## delta G

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

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

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

Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

### delta G

I still don't get why delta G have is zero to find the minimum temperature. Could someone please explain?

Emily Oren 3C
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: delta G

I'm not really sure what you mean by minimum temperature (minimum temperature of what?), but delta G has to be zero at phase changes (ex. at boiling point and melting point) because this is the point at which the process goes from being spontaneous (negative delta G) to not spontaneous (positive).

Michelle Nguyen 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:03 am

### Re: delta G

We set deltaG equal to 0 when we want to find the temperatures at which a reaction is spontaneous. This is because at equilibrium deltaG is 0; neither process, forward or reverse, products or reactants is favored. If we have T for which deltaG is 0, then we know that any slight change in T from that value leads to a favoring of either the forward or reverse reaction (T for which deltaG is 0 is the minimum T for spontaneity for processes where deltaS is positive and deltaH is positive and the maximum T for spontaneity for processes where deltaS is negative and deltaH is negative).

Ya Gao
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: delta G

I think what you are asking what is the minimum temperature when the reaction is spontaneous? The reason that ∆G is 0 because when ∆G is negative, the reaction is spontaneous, therefore, to figure out the minimum temperature, we set the limit:∆G=0, so any temperature above the limit will have a ∆G above 0 and the reaction will be spontaneous.

### Who is online

Users browsing this forum: No registered users and 1 guest