Chapter 5, Problem 143QRT

(a)

Interpretation Introduction

Interpretation:

Enthalpy of formation of Calcium chloride has to be calculated by using Born-Haber cycle.

Concept Introduction:

Born-Haber cycle:

The enthalpy of formation of ionic crystals are calculated by addition of enthalpies of atomization, ionization, affinity and lattice enthalpy, it is given by Born-Haber so it is called as Born-Haber cycle.

Explanation of Solution

Enthalpy of formation of Calcium chloride is calculated as,

  $\begin{array}{l}{\text{Ca}}_{\text{(g)}}^{\text{+}}{\text{+Cl}}_{\text{(g)}}^{\text{-}}\stackrel{}{\to }{\text{CaCl}}_{\text{(s)}}\text{ΔH}{}_{\text{rex}}\text{=lattice}\text{energy}\text{=}\text{-744 kJ}\\ {\text{Ca}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Ca}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}590\text{kJ}\\ {\text{e}}^{\text{-}}{\text{+Cl}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{-349kJ}\\ {\text{Ca}}_{\text{(s)}}\stackrel{}{\to }{\text{Ca}}_{\text{(g)}}^{}{\text{ΔH}}_{\text{sub}}\text{=}178\text{kJ}\\ \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{}\text{ΔH}\text{=}\frac{\text{1}}{\text{2}}\text{BE}\text{=}\frac{\text{1}}{\text{2}}\text{(242kJ)}\\ \overline{{\text{Ca}}_{\text{(s)}}\text{+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{CaCl}}_{\text{(s)}}\text{ΔH}\text{=}\text{-204}\text{kJ}}\end{array}$

Hence, the enthalpy of formation of Calcium chloride is $\text{-204}\text{kJ}$.

(b)

Interpretation Introduction

Interpretation:

The enthalpy change of ${\text{CaCl}}_{\text{2}}$ formation reaction has to be calculated.

Concept Introduction:

Hess's Law:

The enthalpy change of given reaction is calculated by subtraction of sum of enthalpy of formation reactants from sum of enthalpy of formation reactant products.

  ${\text{ΔH}}_{\text{rex}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}$

Explanation of Solution

The enthalpy change of given reaction is calculated as,

  $\text{2}{\text{CaCl}}_{\text{(s)}}\stackrel{}{\to }{\text{CaCl}}_{\text{2(s)}}{\text{+Ca}}_{\text{(s)}}$

Substitute the standard and calculated values in below equation to get enthalpy change of given reaction.

  $\begin{array}{l}{\text{ΔH}}_{\text{rex}}\text{=}{\displaystyle \sum {\text{ΔH}}_{\text{product}}}\text{-}{\displaystyle \sum {\text{ΔH}}_{\text{reactant}}}\\ {\text{ΔH}}_{\text{rex}}{\text{=[ΔH(CaCl}}_{\text{2}}{\text{)+ΔHCa]-2ΔHCaCl}}_{\text{(g)}}\\ \text{=}\text{(-795}\text{.8}\text{kJ/mol)+}\text{(0}\text{kJ/mol)-2(-204}\text{kJ/mol)}\\ \text{=-388}\text{kJ/mol}\end{array}$

Hence, the enthalpy change of given reaction is $\text{-388}\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

Reason for the stability of is ${\text{CaCl}}_2$ higher than $\text{CaCl}$ has to be explained.

Concept Introduction:

Refer part (a).

Explanation of Solution

In part (b), is $\text{CaCl}$ lose energy when reacting to produce ${\text{CaCl}}_2$, it is exothermic reaction.

The stability is higher from the lower energy product so ${\text{CaCl}}_2$ is lower energy than $\text{CaCl}$.

Hence, the ${\text{CaCl}}_2$ is more stable than $\text{CaCl}$.