Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 85

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

(a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using \(\mathrm{CO}_{2}\) as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain.

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0\), and \(w=0\). (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) In thermodynamics, what is the "driving force" for the expansion of the gas?

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix \(\mathrm{C}\). (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2} ?\) (c) If \(P_{\mathrm{SO}_{2}}=P_{\mathrm{H}_{2} \mathrm{~S}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free