/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Many oxides of nitrogen have pos... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 53

Many oxides of nitrogen have positive values for the standard free energy of formation. Using NO as an example, explain why this is the case.

Short Answer

Expert verified
Many oxides of nitrogen, such as NO, have positive values for the standard free energy of formation (螖Gf掳) due to their endothermic formation, which requires energy to break the strong N-N triple bond. The entropy change (螖S) during their formation is not significant enough to compensate for the positive enthalpy change (螖H). Consequently, the formation of these compounds is thermodynamically not favorable under standard conditions, as demonstrated by the calculation of 螖Gf掳(NO) = 27.5 kJ/mol.

Step by step solution

01

1. Definition of Standard Free Energy of Formation

The standard free energy of formation, denoted as 螖Gf掳, is the change in Gibbs free energy when one mole of a compound is formed from its elements in their most stable forms (i.e., standard states) in a given environment, usually 1 atm pressure and 298.15 K temperature.
02

2. Factors Influencing the Value of Standard Free Energy of Formation

The standard free energy of formation for a reaction depends on two factors: enthalpy change (螖H) and the temperature-dependent entropy change (螖S). The relationship between these factors and free energy is given by: \[ 螖G = 螖H - T螖S \] In this formula, 螖G represents the free energy change, 螖H represents the enthalpy change, T represents the temperature in Kelvin, and 螖S represents the entropy change.
03

3. Enthalpy and Entropy Changes for Oxides of Nitrogen

In the case of oxides of nitrogen, the enthalpy of formation (螖Hf掳) is generally positive, indicating that the formation of these compounds requires energy, i.e., they are endothermic reactions. This is because nitrogen is a very stable molecule and breaking the strong triple bond between nitrogen atoms requires a considerable amount of energy. The entropy change (螖S) for the formation of these compounds is not significant enough to compensate for the positive enthalpy change.
04

4. Standard Free Energy of Formation for Nitric Oxide (NO)

Taking nitric oxide (NO) as an example, the standard enthalpy of formation (螖Hf掳) and the standard entropy change (螖S掳) can be looked up in a table or determined experimentally. Values for the formation of NO at 298.15 K and 1 atm are: 螖Hf掳(NO) = 90.37 kJ/mol 螖S掳(NO) = 210.76 J/mol鈥 Using these values, we can calculate the standard free energy of formation (螖Gf掳) for NO at 298.15 K as follows: \[ 螖Gf掳(NO) = 螖Hf掳(NO) - T螖S掳(NO) \] \[ 螖Gf掳(NO) = 90.37 kJ/mol - (298.15 K)(0.21076 kJ/mol鈥) \] \[ 螖Gf掳(NO) = 90.37 kJ/mol - 62.87 kJ/mol \] \[ 螖Gf掳(NO) = 27.5 kJ/mol \] The positive value of 螖Gf掳(NO) indicates that the formation of NO is not spontaneous under standard conditions.
05

5. Conclusion

Many oxides of nitrogen, like NO, have positive values for the standard free energy of formation because their formation is endothermic, requiring energy to break the strong N-N triple bond. The entropy change during the formation is not significant enough to compensate for the positive enthalpy change, resulting in a positive value for 螖Gf. This suggests that the formation of these compounds is thermodynamically not favorable under standard conditions.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

There is evidence that radon reacts with fluorine to form compounds similar to those formed by xenon and fluorine. Predict the formulas of these \(\operatorname{Rn} \mathrm{F}_{x}\) compounds.

What mass of barium is produced when molten \(\mathrm{BaCl}_{2}\) is electrolyzed by a current of \(2.50 \times 10^{5} \mathrm{A}\) for 6.00 \(\mathrm{h} ?\)

Molten \(\mathrm{CaCl}_{2}\) is electrolyzed for 8.00 \(\mathrm{h}\) to produce \(\mathrm{Ca}(s)\) and \(\mathrm{Cl}_{2}(g) .\) a. What current is needed to produce 5.52 \(\mathrm{kg}\) of calcium metal? b. If 5.52 \(\mathrm{kg}\) calcium metal is produced, what mass (in kg) of \(\mathrm{Cl}_{2}\) is produced?

The compound \(\mathrm{Pb}_{3} \mathrm{O}_{4}\) (red lead) contains a mixture of lead(II) and lead(IV) oxidation states. What is the mole ratio of lead(II) to lead(IV) in \(\mathrm{Pb}_{3} \mathrm{O}_{4} ?\)

The inert-pair effect is sometimes used to explain the tendency of heavier members of Group 3 \(\mathrm{A}\) to exhibit \(+1\) and \(+3\) oxidation states. What does the inert-pair effect reference? (Hint:Consider the valence electron configuration for Group 3 \(\mathrm{A}\) elements.)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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.