Problem 51 - Chemistry: Principles and Reactions

Question

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

Step-by-Step Solution

Answer

Based on the given information and thermodynamic relationships, the standard entropy (S⁰) of aqueous hydrogen fluoride (HF) is calculated as -28.01 J/mol·K.

Step 1: Write down the relationship between ΔG, ΔH, and ΔS at equilibrium

At equilibrium, the change in Gibbs free energy (ΔG) is equal to 0. The relationship between ΔG, ΔH, and ΔS is given by the equation: $$\Delta G = \Delta H - T \Delta S$$ Since ΔG=0 at equilibrium, this equation can be rewritten as: $$\Delta H = T \Delta S$$

Step 2: Calculate the ΔH for the dissociation reaction

Using the provided standard enthalpy of formation values for HF(aq) and F⁻(aq), we can calculate the ΔH for the dissociation reaction as follows: $$\Delta H_{\mathrm{rxn}}=\Delta H_{\mathrm{f}}^{\circ} \mathrm{F}^{-}(a q) - \Delta H_{\mathrm{f}}^{\circ} \mathrm{HF}(a q)$$ Substitute the given values: $$\Delta H_{\mathrm{rxn}}=(-332.6 \mathrm{~kJ} / \mathrm{mol}) - (-320.1 \mathrm{~kJ} / \mathrm{mol}) = -12.5 \mathrm{~kJ/mol}$$

Step 3: Calculate the ΔS for the dissociation reaction using the equilibrium constant (Ka)

We are given the equilibrium constant for the dissociation of HF at 25°C, \(K_\mathrm{a}\). This can be used to calculate the change in entropy (ΔS) for the reaction at this temperature. The relationship between K, ΔG, and ΔS is given by: $$K_{\mathrm{a}} = e^{\frac{-\Delta G}{R T}}= e^{\frac{\Delta S}{R}}$$ Here, R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298.15 K). We can calculate the ΔS for the reaction as follows: $$\Delta S = R \ln K_{\mathrm{a}}$$ Substitute the given value for the equilibrium constant, Ka, and the gas constant, R: $$\Delta S = (8.314 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}) \ln (6.9 \times 10^{-4}) = -80.57 \mathrm{J/mol \cdot K}$$

Step 4: Calculate the ΔS⁰ for HF using the relationship ΔH = TΔS and the previously calculated values

Now that we have the ΔH and ΔS values for the dissociation reaction, we can calculate the S⁰ for HF(aq). We know that at equilibrium: $$\Delta H = T \Delta S \Rightarrow S^{\circ} \mathrm{HF}(a q)= \frac{\Delta H - T \cdot S^{\circ} \mathrm{F}^{-}(a q)}{T} - \Delta S_{\mathrm{rxn}}$$ Substitute the values for ΔH, T, ΔS, and \(S^{\circ} \mathrm{F}^{-}(a q)\): \begin{align*} S^{\circ} \mathrm{HF}(a q) &= \frac{-12.5 \mathrm{~kJ/mol} - (298.15 \mathrm{K} \times -13.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K})}{298.15 \mathrm{K}} - (-80.57 \mathrm{J/mol \cdot K}) \\ &= \frac{-12.5 \times 10^3 \mathrm{J/mol} + 4112.67 \mathrm{J/mol}}{298.15 \mathrm{K}} + 80.57 \mathrm{J/mol \cdot K} \\ &= -28.01 \mathrm{J/mol \cdot K} \end{align*}

Step 5: State the final answer for standard entropy of HF(aq)

The standard entropy (S⁰) of aqueous hydrogen fluoride (HF) is -28.01 J/mol·K.

Key Concepts

Equilibrium Constant

In the context of chemical reactions, the equilibrium constant ( \( K \) ) is a key concept. It quantifies the ratio of the concentrations of products to reactants at equilibrium. For the dissociation of HF in water, the equilibrium constant is given as \( K_a = 6.9 \times 10^{-4} \) at 25°C.
Please note that:

• When \( K \) is much less than 1, the position of equilibrium favors the reactants.
• A higher \( K \) value would indicate more products and thus a more complete reaction.

To interpret equilibrium constants, you might consider how conditions like temperature can affect them. Since reaction rates and equilibria are temperature-dependent, changes can shift the position of equilibrium, altering the constant's value.
Understanding \( K \) helps chemists predict the direction and extent of reactions. It is especially beneficial for designing chemical processes or determining reaction yields.

Gibbs Free Energy

Gibbs Free Energy ( \( \Delta G \) ) is a thermodynamic quantity that indicates the spontaneous nature of a reaction. It represents the amount of reversible work a system can perform at constant temperature and pressure.
For a system in equilibrium, such as HF dissociation, \( \Delta G \) is zero, leading to the equation:\[ \Delta G = \Delta H - T\Delta S = 0 \] Here:

• \( \Delta H \) is the change in enthalpy.
• \( \Delta S \) is the change in entropy.

This informs us that the system's enthalpy change is balanced by its entropy change, scaled by the temperature.
By calculating \( \Delta G \) and understanding its implications, chemists can predict whether reactions occur spontaneously:- \( \Delta G < 0 \): Reaction is spontaneous.- \( \Delta G > 0 \): Reaction is non-spontaneous.- \( \Delta G = 0 \): Reaction is at equilibrium.

Enthalpy of Formation

Enthalpy of Formation ( \( \Delta H_f^{\circ} \) ) is the change in enthalpy when one mole of a compound forms from its elements in their standard states. For the HF dissociation equilibrium, we calculated \( \Delta H \) using:\[ \Delta H_{\text{rxn}} = \Delta H_f^{\circ} \text{F\(^-\)}(aq) - \Delta H_f^{\circ} \text{HF}(aq) \]This formula uses standard enthalpy of formation values for both HF and \( \text{F}^- \). It's crucial because it tells us the heat absorbed or released in forming compounds from their elements:

• A negative value indicates exothermic reactions, releasing heat.
• A positive value suggests endothermic reactions, absorbing heat.

Understanding \( \Delta H_f^{\circ} \) is essential for predicting reaction behavior and energy changes, aiding in the development and improvement of various chemical processes.

Entropy Calculation

Entropy ( \( S \) ) measures the disorder or randomness in a system. In thermodynamics, it offers insights into the energy dispersal within a reaction.
To find \( S^{\circ} \) for HF(aq), we rearrange the formula for Gibbs Free Energy:\[ \Delta H = T\Delta S \rightarrow S^{\circ} \text{HF}(aq) = \frac{\Delta H - T \cdot S^{\circ} \text{F\(^-\)}(aq)}{T} - \Delta S_{\text{rxn}} \]Where:

• \( T \) is the temperature in Kelvin.
• \( \Delta S_{\text{rxn}} \) is the change in entropy for the reaction.

Entropy calculations help us understand elemental formation complexity and predict reaction spontaneity. In this context, calculating \( S^{\circ} \) for HF includes adjusting for known \( \Delta H \) and \( \Delta S \) of fluoride, offering deeper insights into thermodynamic equilibria.