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Trouble with Gibbs Free Energy & Equilibrium Constant Calc.

  1. Jun 24, 2016 #1
    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: Jun 24, 2016
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  3. Jun 24, 2016 #2

    Borek

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    Staff: Mentor

    [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]
     
  4. Jun 29, 2016 #3

    James Pelezo

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    Gold Member

    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]
     
  5. Jun 29, 2016 #4

    James Pelezo

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    Gold Member

    Correction... Dividing by 2 => increases K-value by ~780x. SBT
     
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