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 \(K_\text{eq}\):

**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 \(K_\text{eq}\).

**Solution:**

*Step 1: List the known quantities and plan the problem.*

__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.*

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

*Step 3: Think about your result.*

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.

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

**Solution:**

*Step 1: List the known values and plan the problem.*

__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.*

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

*Step 3: Think about your result.*

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.