Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 56: Problem 5

How does temperature affect the equilibrium \(\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ?\) Explain in terms of \(\Delta H^{\circ}\)

Short Answer

Expert verified
Increase in temperature shifts the equilibrium to the right, favoring the endothermic reaction (H2O(l) to H2O(g)), as it absorbs the excess heat. This is because \(\Delta H^{\circ}\) is positive for this reaction. Conversely, a decrease in temperature shifts the equilibrium to the left, favoring the exothermic reaction which releases heat.

Step by step solution

01

Review of Le Chatelier's Principle

Le Chatelier's Principle states that if a stress is applied to a system at equilibrium, the system responds by altering the equilibrium position to relieve that stress.
02

Relation between enthalpy and temperature

For the equilibrium \[\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\] the forward reaction (liquid to gas) is endothermic as it requires heat to go from a liquid state (H2O(l)) to a gaseous state (H2O(g)). This means that \(\Delta H^{\circ}\) is positive.
03

Effect of temperature on the equilibrium

According to Le Chatelier's Principle, if temperature is increased, the system will adjust so as to absorb the excess heat. In this case, the equilibrium will shift to the right to favor the endothermic reaction. Likewise, if temperature is decreased, the reaction will shift left, favoring the exothermic reaction to release heat.

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

Recall that \(\Delta G^{\circ}\) can be written as a function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\). Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not temperature dependent and answer each of the following: a) Derive an expression relating \(\ln K, \Delta H^{\circ}\), and \(\Delta S^{\circ} .\) That is, derive an expression that looks like \(\ln K=\) some function of \(\Delta H^{\circ}\) and \(\Delta S^{\circ} .\) The temperature, \(T\), should appear only once in this equation. b) How are the equilibrium constants for reactions with \(\Delta H^{\circ}>0\) affected by an increase in temperature? c) How are equilibrium constants for reactions with \(\Delta H^{\circ}<0\) affected by an increase in temperature?

Show that the proportionality constant found in CTQ 7 is equal to \(\frac{R T}{n F}\), where the universal gas constant \(R=8.314 \frac{\mathrm{J}}{\mathrm{K} \mathrm{mole}}, F\) is Faraday's constant, 96485 coulombs per mole (of electrons), and "n" is the number of moles of electrons transferred in the balanced chemical reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Organic 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