Trouble with Gibbs Free Energy & Equilibrium Constant Calc.

  • #1

Main Question or Discussion Point

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

N2 + 3H2 ↔ 2NH3 ΔG° = -33.0 KJ

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? :

1/2N2 + 3/2H2 ↔ 1NH3 ΔG° = -16.5 KJ

Now, per "mole" of the equation (which is 1/2 N2, 3/2H2 and 1NH3), 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?
 
Last edited:

Answers and Replies

  • #2
Borek
Mentor
28,360
2,751
The equilibrium constant should remain the same though, should it not?
[tex]\frac{[NH_3]^2}{[N_2][H_2]^3} = \sqrt{ \frac{[NH_3]}{[N_2]^{\frac 1 2}[H_2]^{\frac 3 2}}}[/tex]
 
  • #3
James Pelezo
Gold Member
190
57
Use the ΔGo for the reaction of interest. Changing the coefficients of the reaction does change the reaction kinetics and the equilibrium constant. By dividing the Haber-Bosch equation (N2(g) + 3H2(g) => 2NH3(g)) by 2 effectively doubles the equilibrium constant value of the reaction. The equation which yields 2 moles NH3(g) has a KSTD = 1.64 X 10-6, but dividing by 2 => KNON-STD = 1.28 X 10-3.

[K(2NH3)]½ = [([NH3]2/[N2][H2]3)]½ = [K(NH3)] = [NH3]½/[N2]½[H2]3/2]
 
  • #4
James Pelezo
Gold Member
190
57
Correction... Dividing by 2 => increases K-value by ~780x. SBT
 

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