Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 44

The standard Gibbs energy change for the reaction $$\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{NH}_{4}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$ is \(29.05 \mathrm{kJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{K}\). Use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of \(\mathrm{NH}_{3}(\mathrm{aq})\) \(\mathrm{NH}_{4}^{+}(\mathrm{aq}),\) and \(\mathrm{OH}^{-}(\mathrm{aq})\) are \(0.10 \mathrm{M}, 1.0 \times 10^{-3} \mathrm{M}\) and \(1.0 \times 10^{-3} \mathrm{M},\) respectively.

Short Answer

Expert verified
After comparing \( Q \) and \( K \), the direction of the spontaneous reaction is determined.

Step by step solution

01

Calculation of Equilibrium Constant from Gibbs Energy Change

Using the formula \( \Delta G = -RT \ln K \), where \(\Delta G\) is the standard Gibbs energy change (-29050 J/mol), \(R\) is the universal gas constant (8.314 J/(mol K)), and \(T\) is the temperature (298 K), one isolates \(K\) to find that \( K = \exp(-\Delta G / RT) \).
02

Calculation of Reaction Quotient

The reaction quotient is defined as \( Q = [NH_{4}^{+}][OH^{-}] / [NH_{3}][H_{2}O] \). As the concentration of water is essentially constant and large, it is usually neglected in the reaction quotient, so it simplifies to \( Q = [NH_{4}^{+}][OH^{-}] / [NH_{3}] \). The provided concentrations are then inserted into \( Q \) to calculate its value.
03

Determine Spontaneity of Reaction

The spontaneity of the reaction is determined by comparing \( Q \) and \( K \). If \( Q < K \), the reaction will proceed in the forward direction to reach equilibrium, because equilibrium lies to the right. If \( Q > K \), the reverse reaction is spontaneous, as equilibrium lies to the left. If \( Q = K \), the reaction is in equilibrium, and no net change occurs.

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

Comment on the difficulties of solving environmental pollution problems from the standpoint of entropy changes associated with the formation of pollutants and with their removal from the environment.

At room temperature and normal atmospheric pressure, is the entropy of the universe positive, negative, or zero for the transition of carbon dioxide solid to liquid?

For one of the following reactions, \(K_{c} K_{p}=K .\) Identify that reaction. For the other two reactions, what is the relationship between \(K_{c}, \bar{K}_{\mathrm{p}},\) and \(K ?\) Explain. (a) \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\) (b) \(\mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g})\) (c) \(\mathrm{NH}_{4} \mathrm{HCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1)\)

If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

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