# 20.8: Calculations of Free Energy and \(K_\text{eq}\)

- Page ID
- 53928

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:

\[\Delta G^\text{o} = -RT \: \text{ln} \: K_\text{eq}\]

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 \(25^\text{o} \text{C}\).

\[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO} \left( g \right)\]

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__

- \(K_\text{eq} = ?\)

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.*

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

*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.

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

**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__

- \(\Delta G^\text{o} = ? \: \text{kJ/mol}\)

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

*Step 2: Solve.*

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

*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.

## Summary

- The relationship between \(\Delta G\) and \(K_\text{eq}\) is described.
- Calculations involving these two parameters are shown.

## Contributors and Attributions

CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.