## Delta G not

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

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

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

Leah Thomas 2E
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

### Delta G not

Can someone explain the difference between ∆G and ΔG° conceptually. Why don't we always need to calculate for ΔG°?

Cyianna 2F
Posts: 34
Joined: Wed Nov 16, 2016 3:04 am

### Re: Delta G not

It's just a difference of values, the concept of $\Delta G and \Delta G^{\circ}$ are still the same. I think that it just depends of the conditions of the reaction. I'm not entirely sure though, that's a good question.

Posts: 62
Joined: Thu Jul 27, 2017 3:01 am

### Re: Delta G not

ΔG° is in standard condition, so you would have to calculate ΔH° and ΔS°, just like you would in the regular Gibbs Free Energy equation.

Jakob von Morgenland 2C
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### Re: Delta G not

∆G° is considered the standard free energy change of a reaction, while ∆G is considered the free energy change of a reaction. Remember, standard free energy change implies that the reactants and products are in their standard states at 1 atm and generally 25 celsius.

Ivy Lu 1C
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

### Re: Delta G not

∆G° is the standard free energy of the reaction. This means that the reactants and products are at their standard state: 1 atm for gases and 1 M of aqueous solutions, typically at 25°C. ∆G is just the free energy of the reaction that doesn't have to be under the standard state conditions.

Diego Zavala 2I
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### Re: Delta G not

$\Delta G^{^{o}}$ is the standard free energy change in a reaction at 1 bar for gases or 1M for solutions. Because 1 bar is roughly equivalent to 1 atm, we can use the values of the standard free energy for most reactions

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