Chapter 5, Problem 144QRT

(a)

Interpretation Introduction

Interpretation:

Enthalpy of formation of Silver 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 Silver chloride is calculated as,

  $\begin{array}{l}{\text{Ag}}_{\text{(g)}}^{\text{+}}{\text{+Cl}}_{\text{(g)}}^{\text{-}}\stackrel{}{\to }{\text{AgCl}}_{\text{(s)}}\text{ΔH}{}_{\text{rex}}\text{=lattice}\text{energy}\text{=}\text{-910}\text{kJ}\\ {\text{Ag}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Ag}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{731}\text{kJ}\\ {\text{e}}^{\text{-}}{\text{+Cl}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{-349kJ}\\ {\text{Ag}}_{\text{(s)}}\stackrel{}{\to }{\text{Ag}}_{\text{(g)}}^{}{\text{ΔH}}_{\text{sub}}\text{=}\text{248}\text{kJ}\\ \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{}\text{ΔH}\text{=}\frac{\text{1}}{\text{2}}\text{BE}\text{=}\frac{\text{1}}{\text{2}}\text{(242kJ)}\\ \overline{{\text{Ag}}_{\text{(s)}}\text{+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{AgCl}}_{\text{(s)}}\text{ΔH}\text{=}\text{-123}\text{kJ}}\end{array}$

Hence, the enthalpy of formation of Silver chloride is $\text{-123}\text{kJ}$.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer part (a).

Explanation of Solution

Enthalpy of formation of Silver chloride is calculated as,

  $\begin{array}{l}{\text{Na}}_{\text{(g)}}^{\text{+}}{\text{+Cl}}_{\text{(g)}}^{\text{-}}\stackrel{}{\to }{\text{AgCl}}_{\text{(s)}}\text{ΔH}{}_{\text{rex}}\text{=lattice}\text{energy}\text{=}\text{-787}\text{kJN}\\ {\text{Na}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Na}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{496}\text{kJ}\\ {\text{e}}^{\text{-}}{\text{+Cl}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{-349kJ}\\ {\text{Na}}_{\text{(s)}}\stackrel{}{\to }{\text{Na}}_{\text{(g)}}^{}{\text{ΔH}}_{\text{sub}}\text{=}\text{108}\text{kJ}\\ \frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{Cl}}_{\text{(g)}}^{}\text{ΔH}\text{=}\frac{\text{1}}{\text{2}}\text{BE}\text{=}\frac{\text{1}}{\text{2}}\text{(242kJ)}\\ \overline{{\text{Na}}_{\text{(s)}}\text{+}\frac{\text{1}}{\text{2}}{\text{Cl}}_{\text{2(g)}}\stackrel{}{\to }{\text{NaCl}}_{\text{(s)}}\text{ΔH}\text{=}\text{-}\text{411}\text{kJ}}\end{array}$

Hence, the enthalpy of formation of Silver chloride is $\text{-411}\text{kJ}$.

(c)

Interpretation Introduction

Interpretation:

The common factor for the differences in ${\text{ΔH}}_{\text{f}}$ vales in $\text{NaCl}$ and $\text{AgCl}$ has to be commended.

Concept Introduction:

Refer part (a).

Explanation of Solution

In the ionization and sublimation endothermic process of both compounds are,

  $\begin{array}{l}{\text{Na}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Na}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{496}\text{kJ}\\ {\text{Na}}_{\text{(s)}}\stackrel{}{\to }{\text{Na}}_{\text{(g)}}^{}{\text{ΔH}}_{\text{sub}}\text{=}\text{108}\text{kJ}\\ \\ {\text{Ag}}_{\text{(g)}}^{}\stackrel{}{\to }{\text{Ag}}_{\text{(g)}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}{\text{ΔH}}_{\text{rex}}\text{=}\text{731}\text{kJ}\\ {\text{Ag}}_{\text{(s)}}\stackrel{}{\to }{\text{Ag}}_{\text{(g)}}^{}{\text{ΔH}}_{\text{sub}}\text{=}\text{248}\text{kJ}\end{array}$

From the given vales, $\text{NaCl}$ have smaller values than $\text{AgCl}$ but both them have enthalpy of formation values are in negative, exothermic but $\text{NaCl}$ have higher value than $\text{AgCl}$.