ard free energy G° by 18 kJ/mole, with A* having the higher G°.
Use the table below to find how many more molecules will be in state A* compared with state A at equilibrium.
If an enzyme lowered the activation energy of the reaction by 11.7 kJ/mole, how would the ratio of A to A* change?
Table: RELATIONSHIP BETWEEN THE STANDARD FREE- ENERGY CHANGE, ∆G°, AND THE EQUILIBRIUM CONSTANT
Hint: ∆G° represents the free-energy difference under standard conditions (where all components are present at a concentration of 1 mole/litter). From this table, we see that if there is a favourable free-energy change of –17.8 kJ/mole for the transition Y→ X, there will be 1000 times more molecules of X than of Y at equilibrium (K = 1000).