# 6.9: Relationship Between Free Energy and $$K_\text{eq}$$

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Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and re-deposits on the rock as the carbon dioxide is dissipated into the environment.

## Equilibrium Constant and $$\Delta G$$

At equilibrium, the $$\Delta G$$ for a reversible reaction is equal to zero. $$K_\text{eq}$$ relates the concentrations of all substances in the reaction at equilibrium. Through a more advanced treatment of thermodynamics, we can write the following equation:

$\Delta G^\text{o} = -RT \: \text{ln} \: K_\text{eq}\nonumber$

The variable $$R$$ is the ideal gas constant $$\left( 8.314 \: \text{J/K} \cdot \text{mol} \right)$$, $$T$$ is the Kelvin temperature, and $$\text{ln} \: K_\text{eq}$$ is the natural logarithm of the equilibrium constant.

When $$K_\text{eq}$$ is large, the products of the reaction are favored and the negative sign in the equation means that the $$\Delta G^\text{o}$$ is negative. When $$K_\text{eq}$$ is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative, and so the sign of $$\Delta G^\text{o}$$ is positive. The table below summarizes the relationship of $$\Delta G^\text{o}$$ to $$K_\text{eq}$$:

Relationship of $$\Delta G^\text{o}$$ and $$K_\text{eq}$$
Table $$\PageIndex{1}$$: Relationship of $$\Delta G^\text{o}$$ and $$K_\text{eq}$$
$$K_\text{eq}$$ $$\text{ln} \: K_\text{eq}$$ $$\Delta G^\text{o}$$ Description
>1 positive negative Products are favored at equilibrium.
1 0 0 Reactants and products are equally favored.
<1 negative positive Reactants are favored at equilibrium.

Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.

##### Example $$\PageIndex{1}$$

The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at $$25^\text{o} \text{C}$$.

$\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO} \left( g \right)\nonumber$

The actual concentrations of each gas would be difficult to measure, and so the $$K_\text{eq}$$ for the reaction can more easily be calculated from the $$\Delta G^\text{o}$$, which is equal to $$173.4 \: \text{kJ/mol}$$. Find the $$K_\text{eq}$$.

###### Known
• $$\Delta G^\text{o} = +173.4 \: \text{kJ/mol}$$
• $$R = 8.314 \: \text{J/K} \cdot \text{mol}$$
• $$T = 25^\text{o} \text{C} = 298 \: \text{K}$$
###### Unknown

In order to make the units agree, the value of $$\Delta G^\text{o}$$ will need to be converted to $$\text{J/mol}$$ $$\left( 173,400 \: \text{J/mol} \right)$$. To solve for $$K_\text{eq}$$, the inverse of the natural logarithm, $$e^x$$, will be used.

###### Step 2: Solve.

\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ \text{ln} \: K_\text{eq} &= \frac{-\Delta G^\text{o}}{RT} \\ K_\text{eq} &= e^{\frac{-\Delta G^\text{o}}{RT}} = e^{\frac{-173,400 \: \text{J/mol}}{8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right)}} = 4.0 \times 10^{-31} \end{align*}\nonumber

The large positive free energy change leads to a $$K_\text{eq}$$ that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.

##### Example $$\PageIndex{2}$$

The solubility product constant $$\left( K_\text{sp} \right)$$ of lead (II) iodide is $$1.4 \times 10^{-8}$$ at $$25^\text{o} \text{C}$$. Calculate $$\Delta G^\text{o}$$ for the dissociation of lead (II) iodide in water.

$\ce{PbI_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{I^-} \left( aq \right)\nonumber$

###### Known
• $$K_\text{eq} = K_\text{sp} = 1.4 \times 10^{-8}$$
• $$R = 8.314 \: \text{J/K} \cdot \text{mol}$$
• $$T = 25^\text{o} \text{C} = 298 \: \text{K}$$
###### Unknown

The equation relating $$\Delta G^\text{o}$$ to $$K_\text{eq}$$ can be solved directly.

###### Step 2: Solve.

\begin{align*} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ &= -8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right) \: \text{ln} \left( 1.4 \times 10^{-8} \right) \\ &= 45,000 \: \text{J/mol} \\ &= 45 \: \text{kJ/mol} \end{align*}\nonumber

The large, positive $$\Delta G^\text{o}$$ indicates that the solid lead (II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
• The relationship between $$\Delta G$$ and $$K_\text{eq}$$ is described.
6.9: Relationship Between Free Energy and $$K_\text{eq}$$ is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts.