Problem 90 - Chemistry An Atoms First Approach

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 $$HF(a q) \rightleftharpoons H^{+}(a q)+F^{-}(a q)$$ 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. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 M\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} M\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Step-by-Step Solution

Answer

From the given data and calculations, we find the following ΔG values for each set of conditions: a. ΔG < 0, the reaction proceeds in the forward direction (to the right) b. ΔG < 0, the reaction proceeds in the forward direction (to the right) c. ΔG = 0, the reaction is in equilibrium d. ΔG > 0, the reaction proceeds in the backward direction (to the left) e. ΔG > 0, the reaction proceeds in the backward direction (to the left)

Step 1: Calculate ΔGº

Standard Gibbs Free Energy change (ΔGº) can be calculated using the relationship between dissociation constant (Ka) and ΔGº: \[\Delta G^{\circ}=-R T \ln K_{a}\] At 25°C, T = 298 K, R = 8.314 J/mol K, and Ka = 7.2 × 10⁻⁴. Plug these values into the equation to find ΔG°: \[\Delta G^{\circ}=-((8.314 \, \text{J/mol K})(298 \, \text{K})\ln(7.2 \times 10^{-4}))\]

Step 2: Calculate ΔG for each condition

We can find ΔG for each of the five sets of conditions using the equation \(\Delta G = \Delta G^\circ + RT \ln(Q)\), where Q is the reaction quotient, given by \([H^{+}][F^{-}]/[HF]\). Calculate Q for each set of conditions: a. Q = (1.0)(1.0)/(1.0) b. Q = (2.7 × 10⁻²)(2.7 × 10⁻²)/(0.98) c. Q = (1.0 × 10⁻⁵)(1.0 × 10⁻⁵)/(1.0 × 10⁻⁵) d. Q = (7.2 × 10⁻⁴)(0.27)/(0.27) e. Q = (1.0 × 10⁻³)(0.67)/(0.52) Now, plug in the corresponding Q values into the equation ΔG = ΔGº + RTln(Q) to find ΔG for each condition.

Step 3: Determine the direction of reaction

To find in which direction the reaction will shift under each set of conditions, we can compare the calculated ΔG values to zero: - If ΔG < 0, the reaction proceeds in the forward direction (to the right) - If ΔG > 0, the reaction proceeds in the backward direction (to the left) - If ΔG = 0, the reaction is in equilibrium Compare the calculated ΔG values for each set of conditions and determine the direction of the reaction.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a measure that indicates the relative amounts of products and reactants present in a reaction mixture at any given time. It is similar to the equilibrium constant, but unlike the equilibrium constant, \( Q \) is used when the system is not at equilibrium.

• To calculate \( Q \), we use the concentrations of the reactants and products. For the given reaction \( HF(aq) \rightleftharpoons H^+(aq) + F^-(aq) \), the reaction quotient is given by the expression: \( Q = \frac{[H^+][F^-]}{[HF]} \).
• The value of \( Q \) can help predict the direction in which a reaction will proceed to reach equilibrium. If \( Q < K_a \), the reaction will proceed forward to produce more products. If \( Q > K_a \), the reaction will shift backward to form more reactants.

It is important to measure \( Q \) under the same conditions, particularly temperature, to make direct comparisons to \( K_a \) and use it to predict reaction behavior.

Equilibrium

Chemical equilibrium occurs when a reversible reaction reaches a state where the concentrations of reactants and products no longer change with time. At this point, the forward and reverse reaction rates are equal.

• The concept of equilibrium is described by the equilibrium constant \( K \), which is specific to each reaction and changes only with temperature.
• At equilibrium, the reaction quotient \( Q \) becomes equal to the equilibrium constant \( K \). For the dissolution of hydrofluoric acid (\( HF \)), \( K_a = 7.2 \times 10^{-4} \) at 25°C, representing the acid's strength.

While calculating Gibbs Free Energy changes for a reaction, if \( \Delta G = 0 \), the system is at equilibrium, indicating no net change in the composition of the system over time.

Dissociation Constant

The dissociation constant \( K_a \) is a specific type of equilibrium constant that is used to express the extent of ionization for weak acids in an aqueous solution. It provides valuable information about the strength of an acid.

• A large \( K_a \) value indicates a strong acid, which dissociates completely, while a small \( K_a \) suggests a weak acid, which does not ionize fully.
• For our reaction involving \( HF \), the dissociation constant is \( K_a = 7.2 \times 10^{-4} \), signifying that hydrofluoric acid is a weak acid.
• The \( K_a \) value allows us to write the equilibrium expression and thus calculate \( \Delta G \) using the equation \( \Delta G = \Delta G^\circ + RT \ln(Q) \), where \( Q \) is derived from the initial concentrations of \( HF \), \( H^+ \), and \( F^- \).

Understanding how the dissociation constant works is crucial for predicting the behavior of acids and making decisions based on chemical equilibria.

Aqueous Reactions

Aqueous reactions occur when reactants dissolve in water to form a solution, resulting in various chemical processes. The medium of water often influences these reactions due to its solvent properties.

• In the case of the reaction \( HF(aq) \rightleftharpoons H^+(aq) + F^-(aq) \), it is an example of how acids behave in aqueous solutions. \( HF \) dissociates partially, due to the weak nature of the hydrofluoric acid.
• The concentration of the reactants and products in mol/L (molarity) is particularly significant in aqueous reactions. These concentrations directly impact important calculations like the reaction quotient \( Q \), affecting the direction and extent of equilibrium.
• The solvent properties of water provide a dynamic environment that facilitates dissociation, and relative concentration changes can alter the way the reaction progresses.

Aqueous reactions are central to many processes in chemistry, and understanding them is critical for analyzing chemical behavior in solutions.