## Calculating Gibbs Free Energy from a balanced equation

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

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

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

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Marian 2H
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Joined: Tue Nov 25, 2014 3:00 am

### Calculating Gibbs Free Energy from a balanced equation

I came across the question of balancing equations to calculate Gibbs Free Energy on Midterm 2012 Q6A, in which the answer key had the coefficient 9/2 for oxygen gas in the balanced equation. If I balance the equation with every component having a whole number coefficient, the final answer will be doubled. Is it only for oxygen or gases in their standard states that we do not need a whole number coefficient in a balanced equation? Thank you!

AmirMahmoud_1J
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Joined: Fri Sep 26, 2014 2:02 pm

### Re: Calculating Gibbs Free Energy from a balanced equation

I'm confused as well, I do understand when finding the standard formation enthalpy/gibbs/entropy that you want to form 1 mol of product, but how come in the book examples like 8.55 keep molar coeficients of 3 instead of dividing them (2Fe3O4+.5O2-->3FeO2). Also what if a product includes two molecules with say 3 and 4 moles repectively.

Chem_Mod
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### Re: Calculating Gibbs Free Energy from a balanced equation

Well, textbook Q8.55 is not asking for standard enthalpy/entropy/Gibbs free energy of formation. Instead it askes for delta H/S/G for a specific reaction. Similarly, when given a balanced reaction, you don't have to rebalance it to make every coefficient intergers unless you are told to.

Justin Le 2I
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### Re: Calculating Gibbs Free Energy from a balanced equation

They are equivalent and you will just use different coefficients when calculating H and G of reaction.

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