Problem 51 - Chemistry: An Atoms-Focused Approach

Question

Calculate the free-energy change for the dissolution in water of one mole of \(\mathrm{NaBr}\) and one mole of \(\mathrm{NaI}\) at \(298 \mathrm{K}\) from the values in the following table. $$\begin{array}{lcc}\hline & \Delta H_{\text {solution }}^{\circ}(\mathrm{k} J / \mathrm{mol}) & \Delta S_{\text {solution }}^{\circ}[J /(\mathrm{mol} \cdot \mathrm{K})] \\\\\text { NaBr } & -0.60 & 57 \\\\\hline \text { Nal } & -7.5 & 74 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Answer: The free-energy change for the dissolution of one mole of NaBr in water at 298 K is -17.58 kJ/mol, and for one mole of NaI is -29.52 kJ/mol. Since both values are negative, the dissolution of NaBr and NaI in water are spontaneous processes at 298 K.

Step 1: Review the Formula for Free Energy Change

Recall the formula for free-energy change, which relates free energy to enthalpy change and entropy change: $$\Delta G^{\circ} =\Delta H^{\circ} -T\Delta S^{\circ}$$ where \(\Delta G^{\circ}\) is the change in free energy, \(\Delta H^{\circ}\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S^{\circ}\) is the change in entropy for the system.

Step 2: Calculate the Free Energy Change for NaBr Dissolution

Use the provided values for enthalpy and entropy change and \(T=298\,\text{K}\) to calculate the free-energy change for the dissolution of one mole of NaBr in water: $$\Delta G^{\circ}_{\text{NaBr}} =\Delta H^{\circ}_{\text{NaBr}} -T\Delta S^{\circ}_{\text{NaBr}}$$ $$\Delta G^{\circ}_{\text{NaBr}} = (-0.60\,\text{kJ/mol}) - (298 \,\text{K})(0.057\,\text{kJ}/(\text{mol}\cdot\text{K}))$$ $$\Delta G^{\circ}_{\text{NaBr}} = -0.60\,\text{kJ/mol} - 16.98\,\text{kJ/mol}$$ $$\Delta G^{\circ}_{\text{NaBr}} = -17.58\,\text{kJ/mol}$$

Step 3: Calculate the Free Energy Change for NaI Dissolution

Similarly, use the provided values for enthalpy and entropy change, and \(T=298\,\text{K}\) to calculate the free-energy change for the dissolution of one mole of NaI in water: $$\Delta G^{\circ}_{\text{NaI}} =\Delta H^{\circ}_{\text{NaI}} -T\Delta S^{\circ}_{\text{NaI}}$$ $$\Delta G^{\circ}_{\text{NaI}} = (-7.5\,\text{kJ/mol}) - (298 \,\text{K})(0.074\,\text{kJ}/(\text{mol}\cdot\text{K}))$$ $$\Delta G^{\circ}_{\text{NaI}} = -7.5\,\text{kJ/mol} - 22.02\,\text{kJ/mol}$$ $$\Delta G^{\circ}_{\text{NaI}} = -29.52\,\text{kJ/mol}$$

Step 4: Interpret Results

The free-energy change for the dissolution in water of one mole of NaBr is \(-17.58\,\text{kJ/mol}\) and for one mole of NaI is \(-29.52\,\text{kJ/mol}\). Since the values are negative, the dissolution of both NaBr and NaI are spontaneous processes at \(298\,\mathrm{K}\).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \(\Delta H^{\circ}\), is a measure of the total heat content change in a chemical reaction at constant pressure. It provides insight into whether a process releases or absorbs heat. For the dissolution process, when \(\Delta H^{\circ}\) is negative, heat is released, implying an exothermic reaction.

When considering substances like sodium bromide (NaBr) and sodium iodide (NaI), we observe that their dissolution involves interactions between the water molecules and the solutes. For example, the enthalpy change for NaBr is \(-0.60 \, \text{kJ/mol}\), indicating a slight release of heat. In contrast, NaI, with a \(\Delta H^{\circ} = -7.5 \, \text{kJ/mol}\), releases more heat.

This heat change plays a crucial role in determining the reaction’s spontaneity, as a more negative enthalpy change often makes a process thermodynamically favorable.

Entropy Change

Entropy change, indicated as \(\Delta S^{\circ}\), reflects the randomness or disorder change during a process. In chemical reactions, an increase in entropy, leading to greater disorder, is often favorable.

When substances like NaBr and NaI dissolve, the solid lattice structures break apart, resulting in an increase in the disorder of the system. For NaBr, the entropy change is \(+57 \, \text{J/mol}\cdot\text{K}\), whereas for NaI, it is \(+74 \, \text{J/mol}\cdot\text{K}\). These positive values indicate increased disorder as the ions mingle freely in water.

This concept is critical, as a positive entropy change can drive a reaction toward spontaneity, especially when coupled with an exothermic enthalpic contribution.

Dissolution Process

The dissolution process involves a solute, like NaBr or NaI, dissolving in a solvent such as water, forming a homogeneous solution. This process consists of three primary steps: breaking the solute lattice, breaking solvent-solvent interactions, and forming solute-solvent interactions.

Initially, energy is required to overcome the lattice energy of the solid and some of the solvent's internal attractions. The energy released when new interactions between solute and solvent particles are established can lead to a net release or absorption of energy.

In our example, calculating the free energy changes for the dissolutions indicates that both processes have a net release of energy, suggesting that solute-solvent interactions are favorable, outweighing the energy costs of disrupting initial structures.

Spontaneity of Reactions

The spontaneity of a reaction is determined by Gibbs free energy change, \(\Delta G^{\circ}\). A negative \(\Delta G^{\circ}\) implies that a process happens spontaneously under constant temperature and pressure, a key factor in predicting whether reactions will occur naturally.

The formula \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\) shows how spontaneity is affected by enthalpy and entropy changes. At 298 K, NaBr dissolves with \(\Delta G^{\circ} = -17.58 \, \text{kJ/mol}\), and NaI dissolves with \(\Delta G^{\circ} = -29.52 \, \text{kJ/mol}\). These negative values confirm that both dissolutions are spontaneous.

This concept highlights how thermodynamic parameters work together—if \(\Delta H^{\circ}\) is sufficiently negative, or if \(T\Delta S^{\circ}\) favors increased randomness, a reaction will tend towards spontaneity.