## Relation between Q and Gibbs Free Energy

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

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

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

Danielle Hoekstra 1C
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Joined: Tue Oct 02, 2018 12:15 am

### Relation between Q and Gibbs Free Energy

If the reaction quotient (Q) is greater than the equilibrium constant what does that indicate for Gibbs free energy?

Artin Allahverdian 2H
Posts: 76
Joined: Fri Sep 28, 2018 12:26 am

### Re: Relation between Q and Gibbs Free Energy

Greater than 0 because it will be non spontaneous

Danielle Hoekstra 1C
Posts: 30
Joined: Tue Oct 02, 2018 12:15 am

### Re: Relation between Q and Gibbs Free Energy

Artin Allahverdian 2H wrote:Greater than 0 because it will be non spontaneous

Thanks!

Artin Allahverdian 2H
Posts: 76
Joined: Fri Sep 28, 2018 12:26 am

### Re: Relation between Q and Gibbs Free Energy

To clarify if Q>K the reaction would want to shift in the reverse way to make more reactants in order to reach K. Since this will make the reverse rxn more favorable than the forward rxn, the forward rxn will not be spontaneous, and thus, delta G is 0.

Danielle Hoekstra 1C
Posts: 30
Joined: Tue Oct 02, 2018 12:15 am

### Re: Relation between Q and Gibbs Free Energy

Artin Allahverdian 2H wrote:To clarify if Q>K the reaction would want to shift in the reverse way to make more reactants in order to reach K. Since this will make the reverse rxn more favorable than the forward rxn, the forward rxn will not be spontaneous, and thus, delta G is 0.

That’s very helpful! Thanks again

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