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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 \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\) \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\) and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

Short Answer

Expert verified
(a) The chemical equation for the equilibrium corresponding to Kₐ is: HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq). (b) By using the value of Kₐ, we can calculate ΔG° for the dissociation of nitrous acid using the formula ΔG° = -RT ln(Kₐ), and it gives approximately 29.86 kJ/mol. (c) At equilibrium, the value of ΔG is 0. (d) With the given concentrations, we first calculate the reaction quotient Q and then find ΔG using the formula ΔG = ΔG° + RT ln(Q). The value of ΔG will be approximately -7.80 kJ/mol.

Step by step solution

01

a) Chemical equation for equilibrium corresponding to Kₐ

Firstly, we should understand what Kₐ is. Kₐ is the acid dissociation constant for an acid, which shows the strength of the acid. We're dealing with nitrous acid (HNO₂) as our given acid in this exercise. The chemical equation for the dissociation of HNO₂ in water can be written as: HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq) From the equation, we can express Kₐ: Kₐ = \(\frac{[H^+][NO_2^-]}{[HNO_2]}\)
02

b) Calculate the standard Gibbs free energy change (ΔG°)

Next, we will relate Kₐ with the standard Gibbs free energy change (ΔG°) via the following equation: ΔG° = -RT ln(Kₐ) where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Kₐ is the acid dissociation constant for nitrous acid. First, we must convert the temperature given (25°C) to Kelvin: T = 25 + 273.15 = 298.15 K Now, use the given Kₐ value (4.50 x 10⁻⁴) to calculate ΔG°: ΔG° = - (8.314 J/mol·K) (298.15 K) ln(4.50 x 10⁻⁴) This will give you the value for the standard Gibbs free energy change.
03

c) Value of ΔG at equilibrium

At equilibrium, the reaction quotient Q is equal to the equilibrium constant Kₐ. Therefore, the value of ΔG at equilibrium is 0, as it indicates that the reaction is in equilibrium and there is no net change in the system.
04

d) Calculate the value of ΔG with given concentrations

Now, we are given concentrations for H⁺, NO₂⁻, and HNO₂ in the solution. We first need to calculate the reaction quotient Q: Q = \(\frac{[H^+][NO_2^-]}{[HNO_2]}\) Q = \(\frac{(5.0 \times 10^{-2}M)(6.0 \times 10^{-4}M)}{0.20 M}\) Next, we calculate ΔG based on the reaction quotient Q using the following formula: ΔG = ΔG° + RT ln(Q) By substituting the values of ΔG°, R, and T calculated earlier, and the newly calculated value of Q, we can now find the value of ΔG.

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Key Concepts

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

Acid Dissociation Constant
The acid dissociation constant, commonly represented as \(K_{a}\), is a measure of the strength of an acid in a solution. It specifically pertains to the equilibrium process where an acid donates a proton (\(H^+\)) to water. In the context of nitrous acid (HNO₂), the dissociation into hydrogen ions and nitrite ions (NO₂⁻) is key. When HNO₂ dissociates in water, the chemical equation is:
  • HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq)
This equilibrium expression is quantified by \(K_{a}\), which is calculated using the concentrations of these ions at equilibrium:\[K_{a} = \frac{[H^+][NO_2^-]}{[HNO_2]} \]A larger \(K_{a}\) value indicates a stronger acid as it more readily donates protons to the solution. In practical terms, knowing the \(K_{a}\) helps predict how an acid will behave in various chemical environments.
Gibbs Free Energy
Gibbs Free Energy, denoted as \(\Delta G\), is a fundamental concept that describes the amount of energy available to do work during a chemical reaction. It incorporates both enthalpy and entropy to indicate the spontaneity of a process. In equilibrium processes, particularly those involving acid dissociation, it's related to \(K_{a}\) through the equation:\[\Delta G^\circ = -RT \ln(K_{a})\]where:
  • \(R\) is the gas constant (8.314 J/mol·K)
  • \(T\) is the temperature in Kelvin
  • \(K_{a}\) is the acid dissociation constant
The standard Gibbs Free Energy change (\(\Delta G^\circ\)) reflects how spontaneous the reaction is under standard conditions (1 atm pressure, 298 K). When \(\Delta G^\circ\) is negative, the reaction proceeds spontaneously. At equilibrium, \(\Delta G\) is zero since no net change occurs. By knowing \(K_{a}\), you can directly calculate \(\Delta G^\circ\), providing insight into the thermodynamics of the dissociation reaction.
Reaction Quotient
The reaction quotient, \(Q\), provides a snapshot of the current state of a reaction, much like \(K\) (equilibrium constant) but at any given point in the reaction's progress. For the dissociation of nitrous acid, \(Q\) is expressed similarly to \(K_{a}\):\[Q = \frac{[H^+][NO_2^-]}{[HNO_2]} \]Whereas \(K_{a}\) describes the state at equilibrium, \(Q\) can be calculated at any point to determine if the reaction is at, before, or beyond equilibrium. By comparing \(Q\) to \(K_{a}\):
  • If \(Q < K_{a}\), the reaction will shift right, indicating more products will form.
  • If \(Q > K_{a}\), it will shift left, showing excess products reverting to reactants.
  • If \(Q = K_{a}\), the system is at equilibrium.
Understanding \(Q\) helps predict directionality and assess what needs to change to reach equilibrium.
Nitrous Acid Dissociation
Nitrous acid (HNO₂) dissociates in water to form hydrogen ions (H⁺) and nitrite ions (NO₂⁻). This dissociation process can be depicted by the equation:
  • HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq)
This equilibrium reaction is essential in understanding the behavior of weak acids, as nitrous acid is not completely dissociated in solution. Its \(K_{a}\) value is a testament to its partial ionization, reflecting both its acid strength and its equilibrium in aqueous solution.
To calculate how spontaneity changes in response to specific concentrations at a given time, you employ the Gibbs Free Energy equation:\[\Delta G = \Delta G^\circ + RT \ln(Q)\]This equation bridges the gap between standard conditions and the actual reaction conditions described by \(Q\). By considering both \(\Delta G^\circ\) and \(Q\), you gain a complete picture of the thermodynamic viability and progress of nitrous acid dissociation in real-time conditions.

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Most popular questions from this chapter

Use Appendix \(\mathrm{C}\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g)\), \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q)\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol} \mathrm{P}_{2}(g)\), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values.

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) & \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) & \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from \(300 \mathrm{~K}\) to \(200 \mathrm{~K}\). (a) Describe how this change might be carried out while keeping the volume of the gas constant. (b) Describe how it might be carried out while keeping the pressure of the gas constant. (c) Does the change in \(\Delta E\) depend on the particular pathway taken to carry out this change of state? Explain.

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\). (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix C. Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements Web site (www.webelements.com).

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2} .\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

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