# Trouble with Gibbs Free Energy & Equilibrium Constant Calc.

• Electric to be
In summary, changing the coefficients of the reaction changes the equilibrium constant for the reaction.
Electric to be
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:
Electric to be said:
The equilibrium constant should remain the same though, should it not?

$$\frac{[NH_3]^2}{[N_2][H_2]^3} = \sqrt{ \frac{[NH_3]}{[N_2]^{\frac 1 2}[H_2]^{\frac 3 2}}}$$

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]

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

## What is Gibbs Free Energy?

Gibbs free energy is a thermodynamic property that measures the amount of energy available to do useful work in a system at constant temperature and pressure.

## How is Gibbs Free Energy related to Equilibrium Constant?

The Gibbs free energy change in a chemical reaction is related to the equilibrium constant of the reaction through the equation ΔG = -RT ln K, where R is the gas constant and T is the temperature in Kelvin.

## What is the significance of ΔG being negative or positive?

If ΔG is negative, the reaction is spontaneous and will proceed in the forward direction. If ΔG is positive, the reaction is non-spontaneous and will not proceed without an external energy source. If ΔG is zero, the reaction is at equilibrium.

## How does temperature affect Gibbs Free Energy and Equilibrium Constant?

An increase in temperature will decrease the value of ΔG and may cause an exergonic (spontaneous) reaction to become endergonic (non-spontaneous). It will also change the numerical value of the equilibrium constant, but not the direction of the reaction.

## What are the limitations of using Gibbs Free Energy and Equilibrium Constant in predicting reactions?

Gibbs free energy and equilibrium constant calculations assume ideal conditions and do not take into account factors such as reaction kinetics, concentration changes, and non-ideal behavior of reactants and products. These calculations are also only accurate for reactions at constant temperature and pressure.

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