Problem 68 - Chemistry: Principles and Reactions

Question

Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{HF}(a q)\) is \(-320.1 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}\) for \(\mathrm{HF}(a q)\) is \(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\), find \(K_{\mathrm{a}}\) for \(\mathrm{HF}\) at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Answer: The acid dissociation constant, \(K_{\mathrm{a}}\), for aqueous hydrogen fluoride (HF) at \(25^{\circ} \mathrm{C}\) is approximately \(3.70 \times 10^{63}\).

Step 1: Write the balanced chemical equation for the dissociation of HF

The balanced chemical equation for the dissociation of aqueous hydrogen fluoride (HF) is: HF(aq) \(\rightleftharpoons\) H\({^{+}}\)(aq) + F\({^{-}}\)(aq)

Step 2: Calculate the Gibbs free energy, \(\Delta G^{\circ}\), for the reaction

We can use the following equation to calculate the Gibbs free energy, \(\Delta G^{\circ}\), for the reaction: \(\Delta G^{\circ} = \Delta H_{\mathrm{f}}^{\circ} - T\Delta S^{\circ}\) where \(T\) is the temperature in Kelvin. Since the given temperature is \(25^{\circ} \mathrm{C}\), we need to convert it into Kelvin: \(T = 25 + 273.15 = 298.15\) K Now, plug in the given values to calculate \(\Delta G^{\circ}\): \(\Delta G^{\circ} = -320.1\mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{K})(88.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\) \(\Delta G^{\circ} = -320.1\mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{K})(0.0887 \mathrm{~kJ} / \mathrm{mol} \cdot \mathrm{K})\) \(\Delta G^{\circ} = -320.1 - 26.44 = -346.54\) kJ/mol

Step 3: Use the relationship between \(\Delta G^{\circ}\) and \(K_{\mathrm{a}}\) to find \(K_{\mathrm{a}}\)

We can use the relationship between Gibbs free energy and the equilibrium constant, \(K\), as follows: \(\Delta G^{\circ} = -RT\ln{K}\) where \(R\) is the gas constant (8.314 J/mol⋅K). To find the acid dissociation constant, \(K_{\mathrm{a}}\), we can rearrange the equation and solve for \(K_{\mathrm{a}}\): \(K_{\mathrm{a}} = e^{(-\Delta G^{\circ}/RT)}\) First, convert the Gibbs free energy into J/mol: \(\Delta G^{\circ} = -346.54\mathrm{~kJ/mol} \times 1000\mathrm{~J/kJ} = -346540\mathrm{~J/mol}\) Now, plug in the values and calculate \(K_{\mathrm{a}}\): \(K_{\mathrm{a}} = e^{(-(-346540\mathrm{~J/mol})/(8.314 \mathrm{J/mol⋅K} \times 298.15\mathrm{~K}))}\) \(K_{\mathrm{a}} = e^{146.96}\) \(K_{\mathrm{a}} \approx 3.70 \times 10^{63}\) So, the acid dissociation constant, \(K_{\mathrm{a}}\), for HF at \(25^{\circ} \mathrm{C}\), is approximately \(3.70 \times 10^{63}\).

Key Concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is a crucial concept in chemical thermodynamics. It helps predict whether a reaction will occur spontaneously.
To determine \( \Delta G \) for a reaction, use the formula:

• \( \Delta G = \Delta H - T \Delta S \)

where:

• \( \Delta H \) is the change in enthalpy, or heat energy of the system.
• \( T \) is the temperature in Kelvin.
• \( \Delta S \) is the change in entropy, or disorder, in the system.

A negative \( \Delta G \) indicates a spontaneous reaction under constant temperature and pressure. This means that the reaction can proceed without any external energy input.

Dissociation Constant

The Dissociation Constant, often symbolized as \( K_{a} \), measures the tendency of a larger compound to separate, or dissociate, into smaller components. In acids, it reflects the compound's strength:

• A large \( K_{a} \) value means the acid dissociates completely in water, classifying it as a strong acid.
• A small \( K_{a} \) value indicates a weak acid, as it dissociates incompletely.

This constant provides insights into the chemical equilibrium position of acid-based reactions, essential for understanding reactivity and chemical behavior in solutions.

Enthalpy of Formation

Enthalpy of Formation, \( \Delta H_{f}^{\circ} \), represents the change in energy when one mole of a substance is formed from its elements in their standard states. It is an essential parameter in calculating reaction enthalpies:

• A negative \( \Delta H_{f}^{\circ} \) signifies an exothermic reaction, releasing heat to the surroundings.
• A positive \( \Delta H_{f}^{\circ} \) implies an endothermic reaction, absorbing heat from the surroundings.

Understanding this concept helps in determining the energy efficiency of chemical processes and predicting how reactions evolve.

Entropy

Entropy, symbolized by \( S \), measures the level of disorder or randomness in a system. It plays a vital role in determining reaction spontaneity:

• Higher entropy means more disorder and often a higher chance of the process being spontaneous.
• The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time.

In reactions, the change in entropy, \( \Delta S \), can determine how the reaction and environment interact. Enthalpy and entropy together help predict the reaction's direction at a constant temperature and pressure.

Equilibrium Constant

The Equilibrium Constant, \( K \), characterizes the balance point of a reversible reaction. It's calculated using product and reactant concentrations at equilibrium:

• If \( K \) is much greater than 1, products are favored at equilibrium.
• If \( K \) is much less than 1, reactants are favored at equilibrium.

This constant is crucial for predicting how changing conditions affect a reaction's direction and extent, as it quantifies the relative concentrations of products and reactants.