Gibbs Free Energy in relation to H and S

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Sophia Diaz - Dis 1B
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Joined: Fri Apr 06, 2018 11:02 am
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Gibbs Free Energy in relation to H and S

Hey y'all!

So I'm having trouble understand what Gibbs Free Energy is and how it relates to enthalpy and entropy. I know what the equation is to solve it but why can you solve for $\Delta$ G using $\sum$ G (prod) - $\sum$ G (react) if $\Delta$ G, $\Delta$ H, and $\Delta$ S are all different?

Thanks so much!

bonnie_schmitz_1F
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Re: Gibbs Free Energy in relation to H and S

I think it all comes back to this equation: ∆G = ∆H - T∆S

whitney_2C
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Re: Gibbs Free Energy in relation to H and S

you can solve for $\Delta G$ using $\sum \Delta G(products) - \sum \Delta G(reactants)$ because Gibbs Free Energy is defined by the state functions enthalpy and entropy, and it too is a state function.

105169446
Posts: 32
Joined: Fri Sep 28, 2018 12:16 am

Re: Gibbs Free Energy in relation to H and S

And just as a reminder, a state property is not dependent on the path taken to obtain that state.