Chapter 9, Problem 9.CCP

Interpretation Introduction

Interpretation:

An explanation about the sublimation enthalpy of ice at $0^{\circ }C$ 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.

  $q=ms\Delta t$

Where, $m$ is mass of the substance, $s$ is the specific heat capacity of the substance and $\Delta t$ is the temperature difference.

Explanation of Solution

The heat capacity of liquid water and water vapor are $4.18J{g}^{-1}{}^{\circ }{C}^{-1}$ and $1.84J{g}^{-1}{}^{\circ }{C}^{-1}$ respectively.

The heat energy required to heat up one mole of liquid water by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(18.01g\right)\left(4.18J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=7.54kJ.$

Similarly, the heat energy required to heat up one mole of water vapor by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(18.01g\right)\left(1.84J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=3.32kJ.$

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

  $\begin{array}{l}Meltice:H_{2}O\left(s,0^{\circ }C\right)\to H_{2}O\left(l,0^{\circ }C\right){\Delta }_{r}H=6.02kJ/mol\\ \\ Heatwater:H_{2}O\left(l,0^{\circ }C\right)\to H_{2}O\left(l,{100}^{\circ }C\right){\Delta }_{r}H=7.54kJ/mol\\ \\ Boilwater:H_{2}O\left(l,{100}^{\circ }C\right)\to H_{2}O\left(g,{100}^{\circ }C\right){\Delta }_{r}H=40.7kJ/mol\\ \\ \underset{\_}{Coolvapor:H_{2}O\left(g,{100}^{\circ }C\right)\to H_{2}O\left(l,0^{\circ }C\right){\Delta }_{r}H=-3.32kJ/mol}\\ \text{Sublimation}:H_{2}O\left(s,0^{\circ }C\right)\to H_{2}O\left(g,0^{\circ }C\right){\Delta }_{r}H=50.9kJ/mol\end{array}$

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

The heat capacity of liquid $HBr$ and $HBr$ vapor are $2J{g}^{-1}{}^{\circ }{C}^{-1}$ and $1J{g}^{-1}{}^{\circ }{C}^{-1}$ respectively.  This is same for $HI$ also.

The heat energy required to heat up one mole of liquid $HBr$ by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(80.912g\right)\left(2J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=16.1kJ.$

Similarly, the heat energy required to heat up one mole of $HBr$ vapor by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(80.912g\right)\left(1J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=8.09kJ.$

The heat energy required to heat up one mole of liquid $HI$ by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(127.911g\right)\left(2J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=25.6kJ.$

Similarly, the heat energy required to heat up one mole of $HI$ vapor by ${100}^{\circ }C$ is given below.

  $q=ms\Delta t=\left(127.911g\right)\left(1J{g}^{-1}{}^{\circ }{C}^{-1}\right)\left({100}^{\circ }C\right)\times \frac{1kJ}{1000J}=12.8kJ.$

For $HBr$:

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

  $\begin{array}{l}Melt:HBr\left(s,0^{\circ }C\right)\to HBr\left(l,0^{\circ }C\right){\Delta }_{r}H=2.406kJ/mol\\ \\ Heat:HBr\left(l,0^{\circ }C\right)\to HBr\left(l,{100}^{\circ }C\right){\Delta }_{r}H=16.1kJ/mol\\ \\ Boil:HBr\left(l,{100}^{\circ }C\right)\to HBr\left(g,{100}^{\circ }C\right){\Delta }_{r}H=19.3kJ/mol\\ \\ \underset{\_}{Coolvapor:HBr\left(g,{100}^{\circ }C\right)\to HBr\left(l,0^{\circ }C\right){\Delta }_{r}H=-8.09kJ/mol}\\ \text{Sublimation}:HBr\left(s,0^{\circ }C\right)\to HBr\left(g,0^{\circ }C\right){\Delta }_{r}H=29.8\approx 30kJ/mol\end{array}$

For $HI$:

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

  $\begin{array}{l}Melt:HI\left(s,0^{\circ }C\right)\to HI\left(l,0^{\circ }C\right){\Delta }_{r}H=2.870kJ/mol\\ \\ Heat:HI\left(l,0^{\circ }C\right)\to HI\left(l,{100}^{\circ }C\right){\Delta }_{r}H=25.6kJ/mol\\ \\ Boil:HI\left(l,{100}^{\circ }C\right)\to HI\left(g,{100}^{\circ }C\right){\Delta }_{r}H=49.7kJ/mol\\ \\ \underset{\_}{Coolvapor:HI\left(g,{100}^{\circ }C\right)\to HI\left(l,0^{\circ }C\right){\Delta }_{r}H=-12.8kJ/mol}\\ \text{Sublimation}:HI\left(s,0^{\circ }C\right)\to HI\left(g,0^{\circ }C\right){\Delta }_{r}H=36.9\approx 40kJ/mol\end{array}$

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