Question

Consider the equilibrium system $$ \begin{array}{c} \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) \\ \text { Given } \Delta H_{\mathrm{f}}^{\circ} \mathrm{HF}(a q)=-320.1 \mathrm{k} \mathrm{J} / \mathrm{mol} \\ \Delta H_{\mathrm{t}}^{\circ} \mathrm{F}^{-}(a q)=-332.6 \mathrm{k} \mathrm{J} / \mathrm{mol} ; S^{\circ} \mathrm{F}^{-}(a q)=-13.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} ; \\ \mathrm{K}_{\mathrm{a}} \mathrm{HF}=6.9 \times 10^{-4} \text {at } 25^{\circ} \mathrm{C} \end{array} $$ calculate \(S^{\circ}\) for \(\mathrm{HF}(a q) .\)

Short answer

Using the given standard enthalpy and entropy values, as well as the equilibrium constant at 25°C, it was determined that the standard entropy for HF(aq) is approximately -86.29 J/mol·K.

Step-by-step solution

1. Calculate $$\Delta H_{\text{rxn}}^{\circ}$$ for the dissociation of HF(aq)

First, we will calculate the standard enthalpy of the reaction ($$\Delta H_{\text{rxn}}^{\circ}$$), using the given standard enthalpies of formation of the products and reactants. The equilibrium reaction can be written as follows: $$\text{HF}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{F}^{-}(aq)$$ The $$\Delta H_{\text{rxn}}^{\circ}$$ for this reaction can be calculated by subtracting the standard enthalpy of formation of the reactant from the sum of the standard enthalpies of formation of the products: $$\Delta H_{\text{rxn}}^{\circ} = \Delta H_{\text{f}}^{\circ}(\text{F}^{-}(aq)) - \Delta H_{\text{f}}^{\circ}(\text{HF}(aq))$$ Substitute the given values and find $$\Delta H_{\text{rxn}}^{\circ}$$: $$\Delta H_{\text{rxn}}^{\circ} = -332.6 \; \text{kJ/mol} - (-320.1 \; \text{kJ/mol}) = -12.5 \; \text{kJ/mol}$$

Calculate $$\Delta G_{\text{rxn}}^{\circ}$$ using the equilibrium constant

We will now use the given equilibrium constant, $$K_a$$, to determine the Gibbs free energy change ($$\Delta G_{\text{rxn}}^{\circ}$$): $$\Delta G_{\text{rxn}}^{\circ} = -RT \ln K_a$$ Where, $$R$$ is the gas constant (8.314 J/(mol·K)), $$T$$ is the temperature in Kelvin (298K), and $$K_a$$ is the given equilibrium constant for the dissociation of HF(aq) at 25°C (6.9 × 10^(-4)). Plug in the values and calculate $$\Delta G_{\text{rxn}}^{\circ}$$: $$\Delta G_{\text{rxn}}^{\circ} = -(8.314 \; \text{J/(mol·K)}) \times 298 \; \text{K} \times \ln(6.9 \times 10^{-4}) = 45.41 \; \text{kJ/mol}$$

Calculate the missing standard entropy value for HF(aq) using $$\Delta G_{\text{rxn}}^{\circ}$$, $$\Delta H_{\text{rxn}}^{\circ}$$, and the given values

We can now use the relationship between the Gibbs free energy change, enthalpy change, and entropy change for a reaction: $$\Delta G_{\text{rxn}}^{\circ} = \Delta H_{\text{rxn}}^{\circ} - T \Delta S_{\text{rxn}}^{\circ}$$ We are given the standard enthalpy and entropy for the formation of F- ion, but we need the entropy change for the reaction ($$\Delta S_{\text{rxn}}^{\circ}$$). Since HF(aq) dissociates into H+ and F- ions, we can write: $$\Delta S_{\text{rxn}}^{\circ} = S^{\circ}(\text{H}^{+}(aq)) + S^{\circ}(\text{F}^{-}(aq)) - S^{\circ}(\text{HF}(aq))$$ In this equation, we are given $$S^{\circ}(\text{F}^{-}(aq))$$, and we want to find $$S^{\circ}(\text{HF}(aq))$$. Substitute the known values and rearrange the equation to find $$S^{\circ}(\text{HF}(aq))$$: $$S^{\circ}(\text{HF}(aq)) = S^{\circ}(\text{H}^{+}(aq)) + S^{\circ}(\text{F}^{-}(aq)) - \frac{\Delta H_{\text{rxn}}^{\circ} - \Delta G_{\text{rxn}}^{\circ}}{T}$$ Substitute the given values and calculate $$S^{\circ}(\text{HF}(aq))$$: $$S^{\circ}(\text{HF}(aq)) = (-13.8 \; \text{J/mol·K}) - \frac{-12.5 \; \text{kJ/mol} - 45.41 \; \text{kJ/mol}}{298 \; \text{K}} = -86.29 \; \text{J/mol·K}$$ Therefore, the standard entropy for HF(aq) is approximately -86.29 J/mol·K.

Key concepts

Enthalpy Change

Enthalpy change, denoted as ) or ), is the heat content change during a chemical reaction at constant pressure. It's a measure of the total energy change of a system and includes both internal energy and work done due to volume changes against external pressure. In the context of a chemical equilibrium, such as the dissociation of hydrogen fluoride (HF) in water, understanding the enthalpy change helps predict whether a reaction will be exothermic (releasing heat) or endothermic (absorbing heat).

When HF dissociates in water: $$\text{HF}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{F}^{-}(aq)$$, the enthalpy change ) can be calculated using standard enthalpies of formation. It's key to note that a negative ), as in the given exercise, indicates that the reaction is exothermic—heat is released when HF dissociates.

Gibbs Free Energy

Gibbs free energy, represented as ), is a central concept in thermodynamics and physical chemistry that denotes the maximum reversible work obtainable from a chemical reaction at constant temperature and pressure. A negative ) value implies the reaction is spontaneous, meaning it can occur without an external energy input. Conversely, a positive value suggests non-spontaneity.

For the equilibrium system in our exercise, the equilibrium constant ) is used to find ). The equation ) directly relates the chemical potential landscape of reactants and products. Such insights are crucial when predicting the direction in which a reaction will proceed and assessing reaction feasibility under various conditions.

Equilibrium Constant

The equilibrium constant, expressed as ) for acid disassociation constants, is a dimensionless value that reflects the ratio of concentrations of products to reactants at equilibrium for a particular reaction. It is a valuable indicator of the extent to which a reaction proceeds before reaching a state of balance. In the given example, ) quantifies how much HF dissociates into ) and ) at equilibrium. A larger ) value generally means a higher degree of reaction completion.

In the case of HF dissociation, the ) value allows the calculation of Gibbs free energy change and provides an understanding of the reaction's spontaneity and the conditions under which the system reaches equilibrium.

Standard Entropy

Standard entropy, symbolized as ), defines the degree of disorder or randomness at a standard state of one mole of a substance. It's an absolute measure that captures how energy is distributed among the possible microstates of a system, and it plays a critical role in determining the spontaneity and equilibrium position of a chemical reaction.

In the exercise, we aim to find the standard entropy of HF(aq). Using the relationship between Gibbs free energy, enthalpy, and entropy, we derive the unknown entropy value. This calculated ) of HF, which turns out to be negative, helps to appreciate the energy distribution and the kinetic feasibility of the dissociation process.