Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 77

Consider the reaction $2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$ (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of all gases are \(33.4 \mathrm{kPa}\).

Short Answer

Expert verified
(a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

Step by step solution

01

Calculate standard Gibbs free energy change

Using the ∆G° value for each species provided in Appendix C, we can calculate the standard Gibbs free energy change for the reaction. We have the reaction: 2 NO(g) + O₂(g) ⟶ 2 NO₂(g) Using the formula mentioned above, we get: ∆G° = Σ(ν_product * ∆G°_product) - Σ(ν_reactant * ∆G°_reactant) ∆G° = (2 * ∆G°(NO₂) - (2 * ∆G°(NO) + ∆G°(O₂)) Plug in the values for each species: ∆G° = (2 * 51.3 kJ/mol) - (2 * 86.6 kJ/mol + 0 kJ/mol) Calculate the Gibbs free energy change: ∆G° = -171.2 kJ/mol
02

Calculate reaction quotient

We are given that the partial pressures of all gases are 33.4 kPa. Calculate the reaction quotient Q using the partial pressures: Q = [NO₂]² / ([NO]² * [O₂]) Since the partial pressures are the same for each gas (33.4 kPa), we can write: Q = (33.4)² / ((33.4)² * 33.4) Q = 1 / 33.4
03

Calculate non-standard Gibbs Free Energy Change

Now, we can calculate the non-standard Gibbs free energy change (∆G) using the Van't Hoff equation: ∆G = ∆G° + RT ln(Q) Plug in the values: ∆G = -171.2 kJ/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Don't forget to convert kJ to J: ∆G = -171200 J/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Calculate the value to obtain the final answer: ∆G = -142627.28 J/mol In conclusion, (a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

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 conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)$, is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

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. \((\mathbf{c})\) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

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: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(I_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

Sulfur dioxide reacts with strontium oxide as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s) $$ (a) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ} .\) (b) If you had only standard enthalpy data for this reaction, how would you estimate the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) using data from Appendix \(\mathrm{C}\) on other substances.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

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