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As O 2 ( l ) is cooled at 1 atm, it freezes at 54.5 K to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that ∆H for the I → II phase transition is −743.1 J/mol, and ∆S for the same transition is −17.0 J/K mol. At what temperature are solids 1 and II in equilibrium?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 85AE
Textbook Problem
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As O2(l) is cooled at 1 atm, it freezes at 54.5 K to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that ∆H for the I → II phase transition is −743.1 J/mol, and ∆S for the same transition is −17.0 J/K  mol. At what temperature are solids 1 and II in equilibrium?

Interpretation Introduction

Interpretation: The freezing of O2(l) at 54.5K to form solid I is given. At a lower temperature, the given solid converts into solid II. The values of ΔH for the transition from phase I to II is given. The temperature at which these solids, I and II are at equilibrium is to be calculated.

Concept introduction: The expression for free energy change is,

ΔG=ΔH°TΔS°

Explanation of Solution

Given

The value of ΔH is 743.1J/mol .

The value of ΔS is 17.0J/Kmol .

Formula

The formula of ΔG is,

ΔG=ΔH°TΔS°

Where,

  • ΔH° is the standard enthalpy of reaction.
  • ΔG° is the free energy change.
  • T is the given temperature.
  • ΔS° is the standard entropy of reaction

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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