water) LMA expression, but a modified one instead:

K(Lehninger) = K (LMA) * (55 M)^-1


regards

Ingo

[Borek]

It would mean that's not written but taken into account in the result. It's a surprise...

No surprise here, actually it is a standard procedure in the case of diluted water solutions.

It is so standard most people don't realize they do it all the time.

[GeLe5000]

I understand.

Thank you for your help and the weblinks.