4
$\begingroup$

I have another question similar to this one. I just don't have intuition for why reactions are thermodynamically favorable or unfavorable and I'd like to build it. This time I'd like to ask about the reverse aldol reaction $$\ce{Fructose-1,6-Bisphosphate -> Dihydroxyacetone Phosphate + Glyceraldehyde-3-Phosphate}$$

The standard free energy of this reaction is +23.8 kJ/mol.[1]

I was hoping someone could help me understand why. The larger molecule has lower entropy and charge-charge repulsion. Why is it so thermodynamically unfavorable for it to break apart?

[1] Cornish-Bowden, A. Thermodynamic Aspects of Glycolysis. Biochem. Educ. 1981, 9 (4), 133–137.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Take a look at the bonds being formed and broken. That difference in enthalpy overwhelms the entropic consideration. $\endgroup$
    – Andrew
    May 16 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.