Consider the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Fe}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Fe}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for the reaction. \(K_{\mathrm{sp}}\) for \(\mathrm{Fe}(\mathrm{OH})_{2}\) is \(1.6 \times 10^{-14}\)

Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It can be represented symbolically, with the double arrow \rightleftharpoons indicating that the process can proceed in both directions. At equilibrium, dynamic balance is achieved, and the system's composition remains constant.
To delve deeper, let's consider the dissolution of iron(II) hydroxide, \(\text{Fe}(\text{OH})_2(s)\), which dissociates in water to form \(\text{Fe}^{2+}(\text{aq})\) and hydroxide ions \(\text{OH}^{-}(\text{aq})\). This reaction can achieve equilibrium, meaning that as much iron(II) hydroxide dissociates into ions as it reforms from the ions.
Understanding chemical equilibrium is essential because it governs the concentrations of compounds during a reaction. In the context of the exercise, the solubility product constant, \(\text{K}_{sp}\), is directly related to the equilibrium concentrations of the dissolved ions. Hence, by computing Gibbs free energy change for the reaction at equilibrium, one can predict the spontaneity and direction of a chemical process.

The solubility product constant, designated as \(\text{K}_{sp}\), is a special type of equilibrium constant that applies to the dissolution of slightly soluble salts. It represents the extent to which a compound can dissolve to form ions in a saturated solution and is defined by the concentrations of the ions at equilibrium.
For our exercise, the solubility product constant for \(\text{Fe}(\text{OH})_2(s)\) is given as \(\text{K}_{sp} = 1.6 \times 10^{-14}\). This numeric value signifies how much iron(II) hydroxide can dissociate in water. A lower \(\text{K}_{sp}\) generally implies a less soluble compound. When we calculate \(\text{K}_{sp}\), we multiply the concentrations of the resulting ions raised to the power of their coefficients in the balanced chemical equation.
For students to grasp this concept better, it’s key to understand that a saturated solution is in equilibrium between the undissolved solid and its dissolved ions. Therefore, the relationship between solubility and the solubility product is vital to predicting whether a precipitate will form when different solutions are mixed.

Thermodynamics in chemistry deals with the energy changes that accompany chemical reactions and physical processes. Gibbs free energy, symbolized by \(\text{G}\), is a thermodynamic potential that can predict the direction of chemical reactions and the equilibrium position. It combines the concepts of enthalpy, entropy, and temperature to assess the spontaneity of processes.
For a reaction at constant temperature and pressure, the change in Gibbs free energy, \(\text{\Delta G}\), determines whether a process is spontaneous. If \(\text{\Delta G}\) is negative, the process occurs spontaneously. The relation between Gibbs free energy and the equilibrium constant, \(\text{\Delta G}^\circ = -RT \ln \text{K}\), allows us to calculate \(\text{\Delta G}^\circ\) given \(\text{K}_{sp}\) and vice versa.
To provide more clarity in the context of our exercise, the students should understand that calculating \(\text{\Delta G}^\circ\) using \(\text{K}_{sp}\) and the universal gas constant \(\text{R}\) indicates the thermodynamic favorability of iron(II) hydroxide's solubility at a certain temperature. Thus, linking thermodynamics to solubility enables us to better predict and control chemical processes.