## Gibbs Free Energy and equilibrium concentration

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Semi Yoon
Posts: 60
Joined: Fri Sep 28, 2018 12:27 am

### Gibbs Free Energy and equilibrium concentration

Reaction 1: Fructose-6-phosphate +ATP --> Fructose-1,6-biphosphate + ADP deltaG = -13.7 kJ/mol
Reaction 2: Fructose-6-phosphate + HPO4 2- --> Fructose-1,6-biphosphate +H20 deltaG = 16.5 kJ/mol

Question: If the concentration of ATP is adjusted to four times its equilibrium concentration, what is the value of deltaG?

To find Q, the TA multiplied K by 1/4. Why would you divide K by 4?

EllerySchlingmann1E
Posts: 76
Joined: Fri Sep 28, 2018 12:24 am

### Re: Gibbs Free Energy and equilibrium concentration

Because ATP is a reactant, it goes on the bottom of the K and Q expressions. So if the concentration of ATP increases by 4, you would want divide K by 4 to get Q.

Reva Kakaria 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am
Been upvoted: 1 time

### Re: Gibbs Free Energy and equilibrium concentration

Since K is equal to concentration of products over concentration reactants, if one of the reactant's concentration is quadrupled, then the the original K is divided by 4.