Problem 81 - Chemistry: The Central Science

Question

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a} .\) (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\), \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M\), and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M?}\)

Step-by-Step Solution

Answer

ΔG° = 21.89 kJ/mol #tag_title# (c) Find the value of ΔG at equilibrium#tag_content# At equilibrium, ΔG = 0. #tag_title# (d) Determine the value of ΔG for given concentrations#tag_content# To find the value of ΔG for the given concentrations, we use the following equation: ΔG = ΔG° + RT ln(Q) Where Q is the reaction quotient given by: Q = [H⁺][NO₂⁻] / [HNO₂] Given concentrations: [H⁺] = 5.0 x 10⁻² M, [NO₂⁻] = 6.0 x 10⁻⁴ M, and [HNO₂] = 0.20 M. Q = (5.0 x 10⁻² M) * (6.0 x 10⁻⁴ M) / (0.20 M) Now, calculate ΔG: ΔG = 21.89 kJ/mol + (8.314 J/mol K) * (298 K) * ln(Q) Solve for ΔG: ΔG = 10.52 kJ/mol

Step 1: (a) Write the chemical equation for the equilibrium

The chemical reaction for the equilibrium of nitrous acid (HNO₂) in aqueous solution can be written as: HNO₂ (aq) ⇌ H⁺ (aq) + NO₂⁻ (aq) Ka will be the equilibrium constant for this reaction.

Step 2: (b) Calculate ΔG° for the dissociation of nitrous acid

The relationship between Ka and ΔG° (standard change in Gibbs free energy) is given by the following equation: ΔG° = -RT ln(Ka) Where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (given as 25°C, so we convert it to 298 K), and Ka is the equilibrium constant. First, we need to look up the value of Ka for nitrous acid at 25°C in Appendix D. From Appendix D, we have Ka = 4.5 x 10⁻⁴. Now we can calculate ΔG°: ΔG° = - (8.314 J/mol K) * (298 K) * ln(4.5 x 10⁻⁴) Solve for ΔG°:

Key Concepts

Acid Dissociation Constant (Ka)

The Acid Dissociation Constant, often represented as \(K_a\), is a crucial part of understanding acid strength in an aqueous solution. It specifically measures the extent to which an acid can donate a proton (\(H^+\)) in water. The larger the \(K_a\), the stronger the acid, as it more readily donates protons.
For nitrous acid (\(HNO_2\)), the dissociation process can be represented by the equilibrium equation:

• \(\text{HNO}_2 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{NO}_2^- (aq)\)

To find \(K_a\), consider the concentrations of the resultant ions compared to the initial acid concentration in water at equilibrium. Specifically, \(K_a\) is defined by the equation:\[ K_a = \frac{[\text{H}^+][\text{NO}_2^-]}{[\text{HNO}_2]} \]At a given temperature, such as 25°C for nitrous acid, \(K_a\) remains constant and helps predict how the acid behaves under different conditions.

Gibbs Free Energy (ΔG)

Gibbs Free Energy, denoted as \(\Delta G\), is a measure used to predict the feasibility and extent of chemical reactions at constant temperature and pressure. It interrelates enthalpy, temperature, and entropy, and is essential in understanding chemical equilibria.
The relationship between \(K_a\) and \(\Delta G^\circ\) (standard Gibbs Free Energy change) is given by:\[ \Delta G^\circ = -RT \ln(K_a) \]where \(R\) is the gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin, and \(K_a\) is the equilibrium constant for an acid.
For nitrous acid, inserting \(K_a = 4.5 \times 10^{-4}\) at 25°C (or 298 K) into the equation allows us to calculate \(\Delta G^\circ\). This calculation provides insight into the spontaneity of the dissociation process. A negative \(\Delta G^\circ\) indicates a spontaneous reaction under the given standard conditions.

Nitrous Acid (HNO₂)

Nitrous acid (\(HNO_2\)) is a weak acid noted for its role in various chemical applications including organic synthesis and environmental chemistry. Its aqueous dissociation can be expressed by the equilibrium mentioned previously:

• \(\text{HNO}_2 (aq) \rightleftharpoons \text{H}^+ (aq) + \text{NO}_2^- (aq)\)

Understanding nitrous acid involves calculating its tendencies to dissociate in water. This involves both the equilibrium constant \(K_a\) and Gibbs Free Energy changes which give insights into its chemical behavior.
Nitrous acid's properties make it an intermediate species often of concern in environmental settings, especially given its formation and decomposition in nitrogen-containing compounds. Its weak acidic nature means that in comparison to strong acids, \(HNO_2\) has a lower tendency to ionize completely, thus, its equilibrium shifts more towards the undissociated form.