Question

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}\) Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Short answer

In conclusion, for conditions (a), (b), (c), and (e), the reaction will shift in the reverse direction to reach equilibrium. In condition (d), the reaction is already at equilibrium.

Step-by-step solution

Calculate \(\Delta G^\circ\)

Given: \(K_a = 7.2\times 10^{-4}\) \(T = 25^\circ C = 298 \, K\) \(R = 8.314 J\, mol^{-1}K^{-1}\) \(\Delta G^\circ = -RT\ln(K_a) = -(8.314\, J\, mol^{-1}K^{-1})(298\, K)\ln(7.2\times 10^{-4})\) \(\Delta G^\circ = 1920.85\, J\, mol^{-1}\) Now, we can find \(\Delta G\) for each set of conditions provided.

a. Calculate Q and \(\Delta G\) for the first set of conditions

Given: \([HF]=[H^+]=[F^-]=1.0 M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0)(1.0)}{(1.0)}\) Q = 1 \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1)\) Since \(\ln(1) = 0\), \(\Delta G = \Delta G^\circ\). Hence, \(\Delta G = 1920.85\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

b. Calculate Q and \(\Delta G\) for the second set of conditions

Given: \([HF]=0.98M\), \([H^+]=[F^-]=2.7\times 10^{-2}M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(2.7\times 10^{-2})(2.7\times 10^{-2})}{(0.98)}\) Q = 7.47\times 10^{-4} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(7.47\times 10^{-4})\) \(\Delta G \approx 1898.47\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

c. Calculate Q and \(\Delta G\) for the third set of conditions

Given: \([HF]=[H^+]=[F^-]=1.0\times 10^{-5} M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0\times 10^{-5})(1.0\times 10^{-5})}{(1.0\times 10^{-5})}\) Q = 1.0\times 10^{-5} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1.0\times 10^{-5})\) \(\Delta G \approx 1847.52\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction.

d. Calculate Q and \(\Delta G\) for the fourth set of conditions

Given: \([HF]=[F^-]=0.27 M\), \([H^+]=7.2\times 10^{-4} M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(0.27)(7.2\times 10^{-4})}{(0.27)}\) Q = 7.2\times 10^{-4} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(7.2\times 10^{-4})\) \(\Delta G \approx 0\, J\, mol^{-1}\). Since \(\Delta G = 0\), the reaction is at equilibrium.

e. Calculate Q and \(\Delta G\) for the fifth set of conditions

Given: \([HF]=0.52M\), \([F^-]=0.67M\), \([H^+]=1.0\times 10^{-3}M\) Q = \(\frac{[H^+][F^-]}{[HF]}\) Q = \(\frac{(1.0\times 10^{-3})(0.67)}{(0.52)}\) Q = 1.29\times 10^{-3} \(\Delta G = \Delta G^\circ + RT\ln(Q) = 1920.85 + (8.314)(298)\ln(1.29\times 10^{-3})\) \(\Delta G \approx 402.93\, J\, mol^{-1}\). Since \(\Delta G > 0\), the reaction will shift to achieve equilibrium in the reverse direction. In conclusion, for conditions (a), (b), (c), and (e), the reaction will shift in the reverse direction to reach equilibrium. In condition (d), the reaction is already at equilibrium.

Key concepts

Equilibrium Constants

The equilibrium constant, typically denoted as \(K\), is a crucial value in thermodynamics that reveals the balance point of a chemical reaction at a certain temperature. For reactions in aqueous solutions, such as the dissociation of an acid like hydrofluoric acid (HF), we often use specific constants such as the acid dissociation constant \(K_a\).

The equilibrium constant is calculated from the concentrations of the reactants and products when a reaction reaches equilibrium. It is expressed as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients in the balanced equation. For the HF reaction, it's given by the formula:

• \(K_a = \frac{[H^+][F^-]}{[HF]}\)

This provides insight into the position of equilibrium. If \(K_a\) is large, products dominate at equilibrium, indicating strong acid dissociation, whereas a small \(K_a\) suggests limited dissociation. In our exercise, \(K_a = 7.2 \times 10^{-4}\), indicating HF is a weak acid.

Understanding equilibrium constants helps predict how a system at equilibrium will respond to changes, and is integral to calculating Gibbs Free Energy (∆G), guiding us on how the reaction might proceed.

Reaction Quotient

The reaction quotient \(Q\) plays a vital role in determining the direction a reaction will shift to reach equilibrium. It's calculated in the same way as the equilibrium constant, using the current concentrations of the reactants and products. However, \(Q\) doesn’t require the system to be at equilibrium.

For the reaction of hydrofluoric acid in the original exercise, \(Q\) can be calculated using the equation:

• \(Q = \frac{[H^+][F^-]}{[HF]}\)

Where the concentrations are the current circumstances of the reaction, not necessarily at equilibrium. Comparing \(Q\) to \(K_a\) tells us the system's status:

• If \(Q > K_a\), the system has excess products, suggesting the reaction will shift left towards the reactants.
• If \(Q < K_a\), there are excess reactants, so the reaction will shift right towards the products.
• If \(Q = K_a\), the system is at equilibrium, and no shift occurs.

By calculating \(Q\), we understand where the system stands relative to equilibrium and predict which direction it needs to move to stabilize.

Acid-Base Reactions

Acid-base reactions are a class of chemical reactions that involve the transfer of protons between reactants. In the case of hydrofluoric acid (HF) dissociation, it acts as a Brønsted-Lowry acid by donating a proton (H+) to produce fluoride ions (\(F^-\)) in solution.

The simple dissociation reaction can be written as:

• \(HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\)

This reaction is reversible, and according to Le Chatelier’s principle, it can shift in either direction depending on the conditions such as concentration and temperature. The extent of dissociation of an acid in water is quantified by the acid dissociation constant, \(K_a\), and its resulting effect on \([ ext{H}^+]\), the concentration of hydrogen ions, determines the acidity or pH of the solution.

Understanding this mechanism and its governing rules helps us manage chemical equilibria better in processes involving acids and bases.

Thermodynamics

Thermodynamics is the study of energy changes, particularly the conversion between heat and other forms of energy. In a chemical reaction, thermodynamics helps us understand whether a reaction will occur spontaneously under given conditions by determining the Gibbs Free Energy (∆G).

Gibbs Free Energy combines enthalpy and entropy into a single value that indicates the maximum amount of work obtainable from the reaction at constant temperature and pressure. The formula:

• \(∆G = ∆G^° + RT \ln(Q)\)

relates ∆G to ∆G° (the standard Gibbs Free Energy change) while accounting for the reaction conditions through the reaction quotient \(Q\).

For hydrofluoric acid dissociation, if \(∆G < 0\), the reaction proceeds forward spontaneously; if \(∆G > 0\), it may proceed in reverse to reach equilibrium; and if \(∆G = 0\), it is at equilibrium. This framework allows us to predict and understand the behavior of chemical reactions from a thermodynamic viewpoint, making it essential for practical and theoretical chemistry.