An explanation about the sublimation enthalpy of ice at 0 ∘ C as 51 k J / m o l has to be given. The sublimation enthalpies of H B r a n d H I 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 = m s Δ t Where, m is mass of the substance, s is the specific heat capacity of the substance and Δ t is the temperature difference.

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Chapter 9, Problem 9.CCP

Interpretation Introduction

Interpretation:

An explanation about the sublimation enthalpy of ice at 0∘C as 51 kJ/mol has to be given. The sublimation enthalpies of HBr and HI 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= m s Δt

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

Explanation of Solution

The heat capacity of liquid water and water vapor are 4.18 Jg−1∘C−1 and 1.84 Jg−1∘C−1 respectively.

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

q= m s Δt=(18.01 g)(4.18 Jg−1∘C−1)(100∘C)×1 kJ1000 J=7.54 kJ.

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

q= m s Δt=(18.01 g)(1.84 Jg−1∘C−1)(100∘C)×1 kJ1000 J=3.32 kJ.

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

Melt ice:H2O(s,0∘C )→H2O(l,0∘C ) ΔrH=6.02 kJ/molHeat water: H2O(l,0∘C )→H2O(l,100∘C ) ΔrH=7.54 kJ/molBoil water : H2O(l,100∘C )→H2O(g, 100∘C ) ΔrH=40.7 kJ/molCool vapor: H2O(g, 100∘C )→H2O(l, 0∘C ) ΔrH=−3.32 kJ/mol_Sublimation: H2O(s, 0∘C )→H2O(g, 0∘C ) ΔrH=50.9 kJ/mol

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

The heat capacity of liquid HBr and HBr vapor are 2 Jg−1∘C−1 and 1 Jg−1∘C−1 respectively. This is same for HI also.

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

q= m s Δt=(80.912 g)(2 Jg−1∘C−1)(100∘C)×1 kJ1000 J=16.1 kJ.

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

q= m s Δt=(80.912 g)(1 Jg−1∘C−1)(100∘C)×1 kJ1000 J=8.09 kJ.

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

q= m s Δt=(127.911 g)(2 Jg−1∘C−1)(100∘C)×1 kJ1000 J=25.6 kJ.

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

q= m s Δt=(127.911 g)(1 Jg−1∘C−1)(100∘C)×1 kJ1000 J=12.8 kJ.

For HBr:

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

Melt :HBr(s,0∘C )→HBr(l,0∘C ) ΔrH=2.406 kJ/molHeat : HBr(l,0∘C )→HBr(l,100∘C ) ΔrH=16.1 kJ/molBoil : HBr(l,100∘C )→HBr(g, 100∘C ) ΔrH=19.3 kJ/molCool vapor: HBr(g, 100∘C )→HBr(l, 0∘C ) ΔrH=−8.09 kJ/mol_Sublimation: HBr(s, 0∘C )→HBr(g, 0∘C ) ΔrH=29.8 ≈ 30 kJ/mol

For HI:

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

Melt :HI(s,0∘C )→HI(l,0∘C ) ΔrH=2.870 kJ/molHeat : HI(l,0∘C )→HI(l,100∘C ) ΔrH=25.6 kJ/molBoil : HI(l,100∘C )→HI(g, 100∘C ) ΔrH=49.7 kJ/molCool vapor: HI(g, 100∘C )→HI(l, 0∘C ) ΔrH=−12.8kJ/mol_Sublimation: HI(s, 0∘C )→HI(g, 0∘C ) ΔrH=36.9 ≈ 40 kJ/mol

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

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Answer 2