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### Gibbs Free Energy and equilibrium concentration

Posted: Mon Feb 25, 2019 12:46 am
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?

### Re: Gibbs Free Energy and equilibrium concentration

Posted: Mon Feb 25, 2019 8:48 am
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.

### Re: Gibbs Free Energy and equilibrium concentration

Posted: Mon Feb 25, 2019 8:48 am
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.