Delta S in relation to Gibbs free energy

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

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

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

Katherine Grillo 1B
Posts: 69
Joined: Fri Sep 28, 2018 12:28 am

Delta S in relation to Gibbs free energy

Why does a positive delta S favor the forward process in a reaction?

Camille Marangi 2E
Posts: 60
Joined: Fri Sep 28, 2018 12:26 am

Re: Delta S in relation to Gibbs free energy

This is because of the equation: dG= dH-T*dS. The spontaneity still relies on the change in enthalpy as we see in the equation. If the reaction is exothermic and entropy increases, then no matter the temperature, we will see the forward process being favored as dG will be negative. In general, there is a higher chance the forward reaction will be spontaneous if it generates a positive change in entropy and it all goes back to the aforementioned equation.