Question

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 \(1.6 \times 10^{-14}\)

Short answer

\(\Delta G^{\circ} \approx 77.66 \text{ kJ/mol}\).

Step-by-step solution

Identify the formula

For this reaction, we need to calculate \(\Delta G^{\circ}\) using the relationship between \(\Delta G^{\circ}\) and \(K_{\mathrm{sp}}\), which is:\[ \Delta G^{\circ} = -RT \ln K_{\mathrm{sp}} \] where \(R\) is the universal gas constant and \(T\) is the temperature in Kelvin.

Convert temperature to Kelvin

The given temperature is \(25^{\circ} \text{C}\) which we need to convert to Kelvin using the formula: \[ T = 25 + 273.15 = 298.15 \text{ K} \]

Determine the value of R

The universal gas constant \(R\) has a value of \(8.314 \text{ J/mol⋅K}\). This will be used in the \(\Delta G^{\circ}\) calculation.

Calculate \(\Delta G^{\circ}\)

Substitute \(R = 8.314 \text{ J/mol⋅K}\), \(T = 298.15 \text{ K}\), and \(K_{\mathrm{sp}} = 1.6 \times 10^{-14}\) into the formula:\[ \Delta G^{\circ} = -8.314 \times 298.15 \times \ln(1.6 \times 10^{-14}) \]First, compute \(\ln(1.6 \times 10^{-14})\) which equals approximately \(-31.356\).Then, calculate: \(\Delta G^{\circ} = -8.314 \times 298.15 \times (-31.356)\), resulting in \(\Delta G^{\circ} \approx 77660 \text{ J/mol}\) or \(77.66 \text{ kJ/mol}\).

Key concepts

Solubility Product Constant

The solubility product constant, denoted as \(K_{\mathrm{sp}}\), plays a crucial role in understanding the solubility of sparingly soluble compounds. It is a specific type of equilibrium constant applied to solubility equilibria. Essentially, \(K_{\mathrm{sp}}\) represents the maximum concentration of ions in a saturated solution at equilibrium, which is vital for predicting whether a precipitate will form when two solutions are mixed.

• \(K_{\mathrm{sp}}\) is used primarily for ionic compounds that do not fully dissolve in water, like \(\mathrm{Fe(OH)}_{2}\).
• The expression for \(K_{\mathrm{sp}}\) is derived from the reaction quotient of the solid dissolving into its constituent ions.
• The value of \(K_{\mathrm{sp}}\) indicates solubility; a very low \(K_{\mathrm{sp}}\) means the compound is not very soluble in water.

Understanding \(K_{\mathrm{sp}}\) is essential in various fields, such as chemistry, environmental science, and manufacturing processes, where controlling solubility and precipitation is crucial.

Thermodynamics

Thermodynamics in chemistry provides insights into energy changes within chemical reactions. A principal focus is the Gibbs Free Energy change (\(\Delta G^{\circ}\)), which predicts the spontaneity of a reaction. A negative \(\Delta G^{\circ}\) indicates a spontaneous reaction under standard conditions, whereas a positive \(\Delta G^{\circ}\) suggests non-spontaneity.

To calculate \(\Delta G^{\circ}\) for a reaction involving solubility, the relationship \(\Delta G^{\circ} = -RT \ln K_{\mathrm{sp}}\) is used:

• \(R\) denotes the universal gas constant, valued at \(8.314 \, \mathrm{J/mol\cdot K}\).
• \(T\) is the temperature in Kelvin.
• \(K_{\mathrm{sp}}\) is the solubility product constant.

The calculated \(\Delta G^{\circ}\) aids in determining if the dissolution of a substance like \(\mathrm{Fe(OH)}_{2}\) is feasible under specific conditions.
Studying the thermodynamics of reactions not only helps predict feasibility but also supports the understanding of how energy and matter interact.

Chemical Equilibrium

Chemical equilibrium refers to the state of balance in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time, though not necessarily equal. In the context of dissolving solids, equilibrium concepts apply through the solubility product constant \(K_{\mathrm{sp}}\).

Important points about chemical equilibrium:

• Equilibrium signifies a dynamic process; while concentrations remain stable, particles continuously convert between reactants and products.
• Solubility equilibrium involves the solid (e.g., \(\mathrm{Fe(OH)}_{2}(\text{s})\)) and its ions (\(\mathrm{Fe}^{2+}\) and \(\mathrm{OH}^{-}\)).
• The position of equilibrium can shift with changes in temperature, pressure, or concentration, explained by Le Chatelier's Principle.

Understanding these concepts helps predict the likelihood of precipitation or dissolution under different conditions, critical in various scientific and industrial processes.

Temperature Conversion

Temperature conversion is a simple yet essential skill for many scientific calculations, such as those involving thermodynamic equations. Converting temperatures from Celsius to Kelvin is particularly important when applying formulas involving gases and energy calculations since the Kelvin scale is absolute and directly relates to thermodynamic concepts.

The conversion from Celsius to Kelvin follows the straightforward formula:

• \( T_{\text{(Kelvin)}} = T_{\text{(Celsius)}} + 273.15 \)

For example, to convert \(25^{\circ} \text{C}\) to Kelvin, you add 273.15, resulting in 298.15 K.

Using the Kelvin scale is crucial, as thermodynamic equations require these units for consistent and correct results. Always remember that Kelvin is never expressed in degrees, reflecting its role in scientific contexts.