Chapter 15, Problem 15C.8P

(a)

Interpretation Introduction

Interpretation:

The lattice enthalpy and enthalpy of formation of $CaCl\left(s\right)$ by using Born-Mayer equation and Born-Haber cycle respectively.

Concept Introduction:

Enthalpy of sublimation:

The heat required to change the state of one mole of a substance from solid state to gaseous state at a given combination of temperature and pressure is called as enthalpy of sublimation.

Ionization Energy:

The amount of energy which is required to remove the electron from its outermost orbital is called ionization energy.

Dissociation Energy:

The amount of energy needed to break apart one mole of covalently bonded gases into a pair of atoms is called as the dissociation energy.

Electron Affinity:

The amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion is called as electron affinity.

Lattice Energy:

The energy required to break apart an ionic solid and convert its component atoms into gaseous ions is called lattice energy.

The other definition is the energy released when gaseous ions bind to form an ionic solid is called as lattice energy.

Enthalpy of formation:

The change in enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states is called as the standard enthalpy of formation.

Answer to Problem 15C.8P

The lattice enthalpy and enthalpy of formation of $CaCl\left(s\right)$ has been calculated to be $685kJmo{l}^{-1}$ and $-146kJmo{l}^{-1}$ respectively.

Explanation of Solution

Required data to calculate the lattice enthalpy of $CaCl\left(s\right)$:

  $\begin{array}{l}{r}_{C{a}^{+}}=138pm\\ \\ r_{C{l}^{-}}=181pm\\ \\ d^{*}=34.5pm\\ \\ N_A=6.02214\times {10}^{23}mo{l}^{-1}\\ \\ 1e^{-}=1.602176\times {10}^{-19}C\\ \\ {\epsilon }_0=8.85419\times {10}^{-12}{J}^{-1}{C}^2{m}^{-1}\\ \\ d=\left(138+181\right)pm=319pm\end{array}$

Identification of crystal type:

Radius ratio, $\frac{r_{C{a}^{+}}}{r_{C{l}^{-}}}=\frac{138pm}{181pm}=0.762$, this indicates that it belongs to $CsCl$ structure. Hence, Madelung constant $\left(A\right)=1.763$.

Born-Mayer equation:

  $E_{p,\min }=-\frac{N_A\left|Z_A{Z}_B\right|e^2}{4\pi {\epsilon }_{0}d}\left(1-\frac{d^{*}}{d}\right)A$

Now substituting all the values in the above equation and solving for the lattice enthalpy of $CaCl\left(s\right)$

  $\begin{array}{c}{E}_{p,\min }=-\frac{N_A\left|Z_A{Z}_B\right|e^2}{4\pi {\epsilon }_{0}d}\left(1-\frac{d^{*}}{d}\right)A\\ \\ =-\frac{\left(6.02214\times {10}^{23}mo{l}^{-1}\right)\times \left(1\times 1\right)\times {\left(1.602176\times {10}^{-19}C\right)}^2}{4\pi \times \left(8.85419\times {10}^{-12}{J}^{-1}{C}^2{m}^{-1}\right)\times \left(3.19\times {10}^{-10}m\right)}\times \left(1-\frac{34.5pm}{319pm}\right)\times \left(1.763\right)\\ \\ \simeq 6.85\times {10}^{5}Jmo{l}^{-1}=685kJmo{l}^{-1}.\end{array}$

Therefore, the lattice enthalpy of $CaCl\left(s\right)$ has been calculated to be $685kJmo{l}^{-1}$.

Required data to calculate the enthalpy of formation of $CaCl\left(s\right)$:

1. 1. Sublimation of $Ca\left(s\right)$: $Ca\left(s\right)+\Delta H_{sub}\to Ca\left(g\right)$, $\Delta H_{sub}=+176kJmo{l}^{-1}$.
2. 2. Ionization of $Ca\left(g\right)toC{a}^{+}\left(g\right)$: $\Delta H_{ion}=+589.7kJmo{l}^{-1}$.
3. 3. Dissociation of $C{l}_2\left(g\right)$: $\frac{1}{2}C{l}_2\left(g\right)\to Cl\left(g\right)$, $\Delta H_{diss}=+121.7kJmo{l}^{-1}$.
4. 4. Electron attachment to $Cl\left(g\right)$: $Cl\left(g\right)+e^{-}\to C{l}^{-}\left(g\right)$, $\Delta H_{E.A}=-348.7kJmo{l}^{-1}$
5. 5. Lattice enthalpy, $\Delta H_L=-685kJmo{l}^{-1}$
6. 6. Formation of $CaCl\left(s\right)fromCa\left(s\right)and\frac{1}{2}C{l}_2\left(g\right)$: $Ca\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to CaCl\left(s\right)$, $\Delta H_f$

The Born-Haber cycle for this solid can be made like given below

Figure 1

The sum of the contributions around the cycle is

$\Delta H_f=\Delta H_{sub}+\Delta H_{ion}+\Delta H_{diss}+\Delta H_{E.A}+\Delta H_L$

Then,

$\Delta H_f=\left(176+589.7+121.7-348.7-685\right)kJmo{l}^{-1}=-146kJmo{l}^{-1}.$

Therefore, the enthalpy of formation of $CaCl\left(s\right)$ has been calculated to be $-146kJmo{l}^{-1}$.

(b)

Interpretation Introduction

Interpretation:

The explanation for the non-existence pf $CaCl\left(s\right)$ can be found in the reaction enthalpy for the disproportion reaction of $CaCl\left(s\right)$ has to be verified.

Concept Introduction:

Enthalpy of a reaction:

  $\Delta H^0={\displaystyle \sum \Delta H_{products}^0}-{\displaystyle \sum \Delta H_{reac\tan ts}^0}$

Explanation of Solution

The disproportion reaction of $CaCl\left(s\right)$ can be written as

  $2CaCl\left(s\right)\to Ca\left(s\right)+CaC{l}_2\left(s\right)$

Enthalpy of this reaction is

  $\begin{array}{c}\Delta H^0={\displaystyle \sum \Delta H_{products}^0}-{\displaystyle \sum \Delta H_{reac\tan ts}^0}\\ \\ =\left[0-795.8-2\left(-146\right)\right]kJmo{l}^{-1}\\ \\ =-503.8kJmo{l}^{-1}.\end{array}$

It shows that the process is exothermic in nature. Hence, the thermodynamically unstable $CaCl\left(s\right)$ disproportionate to give $Ca\left(s\right)$ and $CaC{l}_2\left(s\right)$.