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Chapter 19: Problem 81

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 \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M?}\)

Short Answer

Expert verified
螖G掳 = 21.89 kJ/mol #tag_title# (c) Find the value of 螖G at equilibrium#tag_content# At equilibrium, 螖G = 0. #tag_title# (d) Determine the value of 螖G for given concentrations#tag_content# To find the value of 螖G for the given concentrations, we use the following equation: 螖G = 螖G掳 + RT ln(Q) Where Q is the reaction quotient given by: Q = [H鈦篯[NO鈧傗伝] / [HNO鈧俔 Given concentrations: [H鈦篯 = 5.0 x 10鈦宦� M, [NO鈧傗伝] = 6.0 x 10鈦烩伌 M, and [HNO鈧俔 = 0.20 M. Q = (5.0 x 10鈦宦� M) * (6.0 x 10鈦烩伌 M) / (0.20 M) Now, calculate 螖G: 螖G = 21.89 kJ/mol + (8.314 J/mol K) * (298 K) * ln(Q) Solve for 螖G: 螖G = 10.52 kJ/mol

Step by step solution

01

(a) Write the chemical equation for the equilibrium

The chemical reaction for the equilibrium of nitrous acid (HNO鈧�) in aqueous solution can be written as: HNO鈧� (aq) 鈬� H鈦� (aq) + NO鈧傗伝 (aq) Ka will be the equilibrium constant for this reaction.
02

(b) Calculate 螖G掳 for the dissociation of nitrous acid

The relationship between Ka and 螖G掳 (standard change in Gibbs free energy) is given by the following equation: 螖G掳 = -RT ln(Ka) Where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (given as 25掳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掳C in Appendix D. From Appendix D, we have Ka = 4.5 x 10鈦烩伌. Now we can calculate 螖G掳: 螖G掳 = - (8.314 J/mol K) * (298 K) * ln(4.5 x 10鈦烩伌) Solve for 螖G掳:

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳C for nitrous acid, \(K_a\) remains constant and helps predict how the acid behaves under different conditions.
Gibbs Free Energy (螖G)
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路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掳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 (HNO鈧�)
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.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility of manufacturing \(\mathrm{NH}_{3}\) from \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) depends entirely on the value of \(\Delta H\) for the process \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) The reaction of \(\mathrm{Na}(s)\) with \(\mathrm{Cl}_{2}(g)\) to form \(\mathrm{NaCl}(s)\) is a spontaneous process. (c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system.

Using data in Appendix \(C\), calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\). (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(\mathrm{s}\), graphite \()+2 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CCl}_{4}(\mathrm{~g})\) (c) \(2 \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{POCl}_{3}(\mathrm{~g})\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The normal freezing point of 1 -propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\right)\) is \(-127{ }^{\circ} \mathrm{C}\). (a) Is the freezing of 1 -propanol an endothermic or exothermic process? (b) In what temperature range is the freezing of 1 -propanol a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid and solid 1-propanol are in equilibrium? Explain.

Aceticacid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{CO}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}(l) $$ (a) Calculate the equilibrium constant for the reaction at \(25^{\circ} \mathrm{C}\). (b) Industrially, this reaction is run at temperatures above \(25^{\circ} \mathrm{C}\). Will an increase in temperature produce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to \(1 ?\) (You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, and you may ignore any phase changes that might occur.)

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(s) \stackrel{\mathrm{L}}{\longrightarrow} \mathrm{BaO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (d) \(\mathrm{FeCl}_{2}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{HCl}(g)\)
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