Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 62

Consider the reaction $$ \mathrm{NH}_{4}{ }^{+}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NH}_{3}(a q) $$ Use \(\Delta G_{i}^{\circ}\) for \(\mathrm{NH}_{3}(a q)\) at \(25^{\circ} \mathrm{C}=-26.7 \mathrm{~kJ} / \mathrm{mol}\) and the appropriate tables to calculate (a) \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) (b) \(K_{\mathrm{a}}\) at \(25^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Question: Calculate the standard Gibbs free energy change and the acid dissociation constant for the following reaction at 25 °C using the given standard Gibbs free energy value: \(NH_3{(aq)} + H^+_{(aq)} \rightleftharpoons NH_4^+_{(aq)}\) \(\Delta G_{NH-3}^\circ = -26.7 \frac{kJ}{mol}\) Answer: The standard Gibbs free energy change, \(\Delta G^\circ\), for the reaction at 25 °C is 52.74 \(\frac{kJ}{mol}\) (part a), and the acid dissociation constant, \(K_a\), at 25 °C is approximately 0.1209 (part b).

Step by step solution

01

Identifying the Equation

First, identify the equation to be used to find the standard Gibbs free energy change: $$\Delta G^\circ = \sum{(n_i \Delta G_i^\circ)}$$ Step 2: Calculate \(\Delta G^\circ\)
02

Calculate \(\Delta G^\circ\)

Using the equation \(\Delta G^\circ = \sum{(n_i \Delta G_i^\circ)}\), we can calculate \(\Delta G^\circ\) for the given reaction: $$\Delta G^\circ = \Delta G_{H^+}^\circ + \Delta G_{NH_3}^\circ - \Delta G_{NH_4^+}^\circ$$ Now, using the given value for \(\Delta G_{NH_3}^\circ = -26.7 \frac{kJ}{mol}\), and finding the values for \(\Delta G_{H^+}^\circ = 0\) and \(\Delta G_{NH_4^+}^\circ = -79.44 \frac{kJ}{mol}\) from standard tables, we can calculate the standard Gibbs energy change for the reaction: $$\Delta G^\circ = (0) + (-26.7) - (-79.44)$$ $$\Delta G^\circ = 52.74 \frac{kJ}{mol}$$ Step 3: Calculate \(K_a\) using \(\Delta G^\circ\):
03

Calculating \(K_a\)

Next, we need to find \(K_a\) at \(25^\circ \mathrm{C}\). Using the equation \(\Delta G^\circ = -RT\ln{K_a}\), where R is the gas constant (8.314 \(\mathrm{J\,mol^{-1}\,K^{-1}}\)) and T is the temperature in Kelvin (298K): $$K_a = e^{\frac{-\Delta G^\circ}{RT}}$$ Plugging in the values for \(\Delta G^\circ\), R, and T, we get: $$K_a = e^{\frac{-52740\,\mathrm{J\,mol^{-1}}}{8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}(298\,\mathrm{K})}}$$ $$K_a = e^{-2.1157}$$ $$K_a = 0.1209$$ Therefore, the Gibbs free energy change, \(\Delta G^\circ\), at \(25^\circ \mathrm{C}\) is 52.74 \(\frac{kJ}{mol}\) (part a) and the acid dissociation constant, \(K_a\), at \(25^\circ \mathrm{C}\) is approximately 0.1209 (part b).

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

Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

It is desired to produce tin from its ore, cassiterite, \(\mathrm{SnO}_{2}\), at as low a temperature as possible. The ore could be (a) decomposed by heating, producing tin and oxygen. (b) heated with hydrogen gas, producing tin and water vapor. (c) heated with carbon, producing tin and carbon dioxide. Solely on the basis of thermodynamic principles, which method would you recommend? Show calculations.

Given the following data for sodium $$ \begin{aligned} &\mathrm{Na}(s): S^{\circ}=51.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\\ &\mathrm{Na}(g): S^{\circ}=153.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \quad \Delta H_{\mathrm{f}}^{\circ}=108.7 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ estimate the temperature at which sodium sublimes at 1 atm. $$ \mathrm{Na}(s) \rightleftharpoons \mathrm{Na}(g) $$

Which of the following processes are spontaneous? (a) building a sand castle (b) outlining your chemistry notes (c) wind scattering leaves in a pile

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

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