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### Calculating Gibbs Free Energy from a balanced equation

Posted: **Mon Feb 09, 2015 10:41 am**

by **Marian 2H**

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!

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

Posted: **Mon Feb 09, 2015 6:33 pm**

by **AmirMahmoud_1J**

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.

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

Posted: **Tue Feb 10, 2015 12:27 pm**

by **Chem_Mod**

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.

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

Posted: **Wed Feb 11, 2015 1:43 pm**

by **Justin Le 2I**

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