Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 20

Show by calculation whether the reaction $2 \mathrm{HOCl}(\mathrm{aq}) \longrightarrow \mathrm{HClO}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})$ will go essentially to completion as written for standardstate conditions.

Short Answer

Expert verified
No, the reaction \(2 HOCl(aq) \rightarrow HClO_{2}(aq) + H^{+}(aq) + Cl^{-}(aq)\) does not go essentially to completion as written for standard state conditions.

Step by step solution

01

Calculate the standard Gibb's free energy

First, we can use the standard free energy of formation of all species involved in the reaction. We can look up these values in a standard table. The standard Gibbs free energy for a reaction can be calculated with the formula: \(\Delta G° = \Delta G_{products}° - \Delta G_{reactants}°\). From a standard table we obtain that the formation of \(HOCl(aq)\), \(HClO_{2}(aq)\), \(H^{+}(aq)\) and \(Cl^{-}(aq)\) is respectively: -137.3 kJ/mol, -121.9 kJ/mol, 0 kJ/mol and -131.2 kJ/mol. Applying these values into the formula, we get \(\Delta G° = [(-121.9) + 0 + (-131.2)] - [2*(-137.3)] = 58.8 kJ/mol\)
02

Interpretation of the result

Since the Gibbs free energy change \(\Delta G°\) is greater than 0, it means the reaction is not spontaneous. For a reaction to go to completion, it must be spontaneous, which implies \(\Delta G°\) should be negative under standard state conditions.
03

Conclusion

The reaction \(2 HOCl(aq) \rightarrow HClO_{2}(aq) + H^{+}(aq) + Cl^{-}(aq)\) does not go to completion under standard state conditions because it has a nonspontaneous Gibbs free energy change of +58.8 kJ/mol. A negative Gibbs free energy change would suggest otherwise.

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 amide anion \(\mathrm{NH}_{2}^{-}\) is a very strong base. On the basis of molecular orbital theory, would you expect \(\mathrm{NH}_{2}^{-}\) to be linear or bent?

One reaction of a chlorofluorocarbon implicated in the destruction of stratospheric ozone is \(\mathrm{CFCl}_{3}+h v \longrightarrow \mathrm{CFCl}_{2}+\mathrm{Cl}\) (a) What is the energy of the photons ( \(h v\) ) required to bring about this reaction, expressed in kilojoules per mole? (b) What is the frequency and wavelength of the light necessary to produce the reaction? In what portion of the electromagnetic spectrum is this light found?

The text mentions that ammonium perchlorate is an explosion hazard. Assuming that \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) is the sole reactant in the explosion, write a plausible equation(s) to represent the reaction that occurs.

To displace \(\mathrm{Br}_{2}\) from an aqueous solution of \(\mathrm{Br}^{-}\) add \((\mathrm{a}) \mathrm{I}_{2}(\mathrm{aq}) ;\) (b) \(\mathrm{Cl}_{2}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2}(\mathrm{g}) ;\) (d) \(\mathrm{Cl}^{-}(\mathrm{aq})\) (e) \(\mathrm{I}_{3}^{-}(\mathrm{aq})\).

Describe a chemical test you could use to determine whether a white solid is \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) or \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} .\) Explain the basis of this test using a chemical equation or equations.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Nuclear 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