Free energy_biochemistry_deltaG?

Question

Q) The reaction, fructose - 6-phosphate + Pi -----> fructose - 1,6-bisphosphate + H2O,
has a Keq of 0.0001 at pH 7. What is the deltaG for this reaction?

Answer 1

You mean deltaG0, not deltaG. It is important to get this straight; deltaG0 is the change in free energy under standard conditions, while deltaG could mean the change under any conditions.

This difference matters very much. For example, at equilibrium, since the system is in its state of lowest free energy, deltaG = 0

This is how we get to the equation which you now need to use,

deltaG0 = - RT ln Keq

Substitute into this, noticing that you need to use R in J K-1 mol-1 and T in K, and also ln, not log, and do the arithmetic.