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Electric to be

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Hello. A known equation that is useful for calculating equilbrium constants is:

ΔG° = -RT * ln(K)

This is all well and good. Given a standard gibbs free energy of reaction for some given reaction, the equilibrium constant for the reaction can be found.

My trouble is in

N

The ΔG° can be calculated by taking the difference of Gibbs Free Energy of formation of each reactant and product, multiplying by the respective stoichiometric constant, and taking the difference.

Then, when I plug in this value for ΔG° I can find a unique equilibrium constant to describe this reaction.My trouble is the following. What's to stop me from halving all constants of the equation and getting this? :

1/2N

Now, per "mole" of the equation (which is 1/2 N

ΔG° = -RT * ln(K)

This is all well and good. Given a standard gibbs free energy of reaction for some given reaction, the equilibrium constant for the reaction can be found.

My trouble is in

*which*ΔG° to use. For example given the following reaction:N

_{2}+ 3H_{2}↔ 2NH_{3}ΔG° = -33.0 KJThe ΔG° can be calculated by taking the difference of Gibbs Free Energy of formation of each reactant and product, multiplying by the respective stoichiometric constant, and taking the difference.

Then, when I plug in this value for ΔG° I can find a unique equilibrium constant to describe this reaction.My trouble is the following. What's to stop me from halving all constants of the equation and getting this? :

1/2N

_{2}+ 3/2H_{2}↔ 1NH_{3}ΔG° = -16.5 KJNow, per "mole" of the equation (which is 1/2 N

_{2}, 3/2H_{2}and 1NH_{3}), there is one half of the original Gibb's free energy release. However, this still represents a standard gibbs free energy, by following the same process as before of taking the difference between products and reactants.**However, I will certainly get a different value of K.**The equilibrium constant should remain the same though, should it not? It shouldn't matter what multiple of the equation have. This is my confusion.As a side question, should ΔG° describing this type of reaction have units of KJ/mol? I know in the Gibbs free energy of formation it is. However, generally when I've seen free energy of reaction written it has only had units of J or KJ, since the moles were multiplied through in the process of finding ΔG°. If it does have KJ/mol units, would 1 "mol" of reaction essentially be 1 mol * each stoichiometric constant of the products and reactants for a given equation?
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