## Gibbs Free Energy

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

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

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

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

### Gibbs Free Energy

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

To find the standard Gibb's free energy for ATP hydrolysis, my TA multiplied the Gibbs free energy of the first reaction by 1 and the Gibbs free energy by -1. Then to find the standard Gibb's free energy, he added the two and got -30.2 kJ/mol.

I don't really understand why you have to multiply by 1 and -1. Why do you need to multiply one by 1 and the other by -1?

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

### Re: Gibbs Free Energy

This is similar to the Hess's Law approach we used for calculating reaction enthalpy. Basically, you want to make sure that the substances you are not interested in cancel each other out. In this case, you are trying to find the Gibbs Free Energy of ATP hydrolysis, and therefore want to cancel out the fructose compounds. In order for this to happen, Fructose-6-phosphate in the first equation must be on the opposite side of Fructose-6-phosphate on the second equation, and the same goes for Fructose-1,6-biphosphate. Since originally, Fructose-6-phosphate is on the reactants side for both the equations and Fructose-1,6-biphosphate is on the products side for both of the equations, one of the delta Gs must be multiplied by negative one to "flip" the chemical reaction around. Then, the values can be added.