Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 9, Problem 9.CCP
Interpretation Introduction

Interpretation:

An explanation about the sublimation enthalpy of ice at 0C as 51kJ/mol has to be given.  The sublimation enthalpies of HBrandHI at their melting points have to be estimated.

Concept Introduction:

If a transition or process is taking place where the initial and final state of the substance are different, then the formula for calculating the heat required for the process is given below.

  q=msΔt

Where, m is mass of the substance, s is the specific heat capacity of the substance and Δt is the temperature difference.

Expert Solution & Answer
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Explanation of Solution

The heat capacity of liquid water and water vapor are 4.18Jg1C1 and 1.84Jg1C1 respectively.

The heat energy required to heat up one mole of liquid water by 100C is given below.

  q=msΔt=(18.01g)(4.18Jg1C1)(100C)×1kJ1000J=7.54kJ.

Similarly, the heat energy required to heat up one mole of water vapor by 100C is given below.

  q=msΔt=(18.01g)(1.84Jg1C1)(100C)×1kJ1000J=3.32kJ.

Four required thermochemical processes that sum to produce form the sublimation equation of water is given below.

  Meltice:H2O(s,0C)H2O(l,0C)ΔrH=6.02kJ/molHeatwater:H2O(l,0C)H2O(l,100C)ΔrH=7.54kJ/molBoilwater:H2O(l,100C)H2O(g,100C)ΔrH=40.7kJ/molCoolvapor:H2O(g,100C)H2O(l,0C)ΔrH=3.32kJ/mol_Sublimation:H2O(s,0C)H2O(g,0C)ΔrH=50.9kJ/mol

This result is quite close to the 51kJ/mol as given in the question.

The heat capacity of liquid HBr and HBr vapor are 2Jg1C1 and 1Jg1C1 respectively.  This is same for HI also.

The heat energy required to heat up one mole of liquid HBr by 100C is given below.

  q=msΔt=(80.912g)(2Jg1C1)(100C)×1kJ1000J=16.1kJ.

Similarly, the heat energy required to heat up one mole of HBr vapor by 100C is given below.

  q=msΔt=(80.912g)(1Jg1C1)(100C)×1kJ1000J=8.09kJ.

The heat energy required to heat up one mole of liquid HI by 100C is given below.

  q=msΔt=(127.911g)(2Jg1C1)(100C)×1kJ1000J=25.6kJ.

Similarly, the heat energy required to heat up one mole of HI vapor by 100C is given below.

  q=msΔt=(127.911g)(1Jg1C1)(100C)×1kJ1000J=12.8kJ.

For HBr:

Fusion enthalpy of HBr at its melting point is 2.406kJ/mol.

  Melt:HBr(s,0C)HBr(l,0C)ΔrH=2.406kJ/molHeat:HBr(l,0C)HBr(l,100C)ΔrH=16.1kJ/molBoil:HBr(l,100C)HBr(g,100C)ΔrH=19.3kJ/molCoolvapor:HBr(g,100C)HBr(l,0C)ΔrH=8.09kJ/mol_Sublimation:HBr(s,0C)HBr(g,0C)ΔrH=29.830kJ/mol

For HI:

Fusion enthalpy of HI at its melting point is 2.870kJ/mol.

  Melt:HI(s,0C)HI(l,0C)ΔrH=2.870kJ/molHeat:HI(l,0C)HI(l,100C)ΔrH=25.6kJ/molBoil:HI(l,100C)HI(g,100C)ΔrH=49.7kJ/molCoolvapor:HI(g,100C)HI(l,0C)ΔrH=12.8kJ/mol_Sublimation:HI(s,0C)HI(g,0C)ΔrH=36.940kJ/mol

Therefore, the sublimation enthalpies of HBr and HI are 30kJ/mol and 40kJ/mol respectively.

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Chapter 9 Solutions

Chemistry: The Molecular Science

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