Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 91

Consider the reactions $$\begin{aligned}\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)\end{aligned}$$ where $$\text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(K_{\text {reaction } 2}>K_{\text {reaction } 1 .}\) Explain.

Short Answer

Expert verified
The difference in the equilibrium constants (K) for the two reactions is due to the difference in the change in entropy (ΔS). Ethylenediamine (en) is a bidentate ligand and can replace two ammonia ligands, resulting in a greater decrease in entropy (more negative ΔS) in reaction 2. This leads to a more negative Gibbs free energy (ΔG) and a higher equilibrium constant (K₂) for reaction 2 compared to reaction 1.

Step by step solution

01

Recall the equilibrium constant (K) and Gibbs free energy (ΔG) equation:

First, remember the equation that relates the equilibrium constant (K) to the Gibbs free energy change (ΔG): ΔG = -RT ln(K) From this equation, we can tell that as the equilibrium constant (K) increases, the Gibbs free energy change (ΔG) becomes more negative.
02

Analyze the change in enthalpy (ΔH) and change in entropy (ΔS):

We are given that the change in enthalpy (ΔH) for both reactions are quite similar. So, we should focus on the change in entropy (ΔS) to explain the difference in equilibrium constants. Recall the equation that relates ΔG, ΔH, and ΔS: ΔG = ΔH - TΔS When the reactions have similar ΔH values but different ΔG values, it is the ΔS value that accounts for the difference.
03

Compare the reactions in terms of ligands being used:

Reaction 1 involves ammonia (NH₃) as the ligand, while reaction 2 involves ethylenediamine (en) as the ligand. Ethylenediamine is a bidentate ligand, meaning it has two donor atoms that can bind to the central metal atom, in this case, Ni²⁺. Ammonia is a monodentate ligand, meaning it has only one donor atom that can bind to the central metal atom.
04

Relate the difference in ligands to the change in entropy (ΔS):

As ethylenediamine is a bidentate ligand, it can replace two ammonia ligands in the coordination sphere of the nickel ion. This results in a decrease in the number of particles in the reaction. The decrease in the number of particles corresponds to a decrease in entropy (ΔS) for reaction 2. Since reaction 1 involves six ammonia ligands, there is a relatively smaller decrease in particles and thus a smaller decrease in entropy (ΔS) in comparison to reaction 2.
05

Relate the change in entropy (ΔS) to the equilibrium constants (K):

From the equations ΔG = -RT ln(K) and ΔG = ΔH - TΔS, It can be inferred that the reaction 2 with a lower ΔS (more negative) will have more negative ΔG, leading to a higher equilibrium constant (K₂). On the other hand, reaction 1 with a lesser decrease in entropy (less negative ΔS) will have a less negative ΔG, leading to a lower equilibrium constant (K₁). By understanding the relationship between the changes in enthalpy, Gibbs free energy, and entropy, we can explain that K₂ > K₁ is due to the greater decrease in entropy (ΔS) in reaction 2 involving bidentate ligand ethylenediamine compared to reaction 1 involving monodentate ligand ammonia.

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

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix 4, predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature dependent. Predict the sign of \(\Delta S_{\text {sarr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) at \(298 \mathrm{~K}\). Using these values and data in Appendix 4, calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the preceding reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\).

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\right.\) also called ethyl mercaptan) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy of vaporization for this substance?

Consider two reactions for the production of ethanol: $$\begin{array}{l}\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \\\ \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{H}_{2}(g) \end{array}$$ Which would be the more thermodynamically feasible at standard conditions? Why?

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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